OF THE 
 
 ' , 
 
 . . 
 
 UNIVERSITY OF CALIFORNIA. 
 
 
 
 . 
 
 
 Class 
 
ELEMENTARY MANUAL 
 
 ON 
 
 APPLIED MECHANICS 
 
 NINTH EDITION 
 THOROUGHLY REVISED 
 
GRIFFIN'S SCIENTIFIC TEXT-BOOKS. 
 
 By PROFESSOR JAMIESON, M.Inst.C.E., M.Inst.E.E., F.R.S.E., Formerly 
 Professor of Engineering in the Glasgow and West of Scotland Technical College; 
 Consulting Engineer and Electrician, 16 Kosslyn Terrace, Kelvinside, Glasgow. 
 
 INTRODUCTORY MANUALS. 
 
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OF THE 
 
 UNIVERSITY 
 
 OF 
 
ELEMENTARY MANUAL 
 
 ON 
 
 APPLIED MECHANICS 
 
 SPECIALLY ARRANGED TO SUIT THOSE PREPARING FOR THE 
 INSTITUTION OF CIVIL ENGINEERS; ROYAL INSTITUTE OF 
 BRITISH ARCHITECTS ; CITY AND GUILDS OF LONDON INSTI- 
 TUTE ; BRITISH AND COLONIAL BOARDS OF EDUCATION AND 
 ALL KINDS OF FIRST YE1R ENGINEERING STUDENTS. 
 
 BY 
 
 ANDEEW JAMIESON, M. INST. C.E. 
 
 FORMERLY PROFESSOR OF ENGINEERING, THE GLASGOW TECHNICAL COLLEGE 
 
 FELLOW OF THE ROYAL SOCIETY, EDINBURGH; MEMBER OF THE INSTITUTION OF 
 
 ELECTRICAL ENGINEERS; AUTHOR OF "TEXT-BOOKS ON STEAM AND STEAM 
 
 ENGINES," "ADVANCED APPLIED MECHANICS," "MAGNETISM AND 
 
 ELECTRICITY," "ELECTRICAL ENGINEERING RULES AND TABLES." 
 
 NINTH EDITION, THOROUGHLY REVISED 
 
 diagrams, arithmetical Samples, examination 
 anb 
 
 Of THE 
 
 UNIVERSITY 
 
 OF 
 
 |L,ONDON 
 
 CHARLES GRIFFIN' AND COMPANY, LIMITED, 
 
 EXETER STREET STRAND. 
 
 1910 
 [All rights reserved.] 
 
PREFACE TO NINTH EDITION 
 
 THIS Elementary Manual on Applied Mechanics has been care- 
 fully revised, whilst several important alterations and additions 
 have been made to it. 
 
 I have added a drawing and description to the end of Lecture 
 VIII of Butters Brothers and Co.'s new 1909 "Jib Crane 
 Arrester." 
 
 A new Frontis Plate, as well as a new plate for Lecture XXI, 
 showing the latest 1909 improvements in " Alfred Herbert's 
 Electrically Driven Turret Lathe," together with a description 
 thereof, have been duly inserted at the proper places. 
 
 Owing to a very clear and useful hint which was given in The 
 Electrician, by the reviewer of the previous Edition, I have trans- 
 ferred the former Lectures numbered XXIV and XXV, upon 
 " Bending and Shearing Stresses in Beams and Metal Structures," 
 <fec., to my new and more advanced book, Volume III, which deals 
 entirely with the "Theory of Structures." This change has 
 enabled me not only to reduce the total number of Lectures in 
 the book from twenty-eight to twenty- six, but to re-arrange and 
 considerably extend the last Lecture. To it, I have added an 
 account of " Sir Joseph Whit worth's Early Realisations of 
 Mechanical Accuracy " ; as well as a detailed description of one of 
 his " Millionth Measuring Machines." 
 
 Further, I have described by aid of six very clear photo-process 
 diagrams, a " New Set of English Standard Guages." I have 
 specially to thank Mr. H. M. Budgett, of the Crown Works, 
 Chelmsford, England, not only for supplying these good figures, 
 but also for permitting me to copy the Certificate given to him 
 last autumn by The National Physical Laboratory of their very 
 careful tests of these British-made gauges. 
 
 Special attention has been paid to the Appendices A, B, 0, and 
 D, where Teachers and Stage I Students will find, not only the 
 latest General Instructions of the Board of Education (B. of E.) 
 and the City and Guilds of London (C. & G.); but also the concise 
 abstract from the Rules and Syllabus for the examination of 
 Students of The Institution of Civil Engineers (Stud. I.C.B.). 
 Further, all the questions given by these three well-known and 
 respected Examining Bodies, up to and including the year 1908, 
 have been collected and arranged under the numbers of the various 
 
 204114 
 
VI PREFACE 
 
 Lectures to which they naturally belong. Finally, the entire 
 examination papers for 1909 and 1910 have been given as issued, 
 with the view that students should test their knowledge of this 
 book by trying these papers under exact Examination conditions. 
 At all events, I find that this is one of the best methods of pre- 
 paring my C.E. and other Engineering Correspondence Students for 
 these and several other Examinations and Appointments. 
 
 I have again to thank Mr. John Ramsay, Assoc. M. Tnst. C.E., 
 for his assistance with this book. 
 
 ANDREW JAMIESON. 
 
 Consulting Engineer and Electrician. 
 1C ROSSLYN TERRACE, 
 
 KELVINSIDE, GLASGOW. 
 August 1910. 
 
 PEEFACE TO EIGHTH EDITION 
 
 I HAVE taken advantage of the working out of many Questions 
 at the end of the several Lectures, and of various Examination 
 Papers in the Appendices, by my " Science Correspondence 
 Students,'' to obtain a number of checked Answers. These 
 Answers have been inserted at their Questions up to Lecture 
 XXIY ; but thereafter, they have been printed just before the 
 Index, at the end of Appendix C. 
 
 The 1907 Examination Papers for Stage I of the Board of 
 Education (B. of E.), South Kensington ; the City and Guilds 
 (C.and G.) of London Institute, Ordinary Grade, Part I. Sections A 
 and B on Mechanical Engineering; and the Entrance or Students' 
 Examinations of The Institution of Civil Engineers (Stud. I.C.E.) 
 in Elementary Mechanics for 1907 and Feb. 1908 will be found 
 in AppeLdix B, along with those of previous }ears, the Rules, 
 and an Abstract of the Syllabus for these Examinations. 
 
PKEFACE Vll 
 
 These Answers and new up-to date Examination Questions, 
 which have been taken from the best sources, should prove of 
 great interest and assistance to Teachers and Students of Applied 
 Mechanics. 
 
 At the end of Lecture I a new page has been added. It deals 
 with the terms and definitions of words used in measuring Pres- 
 sures, as well as a discussion of the recent endeavour by some 
 authors to introduce the words Thi*ust and Resultant Thrust. I 
 see no necessity for these terms and shall feel obliged if practical 
 Engineering Reviewers, Teachers and Students will give me 
 their opinion after carefully studying my remarks. 
 
 At the end of Lecture XVIII a part of a new page has been in- 
 serted, wherein are given concise distinctions between solids, liquids 
 and gases, with definitions of perfect viscous and elastic fluids. 
 
 A more detailed and clearer explanation of the application and 
 uses of Limit Gauges as well as of the construction and uses of 
 Berridge's Tangentometer have been given in Lecture XXVIII 
 Students should study and apply this instrument to find the 
 sines, cosines, tangents, &c., of different angles and then check 
 their results by the Tables near the end of Appendix C, on 
 " Functions of Angles," before they finish Lecture IV. 
 
 Owing to the necessity for young engineers becoming early 
 acquainted with Electrical Engineering terms and units of 
 measurement, I have added to appendix C what is believed to be 
 the latest and most correct set of definitions for the Nomenclature 
 with their Abbreviations or Symbols of the C.G.S. and of the 
 authorised Practical Electrical Units. 
 
 Finally, the whole book has been most carefully revised and 
 brought up to date. 
 
 I have much pleasure in thanking Forrest Sutherland, B.Sc., 
 Science and Technical Educationist, Natal, South Africa, for his 
 useful suggestions, and John Ramsay, Assoc. M. Inst. C.E., for 
 the interest which he has taken and the help he has given with 
 this new Edition. The Author hopes that this eighth edition 
 will be found to be an advance upon previous issues. He also 
 desires reviewers, teachers and students to kindly let him know 
 in what way they think that he can still further improve this 
 " First Year Manual on Appli ed Mechanics." 
 
 f ANDREW JAM IE SON, 
 
 Consulting Engineer and Electrician. 
 16 KCSSLYB^TEKBACE, 
 
 KELVINSIDE, GLASGOW. 
 
INSTRUCTIONS TO BE FOLLOWED IN THE WRITING 
 OF HOME EXERCISES. 
 
 1. Put the date of handing in each exercise at the right-hand top corner. 
 
 2. Leave a margin an inch wide on th left-hand side of each page ; and 
 in the margin place the number of the question, and nothing more. 
 
 3. Leave * space of *t least three lines between your answers for remarki 
 or corrections. 
 
 4. Be sure yon understand exactly what the question requires you to answer, 
 then give all it requires, but no more. If unable to answer any question, write 
 down its number and the reason why. 
 
 5. Make your answers concise, clear, and exact ; and accompany them, 
 whenever practicable, by an illustrative sketch. 
 
 6. Make all sketches large, open, and in the centre of the page, and do not 
 crowd any writing about them. 
 
 NOTE. The character of the sketches will be considered in awarding the 
 marks to the several questions. Neat sketches and an " Index to 
 Parts," with the first letter of name of Part, will always receive more 
 marks than a bare written description. 
 
 7. Every sketch must be accompanied by an " Index to Parts " written 
 immediately beneath it, and must accompany the answer it is designed ie 
 illustrate. 
 
 NOTE. The initial letter or letters of the name of the Part must be used, 
 and not A, IB, C, or i, 2, 3, &c, 
 
 8. Unless otherwise specially requested by the question, every sketch must 
 be accompanied by a concise, written description. 
 
 9. Every answer which receives less than half of the full marks awarded to 
 It, must be re-written correctly for next evening, before the usual class work f 
 and headed "Re-written." 
 
 REMARKS. Students are strongly reromme? r el to wr te out each 
 answer in scroll first, and then to compare it with the question. After 
 committing it to their book, they should then read it over a second time, 
 to correct any errors they may discover. Reasonable and easily intelli- 
 gible contractions are permitted. Students are invited to ask questions 
 and explanations regarding anything they do not understand. Except in 
 special cases, arrears of Home work will not receive marks. 
 
 N.B. Students who from any cause have been absent from a lecture, must 
 tend a post-card or note of explanation to the teacher. If they miss any exercise 
 or exercises, they must state the reason (in red ink, or underlined) in their 
 exercise books when handing them in next night. If these rules are not com- 
 , then marks will be deducted. 
 
CONTENTS. 
 
 FAOl 
 
 LECTURE I. 
 
 Definition of Applied Mechanics Force Matter Unit of Force 
 The Elements of a Force Graphic Representation of 
 Forces Forces in Equilibrium Action and Reaction 
 Resultant and Components Resultant of Forces acting 
 in a Straight Line Terms and Definitions used in Measur- 
 ing or Calculating Pressures and their Effects Definitions 
 of Scalar, Vector and Rotor Quantities Engineering Cal- 
 culations . 1-5 
 
 LECTURE II. 
 
 Work Unit of Work Examples I. II. III. IV. Work done 
 against a Variable Resistance Example V. Diagrams of 
 Work With Uniform Resistance With a Uniformly In- 
 creasing Resistance With a Uniformly Decreasing Resis- 
 tance With a Combination of Uniform and Variable 
 Loads Example VI. Power or Activity Units of Power 
 The Horse-power Unit To find the Horse-power of any 
 working Agent Example VII. Uses of Squared Paper 
 Clark's Adjustable Curve Example VIII. Questions . 6-ao 
 
 LECTURE III. 
 
 The Moment of a Force Principle of Momenta applied to the 
 Lever Experiments I. II. III. Pressure on and Reaction 
 from the Fulcrum Equilibrant and Resultant of two 
 Parallel Forces Couples Centre of Parallel Forces or 
 Position of Equilibrant and Resultant Centre of Gravity 
 j: Examples of Centre of Gravity The Lever when its 
 
 Weight is taken into Account Examples I. II. Position 
 of the Fulcrum Example III. Questions . . 21-34 
 
 LECTURE IV. 
 
 Practical Applications of the Lever The Steelyard, or Roman 
 Balance Graduation of the Steelyard The Lever Safety 
 Valve Example I. Lever Machine for Testing Tensile 
 
CONTEXTS, 
 
 PAGES 
 
 Strength of Materials Straight Levers acted on by In- 
 dined Forces Bent Levers The Bell Crank Lever Bent 
 Lever Balance Duplex Bent Lever, or Lumberer's Tongs 
 Turkns, or Pincers Examples II. and III. Toggle Joints 
 Questions ......... 35-51 
 
 LECTURE V. 
 
 The Principle of Work Work put in, Work lost, Useful Work 
 Efficiency of a Machine Principle of Work applied to the 
 Lever Experiments I. II. Wheel and Axle The Prin- 
 ciple of Moments applied to the Wheel and Axle The 
 Principle of Work applied to the Wheel and Axle Experi- 
 ment III. The Winch Barrel Example L Ship's Cap^an 
 The Fusee Questions ....,. 52-62 
 
 LECTURE VI. 
 
 Pulleys Snatch Block Block and Tackle Theoretical Advan- 
 tageVelocity Ratio The Principle of Work applied to 
 the Block and Tackle Actual or Working Advantage 
 Work pnt in Work got out Efficiency Percentage 
 Efficiency Example I. Questions . , . . 63-71 
 
 LECTURE VII. 
 
 The Wheel and Compound Axle, or Chinese Windlass The Prin- 
 ciple of Moments applied to the Wheel and Compound 
 Axle The Principle of Work applied to the Wheel and 
 Compound Axle Examples I. II. Weston's Differential 
 Pulley Block The Principle of Work applied to Weston's 
 Differential Pulley Block Experiment I. Cause of the 
 Load not overhauling the Chain Questions . . .72- 79 
 
 LECTURE VIII. 
 
 Graphic Demonstration of Three Forces in Equilibrium 
 Parallelogram of Forces Triangle of Forces Three Equal 
 Forces in Equilibrium Two Forces acting at Right Angles 
 Resolution of a Force into Two Components at Right 
 Angles Resultant of Two Forces acting at any Angle on 
 a Point Resultant of any number of Forces acting at a 
 Point Example I. Stresses in Jib Cranes Jib Arrester 
 Example II., III. Stresses on a Simple Roof Example 
 IV. Questions 80-92 
 
CONTENTS. XI 
 
 LECTURE IX. 
 
 PA6B 
 
 Inclined Planes -The Inclined Plane without Friction When the 
 Force acts Parallel to the Plane Example I. When the 
 Force acts Parallel to the Base Example II. When the 
 Force acts at any Angle to the Inclined Plane Example III. 
 The Principle of Work applied to the Inclined Plane- 
 Example IV. Questions 93-IOO 
 
 LECTURE X. 
 
 Friction Heat is Developed when Force overcomes Friction 
 Laws of Friction Apparatus for Demonstrating First and 
 Second Laws of Friction Experiment I. Example I. 
 Angle of Repose or Angle of Friction Experiment II. 
 Diagram of Angles of Repose Limiting Angle of Resistance 
 Experiment III. Apparatus for Demonstration of the 
 Third Law of Friction Experiment IV. Lubrication 
 Anti-Friction Wheel Ball Bearings Work done on 
 Inclines, including Friction Example II. Questions 101-1x5 
 
 LECTURE XI. 
 
 Difference of Tension in the Leading and Following Parts of a 
 Driving Belt, Brake Horse-Power transmitted by Belts 
 Examples I. II. Velocity Ratios in Lelo Gearing Examples 
 III. IV. Open and Crossed Belts Fast and Loose Pulleys 
 Belt Gearing Reversing Motions Stepped Speed Cones 
 with Starting and Stopping Gear Driving and Following 
 Pulleys in Different Planes Shape of Pulley Face 
 Questions ... 116-129 
 
 LECTURE XII. 
 
 Velocity Ratio of Two Friction Circular Discs Pitch Surfaces 
 and Pitch Circ'es Pitch of Teeth in Wheel Gearing Rack 
 and Pinion Velocity Ratio in Wheel Gearing Example I. 
 Principle of Work applied to Wheel Gearing Examples 
 II. III. Questions 130-139 
 
 LECTURE XIII. 
 
 Sirgle-purchase Winch or Crab Example I. Double-purchase 
 Winch or Crab Example II. Wheel Gearing in Jib-Cranes 
 Questions 140-147 
 
aii CONTENTS. 
 
 LECTURE XIV. 
 
 FAQK8 
 
 Screws The Spiral, Helix, or Ideal Line of a Screw Thread The 
 Screw viewed as an Inclined Plane Characteristics and 
 Conditions to be fulfilled by Screw Threads Different 
 Forms of Screw Threads Whitworth's V-Threads Whit- 
 worth's Tables of Standard V-Threads, Nuts and Bolt 
 Heads Seller's V-Thread The Square Thread The 
 Rounded Thread The Buttress Thread Right and Left- 
 hand Screws The Screw Coupling for Railway Carriages 
 Single, Double and Treble Threaded Screws Backlash in 
 Wheel and Screw Gearings Questions .... 148-159 
 
 LECTURE XV. 
 
 Efficiency, &c., of a Combined Lever, Screw and Pulley Gear 
 Example I. Bottle Screw- Jack Example II. Traversing 
 Screw-Jack Screw Press for Bales Screw Bench Vice 
 Example III. Endless Screw and Worm-WheelCom- 
 bined Pulley, Worm, Worm-Wheel and Winch-Drum 
 Worm- Wheel Lifting Gear Example IV. Questions . 160-173 
 
 LECTURE XVI. 
 
 General Idea of the Mechanism in a Screw-cutting Lathe Motions 
 of the Saddle and Slide Rest Velocity Ratio of the Change 
 Wheels Rules for Calculating the Required Number of 
 Teeth in Change Wheels Examples I. II. Movable Head- 
 stock for a Common Lathe Descriptions of a Screw-cutting 
 Lathe and of an Electrically Driven Hexagon Turret Lathe, 
 with Frontis-plates and complete sets of Detail Drawings 
 Questions 174-206 
 
 LECTURE XVII. 
 
 Hydraulics Definition of a Liquid Axioms relating to a Liquid 
 at Rest Transmission of Pressure by Liquids Pascal's 
 Law "Head" or Pressure of a Liquid at Different Depths 
 Total Pressure on a Horizontal Plane immersed in a 
 Liquid Lord Kelvin's Wire-testing Machine Total 
 Pressure on any Surface immersed in a Liquid Examples 
 I. II. Questions *. 207-213 
 
CONTENTS. Xlll 
 
 < 
 
 LECTURE XVIII. 
 
 tAom 
 
 tJsef ul Data regarding Fresh and Salt Water Example* I. II. III. 
 IV. Centre of Pressure Immersion of Solids Law of 
 Archimedes Floating Bodies Example V. Atmospheric 
 Pressure The Mercurial Barometer Example VI. Low 
 Pressure and Vacuum Water Gauges Example VII. The 
 Siphon Distinction between Solids, Liquids and Gases 
 Definitions of perfect, viscous, and elastic Fluids Cohesion 
 Questions . . 214-226 
 
 LECTURE XIX. 
 
 Hydraulic Machines The Common Suction Pump Example I. 
 The Plunger, or Single-acting Force Pump Example II. 
 Force Pump with Air Vessel Continuous-delivery Single- 
 acting Force Pump without Air Vessel Combined Plunger 
 and Bucket Pump Double-acting Force Pump Example 
 III. Centrifugal Pumps Example IV. Questions . . 227-240 
 
 LECTURE XX. 
 
 Bramah's Hydraulic Press Bramah's Leather Collar Packing- 
 Examples I. II. Large Hydraulic Press for Flanging 
 Boiler Plates The Hydraulic Jack Weem's Compound 
 Screw and Hydraulic Jack Example III. The Hydraulic 
 Bear or Portable Punching Machine The Hydraulic Accu- 
 mulator Example IV. Questions 241-258 
 
 LECTURE XXI. 
 
 Motion and Velocity Uniform, Variable, Linear, and Angular 
 Velocity Unit of Velocity Acceleration Unit of Ac- 
 celeration Acceleration due to Gravity Graphic Re- 
 presentation of Velocities Composition and Resolution 
 
 of Velocities Newton's Laws of Motion Formulae for 
 
 Falling Bodies Formulas for Linear Velocity with 
 Uniform Acceleration Atwood's Machine with Experi- 
 ments Results and Formulae Galileo's and Eater's 
 Pendulum Experiments The Path of a Projected Bodj 
 Centrifugal Force due to Motion in a Circle Cen- 
 trifugal Force Machine Experiments I. II. III. 
 Example I. Balancing High-speed Machinery Cen- 
 trifugal Stress in the Arms of a Fly-wheel Example IL 
 Energy Potential Energy Kinetic Energy Accu 
 mulated Work Accumulated Work in a Rotating Body 
 
XIV CONTENTS. 
 
 The Fly- wheel Radius of Gyration Example III. The 
 Fly Press Example IV. The Energy Stored in a Rotating 
 Fly-wheel Motion on Bicycle and Railway Curves- 
 Momentum Examples VI. to IX. Questions . . 259-296 
 
 LECTURE XXII. 
 
 Some Properties of Materials employed by Mechanics Essen- 
 tial Properties Extension Impenetrability Contingent 
 Properties Divisibility Porosity Density Cohesion 
 Compressibility and Dilatabilifcy Rigidity Tenacity- 
 Malleability Ductility Elasticity Fusibility Load, 
 Stress, and Strain Total Stress, and Intensity of Stress- 
 Tensile Stress and Stress Example I. Compressive Stress 
 and Strain Example II. Limiting Stress or Ultimate 
 Strength Safe Loads and Elasticity Limit of Elasticity 
 Hooke's Law Factors of Safety Modulus of Elasticity 
 Ratio of Stress to Strain Examples III.-V. Resilience 
 or Work Done in Extending or Compressing a Bar within 
 the Elastic Limit Examples VI. -IX. Single Riveted 
 Lap Joints Example X. Questions. . 297-315 
 
 LECTURE XXIII. 
 
 Stresses in Chains Shearing Stress and Strain Example I. 
 Torque or Twisting Moment Torsion of wires Table 
 giving the strength, moduli of Elasticity and Rigidity of 
 various materials Strength of Solid Round Shafts Ex- 
 ample II. Table giving the Horse-Power which steel 
 shafting will transmit at various speeds Strength of 
 Hollow Round Shafts Relation between the Twisting 
 Moment and Horse-Power transmitted by Shafting, as well 
 as the Diameter necessary to transmit a given Horse- 
 PowerExamples III. IV. Questions . . 316-328 
 
 LECTURE XXIV. 
 
 Hooke's Coupling or Universal Joint Double Hooke's Joint 
 Sun and Planet Wheels Cams Heart Wheel or Heart- 
 shaped Cam Cam for Intermittent Motion Quick Return 
 Cam Example Pawl and Ratchet Wheel Reversible 
 Pawl Masked Ratchet Silent Feed Watt's Parallel 
 Motion Parallel Motion Questions .... 329-342 
 
CONTENTS. XV 
 
 LECTURE XXV. 
 
 Reversing Motions Planing Machine Reversing by Friction 
 Cones and Bevel Wheels Whitworth's Reversing Gear 
 Quick Return Reversing Motion Whitworth's Quick 
 Return Motion Whitworth's Slotting Machine Common 
 Quick Return Horizontal Shaping Machine Quick Re- 
 turn with Elliptic Wheels Vertical Slotting Machine 
 Questions . . . 343~354 
 
 LECTURE XXVI. 
 
 Measuring Tools and Gauges Limit Gauges Micrometer Screw 
 Gauge Sir Joseph Whitworth's Early Realisation of 
 Mechanical Accuracy Improved Equivalents Micrometer 
 Gauge A New Set of English Gauges Whitworth's 
 Millionth Measuring Machine Whitworth's Standard 
 Workshop Measuring Machine Questions . . . 355-366 
 
 APPENDICES, Rules and Examination Papers of The Board of 
 Education, City and Guilds of London, and Institution of 
 Civil Engineers for Admission of Students C.G.S. Sys- 
 tem of Units Practical Electrical Units Examination 
 Tables of Useful Constants, Logarithms, Antilogarithms, 
 and Functions of Angles 367-419 
 
 INDEX 421-430 
 
ELEMENTARY MANUAL 
 
 N 
 
 APPLIED MECHANICS. 
 
 LECTURE I. 
 
 CONTENTS. Definition of Applied Mechanics Force Matter Unit of 
 Force The Elements of a Force Graphic Representation of Forces 
 _ Forces in Equilibrium Action and Reaction Resultant and Com- 
 ponents Resultant of Forces acting in a Straight Line Terms and 
 Definitions used in Measuring or Calculating Pressures and their 
 Effects Definitions of Scalar, Vector and Rotor Quantities Engi- 
 neering Calculations. 
 
 Applied Mechanics is that branch of applied science which not 
 only explains the principles upon which machines are designed, 
 made and act, but also describes their construction and applica- 
 tions, as well as how to calculate and test their strength and 
 efficiency. 
 
 Before a student can successfully master any science, he must 
 thoroughly understand the units of measurement that have been 
 adopted in calculating results, and he should also have a clear 
 conception of the exact meaning of the various terms employed. 
 Consequently, we shall commence the study of Elementary Applied 
 Mechanics with definitions and with units of force, work and 
 power.* 
 
 Force is any cause which produces, or tends to produce, motion or 
 change of motion in the matter upon which it acts. 
 
 Matter is anything which can be perceived by one or more of the 
 senses, and which can be acted on by force. 
 
 Matter exists under three conditions : (i) Solids, (2) Liquids, 
 (3) Gases. For example, pieces of wood and of iron are solids ; 
 water and mercury are liquids ; whilst air and oxygen are gases. 
 
 * For the units of length, surface and cubic measure, and for the 
 mensuration of areas and solids, the student is referred to Lectures I. II. 
 and III. of Author's "Elementary Manual on Steam and the Steam 
 Engine," issued by the publishers of this book. *. 
 
 A 
 
2 LECTURE I. 
 
 Bodies are therefore limited portions of matter. When the 
 resistance to motion of a body is equal to or greater than the 
 force applied, so that no motion takes place, the body is said to 
 be subjected to pressure. 
 
 Solids do not yield readily to pressure, for they tend to retain 
 their original shape and size, whereas liquids and gases yield to 
 a very slight pressure, and consequently possess no definite shape. 
 A gas differs from a liquid since it possesses the property of 
 indefinite expansion. A liquid has therefore a definite size, but 
 not a definite shape, whilst a gas has neither definite shape nor 
 definite size. 
 
 Unit of Force. The British unit of force is the pound avoir- 
 dupois, or GRAVITATION UNIT or ENGINEER'S UNIT. The magni- 
 tude of a force is therefore reckoned by the number of pounds of 
 matter which the force would support against gravity. For 
 example, a force of i Ib. means that force which would just lift 
 the weight of a fixed mass at a fixed place if acted on by gravity 
 &lone. But the force of gravity varies at different parts of the 
 earth's surface, being slightly greater at the Poles than at the 
 Equator. Consequently, our engineer's unit of force is only an 
 absolute or invariable one at a standard place, such as at Green- 
 wich sea level. 
 
 The Elements of a Force. When a force acts upon a body, 
 then, in order to fully determine its effect we must know the 
 thre following elements: (i) The point or place of application 
 of the force. (2) The direction in which the force acts. (3) The 
 magnitude of the force. 
 
 (i) Place of Application. In the case of the force of gravity 
 acting on a body, the place of application may be considered to be 
 the whole mass of the body, or we may estimate the whole weight 
 of the body as concentrated at one point, termed the centre of 
 
 * Where great accuracy of measurement is required an absolute or 
 invariable unit offeree must be selected. An absolute unit of force may be 
 defined as that force which, acting for unit time on unit mass, will produce 
 unit change of velocity. If the units of time, mass, and velocity be the 
 second, pound, and foot per second respectively, then we may define the 
 absolute unit of force (called the poundal) as that force which, acting for one 
 second on a mass of one pound would produce a change in velocity of one foot 
 per second. It has been determined experimentally that if a body be let 
 fall freely in vacuo, near the earth's surface, the attractive force of the 
 earth will produce a change of velocity every second of g ( = 32^2 nearly) 
 feet per second. Clearly, then, the gravitation unit is g times the abso- 
 lute unit. Hence the following relation between the gravitation and 
 absolute units of force: A force of one pound = g poundals, or a force 
 of one poundal = ifg pound. 
 
 In this book the gravitation or the engineer's unit of force and work 
 will be used. 
 
ELEMENTS OF A FORCE. ( 3 
 
 gravity of the body. When an extended surface is subjected to 
 pressure, as in the case of a tank containing a liquid, or the piston 
 of an engine subjected to the pressure of a gas, the whole area 
 under pressure may be considered as the place of application. 
 When a body is pulled by means of a rope, or pushed by means 
 of a rod, or supported on a small area, then we consider the force 
 as acting at a point. 
 
 (2) Direction. The direction of a force is the line or path in 
 which it tends to move the body on which it acts. 
 
 (3) Magnitude. The magnitude of the force is the pound* 
 pull or pressure which the force exerts upon the body on whicii 
 it acts. 
 
 Graphic Representation of Forces. When a force acts on 
 a body at a point, its three elements may be conveniently repre- 
 sented as follows : 
 
 , 
 j ' 1\> ' 20 P 
 
 SCALE DIAGRAM OF A FORCE. 
 
 Where represents the point of application, the straight line 
 OP (with the arrow-head), shows the direction in which the force 
 acts, and the length of the divided line OP indicates to scale the 
 magnitude of the force.* (See end of this lecture for further 
 definitions.) 
 
 Forces in Equilibrium. (i) When any number of forces 
 acting upon a body neutralize each other's effects (i.e., leave the 
 body in the same condition as to rest or motion as before th 
 application of the forces), these forces are said to be in equi- 
 librium. 
 
 ,(2) Forces which are in equilibrium may be applied to or 
 removed from a rigid body without altering its condition as to 
 rest or motion. 
 
 (3) Two equal and opposite forces destroy each other's effects ; 
 and, conversely, no two forces can destroy each other's effects un- 
 less they are equal and opposite. 
 
 (4) A force will have the same effect at whatever point in its 
 own direction it may be supposed to act ; and, conversely, if a 
 force have the same effect whether it act at one or other of two 
 given points, then the straight line joining these points (with the 
 suitably directed arrowhead) will be the direction of the force. 
 
 Action and Reaction. (i) Whenever a fixed rigid body is 
 
 * In the case of the above figure the force is represented as equal to 
 22 Ibs. Students will find it convenient to plot down the repiesentation 
 of forces in their exercise books to a scale of ^ of an inch to a pound, 01 
 hundredweight, or ton, according to the values of the forces. 
 
4 LECTURE I. 
 
 acted on by a force, then naturally there is at once set up in that 
 body a secondary force, or a force of reaction, equal and opposite 
 in direction to the primary force. 
 
 (2) Hence action and reaction are equal and opposite, and 
 neutralize each other's effects. 
 
 For Example. Suppose a weight is placed on a rigid horizontal 
 table. In the table there is set up an opposing force or upward 
 reaction which exactly counterbalances the downward force of the 
 weight. If this were not the case, then motion would take place, 
 and either the table would give way, or the weight would sink 
 through the table ! 
 
 Resultant and Components. (i) If any number of forces 
 acting upon a body be replaced by a single force which shall have 
 the same effect, then this force is termed the resultant of these 
 forces, and the forces are called the components of their resultant. 
 
 (2) The operation of finding the resultant of any number of 
 forces is called the composition of forces; and finding the com- 
 ponents is termed the resolution of forces. 
 
 Resultant of Forces acting in a Straight Line. (i) The 
 resultant of any number of forces acting in the one direction 
 along one straight line is equal to their sum, and acts in that 
 direction. 
 
 For Example. Let P 1 P,P 3 P 4 be any four forces acting in one 
 direction along one straight line, then their resultant 
 
 R = P! + P, + P 3 + P 4 
 
 (2) If the forces do not all act in one direction, then the re- 
 sultant is equal to the difference between the resultant of those 
 acting in one direction and the resultant of those which act in 
 the opposite direction, and has the direction of the greater of the 
 two resultants. 
 
 For Example. Let Pfff 4 be any four forces acting along 
 one straight line to the right hand or in a positive direction ; and 
 QiQaQ ^ e an y three forces acting along the same straight line, 
 but in an opposite or left-hand or negative direction, and let 
 
 Qi + Q, + Q 3 be less than p i + p + p i + P 4 
 
 Then the resultant, 
 
 R = (P, + P, + P, * P 4 ) - (Q, + Q, + Q 8 ) 
 and acts in the same direction as P t P 2 P 3 P 4 , and along the same 
 straight line. 
 
 If equilibrium existed between these two sets of oppositely 
 directed forces, then their algebraical sum would be zero, or the 
 resultant would vanish ; t.e., 
 
 (P, + P, + P, + P 4 ) - (Q, + Q, + Q,) - K - O 
 
TERMS, ETC., USED IN MEASURING PRESSURES. C 
 
 A familiar illustration of the above reasoning is the game of 
 " the tug of war," when, say, a batch of sailors are pitted against 
 a corresponding number of soldiers, each batch pulling their 
 utmost at the opposite ends of a rope, and in opposite directions, 
 with the view of obtaining a resultant. 
 
 Terms aid Definitions used in Measuring or in Calcu- 
 lating the Values of Pressures and their JJffects. The only 
 terms (except in special cases) which are herein used for the above pur- 
 poses, are : ^ 
 
 (1) Pressure; (2) Total Pressure ; (3) Intensity of Pressure ; (4) Resultant 
 Pressure; (5) Centre of Pressure. 
 
 DEFINITIONS. (i) Pressure is a general term for the value or amount 
 of force acting between bodies. In certain cases the words push or pull 
 may be used. 
 
 Note. The general term Pressure has hitherto been loosely used t* 
 mean either the Total Pressure (P) or to indicate the Intensity of Pressure 
 (p). Hence it is desirable to clearly state in all problems and writings 
 which of these is meant when the term pressure is alone used. 
 
 (2) Total Pressure (P) means the whole force acting between bodies. It 
 is usual to measure or estimate total pressure in Ibs., thus, P=pA. Ibs. 
 But, when the values are great, it may be given or found in cwts., or in 
 tons. 
 
 (3) Pressure or, more exactly, Intensity of Pressure (p) is the pressure per 
 unit area ; (e.g.) Ibs. per square inch or Ibs. per square fooc. Ex. Let p = 
 10 Ibs. per sq. in. ; A=ioo sq. ins. Then P =pA> = lox 100=1000 Ibs. 
 
 (4) Resultant Pressure (P R ) is the mean pressure per unit area (p) multi- 
 plied by the total area (A) under consi Oration. Here, P R =^A. 
 
 (5) Centre of Pressure (Pc) is the point at which a single force will 
 balance the total pressure. Or, it may be defined as the point at which the 
 resultant pressure acts. 
 
 Differences in Nomenclature. Students may find in "Examina- 
 tion Questions," or in papers read before various Institutions, and in 
 books by different authors, other terms than those just defined. For 
 example, the word Thrust has been recently introduced to mean the total 
 pressure (P); and the simple word pressure to indicate the Intensity of 
 Pressure (p), or Pressure in Ibs. per square inch (p), as defined above. 
 
 Moreover, to show the necessity for a more definite distinction between 
 total and intensity of pressure, if you ask an Engineer or Manufacturer 
 the pressure of steam in his boiler, he will say (e.g.) 100 Ibs. ; when it is 
 actually 100 Ibs. per square inch by the steam gauge, i.e., where^; = ioolbs. 
 
 Resultant Thrust is sometimes used for the aforementioned Resultant 
 Pressure or Pressure Resultant. 
 
 The Author believes that these new terms only confuse 
 students. It has hitherto been the custom amongst Mechanical 
 Engineers and Naval Architects to chiefly confine the words Thrust 
 and Resultant Thrust to problems and cases dealing with the 
 screw propellers and the thrust blocks in connection with the 
 screw shafts of steamships. 
 
DEFINITIONS SCALAR, VECTOR, AND ROTOR. 
 
 NOTE FOB PAGE 3. The word Vector was used in defining the conditions 
 of equilibrium in frames, consequently it may be as well to define the 
 following terms here : 
 
 Scalar. A quantity which has no relation to definite direction in space, 
 or which is considered apart from such direction, is called a "Scalar " or 
 " Scalar-Quantity." 
 
 Vector. A geometrical quantity which is related to a definite direction 
 in space is called a "Vector" or "Vector-Quantity." 
 
 Vector-Quantity. This requires for its complete determination (i) the 
 magnitude, (2) the direction, and (3) the sense to be given. 
 
 A vector-quantity may be geometrically represented by a line, if 
 
 (1) The length of the line represents to scale the magnitude of the quantity. 
 
 (2) The line be placed in the proper direction. 
 
 (3) The proper sense or way be given to the line. 
 
 The sense is usually indicated by an arrowhead on the line. 
 The line itself with its direction and sense is called a Vector-}- 
 Suppose that a force of known magnitude acts along a line from P to 
 ->-Q ; then, the Vector is written down as PQ7 with a bar- line over the two 
 letters P and Q. 
 
 Any quantity, whether scalar or vector (considered as occupying a definite 
 position in space), is t-aid to be localised. Thus the mass of a body in a 
 given position is a localised scalar, and a force acting on a body at a 
 definite point is a localised vector. 
 
 Vector /Sum. The sum of a number of vectors is often called the 
 Kesultant Vector, and in relation to this resultant the other Vectors are 
 called Components. 
 
 To add a number of vectors, place the first anywhere, the beginning of 
 the second to the end of the first, and so on, then the vector from the 
 beginning of the first to the end of the last is the SUM OF THE GIVEN 
 VECTORS (Henrici and Turner}. 
 
 Rotor. A localised vector is called a Rotor (Clifford}. 
 
 EXAMPLE. If A, B, C, D denote four vectors 
 acting at the point to find the sum or result- 
 ant. 
 
 From any convenient point a, draw ab parallel in 
 direction and containing as many units of length as 
 there are units in the vector A. From the end b, 
 draw be parallel to B, and equal in magnitude to 
 it. Similarly, cd parallel and equal to C, and de 
 equal and parallel to D. The resultant in direction 
 and magnitude of the four vectors is the dotted 
 line joining the initial point a to the final point e. 
 ,- ->a If a vector E, equal in magnitude but in the 
 
 /.'' \ c contrary direction, acts at O, then the five vectors 
 
 S \ A, B, C, D, E will have a sum equal to zero, or the 
 
 five vectors are in equilibrium. 
 
 The polygon abcdea, when applied to forces is 
 called the polygon of forces. Hence, if any number 
 of vectors act at a point and can be represented 
 by the sides of a closed polygon taken in order, the 
 
 SUM OP VECTORS 
 
 ACTING AT A POINT, vectors are in equilibrium. 
 
NOTES ON ENGINEERING CALCULATIONS. 5& 
 
 We shall return to the graphic representation of forces, &c., 
 when we come to deal with the parallelogram and triangle of 
 forces and their application to ascertaining the stresses on simple 
 structures.* 
 
 Note Regarding Engineering Calculations. Engineering 
 
 students should clearly understand, that there is no necessity or advantage 
 to be gained in working out their arithmetical results to a greater nicety, 
 than the tool.-', rules, gauges, and instruments placed at their disposal 
 will enable them to measure with accuracy. 
 
 For example, a skilled mechanic who is furnished with a steel footrule 
 and callipers may express his ideas of length to the Vt r a t best to the 
 rH (' OI ) of an inch, which is equivalent to i part in 1200. It would be 
 ridiculous, therefore, to ask him to calculate such lengths to the third 
 decimal place. 
 
 A carpenter who uses the well-known 3 ft. rule may be perfectly 
 satisfied if he measures to the ^ of an inch, i.e., to I part in 500. 
 
 A mason will usually be satisfied if he can measure to within of an inch 1 
 
 The captain of a sailing vessel could not be expected to spot the position 
 of his ship at sea, to within a couple of nautical miles ; so it is no use to 
 ask him to place his boat on a particular meridian 1 
 
 Note on Questions in Proportion. When dealing with all such 
 questions, it is best for the student to ask himself 
 
 (1) What is required? Then to put the corresponding given value in 
 
 the 3rd term, with x, y, or 2 in the 4th for the value to be found. 
 
 (2) Will the answer be greater or less than the value in the 3rd term T 
 
 Then to put the greater or the less given value (according to this 
 answer) into the 2nd term, and the remaining known quantity 
 in the ist term. 
 
 The answer for x, y, or z is then equal to the product of the 2nd and yd 
 terms, divided by the ist. 
 EXAMPLES : 
 
 ist : 2nd : : $rd : 4th (Term*). 
 (a) If, 10 : 100 :: 1000 : x. 
 
 Then , 
 (6) If, loo 
 
 This method is more convenient to the elementary student than dealing 
 directly with fractional ratios. 
 
 * We have intentionally made this Lecture a short one, and have not 
 appended any question*, "because at the first meeting of a session the 
 Lecturer has to give a series of general instructions to his students, and 
 the class is seldom so completely formed as to make it worth while setting 
 any home work until the second meeting. 
 
LEOTTJBE II. 
 
 CONTENTS. Work Unit of Work Examples I. II. III. IV. Work done 
 against a Variable Kcsistance Example V. Diagrams of Work 
 With Uniform Resistance With a Uniformly Increasing Resistance 
 With a Uniformly Decreasing Resistance With a Combination of 
 Uniform and Variable Loads Example VI. Power or Activity 
 Units of Power The Horse-power Unit To find the Horse-power of 
 any working Agent Example VII. Uses of Squared Paper Clark's 
 Adjustable Curve Example VIII. Questions. 
 
 Work. If a force acts upon a body and causes that body to 
 move through a distance, then the force is said to have done work. 
 It does not matter how long the operation takes, whether a second, 
 a minute, an hour, or a day, or even a year, the same amount of 
 work is done by the force acting through the distance. Time, 
 therefore, does not come into the question of estimating work 
 done, but we must have a force overcoming a resistance through 
 a definite distance. If the force applied be inadequate to over- 
 come the resistance of the body to motion, then no work is done. 
 The amount of work done therefore depends solely upon the pro- 
 duct of the force applied (or the resistance overcome) and the 
 distance through which it acts in its own direction. 
 
 Or, Work = Force x Distance. 
 
 Unit of Work.* The unit of work, is the work done in over- 
 coming unit force through unit distance. Tow, since the British 
 unit of force is th^ pound, and unit distance the foot, the British 
 unit of work is called the foot-pound, and is therefore the work 
 done when a resistance of i Ib. is overcome through a distance of 
 i foot. 
 
 EXAMPLE I. If a weight of i Ib. be elevated a vertical dis- 
 tance of i ft. against the force of gravity, then i foot-pound of 
 
 * In the case of heavy work the unit foot-ton is sometimes used in this 
 country. A foot-ton simply means one ton raised one foot high against 
 gravity, or a force of one ton exerted through a distance of one foot, or a 
 resistance of one ton overcome for a distance of one foot. In Electrical 
 Engineering the unit of work is the work done in overcoming a resistance 
 of one dyne through a distance of one centimetre. It is called the Erg. 
 Since the weight of i gramme is = 981 dynes, the work done in raising 
 i gramme through a vertical height of i centimetre against the force of 
 gravity is 981 ergs or (g) ergs. One foot-pound i '356 x io 7 ergs. 
 
UNIT OF WORK. 
 
 work has been performed. If 10 pounds be elevated vertically 
 through a distance of 10 ft., then result is (10 x io)= 100 ft.-lbs. 
 of work. 
 
 Fig. for Example III. 
 
 UNIT OF WORK. 
 
 W = i lb. weight. 
 R = i Ib. reaction. 
 
 ILLUSTRATING WORK DONE. 
 
 W = Weight in Ibs. 
 P = Pull in Ibs. 
 L = Length in feet. 
 
 EXAMPLE II. If a body offers a constant resistance to motion 
 in any direction of P Ibs., and if it be forced along a distance 
 of L ft., in that direction, then tbo work done is P x L ft.-lbs. 
 Or, Work done = Force x Distance 
 
 i.e. Foot pounds = P Ibs. x L feet. 
 
 Suppose a cart with its load weighs W Ibs. and offers a con- 
 stant resistance of P Ibs. to traction along a road, and that it is 
 pulled through a distance of L feet ; then, 
 
 The work done = P x L (ft.-lbs.) 
 
 EXAMPLE III. In drawing a loaded cart along a level road, a 
 horse has to exert a constant pull of 100 Ibs. ; how much work 
 will be done in 10 minutes supposing the horse to walk at the 
 rate of 6000 yards an hour ? 
 
 Distance in feet through 
 which the resistance of 
 100 Ibs. is overcome in 10 
 minutes. 
 
 _6ooo (yds.) x 10 (m.) 
 60 (m.) 
 
 = 3000 ft. 
 
 3 (ft-) 
 
 Work done in 10 minutes = P x L. 
 
 = 100 x 3000. 
 
 = 300,000 ft.-lbs. 
 
 EXAMPLE IV. A traction engine is employed to draw a loaded 
 waggon along a level road where the resistance to be overcome is 
 
8 
 
 LECTURE II. 
 
 100 Ibs. per ton. How many foot-pounds of work are expended 
 in drawing 10 tons over 100 yards ? 
 
 TRACTION ENGINE AND LOAD. 
 
 1. Tractive force = 100 Ibs. per ton. 
 
 2. Total pull, P, = 100 (Ibs.) x 10 (tons) 
 
 3. Distance, L, - 100 (yds.) x 3 (ft.) 
 
 4. Work done = P x L. 
 
 = IOO X IO X 100 X 3. 
 
 300,000 ft.-lbs. 
 
 Work Done against a Variable Kesi stance. If the resist- 
 ance varies whilst the force overcom- 
 ing it acts through a known distance, 
 then the work clone will be measured 
 by the product of the average resist- 
 ance and the distance. If the resist- 
 ance varies uniformly, its average can 
 be found by adding its values at the 
 commencement and end of the motion, 
 and dividing by two. 
 
 EXAMPLE V. Explain the method of 
 estimating the work done by a force, 
 awid define the unit of work. The sur- 
 face of the water in a well is at a depth 
 of 20 feet, and when 500 gallons have 
 been pumped out, the surface is lowered 
 to 26 feet. Find the number of units 
 of work done in the operation, the 
 weight of a gallon of water being 10 Ibs. 
 (S. and A. Exam. 1887.) 
 For an answer to the first part of this question refer to the 
 previous part of this lecture. 
 
 WORK VARYING 
 UNIFORMLY. 
 
 i. Weight of water raised 
 
 = weight of 500 gallons. 
 = 500 x 10 Ibs. 
 
 Or, 
 
 P, =5000 Ibs. 
 
DIAGRAMS OF WORK. 
 
 [Distance through which the 
 
 2. Mean height water is lifted = -j centre of gravity, G (of raised 
 
 [ water), has been elevated. 
 
 Or, 
 
 3. Work done 
 
 20 + 26 
 2 
 
 21 ft. 
 
 ft. 
 
 5000 x 23. 
 115,000 ft.-lbs. 
 
 " 
 
 , 10 ft. M 
 
 Plbs. 
 I 
 
 \ 
 Bibs. 
 
 DIAGRAM OF UNIFORM WORK. 
 
 Diagrams of Work. (i) Against a Uniform Resistance. If 
 the resistance overcome is uniform, then the work done may be 
 graphically represented by the area 
 of a rectangle. 
 
 To find the work done in over- 
 coming a uniform resistance of 
 5 Ibs. through a distance of 10 ft. : 
 Plot down a vertical line to any 
 convenient scale to represent P (or 
 5 Ibs.) and a horizontal line to the 
 same scale to represent L (or 10 ft.) Then complete the rectangle. 
 
 The area P x L or 5 x 10 = 50 ft.-lbs. of work. 
 
 In the accompanying figure a scale of -^ inch has been used 
 to represent both i Ib. and i ft., consequently each of the SHIM 11 
 squares represents to scale one foot-pound of work. 
 
 (2) With a Uniformly Increasing Resistance, If the resist- 
 ance uniformly increases for ex- 
 ample, in the raising of a length of 
 rope or chain vertically by one end 
 from the ground, then the work 
 done may be graphically represented 
 by the ai ea of a right-angled triangle, 
 where P represents the total weight 
 of chain in Ibs., and L its total 
 length in feet. 
 
 Plbs. 
 
 DIAGRAM OF WORK FOR AN 
 INCREASING RESISTANCE. 
 
 The Total Work done = 
 
 PxL 
 
 ft.-lbs. 
 
 Here the work done per foot of length of chain lifted, uniformly 
 increases from a minimum to a maximum, until the whole rope 
 or chain is off the ground. When any known length, I, has been 
 lifted, then the area enclosed by the triangle whose horizontal 
 side is I, and vertical side p represents the work done. 
 
IO 
 
 LECTURE 
 
 (3) With a Uniformly Decreasing Resistance. If the resistance 
 uniformly decreases, as in the case of 
 winding a rope or chain upon the 
 barrel of a winch or crane, then the 
 work done will also be represented 
 graphically by the area of a right- 
 angled triangle, where P represents 
 the total weight of rope or chain in 
 pounds being lifted at the start, and 
 L its length in feet. 
 
 DIAGRAM OF WORK FOR A 
 DECREASING KESISTANCE. 
 
 Px L 
 
 .. The Total Work done = - ft.-lbs. 
 
 Here the work done per foot of length of chain lifted, gradually 
 diminishes from a maximum at the start to a minimum, when the 
 last foot is being lifted. 
 
 As in Case (2), you can at any time know the work done or still 
 to be done from the scale diagram, if you know the length of chain 
 lifted or to be lifted. 
 
 For example, if I feet have still to come on to the barrel, then 
 the vertical ordinate p on the scale diagram will represent the 
 
 pull being exerted at the time, and consequently i represents 
 
 the work still to be done. 
 
 Or, generally, with any gradually 
 increasing or decreasing resistance 
 the work done is equal to the mean 
 of the average resistance in Ibs. 
 x the distance through which it 
 acts in feet. 
 
 (4) With a Combination of Uni- 
 form and Variable Loads. "When 
 one part of a load is uniform and 
 another part variable, as in the case 
 of lifting a weight with a chain, 
 by winding the chain on the barrel 
 of a winch or crane, the diagram 
 of work for the uniform load is 
 naturally a rectangle, and for the 
 chain a triangle if the chain is 
 completely wound on to the barrel, 
 or a trapezoid if there is still some 
 portion of it to be lifted.* 
 
 * See p. 5 of the Author's " Elementary Manual on Steam and the Steam 
 Engine" for how to find the Area of a Trapezoid. 
 
 ^ 
 
 1 
 
 X 
 
 N. 
 
 n 
 
 \1 
 x x 
 
 (1 
 
 It 
 
 m 
 
 N. 
 
 
 L 
 
 
 -> 
 
 
 
 
 
 Q 
 
 K 
 
 M 
 
 DIAGRAM OF WORK FOR A 
 COMBINATION OF UNI- 
 FORM AND VARIABLE 
 LOADS. 
 
DIAGRAMS OF WORK. II 
 
 Let P . the uniform pull required to lift the load *r over- 
 
 come the uniform resistance. 
 L . = the distance the weight is lifted. 
 Pv Pi = t^e weights of chain hanging at the commence- 
 
 ment and at the finish of the lift. 
 Work done in lifting the uniform load = P x L 
 
 Work done in lifting the variable load =*- ^? x L 
 
 .-. Whole work = P x L+Li x L= 
 
 2 2 
 
 = Area of the figure, DEFC. 
 
 The diagram DEFC represents the work done and also 
 the variation of the resistance during the lift. The rectangle 
 ABCD represents the work done in overcoming the uniform load, 
 and the trapezoid ABFE the work done in overcoming the 
 variable load. The resistance at any instant of the lift will be 
 represented by the vertical line drawn from the horizontal base 
 DC to the inclined line EF through the point on DC or AB 
 which represents the position of the load at that instant. The 
 part of this vertical fine intercepted between AB and EF will 
 represent the resistance offered by the variable part of the load 
 at the instant considered. Thus, at the commencement of the 
 lift the total resistance is P +p l and represented by DE, at tho 
 end of the lift the total resistance is P + p, and represented by 
 CF. At , J, and j of the lift the total resistances are repre- 
 sented by the vertical lines GH, KL, and MN respectively, whil i 
 the resistances due to the variable part of the load at these poh-ts 
 are represented by the lengths grH, L, and rriN respectively 
 If the final resistance due bo the variable part of the load was 
 zero (as would be the case if the whole of the chain were wound 
 on to the barrel) then the diagram of work for this part of the 
 load would be the triangle AEB. 
 
 EXAMPLE VI. Explain fully the mode of measuring the work 
 done by a force. What unit is adopted ? A weight of 2 cwts. 
 is drawn from a mine, 30 fathoms deep, by a chain weighing i lb. 
 per linear foot ; find the number of units of work done. (S. and 
 A. Exam. 1893.) Also find the resistance offered when the 
 weight has been raised through J, ^, and j of the whole depth 
 respectively. 
 
 ANSWER. (i) The work done by a force is measured by the 
 
 product of the force into the distance throagh whieh that force 
 
 moves in its own direction. If P be the force in pounds, and L the 
 
 distance in feet through which it moves in its own direction, then 
 
 Work done- Px L f t, Ibi, 
 
12 LECTURE II. 
 
 (2) The unit of work adopted in this country is the work done 
 when a force of one pound is moved through a distance of i foot, 
 and is called the foot-pound (ft.-lb.). 
 
 (3) Referring to the previous figure, make AB to represent 
 30 x 6 = 1 80 ft., the depth of the mine, AD to represent 2x112 
 = 224 Ibs. the weight of material raised, AE to represent 
 i x 1 80= 1 80 Ibs. the weight of chain at beginning of lift. Then 
 assuming the whole of the chain to be wound up, complete the 
 rectangle ABCD, and join E and B. The area of the figure 
 DEBO then represents the work done. 
 
 . . Work done = area DEBC - \ (DE + CB) x AB. But DE = 
 DA + AE = 224 + 180 = 404 Ibs. 
 CB = DA =224 Ibs., AB = 1 80 ft. 
 .".Work done = \ (404 + 224) x 180 ft.-lbs. = 56,520 ft.-lbs. 
 
 (4) The resistance at \ lift, or when the weight has been raised 
 45 ft., is Gh = Gg + gh = 224 + $x 1 80 = 359 Ibs. 
 
 At ^ lift the resistance is K =K& + kl = 224 + Jx 180 = 
 314 Ibs. 
 
 At f lift the resistance is M?i = Mm4- mn= 224 + | x 180 = 
 269 Ibs. 
 
 Power or Activity is the rate of doing work* In estimating 
 or testing the power of any agent the time in which the work is 
 done must be noted and taken into account. Consequently, we 
 speak of the activity or power of a man, of a horse, or of an 
 engine, as capable of doing so many foot-pounds of work per 
 minute. 
 
 Units of Power, f The unit of power adopted in this country 
 is called the horse-power. It is the rate of doing work at 33,000 
 ft.-lbs. per min., or 550 ft.-lbs. per sec., or 1,980,000 ft.-lbs per 
 hour. 
 
 The Horse-power Unit was introduced by James Watt, the 
 great improver of the steam engine, for the purpose of reckoning 
 the power developed by his engines. He had ascertained by 
 experiment that an average cart-horse could develop 22,000 foot- 
 pounds of work per minute, and being anxious to give good value 
 to the purchasers of his engines, he added 50 per cent, to this 
 amount, thus obtaining (22,000+ 11,000) the 33,000 foot-pounds 
 per minute unit, by which the power of steam and other engines 
 has ever since been estimated. 
 
 * The vrordpoweris very frequently misapplied by writers and students, 
 for they often call the mere pull, pressure, or force exercised on or by an 
 agent the power. Students should strenuously avoid this misuse of the 
 word power, and never employ it in any other sense than as expressing a 
 rate of doing work, or activity. 
 
 t In Electrical Engineering the Unit of Power is called the Watt, and it 
 equals lo 7 ergs per second, or 746 Watts i horse-power. 
 
TO FIND THE HORSE-POWEK OF ANT WOKKTNO AGENT. 1 3 
 
 To find the Horse-power of any Working Agent. Divide 
 the number of foot-pounds of work which it does in one minute by 
 33 000 - 
 
 Let P = Pull exerted or resistance overcome in pounds. 
 L = Length or distance through which P acts. 
 M = Minutes the agent is at work. 
 H.P. = Ilorse-power. 
 Then, 
 
 H P - FxL p = II. P. x 33000x11 . L _ H. P. x 33000 x M 
 33000 x M ' L P 
 
 EXAMPLE VII In what way is the rate of doing work measured 
 hi horse-power ? 
 
 If 40 cubic feet of water be raised per minute through 330 feet, 
 what horse-power of engine will be required, supposing that there 
 is no loss of friction or other resistances ? Note. I cubic foot of 
 water weighs 62 J Ibs. (S. and A, Exam. 1892). 
 
 ANSWER. The rate of doing work, as measured in horse-power, 
 is equivalent to 33,000 foot-pounds of work done per minute. 
 
 ist. i cubic foot of water weighs 62 J Ibs. 
 
 .40 cubic feet of water weigh 40x62^ = 2500 Ibs. 
 
 2nd. Work done per minute = ^ c 2500 (Ibs.) x 330 ft. 
 
 M I 
 
 3 rd. .-. H.P.= PxL = 2500 x 330 = 825000 = 25> 
 
 33000 x M 33000 x i 33000 
 
 Note. Students will find it a great advantage, as well as a sav- 
 ing of time not to multiply figures together until the last stage of 
 the answer has been reached, and then to cancel all common 
 factors in numerator and denominator. For example, in the 
 answer to the above question we might proceed thus 
 i st. 40 cubic feet of water = 40 x 62 \ Ibs. 
 2nd. Work done per minute = 40 x 62! x 330 ft. -Ibs. 
 
 4 i 
 
 40x62^x^0 
 3rd. .. Horse-power of engine = - 
 
 10 
 
 10 10 
 
 The process consists in this the factor, 330, can be cancelled 
 from numerator and denominator, leaving 100 as the denominator. 
 The factor, 10, can then be divided out of 40 in the numerator 
 and from the 100 in the denominator, thus leaving 4 x 62 J as the 
 numerator and 10 as the denominator. The remainder of the 
 work is evident. 
 
LECTURE H. 
 
 Uses of Squared Paper. Squared paper is made by drawing 
 a number of equally spaced horizontal lines and crossing these by 
 vertical ones at the same distance apart. The paper is conse- 
 quently covered with a large number of little squares, the sides of 
 which are usually one-tenth of an inch in length. In order to 
 facilitate the measurement of distances, every tenth line, and some- 
 times every fifth, is heavier or of a different colour to the others. 
 
 By the aid of squared paper we can graphically represent how 
 two varying quantities depend upon each other. For example, 
 take the case of a chain being gradually lifted from the ground, as 
 already considered in connection with diagrams of work and 
 shown by the first of the following figures. The length and 
 weight of chain lifted will alter as the upper end is raised ; but 
 the suspended weight will always depend upon the length which 
 hangs freely from the spring balance. In fact, any change in the 
 length lifted will produce a corresponding change in the load 
 registered by the balance. If we note the pull indicated by the 
 balance for different lengths of hanging chain, we shall be able to 
 obtain a line or curve which will show to the eye how these two 
 quantities depend upon one another. Suppose we obtain the 
 following results : 
 
 Length of "^ 
 chain lifted V 
 in feet. J 
 
 I 
 
 2 
 
 3 
 
 3'5 
 
 4 
 
 S 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 Weight on 
 
 
 
 
 
 
 
 
 
 
 
 
 chain lifted V 
 
 0-8 
 
 i '4 
 
 2'0 
 
 2 '5 
 
 27 
 
 3'S 
 
 4 '3 
 
 .TO 
 
 5 '6 
 
 6'2 
 
 7-0 
 
 in Ibs. J 
 
 
 
 
 
 
 
 
 
 
 
 
 Corrected ^ 
 values in > 
 Ibs. J 
 
 07 
 
 I '4 
 
 ,, 
 
 - 
 
 2'8 
 
 3'5 
 
 4*2 
 
 4 '9 
 
 , 
 
 6-3 
 
 7-0 
 
 We shall represent the lengths of the hanging chain by horizon- 
 tal distances which are termed abscissa, and the corresponding 
 weights by vertical distances called ordinates. We choose such a 
 scale for these quantities as will enable us to get them all upon 
 the squared paper ; at the same time we keep the scales as large 
 as possible. In this case we have chosen five divisions horizontally 
 to represent one foot, and five divisions vertically for one pound. 
 It is, however, not necessary to adopt equal scales for abscissae 
 and ordinates, but we should select the most convenient scale for 
 each according to circumstances. 
 
USES OF SQUARED PAPER. I 5 
 
 To find the point corresponding to the fourth column in the 
 table, take B at 3-5 on the horizontal scale, and C at 2-5 on the 
 vertical one. Draw B A vertically and C A horizontally. Then, 
 the intersection of these two lines is the point required. In 
 practice, these lines are not actually drawn, but the point is 
 found by the eye with the assistance of the lines on the paper, and 
 a x or is placed to mark its position. When all the points have 
 been thus plotted from the table, we draw a mean line or curve 
 between them. In this case, it is a straight line passing through 
 
 Abscissae or Length 
 
 Femt 
 
 HANGING CHAIN. SQUARED PAPER. 
 
 RELATION BETWEEN LENGTH AND WEIGHT OP CHAIN. 
 
 the origin 0. It will be seen that this line does not pass through 
 all the points, but that some of them are on one side and some on 
 the other. This may be due to errors of observation or to 
 irregularities in the chain. If we know that the chain is uniform, 
 the points ought to lie along the straight line we have drawn, 
 and we can correct for errors of observation. Thus, the point A 
 should have been at D, and the correct weight for that length of 
 chain is represented by BD, which is 2*45 Ibs. We can correct 
 the other values in the same way, and so obtain the numbers 
 shown in the third line of the table. 
 
 When the points lie approximately in a straight line, the 
 nearest mean straight line is best found with the help of a fine 
 thread which is stretched and moved among the points until it 
 
1 6 LECTURE II. 
 
 lies most evenly amongst them. The positions of its ends are 
 then marked and a line drawn with a straight-edge through these 
 marks. When the points do not lie near a straight line a smooth 
 curve may be drawn through them, either freehand, or by aid of 
 French curves, a thin strip of wood or steel, or Clark's paiant 
 
 It ~ 
 
 CLARK'S PATENT ADJUSTABLE CURVE, 
 
 adjustable curve. This consists of a flexible strip of celluloid 
 with a brass loop A, for the thumb of the left hand, and another 
 B, for the second or third finger, as shown by the accompanying 
 figures. Now, if these two loops are drawn together the celluloid 
 will be formed into some curve, the shape of which can be ad- 
 justed by moving the sliding rod C with the right hand and 
 fixing this curve by means of a cord (not shown in the figure) 
 joining B with a V-shaped groove in A. It may also be used as 
 
USES OF SQUARED PAPK*. 
 
 a set curve for the purpose of transferring a curve from one 
 drawing to another. 
 
 When we represent distances along one scale of the squared 
 paper and forces along the other, then the areas such as O D B 
 indicate to scale the work done in raising the length of chain O B. 
 This is true, whether the line D is straight or curved. We 
 can, however, represent any two quantities which depend on one 
 another by a curve on squared paper. 
 
 EXAMPLE VIIT. A body is being acted upon by a variable lifting force. 
 When the body is lifted x feet the force F Ib. is observed. 
 
 X 
 
 
 
 15 
 
 25 
 
 50 
 
 70 
 
 100 
 
 125 
 
 150 
 
 180 
 
 210 
 
 F 
 
 530 
 
 525 
 
 516 
 
 490 
 
 425 
 
 300 
 
 210 
 
 160 
 
 110 
 
 90 
 
 Plot on squared paper and find the average value of .Ffrom x = o to 
 *=2io. What is the work done by F when the body has been lifted 
 210 feet? (B. of E. 1904.) 
 
 ANSWER 
 
 
 
 r~ 
 
 
 
 
 
 
 
 
 
 
 
 i 
 
 
 
 
 
 
 
 
 
 500 
 
 SK 
 
 f : 
 
 si^ij 
 
 
 
 
 i:::::i 
 
 
 
 . 
 
 
 
 i- 
 
 F : 
 
 
 
 
 - 4- 
 
 
 
 
 400 
 
 -^ 
 
 
 | :: 
 
 
 
 
 
 
 
 
 
 r JI 
 
 T 
 
 rTrrHrr 
 
 5p 
 
 
 
 
 
 
 
 
 *- 
 
 t: 
 
 :l::::: 
 
 
 ^ *vj 
 
 
 
 
 
 
 300 
 
 
 
 . t : ...: 
 
 
 Tt-f 1 
 
 
 
 
 
 
 
 i 
 
 w 
 
 
 
 |i 
 
 ^p-:::::::: 
 
 
 
 
 
 ?00 
 
 
 ft 
 
 
 
 
 : jK:i; : :: 
 
 
 
 
 
 
 "* 
 
 
 :|::::: 
 
 - ---[- 
 
 L 
 
 - ^* s 
 
 
 
 
 . , 
 
 100 
 
 k 
 
 
 1 
 
 ::-:(- 
 
 t 
 
 H -^ 
 
 F:|S::| 
 
 a 
 
 ""] ' 
 
 = B 
 
 
 
 
 1 
 
 
 
 7 5^y?c 
 
 es { 
 
 |^S 
 
 S :::: i 
 
 
 
 : C 
 
 20 40 60 80 100 120 140 160 180 - 200 210 
 
 DIAGRAM OP WORK DONE BY A VARIABLE FORCE. 
 
 In order that we may obtain the average value of F, we must nse squared 
 paper since the lifting force is variable. Mark off the distances x in feet 
 (to gcale) along the abscissa C, and at each of the values of , plot 
 ordinates corresponding to the values of F in Ibs. Join all the points by 
 a curve as shown in the above diagram. The area of this diagram below 
 the curved line A B, represents the total work done by the variable force Fibs., 
 in passing through a distance of 210 feet. 
 
 We can obtain the area of the diagram O A B C either by aid of aplanimeter 
 or by inding the average value of F and multiplying this by the distance 
 
1 LECTURE H. 
 
 through which Fhas moved, viz., 210 feet. But the ordinary process used 
 by engineers for finding the area of a diagram such as A B C is to divide 
 9-0 into 10 equal parts and then to measure the value of F at the centre 
 of each part. The sum of these values divided by 10 gives us the average 
 value of F. If this number be now multiplied by the value of C the 
 final result will be the area of the diagram in foot-pounds. 
 
 Thus, by the latter method we find the average value of F=305 Ibs. 
 
 Hence the total work = average value of Fx distance it acts. 
 
 Or, the total work done = 3o$ x 2io-64,050 ft.-lbs. 
 
LECTURE H. QUESTIONS. Ip 
 
 LECTURE II. QUESTIONS. 
 
 1. Define tbe unit of work. What name is given to this unit? In 
 drawing a load a horse exerts a constant pull of 120 Ibs. ; how much work 
 will be done in 15 minutes, supposing the horse to walk at the rate of 
 3 miles an hour 1 Ans. 475,200 ft.-lbs. 
 
 2. How is the work done by a force measured ? The resistance to 
 traction on a level road is 150 Ibs. per ton of weight moved ; how many 
 foot-pounds of work are expended in drawing 6 tons through a distance 
 of 150 yards 1 Ant. 405,000 ft.-lbs. 
 
 3. Distinguish between force and work done by a force. How is each 
 respectively measured ! A traction engine draws a load of 20 tons along 
 a level road, the tractive force on the load being 150 Ibs. per ton. Find 
 the work done upon the load in drawing it through a distance of 500 yards. 
 Ans. 4,500,000 ft.-lbs. 
 
 4. Find the number of units of mechanical work expended in raising 
 136 cubic feet of water to a h/.ight of 20 yards. The weight of a cubio 
 foot of water is 62^ Ibs. Ans, 510,000 ft.-lbs. 
 
 5. A weight of 4 tons is raised from a depth of 222 yards in a period of 
 45 seconds ; calculate the amount of work done. Ant. 5,967,360 ft. -Ibs. 
 
 6. A hole is punched through a plate of wrought-iron one-half inch in 
 thickness, and the pressure actuating the punch is estimated at 36 tons. 
 Assuming that tbe resistance to the punch is uniform, find the number of 
 foot-pounds of work done. Ans. 3360 ft. -Ibs. 
 
 7. How is work done by a force measured ? Give some examples. Set 
 out a diagram of the work done in drawing a body weighing 10 Ibs. up a 
 smooth incline 4 feet high, marking dimensions. 
 
 8. A train of 12 coal waggons weighing 133 tons is lifted by hydraulic 
 power (two waggons being raised at a time) through 20 feet in 12 minutes. 
 Estimate the work done in foot-tons. Taking the average of work done, 
 how many foot-pounds are done per minute? Ant. 2660 ft. -tons ; 
 496,533*3 ft.-lbs. per minute. 
 
 9. The plunger of a force-pump Is 8f inches in diameter, the length of 
 the stroke is 2 feet 6 inches, and the pressure of tbe water is 50 Ibs. per 
 square inch ; find the number of units of work done in one stroke, nd 
 plot out a diagram of work to scale. Ans. 7517 ft.-lbs. 
 
 10. A chain 30 feet long, and weighing 100 pounds per yard, lies coiled 
 on the ground. Find by calculation and by a scale diagram of work how 
 many units of work would be expended in just raising it by the top end 
 from the ground. Ans. 15,000 ft.-lbs. 
 
 11. A chain, weighing 30 Ibs. to the fathom, is employed to lift i ton to 
 a height of 30 ft. by winding the chain on a barrel. Find by calculation 
 and by a scale diagram of work, how many units of work will be expended 
 (a) when the outer end of the chain is brought home to the barrel ; 
 (b) when 18 feet of it are still hanging free with the weight at the end of 
 
 it. Ans. (a) 69,450 ft.-lbs. ; (b) 28,320 ft.-lbs. 
 
 12. Define the following mechanical terms : Force, work, unit of work, 
 power, activity, and horse-power. A horse drawing a cart at the rate of 
 2 miles per hour exerts a traction of 156 Ibs. ; find the number of units of 
 work done in one minute. Ans. 27,456 ft. -Ibs. 
 
 13. In what way is the rate of doing work measured in horse-power ? 
 If loo cubic feet of water be raised per minute through 330 feet, what 
 horse-power of engine will be required, supposing that there is no loss by 
 friction or other resistances 1 A<M. 62'$ h.p. 
 
 14. If a horse, walking at the rate of 2$ miles per hour, draws 104 Ibs. 
 out of a well by means of a cord going over a wheel, how many unite of 
 work would he perform in one minute t Ant. 22,880 fc-lbi. 
 
LECTURE II. QUESTIONS. 
 
 15. What unit do you employ in measuring force, and what unit in 
 measuring the work done by a force ? A horse exerting a pull of 40 Ibs. 
 per ton draws a load of 15 cwt. along a level road ; how far will the horse 
 travel in 10 minutes if he does work on the load at the rate of horse- 
 power ? Ans. 5500 ft. 
 
 16. Distinguish between the expressions "foot-pound" and "horse- 
 power " by giving a clear definition of each. A bucket when filled with 
 water weighs 180 Ibs., and is raised at a uniform rate from a depth of 
 150 feet in eight minutes. Find the work done in one minute. 
 
 An*- 3375 ft.-lbs. 
 
 17. What work in foot-pounds is done in raising the materials for 
 building a brick wall 50' high, 12' long, and 2' 3" in thickness, if one cubic 
 foot of brickwork weighs 112 Ibs. ? Ans. 3,780,000 ft. -Ibs. 
 
 1 8. A man of 150 Ibs. climbs a hill regularly 1200' vertically per hour ; 
 at another time he climbs a staircase at 2^' per second. Find in each case 
 the horse-power expended by the man. Ans. '69 h.-p. ; '68 h.p. 
 
 19. An express train going at 40 miles per hour weighs 150 tons ; 
 the average pull on it is 12 Ibs. per ton, what is the horse-power exerted? 
 This power is only 40 per cent, of the total indicated power of the engine ; 
 find the indicated power. Ans. 192 h.-p.; I. H. P. = 480. 
 
 20. Water at 750 Ibs. per square inch pressure acts on a piston one 
 square foot in area, through a stroke of I foot ; what is the work that 
 such water does per cubic foot ? and per gallon ? If an hydraulic com- 
 pany charges 18 pence for a thousand gallons of such water, how much 
 work is given for each penny ? Ans. 108,000 ft.-lbs. ; 17,280 ft.-lbs. ; 
 960,000 ft.-lbs. 
 
 21. Explain how squared paper is used, and mention a few of the 
 purposes to which it is applied. 
 
 22. Plot out a curve from the following data showing the pressure on a 
 piston at various distances from the commencement of the stroke : 
 
 Distance in feet. 
 
 o 
 
 I 
 
 '2 
 
 '3 
 
 '4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 '9 
 
 IX) 
 
 Pressure in \ 
 Ibs. per 
 square inch. J 
 
 20 
 
 21 
 
 21 
 
 20 
 
 19 
 
 18-5 
 
 18 
 
 I3-5 
 
 9 
 
 4'5 
 
 o 
 
 23. A chain weighing 10 Ibs. per foot of its length is 240 feet long, and 
 hangs vertically ; what work is done in winding the chain upon a drum ? 
 Am. 288,000 ft.- Ibs. 
 
C -21 ) 
 
 LECTURE III. 
 
 CONTENTS. The Moment of a Force Principle of Moments applied to the 
 Lever Experiments I. II. III. Pressure on and Reaction from the 
 Fulcrum Equilibrant and Resultant of two Parallel Forces Oouples 
 Centre of Parallel Forces or Position of Equilibrant and Resultant 
 Centre of Gravity Examples of Centre of Gravity The Lever when 
 its weight is taken into Account Examples I. II. Position of the 
 Fulcrum Example III. Questions. 
 
 The Moment of a Force, with r expect to a point, is equal to the 
 force Multiplied % the perpendicular distance from the point to its 
 line of action. 
 
 For example, suppose a body to be resting on the point O, and 
 a force, P, to be applied to the body in the direction PA. Then, 
 if the perpendicular distance from 
 to the line of action of the force be 
 UA, the moment of the force P, tending 
 fo turn the body about the point O, is 
 P x AO. If the force be reckoned 
 in pounds, and the perpendicular dis- 
 tance in feet, the product will be in 
 pounds-feet. The student must there- 
 fore avoid confusing the answer with 
 ft.-lbs. of work. 
 
 Principle of Moments. If any number of forces act in one 
 plane on a rigid body, and if these forces are in equilibrium ; then 
 the principle of moments asserts that the sum of the moments of 
 those forces which tend to turn the body in one direction about a 
 point, is equal to the sum of the moments of the forces which t#nd to 
 turn the body in the opposite direction about the same point. 
 
 Or, to state the principle more concisely, the opposing moments 
 about the point are equal when equilibrium exists. 
 
 If the moments of those forces which tend to turn the body to 
 the right hand (i.e., in the direction of the motion of the hands 
 of a clock) be called positive ( + ), and the moments of the remain- 
 ing set of forces which tend to turn the body to the left hand (".., 
 in the opposite direction to the movement of the hands of a clock) 
 be called negative ( ), then the algebraical sum of the moments of 
 the forces which act in one plane, and which are in equilibrium 
 about a point, is zero. 
 
22 
 
 LECTURE III. 
 
 Principle of Moments applied to the Lever. A lever is 
 simply a rigid rod, bar, or beam, capable of turning about a fixed 
 
 point called the fulcrum (F). Act- 
 ing on the lever in one direction is a 
 force or set of forces which we shall 
 term the pull or pressure (P), and 
 in the other direction there is the 
 resistance or set of resistances to be 
 overcome, which we shall term the 
 weight (W). The pressure, P, and 
 the weight, W, produce a reaction 
 at the fulcrum, which is called the 
 equilibrant (E). 
 
 The parts of the lever between 
 the fulcrum and the pressure and 
 between the fulcrum and the weight 
 are called the arms of the lever. 
 w The accompanying three figures 
 
 LEVERS IN EQUILIBRIUM. show three ways in which F, P and 
 
 W may be arranged with a straight 
 
 lever.* In each case, the opposing moments about the julcrum are 
 equal, when the lever is in equilibrium. 
 
 Or, . . P x AF = W x BF 
 
 satisfies the conditions for equilibrium in the case of a lever. 
 
 EXPERIMENT I. To prove the foregoing statements, take a 
 rigid homogeneous bar, AB, of uniform section. Let the bar be 
 of yellow pine, i inch deep, ^ inch broad, and 32 inches long. 
 Attach to the ends, A and B, light flexible cords with small hooks 
 at their lower ends, and attach to the middle of the bar at F 
 another light flexible cord, and pass this cord over a pulley having 
 a minimum of friction at its bearings. Fix such a weight to the 
 free end of this middle cord as will just counterpoise the bar and 
 cords. Test the accuracy of this preliminary adjustment by 
 
 * The levers represented by the above three figures are assumed to be 
 without weight. A force, P, acts through a perfectly flexible, weightless 
 cord at A, and another force, W, acts also through an exactly similar 
 cord at B, with the fulcrum at F in each case. In the second and third 
 case the cord attached at A passes over a frictionless pulley in order to 
 give the necessary direction to the force P. These three relative positions 
 of P, W and F used to be termed the first, second and third order of levers; 
 but there is no necessity for any such distinction, since all the student has 
 to remember is this, that when equilibrium exists the opposing moments 
 
 about the fulcrum are equal, i.e., (P x AF = W x BF), or, _ = _, or = 
 
 W AF P BF 
 The ratio W to P is termed the theoretical advantage of the lever. 
 
PRINCIPLE OF MOMENTS APPLIED TO THE LEVER. 
 
 observing whether the bar hangs horizontal, and, if pulled down 
 or up a little, whether the weight balances the bar and cords, 
 Now affix equal weights, P, of, say, 4 oz., to the cords hanging 
 
 P=4 oz. 
 
 EXPERIMENT I. ON PARALLEL FORCES. 
 
 from the ends A and B, and add an equilibrating weight, E, of 
 8 oz. to the end of the central cord. You will find that the bar 
 will come to rest in a horizontal position, thus proving that 
 
 i.e., 
 Or, 
 i.e., 
 
 P 
 
 (oz.) 
 
 P x BF 
 
 
 4 (oz.) x 1 6" 
 
 
 BF : AF 
 
 
 BF 16 4 
 
 i 
 
 AF 76 
 
 i 
 
 AF = 
 
 1 6" = 
 
 P:P :: 
 P 
 P 
 
 Also, that the equilibrant, 
 
 E = P + P 
 8 oz. = 4 oz. + 4 oz. 
 
 If P and P are now removed from the ends A and B, and a 
 single weight, R, of 8 oz. be hung from F (as represented by the 
 vertical dotted line and arrow), the result as far as the balancing 
 of the system is concerned will be unaffected. 
 
 Consequently, R E P + P 
 
 i.e., 8 oz. = 8 oz. 4 oz. + 4 oz. 
 
 Or, the resultant of two equal parallel forces acting in the same 
 direction is equal to the sum of the two forces, and acts midway 
 between them and parallel to them i.e., at the same point as the 
 equilibrant, and in the same line therewith, but in the opposite 
 direction. 
 
LECTURE IIT. 
 
 EXPERIMENT II. Take another rigid homogeneous bar, AB, of 
 the same uniform section as the previous bar, but let its length 
 be 24 inches. Attach cords with hooks to the ends A and B, and 
 t/o a point F, say 8 inches from A and 16 inches from B. Pass 
 this latter cord over the guide-pulley, and fix it there until you 
 
 8'-' 
 
 
 =24 oz. 
 
 Q= ti oz. 
 
 =16 oz. 
 
 EXPERIMENT II. ON PARALLEL FORCES. 
 
 have just added sufficient weight to the end At) balance the 
 longer end BF; then unfix the end of the middle cord, and 
 attach such a weight to it as will counterpoise the whole system. 
 Now attach to the cord at A a weight P = 1 6 oz. ; to the other 
 end, B, a weight Q ; and a weight at E, so as to again balance 
 the whole system. It will be found that Q equals 8 oz. and 
 E equals 24 oz., thus proving that 
 
 Or, 
 
 i.e., 
 
 P x AF = Q x 
 
 1 6 (oz.) x 8'' =8 (oz.) x 
 
 P :Q :: BF : AF 
 
 P _BF 16 2 
 
 Q = "AF 8~ I 
 
 BF 
 
 i6 v 
 
 Or, the point F is twice the distance from the end B that it is 
 from the end A, and P has twice the value of Q. 
 Also, that the equilibrant, 
 
 E = P + Q 
 
 For, 24 oz. = 1 6 oz. + 8 oz. 
 
 If P and Q be now removed from the cords at A and B, and a 
 single weight, R, of 24 oz., be hung from F (as represented by 
 the vertical dotted line and arrow), the result, as far as the balanc- 
 ing of the system is concerned, will be unaffected. 
 
PRESSURE ON, AND REACTION FROM, THE FULCRUM. 25 
 
 Consequently, R - E = P + Q 
 For, 24 oz. = 24 oz. = 1 6 oz. + 8 oz. 
 
 Or, the resultant of any tvto parallel forces acting in the same 
 direction is equal to the sum of the two forces, and acts parallel to 
 them and at a point between them, so that the ratio of the forces is 
 inversely proportional to their distances from the point ; or so that 
 P : Q :: BF : AF 
 
 Pressure on, and Reaction from, the Fulcrum. You may 
 
 also conclude from these two experiments, if the lever had 
 been balanced on a knife-edge or journals, that the pressure on 
 the fulcrum due to the forces P and Q would have been equal to 
 and act in the direction of the resultant R, and that the reaction 
 from the fulcrum would have been equal to and act in the 
 direction of the equilibrant E. 
 
 EXPERIMENT III. Supposing that in the last experiment, after 
 adjusting the lever by placing a counterpoise weight at A, in order 
 to bring the beam to a horizontal position, and after balancing 
 the weight of the beam and cords by an equivalent weight at 
 position E, you added a weight Q, of 8 oz., to the cord at B, and 
 a weight E, of 24 oz., to the cord attached at F, the beam would 
 turn, and would only be brought to a horizontal position by 
 attaching a weight at A of 16 oz. Hence you observe that P 
 acts at A as the equilibrant both in direction and magnitude to 
 the two unequal parallel wdike forces Q and E. Consequently 
 a force equal and opposite in direction to P would be the resultant 
 of the two forces Q^and E ; and it would replace their combined 
 effect on the balanced beam. 
 
 Further, P = E - Q; 
 
 for, 1 6 oz. = 24 oz. - 8 oz. 
 
 And the moments about the position where the equilibrant acts 
 are equal, 
 
 for Q x BA = E x FA 
 
 i.e., 8 (oz.) x 24" = 24 (oz.) x 8" 
 
 Equilrbrant and Resultant of Two Parallel Forces. 
 
 From the above experiments you conclude that the equilibrant 
 and the resultant of any two like parallel forces are equal to 
 their sum, and any two unlike parallel forces are equal to their 
 difference. , . 
 
 Couples. When the two parallel forces are equal and act in 
 opposite directions upon a body, they are termed a couple. Ihe 
 perpendicular distance between the two forces is termed iM 
 arm of the couple," and the " moment of the couple" is the product of 
 
LECTURE III. 
 
 -me of the forces and the arm. A couple simply tends to cause 
 rotation of the body upon which it acts, for it has no resultant, 
 
 snce 
 
 = P - P = 0. 
 
 One couple can, however, be equilibrated or balanced by another 
 oouple of an equal moment, acting x m the.same plane, and tending to 
 
 A COUPLE. 
 
 TWO BALANCING COUPLES. 
 
 turn the body in the opposite direction. In the accompanying 
 figure the couple P,AB,P will be balanced by the couple Q,CD,Q 
 if their moments are equal ; i.e., if 
 
 P x AB = Q x CD.* 
 
 We shall frequently have to refer to practical examples of 
 couples, such as in a ship's capstan, the screw press used in copy- 
 ing manuscript, pressing bales of goods, and the fly press for 
 punching holes in thin plates, or for stamping or embossing 
 metals, &c. 
 
 Centre of Parallel Forces, or Position of Equilibrant 
 and Resultant. From Experiment III. and the accompanying 
 figure to Experiment II., you conclude that the position where 
 the equilibrant and resultant act is such, with respect to the 
 positions where the forces act, that the moments of the forces 
 about that position are equal and opposite in effect upon the 
 lever. 
 
 For, Q x BA = E (orK) 
 
 FA; or, = 
 
 E BA 
 
 * Let P = -8 Ibs., and Q = 10 Ibs. : AB = 10 ft. and CD = 8 ft. 
 Then, P x AB = Q x CD 
 Or, 8 x no =: 10 x 8 
 
 80 = So 
 
CENTRE OF PARALLEL FORCES. 
 
 27 
 
 
 
 
 8 
 
 8 
 
 i 
 
 X 
 
 8fl 
 j 
 
 or, 
 
 = 
 
 
 
 
 
 
 24 
 
 24 
 
 3* 
 
 X 
 
 FB; 
 
 or, 
 
 P 
 
 E~~ 
 
 FB 
 AB 
 
 
 X 
 
 1 6"; 
 
 or, 
 
 16 _ 
 
 16 
 
 o 
 
 
 
 
 24 
 
 24 
 
 *> 
 
 j 
 
 i.e., 8 (oz.) x 24" = 24 (oz..) 
 And P x AB = E (or R) 
 
 i.e., 1 6 oz. x 24" = 24 (oz.) 
 
 The fulcrum F, where the equilibrant and resultant act, is 
 termed the centre of the two parallel forces, and it is J of the 
 length of the lever from one end, and f from the other end. 
 
 Reasoning generally trom this particular case, if you have 
 any two unequal unlike parallel forces, P and Q, acting on a 
 body in the directions AP and BQ respectively, and of which Q 
 
 CENTRE OP PARALLEL UNLIKE FORCES. 
 
 is the greater force, then if the line AB be drawn perpendicular 
 to the directions of these forces, and prolonged, a single force E 3 
 parallel to P and Q and equal to Q P, will balance these two 
 forces at a point C, so thai the moments about C are equal and 
 opposite ; or, 
 
 P x AC = Q x BC 
 
 Further, a force R, equal and opposite to E, acting at C, will 
 represent the resultant of P and Q. This point, C, is termed the 
 centre of the parallel forces. 
 
 The position of the point C, which is determined by the above 
 equation, is not affected by the directions of the forces so long as 
 they act at the same points A and B, and have the same mag- 
 nitudes. 
 
 You may imagine any number of parallel forces acting in one 
 plane being replaced by a single force. For in the above case 
 you have formed a resultant, R, for the two forces P and Q 
 consequently you could find a resultant, R I} for R and any other 
 parallel force say S ; and so on for any number. 
 
28 LECTURE HI. 
 
 The fined resultant of the whole of the forces would act at a point 
 which would be the centre of the system of the whole of the parallel 
 forces acting on the body. 
 
 Centre of Gravity. Since gravity attracts towards the earth 
 each particle of matter of which a body is composed, the weight 
 of a body may be considered as the sum of a system of parallel 
 forces. The centre of these parallel forces is called the centre of 
 gravity of the body, and is the point where the resultant of the 
 weights of all the particles composing the body acts. 
 
 The following statements in small type, which are generally 
 proved as propositions and corollaries in books on Elementary 
 Theoretical Mechanics, should be remembered by the student : 
 
 1. If a body is symmetrical, the centre of gravity (or e.g. of the body) coin- 
 cides with the centre of the mass. 
 
 2. If a body be uniform, the c.g. coincides with the centre of volume. 
 
 3. In a very thin plate of uniform density the c.g. coincides with the 
 centre of surface. 
 
 4. If the c.g. of a body be determined for any one position of the body, 
 the same point is the c.g. for every other position. 
 
 5. If a body be supported on its centre of gravity, the body will balance 
 in any position. Or, a body will balance about its c.g. in all positions. 
 
 6. If a body balance in all positions about a straight line through it, the 
 c.g. lies in that line. 
 
 7. If the c.g. be vertically above or below the point of support, the body 
 will rest in that position. Hence, if you balance or support a body from 
 two different points, the c<g. lies in the intersection of the two vertical 
 lines from the two points respectively. Or, if you balance a body on an 
 edge, the &jg. is In the vertical plane passing through that edge. Balance 
 it again on a different edge, thus finding another plane which passes 
 through the c.g. Then the c.g. lies in the straight line constituting the 
 intersection of the two planes. Balance the body for a third time in 
 another position, then the point where this third vertical plane intersects 
 with the straight line will be the c.g. of the body. 
 
 8. The c.g. of regular geometrical bodies may easily be found by mere 
 inspection when they are of uniform density. 
 
 For Example. The c.g. of a line is at the middle of the line ; of a circle 
 at its centre ; of a sphere at its centre ; of the surface of a uniform cylinder 
 and of a solid cylinder at the centre of the axis ; of a parallelogram at t>he 
 intersection of its diagonals ; of a triangle at the intersection of straight 
 lines drawn from two of the angles to the middle points of the opposite 
 sides i.e., at a distance from one of the angles along one of these lines 
 equal to | of the line ; of the perimeter of a triangle (i.e., of three uniform 
 rods forming a triangle) at the intersection of the two straight lines which 
 bisect two of the angles of the triangle formed by joining the centres of 
 the three uniform rods ; of a polygon at the point of application of the 
 resultant of the parallel forces represented by the areas of the respective 
 triangles into which the polygon may be formed, and where each of these 
 forces is considered to act at the c.g. of its own triangle ; of a pyramid at 
 f of the line from the vertex to the c.g. of the base ; of a cone at of the 
 axis from the vertex ; of the curved surface of a cone at f of the axis from 
 the vertex ; of a. prism at the middle of the line connecting the c.$r.'s of its 
 ends. 
 
THE LEVER. 2Q 
 
 The Lever when its Weight is taken into Account. 
 
 In this case we have to add the moment due to the weight of the 
 lever, to the moment of P or of W according as it acts along 
 with the one force or with the other ; i.e., according as the e.g. of 
 the lever is on the same side of the fulcrum as P or W. When 
 the lever is of uniform section and density throughout, then the 
 e.g. of the lever is at its middle point, and consequently the 
 whole weight of the lever may be considered as concentrated and 
 acting at that point. 
 
 A 
 
 t 
 
 WEIGHT OF LEVER CONSIDERED. 
 
 Let AB be a uniform lever, of weight w, acting at its e.g. or 
 middle point C, let a weight, IF, be attached to the end B, then 
 the force P, which will have to be applied to the other end A, in 
 order to balance the whole about the fulcrum F, will be found 
 by taking moments about F. 
 
 Thus, P x AF 4- w x CF = W x BF 
 
 Or* p_WxBF-wxCF 
 
 EXAMPLE I. A uniform lever, 5 ft. long, of 30 Ibs. weight, is 
 placed on a fulcrum 10 in. from one end, and has a weight of 
 100 Ibs. attached to the short end. What force must be applied, 
 and in what direction, in order to produce equilibrium ? Also, 
 what is the pressure on the fulcrum, and in what direction does 
 the reaction from the fulcrum act ? 
 
 1 . Referring to 'the above figure, we find from the question that 
 
 AB = 5 ft. = 60 in.; BF= 10 in. .*. AF = 50 in. and CF = 20 in. 
 - W = 100 Ibs. and w = 30 Ibs. 
 
 2. By the principle of moments 
 
 The Opposing Moments about the Fulcrum are equal. 
 
 Consequently, P x AF + w x CF = W x BF 
 
 p_WxBF-tpxGF 
 AF 
 
 Substituting the numerical values 
 
 P = 100x10-30x20 _ 8 lbg 
 
 5 
 
 * If the e.g. of the lever was on the opposite side of the fulcrum, i.e., 
 on the side of W, then P x AF=* W x BF + to x CF. 
 
3O LECTURE III. 
 
 3. P acts vertically downwards, since the moment due to the 
 weight of the lever is not sufficient to equalise the moment due to 
 the weight W about the point F. 
 
 4. The pressure on the fulcrum is evidently equal to the sum of 
 all the forces, since all the forces act in one direction, or vertically 
 downwards. It is therefore equal to 
 
 W + w + P ^ soo 4- 30 + 8 = 138 Ibs. 
 
 5. The reaction from the fulcrum is equal and opposite in direc- 
 tion to this resultant. It therefore acts vertically upwards, and 
 is the equilibrant of the whole of the forces, for a vertical force of 
 138 Ibs. applied to the lever at F would counterpoise or just lift 
 the whole bar with the attached weights P and W. 
 
 EXAMPLE II. Suppose everything the same as in the previous 
 example but the weight of the lever, which you may consider as 
 now equal to 60 Ibs. ; what force P would be required, and in 
 what direction would it have to act, in order to produce equili- 
 brium ? Also, what would be the resultant or downward pressure 
 at F. 
 
 1. You observe at once that the moment of the weight of the 
 lever is greater than the moment of W about the fulcrum. 
 
 For, w x OF > W x BF 
 
 Since, 60 x 20 > 100 x 10 
 
 Consequently by the principle of moments P must act against 
 w, or vertically upwards, so as to assist W, in order that the oppos- 
 ing moments about the fulcrum may be equal. 
 
 2. The formula therefore becomes 
 
 !0xCF-PxAF 
 Or, >xCF 
 
 . u?xCF-WxBF -D 
 
 " 
 
 Substituting the numerical values, we have 
 
 6ox20-iGoxio_ p _ 1 lbs 
 
 5o 
 
 3. The resultant pivssure at F is equal to the algebraical sum of 
 the forces, or 
 
 W + w - P = 100 + 60-4 = 156 lbs. 
 
 And acts vertically downwards. The equilibrant would there- 
 fore be 156 lbs. acting on the lever at Fand vertically upwards. 
 
 Position of the Fulcrum. In answering questions which give 
 the magnitude of the forces with which they act, and require only 
 an answer for the position of the fulcrum, the student has 
 
THE LEVER. 3 1 
 
 simply to employ the general formula for the principle of mo- 
 ments, and then to substitute the known numerical values in 
 order to get the unknown. Or, he may reason out the formula 
 into the following shape, and then interpolate the numerical 
 values. Referring to the last figure, suppose that the distance 
 AF is required : 
 
 Then, neglecting the weight of the lever, we have by the principle 
 of moments 
 
 Or, PxAF + Wx AF = AF(P + W) =WxBA 
 
 WxBA 
 AF = P + W 
 
 Now, taking the weight of the lever into account , we have by thd 
 principle of moments : 
 
 PxAF + wxCF = Wx BF. 
 Or, 
 
 r BA 
 
 PxAF + wxAF + Wx AF = WxBA + M?x 
 
 W 
 
 EXAMPLE III. Where should the fulcrum be placed under a 
 uniform lever in order to produce equilibrium, if the lever is 5 ft. 
 long, weighs 30 Ibs., and has weights of 100 and 8 Ibs. respectively 
 hung at its ends. 
 
 From the above general equation for equilibrium viz. : 
 
 
 8 + 3Q-I- TOO 
 Which proves the data given in Example I. to be correct, 
 
32 LECTURE III. QUESTIONS. 
 
 LECTURE III. QUESTIONS. 
 
 1. Define what is meant by " the moment of a force,* 1 and give an ex 
 ample with a sketch. 
 
 2. State " the principle of moments," and apply it to the case of a 
 simple straight lever. 
 
 3. A weight of 10 Ibs. on the end of a lever 100 inches from the centre 
 of motion is found to balance a weight of 100 Ibs. at a distance of 10 inches. 
 Explain the natural law which governs matter and motion, upon which the 
 above mechanical fact depends. (Answer this by giving the definition of the 
 principle of moments.) 
 
 4. Describe an experiment to prove the equality of the moments when 
 the pull is between the weight and the fulcrum and acts in the opposite 
 direction to the weight. 
 
 5. In the case of a straight lever, how would you ascertain the pressure 
 on and the reaction from the fulcrum ? 
 
 6. Three forces, of 12, :o and 2 Ibs., act along parallel lines on a rigid 
 body ; show by a sketch how they may be adjusted so as to be in equili- 
 brium 1 Ans. The force of 12 Ibs. must act as the equilibrant to the forces 
 2 and 10 Ibs. i.e., in a line with their resultant, but in the opposite direc- 
 tion. 
 
 7. Two parallel forces of 10 and 12 Ibs. act in opposite directions on a 
 rigid body, and at 2 fe^t apart. Where is the centre of the two forces, 
 and what is their resultant 1 Ans. 10 feet from the force of 12 Ibs., 2 Ibs. 
 
 8. Define the " centre of gravity " of a body, and show how you would 
 find it experimentally in the case of any irregular body. Give an example. 
 
 9. State the rule which applies when two unequal forces balance on 
 opposite sides of the fulcrum of a straight lever, the weight of the lever 
 being neglected. A uniform straight lever, 4 feet long, weighs 10 Ibs., the 
 fulcrum is at one end ; find what upward force acting at the other end 
 will keep the lever horizontal when a weight of 10 Ibs. is hung at a dis- 
 tance of i foot from the fulcrum. Find also the pressure on the fulcrum 
 and the direction in which it acts. Ans. 7'5 Ibs. ; 12-5 Ibs. downwards. 
 
 10. A uniform bar, 4 feet long and weighing 4 Ibs., can turn about a 
 fulcrum at one end, and a weight of 10 Ibs. is hung upon the bar at a dis- 
 tance of i foot from the fulcrum. Find the upward force at the free end 
 which will keep the bar horizontal. Ans. 4-5 Ibs. 
 
 11. A uniform bar of metal 10 inches long weighs 4 Ibs., and a weight 
 of 6 Ibs. is hung from one end. Find the fulcrum or point upon which the 
 bar will balance. Ans. 2 inches from the 6 Ibs. weight. 
 
 12. Two parallel forces whose magnitudes are 8 and 12 Ibs. respectively, 
 act in the same direction on a rigid body at points 10 inches apart. Find 
 the magnitude and line of action of the resultant of the two forces. 
 
 Ans. 20 Ibs. at a point 6 inches from the force of 8 Ibs. 
 
 13. A uniform lever is 5 feet long, and weighs 10 Ibs., the fulcrum being 
 at one end. A weight of 30 Ibs. is hung at a distance of 4 feet from the 
 fulcrum ; what upward force acting at the middle point of the lever will 
 keep it in a horizontal position 1 Ans. 58 Ibs. 
 
 14. Define " moment of a force." How is it measured ? A bar of metal 
 of uniform section weighs 5 Ibs., and a weight of 10 Ibs. hangs from one 
 end. It is found that the bar balances on a knife edge at 9 inches from 
 the end at which the weight hangs ; what is the length of the bar I 
 
 4 ft. 6 in. 
 
LECTDBE HI. QUESTIONS. JJ 
 
 15. State the principle of the lever, and prove it when F and W act on 
 opposite aides of the fulcrum. A weight of 5 Ibs. is hang at one end of 
 a uniform bar, which is balanced over a knife edge at a point 14 inches 
 from the end at which the weight hangs. The bar weighs 30 Ibs. ; find its 
 length. Ant. 32$ inches. 
 
 16. State the principle of the lever. A uniform straight bar, 14 inches 
 long, weighs 4 Ibs. ; it is used as a lever, and an 8 Ib. weight is suspended 
 at one end. Find the position of the fulcrum when there is equilibrium. 
 Ant. 2& inches from the 8 Ib. weight. 
 
34 NOTES AND QUESTIONS. 
 
( 35 ) 
 
 LECTURE IV. 
 
 CONTENTS. Practical Applications of the Lever The Steelyard, or Roman 
 Balance Graduation of the Steelyard The Lever Safety Valve 
 Example I. Lever Machine for Testing Tensile Strength of Materials 
 Straight Levers acted on by Inclined Forces Bent Levers The 
 Bell Crank Lever Bent Lever Balance Duplex Bent Lever, or Lum- 
 berer's Tongs Turkus, or Pincers Examples II. and 111. Toggle 
 Joints Questions. 
 
 IN this Lecture we shall give a number of examples of the 
 application of the lever. 
 
 STEELYARD, OB ROMAN BALANCE. 
 INDEX TO PARTS. 
 
 F represents Fulcrum. 
 GA Graduated arm. 
 SW Sliding weight. 
 
 P Pull due to SW 
 SP Scale pan. 
 
 W represents Weight in SP. 
 AF Distance of P 
 
 from F. 
 
 BF Distance of W 
 fromF. 
 
36 LECTURE IV 
 
 The Steelyard, or Roman Balance, is a straight lever with 
 unequal arms, having a movable or sliding weight on the longer 
 arm. It is very much used by butchers for weighing the carcasses 
 of cattle and sheep, and in such cases it generally has two fulcra 
 and two scales of division corresponding to them, the one set 
 being, say, for hundredweights and the other for pounds. 
 
 Graduation of the Steelyard. The practical method of gradu- 
 ating the steelyard is to put unit weight (say i Ib.) into the scale 
 pan, SP (or attach it to the hook on the shorter arm if there 
 should be no such pan), and mark the position where the sliding 
 weight, SW, has to be placed in order to cause equilibrum. Mark 
 this position i on the scale. Then put in two units (say 2 Ibs.) 
 into SP, and adjust SW as before, marking its new position as 
 2 on the scale ; and so on until SW is at the end of the longer 
 arm. 
 
 In tnis form of steelyard, if the differences of the weights W, 
 corresponding to successive distances, i to 2, to 3, <fcc., be the same, 
 the graduations will be equal to each other. This may be proved 
 in the following manner : First of all, it is clear that the instru- 
 ment can be so constructed that the centre of gravity of the beam 
 and scale pan may occupy one or other of three different positions. 
 The centre of gravity may coincide with F, or it may be on the 
 longer arm, or it may be on the shorter arm. 
 
 Suppose the centre of gravity to coincide with F, the fulcrum. 
 
 Let the scale pan be loaded to the extent of W units, and 
 suppose that the sliding weight of P units has to be placed at A 
 in order to keep the beam horizontal. 
 
 B F . . , A At t r . . A 
 
 8 9 1 11 12 
 
 A 1 2 3 4 5 tf 7 
 W P 
 
 STEELYARD WITH THE CENTRE OP GRAVITY COINCIDING 
 WITH THE FULCRUM. 
 
 Then, PxFA = WxFB . . (i) 
 
 Increase W by one unit, and to restore equilibrium, let P be 
 placed at A r Then, for equilibrium we must have 
 
 PxFA 1= =(W+i)xFB . (2) 
 
 Subtracting corresponding members of equations (i) and (2) 
 we get P(FA t - FA) = ( W + i - W) x FB, 
 
 Or, P x AAj = FB, 
 
 FB 
 = ^p- (3) 
 
THE LEVER. J7 
 
 Increase W by n units, and let P occupy the position An. 
 Then, for equilibrium, we must have 
 
 PxFAra = (W + n)xFB . (4) 
 
 As before, subtract the corresponding members of (i) and (4), 
 when we get P x A A* = n x FB, 
 
 FB 
 
 Or, AA n = n x A A t by equation (3) 
 
 Thus we see, that the graduations are all equal for equal incre- 
 ments of W. 
 
 The student will readily observe that the zero of the scale is at 
 F, and by putting W= i in equation (i) we can fix the position 
 of the first number on the scale 
 
 i.e. t P x Fi = i x FB, 
 
 Or, Fi=. 
 
 Nexti suppose the centre of gravity to lie in the longer arm at G. 
 
 B | .F G . A Ai t _ | , l l A* 
 
 | / ^>3\ 4 5 6 7 ]s 9 10 11 12 13 14 15 
 I j 
 
 W P 
 
 STEELYARD WITH THE CENTRE OF GRAVITY IN THE' 
 LONGER ARM AT G. 
 
 Let w = weight of beam and scale pan, and suppose P at A 
 and to at G to balance W units at B. Then, for equilibrium, 
 we have 
 
 P x FA + 10 x FG = W x FB . . (5) 
 
 As before, increase W by one unit, and let P be shifted to A, 
 in order to restore equilibrium, then we must have 
 
 Subtracting (5) from (6) we get 
 
 PxAA 1 = FB, 
 
 _FB 
 Or, AA t p . . , / \ 
 
 Now increase W by n units, and let P occupy the position A*, 
 then 
 
 P x FA,. + w x FG = (W + n) x FB . (8) 
 Subtracting (5) from (8) we get 
 
 PxAA M = nxFB, 
 
 FB 
 Or, AA,. = n x -p-, 
 
 That is, AA,, = n x AA, by equation (7) 
 
38 LECTURE IV. 
 
 Thus we again see that the graduations are equal for equal 
 increments of the weight W. 
 
 To find the zero of the scale in this case : 
 In equation (5) put W = 0, then 
 PxFO 
 
 That is, the zero is in the shorter arm at a distance from F, 
 represented by - J p units of distance, the units in this case 
 
 being the same as those measuring FG. 
 
 By making W= i in equation (5) the position of the first figure 
 on the scale can be fixed, and then the whole beam graduated, 
 since all the divisions are of the same size. 
 
 One important point to be observed in this arrangement is, 
 that when the centre of gravity lies in the longer arm, there is a 
 limit to the smallness of the weight which can be weighed in the 
 scale pan, since the sliding weight moves along the longer arm 
 only. 
 
 Let P coincide with F, then the weight W which must be 
 placed in the scale pan in order to just balance the weight of the 
 beam and scale pan at G is 
 
 . 
 
 Any weight less than this cannot be weighed. This is not an 
 objection to the instrument where the weights to be measured 
 are great, as in the case of the butcher's steelyard used for 
 weighing heavy carcasses. 
 
 When the centre of gravity lies in the shorter arm the graduations 
 will still be equal. The reasoning is the same as in the last two 
 cases. The student can also prove that the zero of the scale is 
 on the longer arm at the point given by the equation 
 
 In this case all weights, however small, can be weighed. 
 
 The Lever Safety Valve.* The lever safety valve is a simple 
 
 * For a more detailed description of safety valves and their action 
 refer to Lecture XXVII. of the author's Elementary Manual on ' Steam 
 and the Steam Engine." 
 
THE LEVER SAFETY VALVE. 
 
 39 
 
 contrivance fixed on the top of a boiler for the purpose of auto- 
 matically preventing the steam exceeding an agreed-upon working 
 pressure. 
 
 Referring to the next figure, YC is a cast-iron valve chest, 
 containing a tightly-fitted gun-metal valve seat, VS, on which 
 rests a steam-tight gun-metal valve, V. On the centre of the 
 upper side of this valve rests a conical steel pin attached to a 
 straight lever by an eye and bolt. One end of this lever is free 
 to turn en a fulcrum fixed to the upper flange of the valve chest, 
 and a lock-fast cast-iron weight is placed near the other end, so 
 
 LOCKFAST LEVEB SAFETY VALVE. 
 INDEX TO PARTS. 
 
 VC represents Valve chest. 
 
 VS 
 
 Valve seat. 
 
 V represents Valve. 
 
 P i Locking pin. 
 
 that the downward moment of the weight about the fulcrum 
 balances the upward moment of the steam pressure on the valve 
 about the same fulcrum. 
 
 Let L = length of lever in inches from fulcrum to the e.g. of 
 
 the weight, W. 
 F = Distance in inches from fulcrum to centre line of 
 
 valve, V. 
 
 ^ = to e.g. of the lever. 
 
 W = "Weight in Ibs. of the cast-iron counterpoise block. 
 w / = lever. 
 
 w * = valve. 
 
 P =: Pressure of steam in Ibs. per square inch* 
 
4<3 LECTURE IV, 
 
 d = Diameter of valve in inches. 
 A ='Area of valve in square inches = -d 2 . 
 P x A = Total pressure in Ibs. on the valve. 
 
 Then, by taking moments about the fulcrum, we find the pressure 
 of steam per square inch which will balance the several forces. 
 
 for the upward moment = the sum of the downward moments. 
 
 (P x A- W,)F = (W x L) + (W, x G). 
 Or, (PxA)xF = (WxL) + (W,xG) + (W.xF) 
 
 AxE 
 
 If we neglect the weight of the lever and the valve 
 Then, (PxA)xF = WxL 
 
 -?s 
 
 EXAMPLE I. A valve, 3 inches in diameter, is held down by a 
 lever and weight, the length of the lever being 10 inches, and the 
 valve spindle being 3 inches from the fulcrum. You are to dis- 
 regard the weight of the lever and to find the pressure per square 
 inch which will lift the valve when the weight hung at the end 
 of the lever is 25 Ibs. 
 
 -10 
 
 W =25 Ibs. 
 
 Referring to the previous figure as well as to the accompanying 
 one, we see from the question that 
 
 <* = 3" ' A = ^2 = .7854x3x3 = 7.07 sq. ins. ; 
 
 BF= 3', AF= 10" and W = 25 Ibs. 
 Taking moments about F, we get 
 
 (PxA)xBF = Wx AF 
 
 P x 7.07 x 3 = 25 x 10 
 
 P = H'8 Ibs. per square inch. 
 
 Testing Machine. The following figures illustrate a machine 
 which is used for testing the tensile strength of iron, steel and 
 such like materials. It consists of a combination of levers. After 
 
LEVER TESTING MACHINE. 
 
 >HW 
 
 LEVEB MACHINE FOR TESTING TENSILE STRENGTH OF MATERIALS. 
 
 A, 
 
 L, 
 
 7 
 
 MW 
 
 f 
 
 W 
 
 n 
 
 R BW a 
 
 A 2 
 
 DIAGRAM OF THE LEVERS. 
 
 INDEX TO PARTS. 
 
 LjL, represent Levers. 
 F,F 2 Fulcra. 
 
 P Pull, or dead 
 weights. 
 
 A 
 
 i 
 
 Where P acts on L,. 
 Specimen under te.->t. 
 
 BpA,, represents Where R acts on 
 
 L,,L 2 . 
 
 B., Where W acts on S. 
 M\V Movable weight. 
 BW Balance weight. 
 
 HW Hand-wheel and 
 screw for elevat- 
 ing F lf &c. 
 
2 LECTURE IV. 
 
 mastering the general arrangement of the machine by comparing 
 the index to parts with the side elevation, the student should 
 refer to the accompanying skeleton diagram (where the same index 
 letters have been used), from which he will readily understand how 
 the stresses are transmitted and magnified. 
 
 Looking at the second of the above figures, or skeleton diagram 
 of the levers, it will be seen that when equilibrium exists between 
 the stress W on the specimen S, and the pull P, applied at A v 
 
 P x A^\ = E x B^j, and R x A 2 F 2 = W x B 3 F 2 
 PxA,F, Wx 
 
 *i*\ ~ A 
 
 P x A 1 F 1 x A 2 F 8 
 -L 
 
 PxA,F, WxB,F 2 
 
 ~ 
 
 . . _ 
 Consequently, W= 
 
 Straight Levers Acted on b3 Inclined Forces. In the 
 previous Examples and in Leeeure III. we have considered the 
 forces P and W as acting at right angles to the straight levers. 
 In such cases the forces had the greatest advantage, or their 
 turning moments were a maximum. But the principle of moments 
 is equally applicable to inclined forces acting on straight levers and 
 to bent levers. 
 
 X s 
 / \ 
 
 W 
 
 STRAIGHT LEVEES WITH INCLINED FORCES. 
 
 For, let AjBj be a straight lever acted on by inclined forces, 
 P and W. "Draw from the fulcrum, F, lines at right angles to 
 the produced directions of the forces as shown by the dotted lines 
 in the above figure. 
 
 Then, the effective arms for the forces P and W are respectively 
 A 8 F and B,F ; and equilibrium takes place when their momenta 
 about F are equal ; 
 
 i.e., when P x A,F = W x B F 
 
 Or, P: W::B,F: A,F. 
 
 Bent Levers. The BeU Crank Lever. The same principle 
 and action hold good in the case of bent levers. Take an ordi- 
 
BENT LEVERS. 
 
 43 
 
 nary right-angle bell crank lever, as shown by the accompanying 
 figure. Here the effective arms are equal to the actual arms of 
 the lever, because the forces have been shown as acting at right 
 angles to their respective arms, or with maximum turning 
 moments. 
 
 Therefore, P x AF = W x BF. 
 
 But, if the lever be turned round through any angle by, say, an 
 extra pull at P, then, in order to ascertain the virtual moments 
 we should have to draw lines at right angles from F on the 
 directions of P and W in order to calculate their effective 
 arms. 
 
 S.P. 
 
 w 
 
 BELL CRANK LEVER 
 
 BENT LEVER BALANCE. 
 
 Bent Lever Balance. Examine an ordinary bent lever 
 balance, such as is frequently used for weighing letters and light 
 parcels, where the force P is a constant quantity, and the variable 
 force W is represented by the article to be weighed. As showu 
 by the accompanying figure, the effective arms change with each 
 weight to be ascertained, and consequently the scale S of this 
 balance has to be graduated by trial, or by introducing standard 
 pounds, such as SP, or other units, and marking the values on 
 the scale opposite the position where the end of the pointer on 
 P comes to rest. Or, the graduation might be done by plotting 
 the various positions of the arms and values of the forces to scale, 
 In the illustration we have evidently got equilibrium when 
 
 P x AF = W x BF. 
 
 Duplex Bent Lever, or Lumberer's Tongs. The accom- 
 panying illustration shows a very useful and simple application of 
 the bent lever, which is used at the end of a winch or crane chain, 
 
44 
 
 LECTURE IV. 
 
 for affixing to and holding fast stones, logs of wood, blocks of ice, 
 or other heavy articles when they have to be lifted. 
 
 P, P indicate the directions of the pulling forces on the short 
 chains between the ends of the shorter arms and the common link 
 which is attached to the crane chain. F is the common fulcrum, 
 and W, W show the directions of the forces with which the article 
 is gripped. The student will be able to draw a diagram of the 
 forces and calculate their effective moments for himself for any 
 particular case. 
 
 The Turkus, or Pincers. The ordinary carpenter's turkus, 
 or pincers, which is frequently used for extracting nails from wood, 
 is another familiar illustration of the duplex bent lever. As 
 shown by the accompanying figure, the forces P^Pj, represent the 
 forces with which the pincers is gripped by the hand after the 
 jaws have been closed on the neck of the nail, and the force B 
 
 TONGS OE DUPLEX BENT LEVEE. 
 
 TUKKUS, OE PlNCEES. 
 
 the pressure which has to be exerted by the arm and body in 
 order to extract the nail from the wood i.e., to overcome the 
 frictional resistance, W, between the wood and the nail. As 
 shown by the separate diagram of forces in dotted lines, straight 
 lines have been drawn, not from the joint of the pincers, but 
 from a position representing the fulcrum F (or point where the 
 nose of the pincers rests on the wood), perpendicular to the direc- 
 tions of the forces P and W, in order to obtain the lengths AF 
 and BF of the effective arms of the bent lever. 
 
 Here again, P x AF = W x BF. 
 
 EXAMPLE II. The handle of a claw-hammer is 15 inches 
 long, and the claw is 3 inches long. What resistance of a nail 
 would be overcome by the application of a pressure of 50 Ibs 4 at 
 the end of the handle ? 
 
BENT LEVERS. 45 
 
 You are required to show, by a diagram, the manner in which 
 you arrive at your result. (S. and A. Exam. 1892.) 
 
 ANSWER. Here we have a simple case of a bent lever, with 
 fulcrum at F, and effective arms, AF, BF, 1 5 and 3 inches long 
 
 w 
 
 P-50 LBS. 
 
 EXAMPLE OP A BENT LEVER. 
 
 respectively. Let W represent the resistance in Ibs. offered by 
 the nail at B. Then, by taking moments about F, we get 
 
 W x BF = P x AF 
 
 Or, W x 3 = 50 x 15 
 
 ... W = 5 x I5 = 250 Ibs. 
 3 
 
 EXAMPLE III. State the mechanical law known as the Principle 
 of the Lever. In a pair of pincers the jaws meet at i J inches from 
 the pin forming the joint. The handles are grasped with a force 
 of 50 Ibs. on each handle at a distance of 8 inches from the pin. 
 Find the compressive force on an object held between the jaws, 
 and also the pressure upon the pin. (S. and A. Exam. 1888.) 
 
 Let P denote the force of 50 Ibs. with which the handles are 
 grasped at a distance of 8 inches from F, the pin. Let W denote 
 
 P 7 =50/6s. 
 
 R=316-6lb*. |P,-50/to. 
 
 PINCERS OR NIPPERS. 
 
 the compressive force on the object O, and R the resultant reaction 
 or pressure on the pin or fulcrum F. Although there are two 
 levers here, each having a common fulcrum, F, it is best to con- 
 
 NoU. It is a mistake to speak of the " Principle of the Lever " ; what 
 is evidently meant is the Principle of Moments as applied to the lever. 
 
4< LECTURE IV. 
 
 sider the action of one lever only. Suppose the lower handle, H,, 
 to be fixed, and consider the action of the upper handle, H t . It 
 then becomes a simple question on the lever. 
 
 (1) To find W, take moments round F, then 
 
 W x ij" = 50 x 8" 
 .. W = 266-6 Ibs. 
 
 (2) To find R, the pressure on the pin F, take moments round O 
 then 
 
 R x ij" = 50 x (ij + 8") - 50 x 9 J" 
 
 ,'. R = 316 8 Ibs. 
 
 Or, since R must be the resultant of 3? and W, we get 
 R = P + W = $o + 266-6 = 316-6 Ibs. 
 
 The M Toggle," or " Knuckle Joint," consists of a well- 
 known combination of levers. It is characterised by its capability 
 of exerting an enormous force through a short distance by means 
 of very compact and simple elements. This device has been 
 applied in many well-known cases, such as in the stone-crushing 
 machine, certain brakes, printing and several forms of packing 
 presses, and in the familiar frame by which carriage hoods are 
 held in position. 
 
 The accompanying diagrams illustrate the principle of the 
 " Toggle," and the plan of its application to a cane mill. 
 
 In Fig. i, A A' are two links jointed at B, G, and D. The 
 point G is fixed, whilst the point 7? is free to move vertically along 
 the line E C. The centre D is connected with the point F by 
 means of a third link, formed partly of a spring S. The point F 
 is supposed to be fixed in relation to its distance from the vertical 
 line E C, but is free to move in a line parallel to E C. Under 
 these conditions, a force acting in the direction of the arrow along 
 the line G H will resist a much greater force acting along the 
 line E (7, and tending to move the point B upwards. The force 
 along G H acts with a leverage equal to the distance from J to 
 C, whilst the force along B D has only a leverage equal to the 
 short length K C. The nearer the centre D approaches the 
 vertical line, the greater will be the leverage of the force acting 
 on the centre D in the direction G H, and the less will be the 
 leverage of the force acting in the direction E G through the 
 link A'. In practice, a few railway buffer springs, corresponding 
 to S in Fig. i, combined with simple links in a compact arrange- 
 ment, occupying little space, are sufficient to resist forces amount- 
 ing to several hundreds of tons. The line E G corresponds to 
 Wie centre line of one of the top cover bolts of a cane mill, G 
 
"THE TOGGLE," OR "KNUCKLE JOIflT." 
 
 47 
 
 corresponding to the nut at the top of the bolt, and therefore 
 one of the fixed points in the system ; whilst B may be taken as 
 representing a point in the top cover itself, which is supposed to 
 be free to move up and down. 
 
 Fig. 2 illustrates the operation of the system, and directs 
 attention to a feature of great practical interest and value. The 
 
 FIG. i. FIG. 2. 
 
 Iti 
 
 y 4 ^MOCGr 
 
 E 
 
 THE PEINCIPLB OP THE TOGGLE JOINT. 
 
 parts in this figure correspond with those in Fig. i, but it is 
 supposed that a force acting in the direction E C has raised the 
 point B from its original position, as in Fig. i, to the position B'. 
 The point D has consequently been moved further away from 
 the vertical line EC to the position D'. In this position, the 
 leverage of the force acting along G H is reduced, whilst the 
 leverage of the force acting through the link A' is increased, 
 as compared with the previous conditions in Fig. i ; but it will 
 be seen that Z>. in moving to the position />', has compressed the 
 spring S, and therefore increased its resistance. 
 Hence, by the principle of moments : 
 
 (Force along B D) x K C = ( Force along G F) x J 0. 
 /. Force along B D = (Force along G F) 
 
 NOTE. See the Tangentometer at end of Lecture XXVIII. 
 
LECTURE IV. 
 
 T> J 
 
 But the vertical force E B = (Force aloog B D) x ^-^. 
 
 J C 1 R T 
 . = (Force along G F) x ^-^ x g-j-j. 
 
 But if the links B D and D C are equal, then B J = J C. 
 
 J Q 2 
 
 Hence, force along E B = (force along G F) x 7^^ TT^. 
 
 If the movement along B C is very small compared with the 
 length of the links B D and D C, then J C may be considered 
 as a constant length. Hence, when the spring is so adjusted that 
 
 its resistance to compression 
 is proportional to K C, the 
 pressure E B upon the rollers 
 will be nearly constant. This 
 arrangement of the " toggle 
 joint," therefore, automati- 
 cally permits of light or heavy 
 feeding of the canes in a sugar 
 mill without bringing undue 
 stresses upon the various 
 parts, and thus diminishes 
 the chance of breakdowns. 
 
 The following illustration 
 shows the actual construction 
 of the " Toggles " as applied 
 to the top roller of an ordinary 
 3-roller cane-crushing mill. 
 The left-hand half of the 
 illustration shows the posi- 
 tions of the parts when the 
 mill is empty. In this con- 
 dition the top roller rests in its 
 bearings and gix'es the mini- 
 mum openings between the 
 rollers. The right-hand half 
 of the illustration shows the 
 condition of the system when 
 
 PATENT TOGGLE APPARATUS. the mill is taking a heavy 
 
 feed. It will be seen flint 
 as the roller rises it lifts the top caps, which are under the contiol 
 
THE TOGGLE," OR "KNUCKLE JOINT.'' 
 
 49 
 
 of the " Toggles," until the upward pressure is balanced by the 
 resistance of the " Toggles." 
 
 v. 
 
 TOGGLE JOINT AS APPLIED TO A SUGAR-CANE MILL. 
 
 The above is an illustration of a mill made by the Messrs. 
 Mirrlees, Watson & Co., Ltd., of Glasgow, with rollers 32 inches 
 in diameter by 60 inches long. The arrangement of their patent 
 " Toggles " in this case is of a special kind. The vertical bolts 
 of the " Toggles " are formed by continuations of the top cover 
 bolts, but the mill covers or caps are fixed by nuts screwed directly 
 down upon them, and do not lift with the roller. The communi- 
 cation between the " Toggles " and the top roller is made by means 
 of plungers formed on the under side of the bottom plates of the 
 " Toggles," and working through the top caps. The only advantage 
 of this arrangement is, that it- permits the top cap to be used as 
 a brace to bind the upper jaws of the mill cheek together, and 
 thus adds in some measure to the strength of the cheek. 
 
5O LECTUEB IV. QUESTIONS. 
 
 LECTUBE IV. QUESTIONS. 
 
 1. Sketch and describe the steelyard, or Roman balance, and explain 
 fully how the graduations on the scale are equal for equal differences in 
 the weights applied to the shorter arm. 
 
 2. Sketch and describe a lockfast lever safety valve. A valve, 3 
 inches in diameter, is held down by a lever and weight, the length of the 
 lever being 30 inches, and the valve spindle being 4 inches from tho 
 fulcrum. You are to disregard the weight of the lever and to find the 
 pressure per square inch which will lift the valve when the weight hung 
 at the end of the lever is 56 Ibs. Ana. 59*4 Ibs. 
 
 3. The diameter of a safety valve is 3", its weight 3^ Ibs. ; length of 
 lever is 30", and its weight 16 Ibs. ; the distance from fulcrum to centre 
 of valve is 3", and to e.g. of lever 12". Find where a weight of 50 Ibs. 
 must be placed on the lever in order that steam may just blow off at 70 Ibs. 
 per square inch by gauge. Ans. 25-65 inches from the fulcrum. 
 
 4. The safety valve of a boiler is required to blow off steam at 100 Ibs. 
 per square inch by gauge. The dead weight is 100 Ibs., weight of lever 
 10 Ibs., and of valve 5 Ibs. ; diameter of valve 3^", distance from centre of 
 valve to fulcrum 4", from e.g. of lever to fulcrum 15". Where should you 
 place the weight on the lever? Ans. 36-9 ins. from fulcrum. 
 
 5. Sketch and describe a lever machine for testing the tensile strength 
 of materials. If the advantage, or ratio of resistance R to pull P in the first 
 lever, is 56 to I, and of the second lever 40 to I, what stress will be produced 
 on the test specimen when P= 100 Ibs. 1 Ans. 100 tons. 
 
 6. A force of 100 Ibs. acts at one end of a straight lever, but at an angle 
 of 60 to it. What force acting at the other end of the lever, at an angle of 
 45 to it, will keep the lever in equilibrium if the fulcrum be placed half 
 the distance from the first force that it is from the second ? Draw a dia- 
 gram of the forces and their effective arms. Ans. 61.25 lbs - 
 
 7. Sketch a bell crank lever, to convey a small movement from one 
 line to another, cutting each other at 60 ; the distances moved through 
 to be as i to 2. 
 
 8. The handle of a claw-hammer is 12 inches long, and the claw is a 
 inches long. What resistance of a nail would be overcome by the appli- 
 cation of a pressure of 40 lbs. at the end of the handle ? Show, by a diagram, 
 the manner in which you arrive at your result. Ans. 240 lbs. 
 
 9. In a pair of pincers the jaws meet at i^ inches from the pin forming 
 the joint. The handles are grasped with a force of 30 lbs. on each handle 
 at a distance of 7^ inches from the pin. Find the compressive force on 
 an object held between the jaws, and also the pressure upon the pin. 
 Sketch the apparatus and show the direction and values of all the forces. 
 Ans. 1 80 lbs ; 210 lbs. 
 
 10. There is a contrivance for obtaining great pressure through a small 
 distance, commonly termed the toggle or toggle joint. Will you explain 
 it, and show wherein its peculiar action and efficiency consist ? 
 
 11. Explain the mechanical advantage of the combination known as a 
 toggle joint. Show its application in printing machinery, or in stone 
 crushing machines, or in any other instances with which you are 
 acquainted. 
 
NOTES AND QUESTIONS 
 
 12. Sketch any one form of toggle joint with which you are acquainted, 
 and poinc out its object. (C. & G., 1903, O., Sec. A.) 
 
 13. The figure shows a bent lever AOB with a frictionless fulcrum 0. 
 AOis 12", BO is 24". The force Q, of 1000 Ib. acts at A, what force P 
 acting at B will produce a balance ? Work the question graphically or in 
 any other way. Neglect the weight of the lever. (B. of E. 1904.) 
 Ant. Force P = 550 Ibs. 
 
 NOTE Before answering this question and any future questions in- 
 volvirxr angles, or the ratios of the sides of a right angled triangle, 
 students should refer to the construction, action and uses of the Tan- 
 gentometer at the end of Lecture XXVIII. 
 
LECTURE V. 
 
 CONTENTS. The Principle of Work Work put in, Work lost, Useful 
 Work Efficiency of a Machine Principle of Work applied to the 
 Lever Experiments I. II. Wheel and Axle The Principle of 
 Moments applied to the Wheel and Axle The Principle of Work 
 applied to the Wheel and Axle Experiment III. The Winch Barrel 
 Example I. Ship's Capstan The Fusee Questions. 
 
 The Principle of Work.* The principle of work is applicable 
 to all machines, and may be stated as follows : 
 
 The work put into a machine is equal to the work absorbed by 
 the machine plus the work given out by the machine. 
 
 Or, WORK PUT IN = LOST WORK + USEFUL WORK. 
 
 This is an aociom. But, nevertheless, many deluded would-be 
 inventors have spent much time and money in devising "perpetual 
 motion " appliances, or machines which should turn out as much 
 work as, or even more than, was put into them ! 
 
 1. When a machine is employed to perform mechanical work, 
 a certain force must be applied to one part of it in order to move 
 the machine and to perform work at another part. 
 
 The product of this applied force and the distance through 
 which it acts constitute tlie whole work put into the machine. 
 
 2. Some of this work must be expended in merely keeping the 
 different parts in motion, against natural resistances due to fric- 
 tion at the fulcra or journals, and friction between moving parts 
 and the air or water in the case of an hydraulic apparatus. The 
 work so absorbed is termed lost work. 
 
 The mean value of the frictional resistances, multiplied by the 
 mean distance through which they are overcome, constitute the 
 work lost in the mechanism. One great object to be kept in view, 
 in designing most machines, is to minimise this lost work by 
 minimising the internal resistances to motion in the machine 
 
 * The Principle of Work is usually stated as follows in books on Mecha- 
 nics, but I find that engineering students much prefer the above definition. 
 " If a system of bodies be at rest under the action of any forces, and be 
 moved a very little, no work will be done." " Conversely : If no work is 
 done during this small movement, the forces are in equilibrium." Prof. 
 Goodeve's " Manual of Mechanics," p. 73. 
 
PRINCIPLE OF WORK APPLIED TO THE LEVER. S3 
 
 itself ; but you must remember that these can never be entirely dis- 
 posed of, as has only too often been conjectured by " perpetual 
 motion " faddists. 
 
 3. The remainder goes to do the useful work for which the 
 machine was designed, and therefore 
 
 4. The efficiency of a machine = the work S ot out ' 
 
 the work put in. 
 
 To impress these facts on the mind of the student we present 
 them in the following condensed form : 
 
 1. Work put in = force applied x the distance it acts. 
 
 2. Work lost = force absorbed in overcoming internal resistances 
 
 x the distance it acts. 
 
 3. Useful work = force given out x the distance it acts. 
 
 4. Efficiency [ = ratio of work got out to ivork put in. 
 
 5. Work put in = lost work + useful work. 
 
 Principle of Work applied to the Lever. In applying the 
 above " principle of work " to the lever, we will take the liberty 
 of neglecting the lost 
 work. We shall therefore 
 assume that the friction 
 at the fulcrum is so small 
 that it may be neglected 
 for the purpose we have 
 in view. (^ 
 
 EXPERIMENT I. Let F fr A| 
 
 AjF be a straight lever 3/| 
 
 without weight, having yy 
 
 its fulcrum at Fa force, PmNCIPLE OP WoRK APPLIED T0 A 
 W , acting vertically down- T __ 
 
 j p AI T~ iMByJoLK. 
 
 wards from the point B 1? 
 
 and a force, P, acting vertically upwards at the end A lt keeping 
 W in equilibrium. Now imagine the lever elevated to the posi- 
 tion A,F. 
 
 The work put in at A : = P x the vertical distance from A t to A^. 
 
 The work got out at B x = W x the vertical distance from B x to B,. 
 
 Therefore, since we neglect all frictional resistances 
 
 The work put in = the work got out 
 Or, P x A>A, = W x B^, 
 
 ie P B ' B > 
 
 W AA 
 
 But by Euclid the triangles A 4 FA, and BjFB a are similar in every 
 respect. 
 
54 
 
 LECTURE V. 
 
 Therefore, 
 
 Hence, 
 Or, 
 
 W ' 
 Af = 
 
 BF 
 
 But this is the equation we proved in Lecture III. with respect 
 to the lever as complying with the " principle of moments." 
 Hence the "principle of work " and the " principle of moments " are 
 in agreement. 
 
 In the accompanying figure the force P has been shown as 
 elevated through 12", and the force W as elevated through 6", 
 
 Therefore, P x 12" = W x 6" 
 
 4 - T, ' 7 
 
 P being half the magnitude of W, it has to be elevated through 
 double the distance in order that the same amount of work may 
 be done in the same time. 
 
 EXPERIMENT II. Consider the case of a simple lever, where a 
 weight, W, at B is balanced by another weight, P, at A, around 
 
 PRINCIPLE OP WORK APPLIED TO JL LEVEB. 
 
 a fulcrum at F, without friction. Let the lever be turned through 
 90, or a quarter of a revolution i.e., from a horizontal position, 
 AB, to a vertical position, A'B'. 
 Then by the definition of work 
 
 The work put in at A = P x A'F, and 
 The work got out at B = W x B'F. 
 
THE WHEEL AND AXLE. 
 
 55 
 
 It does not matter in the slightest degree how circuitous the 
 paths P and "W take in passing from their original to their new 
 positions in this case, since all we require to know is the vertical 
 distances through which P is depressed and "W elevated. 
 
 Consequently, by the " Principle of Work" 
 
 P x A'F = W 
 
 But, A'F = AF, and B'F = BF, 
 
 .-. Substituting AF for A'F, and BF for B'F, 
 We get, P x AF = W x BF 
 
 But this is the equation for the "principle of moments" which 
 we have again deduced from the " principle of tuork " by another 
 and simpler form of reasoning. We find that this latter method 
 appeals more directly to the minds of young engineering students 
 than the proofs usually found in books on Mechanics. 
 
 The Wheel and Axle. The wheel and axle has been used for 
 centuries for drawing water by a bucket from a well. It is used 
 by the navvy for lifting the 
 material which he excavates 
 from the earth, by the mason 
 for raising stones, bricks and 
 mortar, and by many other 
 tradesmen for a variety of 
 purposes ; as well as by the 
 quartermaster as a steering- 
 gear, and the able seaman as 
 a capstan. The accompanying 
 illustration shows the form it 
 takes when used for elevating 
 goods in a store or mill.* It 
 is simply a practical arrange- 
 ment for continuing the action 
 of the lever as long as re- 
 
 quired. So long as a sufficient WHEEL AND AXLE. 
 
 pull is applied to the rope, 
 
 which fits into the grooved wheel, to overcome the resistance of 
 the load attached to the chain hook, the weight will be raised. 
 The wheel and axle is therefore a form of lever by which a weight 
 may be raised through any desired height. 
 
 The Principle of Moments applied to the Wheel and 
 Axle. In the diagram let the larger circle represent the circum- 
 ference of a wheel of radius, R, to the periphery of which a force, 
 
 * The above figure represents a wheel and axle as supplied by Messrs. 
 P. & W. MacLellan, of Glasgow. 
 
LECTURE V. 
 
 P, is applied. Let the smaller circle represent the circumference 
 of the axle or barrel of radius, r, to the periphery of which is 
 applied a resistance W. Let the forces P 
 and W act in the same direction and verti- 
 cally downwards. Join the points where the 
 lines of action of the forces are tangents to 
 the wheel and axle by a straight line, AB. 
 Then, AB passes through the common centre 
 of the circles i.e., through their common 
 centre of motion or fulcrum F, and AF is 
 the effective arm for the force P, whilst BF 
 is the effective arm for the force W. In 
 fact, AFB is a straight lever in equilibrium, 
 with the fulcrum at F. 
 Therefore, taking moments about F, we have 
 
 P x AF = W x BF 
 Or, P x R, = W x r. 
 
 The Principle of Work applied to the Wheel and Axle. 
 
 EXPERIMENT III. Take a model of the wheel and axle as illus- 
 trated by the accompanying figure. Let, forces, P and W, act in 
 equilibrium, as in the previous case, at radii R and r respectively. 
 
 w 
 WHEEL AND AXLE. 
 
 W 
 
 MODEL TO TEST THE PEINCIPLE OF WOEK APPLIED TO THE 
 WHEEL AND AXLE. 
 
 Now mark carefully with a piece of coloured chalk or ink the exact 
 positions where the tape supporting P is a tangent to the wheel, 
 
THE WINCH BARREL. 
 
 and where the cord supporting W is a tangent to the barrel. Pull 
 P until the wheel and barrel have just made one complete revolu- 
 tion. Then, neglecting any force required to overcome friction at 
 the bearings of the spindle 
 
 The work put in by P = P x 27rB, 
 The work got out in raising W = W x 2irr 
 
 But the work put in = the work got out 
 . P x 27rR = W x 2rrr 
 
 Cancelling 2n- from each side of the equation 
 We have P x R = W x r. 
 
 But this is the same equation as we obtained above by applying 
 the "principle of moments." Therefore, we see that the "prin- 
 ciple of moments" and the "principle of work" harmonise. 
 
 The Winch Barrel. The wheel may be replaced by a handle 
 H, and the mere axle by a barrel or drum D, of any desired size. 
 
 W 
 
 SIDE VIEW. END VIEW. 
 
 WINCH BARREL AND HANDLE. 
 
 EXAMPLE I. A man exerts a constant force of 30 Ibs. on a 
 winch handle of 15" radius ; what weight will he be able to lift 
 attached to a rope hanging from a barrel of 5" radius ? 
 
 By the principles of moments and of work ; and interpolating 
 the numerical values 
 
 Px R-Wxr 
 30 x i5=Wx 5 
 
 ... w = 3^115 = BO Ib.. 
 
 Ship's Capstan. A partly sectional, partly outside view of 
 this useful machine is illustrated by the following figure : 
 
 A capstan is generally fixed upon the forecastle of a ship, or 
 near to the side of a quay or dock, for the purposes of warping and 
 
LECTURE V. 
 
 7 
 
 !BF - 
 
 SHIP'S 
 
 CAPSTAN. 
 
 INDEX 
 
 TO PAETS. 
 
 OH represents Capstan head. 
 SS ,, Spokes or arms. 
 
 E Eadius of S. 
 
 B Barrel. 
 
 r Radius of B. 
 
 PR represents Pall and Ratchet. 
 
 F Frame. 
 CP Capstan pillar. 
 FS Footstep of CP. 
 
 berthing the vessels. The above illustration shows a capstan as 
 built into a forecastle, where the round turned footstep, FS, of 
 the vertical cast-iron capstan pillar, CP, bears in a cast-iron or 
 cast-steel shoe fitted upon the steel or wrought beams of the main 
 deck. The frame F, which supports the casing for the pall and 
 ratchet gear, may be the beams of the upper or forecastle deck. 
 A strong rope made fast on shore is passed several times round 
 the capstan barrel B, and the slack end of the rope is coiled on 
 deck. The addition of the rope to the barrel increases the effective 
 arm or radius r, at which the resistance of the ship acts by half 
 the diameter of the rope. Eight or any desired less number of 
 wooden spokes, S, S, having their inner ends squared and tapered, 
 are fixed into hollow square holes in the cast-iron capstan 
 head CH. Then, just as many sailors as may be required to 
 
SHIP'S CAPSTAN. 59 
 
 overcome the resistance of the ship apply themselves to the outer 
 rounded ends of the spokes, and push away as hard as they can. 
 
 It will be observed that, calling, p, the force applied by each 
 sailor at radius R ; then, when we have two saiJors acting on 
 diametrically opposite spokes /?, 2R, p forms a couple tending to 
 cause rotation of the capstan in one direction. Consequently 
 from the property of couples (as we showed in Lecture III.) this 
 couple can only be balanced by another couple acting in the 
 opposite direction and having an equal moment. Such another 
 couple exists, when the resistance of the ship, W, acting with an 
 arm, r (equal to the distance from centre of capstan to centre of 
 rope), balances the corresponding reaction at the centre of the 
 capstan barrel. Hence, when the force applied by the two sailors 
 is balanced by the resistance to motion of the ship, we have the 
 one couple just balancing the other one. 
 
 Or Couple p, 2R, p balancing couple W, r, W 
 
 i.e., p x 2B = W x r 
 
 In the same way, with two, three, or four pairs of sailors, each 
 pair being supposed to act on diametrically opposite spokes, we 
 have two, three, or four couples acting in one direction, balanced 
 by one couple, viz., the resistance of the ship into the distance 
 from the centre of the capstan barrel and the reaction from that 
 centre.* 
 
 In the case of four sailors just being able to move the ship, two 
 couples, p, 2R, p+p, 2R, p, balance one couple, W, r, W; 
 
 *'.., p x 2R.+J9 x 2R = W xr 
 
 The Fusee. As an illustration of the lever action and of 
 work put into and got out of a machine, we cannot do better 
 than finish this lecture by a description of the construction 
 and action of the simple yet most ingenious contrivance termed 
 the fusee. In good watches and clocks, where the elastic force 
 of a coiled spring is used to drive the works, the fusee is used 
 for the purpose of compensating the gradually diminishing pull 
 of the uncoiling spring. The driving of the works at a constant 
 rate is the object for which a watch or clock is designed. This 
 naturally entails a constant resistance to be overcome, but since 
 one of the most compact and convenient forms of mechanism into 
 
 * The student should draw a plan of the capstan barrel, and show radial 
 lines to indicate one, two, or more pairs of diametrically opposite spokes 
 with forces, p, acting at their ends, all tending to turn the barrel in one 
 direction. He will then see that a couple formed by resistance to the stress 
 on the rope, and an equal reaction from the centre of motion, will be required 
 to act in the other direction in order that equilibrium may take place. 
 
6o 
 
 LECTURE V. 
 
 which mechanical force can be stored is that of a coiled spring, 
 and since the very nature of the spring is such that its force 
 decreases as it uncoils, we must employ some compensating 
 device between this variable driving force and the constant 
 resistance. The fusee does this in a most accurate and complete 
 
 THE FUSEE FOE A CLOCK OR WATCH. 
 
 INDEX TO PARTS. 
 
 F represents Fusee. 
 B Barrel. 
 BW Ratchet wheel. 
 
 TW represents Toothed wheel. 
 WS ,, Winding square. 
 
 manner. Looking at the accompanying figures and index to parts, 
 we see that the barrel B, which contains the watch or clock 
 spring, is of uniform diameter, and that between the outside of 
 this barrel and the fusee, or spirally grooved cone, there passes a 
 cord or chain. When the winding key is applied to the winding 
 square WS, and turned in the proper direction, a tension is 
 applied to the cord, and it is wound upon the spiral cone, thus 
 coiling up the spring inside the barrel B ; for the outer end of 
 this spring is fixed to the periphery of the barrel, and the inner 
 end to its spindle or axle, is in direct gear with the works of the 
 clock. When the spring is fully wound up it has the greatest 
 force, but it acts with the least advantage, since then the cord is 
 on the smallest groove of the cone pulley. When the spring is 
 almost uncoiled it acts with the greatest advantage, for then the 
 eord is on the largest groove of the cone. Consequently the radii 
 
THE FUSEE. 6 1 
 
 of the grooves of this cone are made to increase in proportion as 
 the force applied to the cord decreases in order that there shall 
 be a constant turning effort on the works of the clock. 
 
 The work put in when winding up the coiled spring, is given up 
 by it in overcoming the frictional resistances of the different parts 
 of the mechanism. 
 
 Or the work put in = lost work, for the whole of the work put 
 in is devoted to simply keeping the parts of the machine in 
 motion, thus leaving nothing for other work, unless the clock is 
 used to strike a bell or do some other kind of work. 
 
62 LECTURE Y. QUESTIONS. 
 
 LECTURE V. QUESTIONS. 
 
 1. State the " Principle of Work," and explain the manner in which it 
 is applied in determining the relation of a P to W in the lever. A lever, 
 centred at one end, is 15 feet long, and a weight of W Ibs. hangs from the 
 opposite end. The weight W is supported by an upward pressure of 
 28,270 Ibs. at 13 feet from the fulcrum. Find W. Ans. 24,5006 Ibs. 
 
 2. Define work put in, lost in, and got out of a machine, and prove that 
 the work put in = lost work plus the useful work. How are the "advan- 
 tage " and the efficiency of a machine reckoned ? 
 
 3. Sketch and describe the wheel and axle. Apply both the "principle 
 of moments" and the "principle of work" to find the relation between 
 the force applied and the weight raised by aid of this machine. A wheel 
 and axle is required so that the force applied at the circumference of the 
 wheel in moving through a distance of 10 feet shall raise a weight of 
 4 cwts. through a height of 2 feet. If the diameter of the axle is 10 inches, 
 find the force applied in Ibs., and the radius of the wheel in feet. Ans. 
 89-6 Ibs. ; 2 feet I inch. 
 
 4. The crank or handle which turns a windlass is 14 inches in length j 
 what must be the diameter of the axle when a man exerting a force of 
 60 Ibs. upon the handle raises a tub of coals weighing 2 cwt. 1 Ans. 
 7^ inches. 
 
 5. In a windlass the barrel is 8 inches diameter, the rope is I J inches 
 diameter, and the crank handle 15^ inches long. What force must b 
 applied at the handle to raise 2 cwt. 1 Also, what weight would be raised 
 by a constant force of 30 Ibs. applied at the handle 1 Ans. 66-8 Ibs. ; 
 100-5 Ibs. 
 
 6. A capstan is worked by four men ; each man exerts a constant force 
 of 30 Ibs. at a distance of 4 feet from the axis. A rope of f -inch diameter 
 is wound round the drum, of 5| inches radius. Find the pull on the rope 
 which balances the pressure on handles. Make a diagram showing the 
 action of the forces, and find the pressure on the central shaft of the cap- 
 stan. Ans. 921 6 Ibs. ; 921-6 Ibs. 
 
 7. Describe, with a sketch, the spring-barrel and fusee of a clock or of a 
 watch. Explain its action by reference to the principle of moments. 
 
 8. A ship's capstan has a ratchet-wheel with two detents or pawls, 
 arranged so that when one is engaged with a tooth of the wheel the point 
 of the other is midway between two teeth. Sketch the arrangement, and 
 say why in this case two detents or pawls are better than one. 
 
 9. Find the average horse-power exerted by a winding engine to lift 
 3 tons from a pit f mile deep at a uniform speed in two minutes, supposing 
 that 30 per cent, of the total work done is lost in friction. 
 
 (0. & G., 1905, O., Sec. A.) Ans. 576 H.P. 
 
LECTURE VI. 
 
 CONTENTS. Pulleys Snatcn Block Block and Tackle Theoretical Ad- 
 vantage Velocity Ratio The Principle of Work applied to the Block 
 and Tackle Actual or Working Advantage Work put in Work 
 got oat Efficiency Percentage Efficiency Example I. Questions. 
 
 Pulleys. Suppose you had to elevate a sack of flour from the 
 ground to an upper storey of a mill or store, you might place it 
 upon your back and carry it up the stairs. In doing so, you 
 would expend so many foot-pounds of work. Let the sack of 
 flour be 100 Ibs., your own weight 150 Ibs., and the height to 
 which it is raised be 30 feet. Then the 
 
 Work done in elevating the flour = 100 Ibs. x 30' = 3000 ft.- Ibs. 
 yourself =150 x 30' = 4500 
 
 Total work done a 250 x 30' = 7500 
 
 And your efficiency as a machine would be found thus 
 
 Mechanical efficiency = sefulwork work got out = 3000 ft-lba. 
 total work work put in 7500 ft.-lbs. 
 
 Or, your percentage efficiency would be found from the propor- 
 tion 
 
 7500 : 3000 :: 100 : x 
 
 7500 
 
 In other words, 60 per cent, of the total work done is lost work, 
 and only 40 per cent, is useful work. 
 
 If instead of carrying the sack upstairs, you found ready to 
 hand a long rope (with its two ends close to the ground) that had 
 been passed over a smooth iron hook fixed to the outside wall 
 above an outside landing for the particular storey of the building, 
 and, if you attached one end of this rope to the sack and found 
 that by pulling with all your strength (or say with a force of 
 150 Ibs., i.e., equal to your weight) on the other end, you could 
 just lift the sack. Then, if by this means you elevated the sack 
 to the landing, you would have expended less work than by the 
 iormer method ; for, 
 
6 4 
 
 LECTURE VI. 
 
 Work done in elevating flour = loolbs. x 30'= 3000 ft.-lbs. 
 against friction, &c. = 50 ,, x 30' = 1500 
 
 Total work done 
 
 =150 x 30 = 4500 
 
 .-. Mechanical efficiency = Hi 6 *" 1 
 
 or, ut^ 3000^.5 
 
 work put in 4500 
 
 total work 
 And the percentage efficiency is therefore 66-6. 
 
 For, 4500 : 3000 :: 100 
 3000 x 100 
 45 
 
 66-6 % 
 
 Hence 33.3 per cent., or ^ of the total 
 work put in by you in pulling at one 
 side of the rope, is spent in overcoming 
 the friction between the rope and the 
 hook and bending the rope over the hook, 
 whilst only 6 6' 6 per cent., or |, remain 
 for elevating the sack of flour. 
 
 If, instead of the iron hook you had 
 found a double-flanged deep V-grooved 
 pulley with a rope over it, as in the ac- 
 companying illustration, and that this 
 pulley revolved so easily on its bearings 
 that you had only to pull with a constant 
 
 force of no Ibs. in order to lift the sack of flour from the ground 
 
 up to the 3o-feet level, then 
 
 Work done in elevating flour = 100 Ibs. x 30' = 3000 ft.-lbs. 
 against friction, &c. = 10 x 30'= 300 
 
 PULLEY AND WEIGHTS. 
 
 Total work done 
 
 no 
 
 30 =3300 
 
 .*. Mechanical efficiency = lisefnl work . O r ^J^ out = 3ooo = .^ 
 total work ' ' work put in 3300 
 
 And the percentage efficiency is 90-9 
 For 3300 : 3000 :: 100 : x 
 
 x = 300Q * IPO sf 9Q . 9 % 
 
 Hence only 9-1 per cent, of the total work put in is lost work 
 In overcoming friction at the pulley bearing and in bending the 
 rope over the pulley. 
 
 You see, therefore, what a useful machine a pulley is, not only 
 for enabling you to change the direction of a force, but aho for 
 the saving of labour. 
 
SNATCH BLOCK, , 65 
 
 A pldley is simply a wfael and axle wherein their radii are one 
 and the same, or a lever with equal arms. Hence the principles of 
 moments and of work may be applied to it in the same way ag 
 we applied them to the lever and to the wheel and axle. 
 
 Snatch Block. If you should require to put the bend of a 
 rope on a pulley, and at the same time prevent the possibility of 
 the rope coming out of the groove, without having to reeve the 
 end of the rope between its cheeks, you would use what is called a 
 snatch block. One form of snatch block is illus- 
 trated by the accompanying figure, where on the 
 side of one cheek there is a sneck or snatch, 
 which is turned to one side, to enable the bend of 
 the rope to be placed around the U groove of the 
 pulley. The snatch then falls down and closes 
 upon the central pin. Another form has a 
 hinged snatch which can be lifted up at right 
 angles to the face of the cheek, and after the 
 rope has been put on the pulley the snatch is 
 closed down and locked by a pin attached to a 
 short chain fixed to the side of the cheek, just 
 like an ordinary front hinge for closing a chest. 
 The single movable pulley, which is used for sup- 
 porting the load to be lifted by a Chinese wind- 
 lass or by a jib crane, is sometimes called a * u '. J"" 
 snatch block (see the illustration of the wheel and ' 
 compound axle in next Lecture, and of jib cranes in Lectures 
 VIII. and XIII.). In the latter case the chain passes from 
 the barrel of the crane over the pulley at the point of the jib, then 
 vertically down, underneath the snatch-block pulley, and vertically 
 upwards to a point on the under side of the jib where it is fixed 
 by an eye-shackle with a bolt and nut. If the load, including the 
 weight of the snatch-block, be W, then, neglecting friction, the 
 
 W 
 
 pull P on the chain will be ; f or W is supported by two ver- 
 tical or parallel parts of the chain, each part carrying half the 
 load, or W = 2?. If the load be elevated any distance L, then 
 the chain will have to be pulled in on the barrel a distance of 2L, 
 for by the principle of work 
 
 The pull x its distance = the load x its distance. 
 Or, P x 2L = W x L. 
 
 The theoretical advantage is therefore 2 to i, or a certain force 
 would lift double the weight, neglecting friction. 
 
 Block and Tackle. Passing over the various arrangements 
 of pulleys for lifting weights which are treated of in theoretical 
 
66 
 
 LECTURE VI. 
 
 mechanics, we come to this well-known and Useful contrivance. 
 As will be seen from the accompanying sketch, it consists of a 
 number of pulleys (or sheaves as they are technically termed) free 
 
 to run round on a turned central 
 iron or steel spindle, and inserted 
 in a block, having their iron divi- 
 sions between each pulley, and strong 
 iron cheeks fixed to a swivel joint- 
 terminating in an iron hook hung 
 from an eye bolt. Three sheaves 
 are shown in this block, but the 
 number may range from one up- 
 wards, according to the size and 
 work to be done. There is a simi- 
 larly constructed block with two 
 sheaves, to which the weight to 
 be raised, or the body to be pulled, 
 is attached, and this is called the 
 movable block, whereas the upper 
 or home one is termed the fixed 
 block. Around the pulleys of both 
 blocks there is reeved a rope with 
 the inner end made fast to an eye 
 on the movable block, whilst the 
 free end hangs from one of the 
 outside sheaves ; but this arrange- 
 ment is frequently reversed, for the 
 inner end of the rope may be at- 
 tached to an eye on the fixed block, 
 and the free end may spring from 
 the other one (see the figure in con- 
 nection with Example I. of this 
 Lecture). The free end of the rope 
 is then ready to be pulled by the 
 hands or by aid of a winch. 
 
 Now, neglecting friction, and 
 
 supposing the rope to be perfectly flexible, a force, P, applied to 
 the free end of the rope would be transmitted throughout it to 
 the other end at the movable block. Hence the effect of this 
 force in overcoming a resistance, W, is multiplied by the number, 
 n > of parts of tfie rope which spring from the movable block. 
 
 BLOCK AND TACKLE. 
 
 Or, 
 
 W 
 
 And (i) The theoretical advantage 
 
 W 
 P 
 
I UNIVERSITY I 
 
 X^LIFOR!*^^ 
 
 THB BLOCK AND TACKLE/ 67 
 
 (2) The velocity ratio, or ratio of the distance through which 
 P acts, to that through which W is overcome in the same time. 
 
 Or, Velocity ratio = P ' 8 ^ tapce = ^ 
 
 W s distance i 
 
 In the figure there are shown three pulleys in the upper block 
 and two in the lower, with five parts of rope springing from the 
 latter ; therefore in this case n = 5. 
 
 Here W = nP = 5 P; or, P * ^ - ^ 
 
 n 5 
 
 since P must pass through five times the distance that W does 
 in the same time. 
 
 The velocity ratio = ^ d ^ stapce = = * 
 W's distance i 1 
 
 So that the theoretical advantage and the velocity ratio have the 
 same algebraical expression and numerical value. - (See note, p. 68.) 
 The Principle of Work applied to the Block and Tackle. 
 Using the very kind of block and tackle represented by the 
 previous figure, attach a light Salter's spring balance by its hook 
 to the rope where the hand is shown. Fix such a weight to the 
 lower block that the weight of rope between the blocks, the 
 movable block, and the load are 60 Ibs. Call this W. Now 
 pull the ring of the spring balance until the load rises slowly and 
 uniformly, and note the reading on the balance ; let it be 18 Ibs., 
 and let the weight of this balance and the hanging free end of 
 the rope, which is assisting the arm, be 2 Ibs. Call this total 
 pull of 20 Ibs. P ; then : 
 
 (3) 27U actual or Dicing advance = wei 8 ht raised = W = 6olbs. = 8 
 
 pull applied P 20 Ibs. 1 
 
 Lift W up through one foot exactly, and measure the length of 
 rope which you have pulled out from the upper block, and you 
 will find that it is five feet ; hemce,* 
 
 (4) The work put in = P x n= 20 Ibs. x 5 ft. = 100 ft.-lbs. 
 
 (5) The work got out = W x i = 60 Ibs. x i ft. = 60 ft.-lbs. 
 
 Work got out 60 ft.-lbs. 
 
 (6) The efficiency =- -J _ =8-3 
 
 Work put in I00 ft.-lbs. 
 
 (7) The percentage efficiency = '6 x 100 = 60 % 
 
 In the same way the efficiency of uny other block and tackle 
 may be found, and the student should carry out a series of 
 
 * The above results were obtained by the Author from a block and 
 tackle of the same kind as that shown by the previous figure, at a demon- 
 stration in his Junior Applied Mechanics class. 
 
68 
 
 LECTUEE VI. 
 
 experiments in a laboratory or workshop so as to impress the 
 various measurements and the results on his memory. He will 
 find that if the efficiency is over 50 per cent, a comparatively 
 small load will run down and overhaul the free end of the rope, 
 unless it has some restraining force applied to it, or be fixed to 
 some rigid body. It is for this reason that 
 sailors, who work very much with ordinary 
 block and tackle, always " belay " tne free 
 end of the rope when they have adjusted 
 their sails or have heaved up a body to 
 the required height. 
 
 EXAMPLE I. A tackle, consisting of an 
 ordinary double and treble block, is em- 
 ployed for lifting a weight of 600 Ibs. 
 attached to the double block. What 
 force is required, - neglecting friction ? 
 If the tackle is reversed, so that the 
 weight is attached to the treble block, 
 the free end of the rope being pulled 
 upwards, what force would now be re- 
 quired to lift the weight? (S. and A. 
 Exam. 1892.) 
 
 BLOCK AND TACKLE ANSWER. First Case. By an inspec- 
 
 2ND CASE, EXAMPLE I. ,. - ,. . J .. . T 
 
 tion or the previous figure in this Lecture, 
 
 it' will be apparent that the weight W is supported by Jive parts 
 of the rope, or n = 5. 
 
 *,- -- 
 
 n 
 
 W 
 T 
 
 = 120 Ibs. 
 5 
 
 /Second Case. Here the system is inverted, so that the block 
 with the three pulleys is lowermost, as shown by the accompany- 
 ing figure. In this case it is evident that there are six parts of 
 the rope supporting W, or n = 6. 
 
 n 
 
 Note. If a machine be supposed to work without friction, then the ratio 
 of the resd&tanoe overcome to the effort applied is termed the theoretical 
 or hypothetical mechanical advantage. If, however, friction be taken into 
 account and an effort P be able to overcome a resistance W, then the ratio 
 W 
 is termed the mechanical advantage. 
 
LBCTURB VI. QUESTIONS. v 69 
 
 LBCTUBB VI. QUESTIONS. 
 
 1. Suppose that your weight is 10 stone 10 Ibs., and that you lift a 
 weight of \ cwt. on your shoulder r and walk upstairs with it to a height of 
 20 ft. ; what work have you expended, and what will be your efficiency as 
 a machine 1 Am. 4120 ft-lbs. ; 27 per cent. 
 
 2. Suppose that you had a rope passed round a beam of wood, and that 
 you attached 4 cwt. to one end and pulled with a force of 84 Ibs. on the 
 other end and then elevated it 10 ft. : (a) what work have you put in ? 
 (b) what is the percentage efficiency of the arrangement ? (c) what is the 
 percentage of lost work ? ATM. (a) 840 ft. -Ibs. ; (b) 66-6 ; 33-3. 
 
 3. Suppose that a weight of 4 cwt. is attached to one end of a rope passed 
 round a pulley, and that you lift it 10 ft. by pulling on the other end of 
 the rope with a force of 70 Ibs. : what percentage of the work done is lost 
 in overcoming the friction at the pulley ? Ans. 20 per cent. 
 
 4. What will be the difference, and why, in the tension on the chain of 
 a crane when a snatch-block is used, and when the weight is lifted directly 
 Sketch a snatch-block, and describe its construction and action. 
 
 5. In a rope and pulley lifting block with three sheaves in the upper 
 block, and two sheaves in the lower block, find the theoretical advantage 
 gained. Give the reason for your answer, and sketch the arrangement, 
 showing where the rope is to be attached. Arts. W : P : : 5 : I. 
 
 6. Sketch an arrangement of 5 equal pulley sheaves for lifting a weight 
 of i ton. What force is exerted on the rope in your arrangement ? Ex- 
 plain the mo^e of arriving at this numerical result by the principle of 
 work. An. With 3 pulleys in upper block and 2 in lower block, 
 P = 448 Ibs. 
 
 7. A tackle is formed of two blocks, each weighing 15 Ibs., the lower 
 one being a single movable pulley, and the upper or fixed block having two 
 sheaves ; the parts of the cord are vertical, and the standing end is fixed 
 to the movable block ; what pull on the cord will support 200 Ibs. hung 
 from the movable block, and what will then be the pressure on the point 
 of support of the upper block ? Give a sketch. Ant. 71*6 Ibs. ; 301-6 Ibs. 
 
 8. A weighf of 400 Ibs. is being raised by a pair of pulley blocks, each 
 having two sheaves. The standing part of the rope is fixed to the upper 
 block, and the parts of the rope, whose weight may be disregarded, are 
 considered to be vertical. Each block weighs 10 Ibs.; what is the pres- 
 sure at the point from which the upper block hangs 1 An*. 522-5. 
 
 9. A tackle, consisting of an ordinary double and treble block, is em- 
 ployed for lifting a weight of 1000 Ibs. attached to the double block. What 
 force is required, neglecting friction ? If the tackl* is reversed, so that 
 the weight is attached to the treble block, the free nd of the rope being 
 pulled upwards, what force would now be required to lift the weight ? 
 Sketch the two arrangements. Ans. 200 Ibs.; 166-6 Ibs. 
 
 10. Apply the " principle of work " to find the relation between the force 
 applied and the weight raised by an ordinary set of block and tackle. 
 State what is meant by the following terms : {i) velocity-ratio ; (2) theo 
 retical mechanical advantage ; (3) actual or working advantage ; (4) work 
 put in ; (5) work got out ; (6) efficiency of an apparatus or machine ; 
 (7) percentage efficiency. 
 
 11. With an ordinary block and tackle haying 3 pulleys in upper block 
 and 2 in lower block i.e., 5 ropes attached to lower block it is found that 
 a pull of 50 Ibs. is required to raise a weight of 165 Ibs. Find (i) Theo 
 retical advantage and velocity ratio = 5 : i ; (2) Actual advantage =3- 3 : I ; 
 (3) Efficiency of apparatus =-66 ; (4) Percentage efficiency of apparatus =66. 
 
70 LECTURE VI. QUESTIONS. 
 
 12. If the upper block of a set of pulleys and tackle has four equal 
 sheaves, and the lower block three equal sheaves, and if a weight of one 
 ton is hung on the lower block, one end of the rope being fixed to the 
 ground and the other end free, what pull upon the free end will raise the 
 weight, and what distance will the weight rise for every yard of increase 
 of length in the free end? If the rope be fastened to the lower block 
 instead of to the ground, what pull will raise the weight ? 
 
 Ans. 373 Ibs. ; 6 inches ; and 320 Ibs. 
 
 13. A machine is concealed from sight except that there are two 
 vertical ropes ; when one of these is pulled down the other rises. How 
 would you find the efficiency of this lifting machine? What do you 
 mean by velocity ratio, and by mechanical advantage ? (S. E. B. 1901.) 
 
 14. In a lifting machine an effort of 26*6 Ibs. just raised a load of 
 2260 Ibs. ; what is the mechanical advantage 1 If the efficiency is 0755, 
 what is the velocity ratio ? Ans. 85 ; 113. (B. of E., 1902.) 
 
 15. Distinguish between force, work and rate of work. Find the pull on 
 the draw bar exerted by a locomotive which develops 600 horse-power 
 when travelling at 60 miles an hour the mechanicaf efficiency of the 
 locomotive being taken as 60 per cent. Ans. 2250 Ibs. (.0. & G., 1903, 0.) 
 
 16. Define the terms mechanical advantage, velocity ratio, and efficiency, as 
 applied to lifting tackle. 
 
 In a lifting machine an effort of 26 Ibs. just raises a load of 2200 Ibs.. 
 and the efficiency is 075. Find the values of the mechanical advantage 
 and velocity ratio. If, with the same machine, a load of 12 Ibs. lifts a 
 load of 600 Ibs., what is the new efficiency? (C. & G., 1905, 0., Sec. A. 
 Ans Mech. Adv. = 84-6 ; Vel. Ratio = 112-5 : 1 ; Efficiency = '44. 
 
KOTES AND QUESTIONS. 
 
LECTURE VII. 
 
 CONTENTS. The Wheel and Compound Axle, or Chinese Windlass The" 
 Principle of Moments applied to the Wheel and Compound Axle The 
 Principle of Work applied to the Wheel and Compound Axle Ex- 
 amples I. II. Weston's Differential Pulley Block The Principle of 
 Work applied to Weston's Differential Pulley Block Experiment I. 
 Cause of the Load not overhauling the Chain Questions. 
 
 The Wheel and Compound Axle, or Chinese Windlass. 
 
 This ingenious contrivance was first devised by the Chinese 
 for the purpose of lifting weights. The theoretical mechanical 
 advantage is very great, but it possesses the disadvantage of re- 
 quiring a long length of rope to lift the weight a small height. 
 
 Its construction and action will be easily understood from the 
 accompanying side and end views, which are taken from a model 
 
 SIDE VIEW. END VIEW 
 
 (Without End Bearing). 
 
 THE WHEEL AND COMPOUND AXLE. 
 
THE WHEEL AND COMPOUND AXLfc. 73 
 
 made in the author's engineering workshop for the purpose of 
 demonstrating its action and efficiency to his students. 
 
 The Principle of Moments applied to the Wheel and 
 Compound Axle. Taking moments about the axle, we have, 
 when there is equilibrium between P and W, 
 
 The Principle of Work applied to the Wheel and Com- 
 pound Axle. Neglecting friction, and supposing the rope to be 
 perfectly flexible, cause the wheel to make one complete revolu- 
 tion in the direction shown by the arrow near its circumference 
 on the end view. 
 
 Then, by the principle of work, 
 
 The work put in = the work got out. 
 Or, P x its distance = W x its distance ; * 
 i.e., P x circumference = W x J of the difference of the cir- 
 of wheel cumferences of the larger and 
 
 smaller axles.* 
 
 Or, P x 27rR = W x (27^ - 2irrJ 
 
 (Dividing both sides of the equation by 2r) 
 
 Which is the same result as the one above ; consequently the 
 principle of moments and the principle of work agree. 
 
 EXAMPLE I. In a compound wheel and axle, where the weight 
 hangs on a single movable pulley, the diameters of the two por- 
 tions of the axle are 3 and 2 inches respectively, and the lever 
 handle which rotates the axle is 12 inches in length. If a force 
 
 \v 
 
 * If iff raised the circumference of the larger circle on one side, 
 
 W 
 
 then is lowered at the same time on the other side, the circumference 
 
 of the smaller axle ; consequently W will be elevated a distance equal to 
 fto//the difference of the circumferences of two axles, or= (2Trr l -2irr^. 
 
74 LECTURE vn. 
 
 of 10 Ibs. be applied to the end of the lever handle, what weight 
 can be raised ? 
 
 ANSWER. Here P= 10 Ibs. ; R= 12" ; r v = 1.5" and r,= i". 
 By the principles of moments and of work 
 
 .. \V= 10 x 12 x 4 = 480 Ibs. 
 
 EXAMPLE II. In a compound wheel and axle, let the diameter 
 of the large axle be 6 inches, and that of the smaller axle 4 inches, 
 and the length of the handle 20 inches ; find the ratio of the 
 velocity of the handle to that of the weight raised. 
 
 ANSWER. Here R = 20" ; r, = 3" ; r, = 2". 
 By the principles of moments and of work 
 
 ... p-ifri-v; 
 
 W R 
 
 P 1(3-2) 
 
 W~ 20 
 
 .Stttf 6y the principle of work 
 
 P x its distance = W x its distance 
 i x P's distance = 40 x Ws distance 
 .. The velocity ratio, 
 
 Q P's distance _ 40 
 
 Ws distance 1 
 
 Weston's Differential Pulley Block. This practical appli- 
 cation of the Chinese windlass is simply a compound axle without 
 the wheel. Or, where R = r r 
 
 Hence, P x R = (R _ r) 
 
 2 
 
 where R is the radius of the larger axle- or pulley, and r the 
 radius of the smaller one. A fter describing Weston's differentia] 
 pulley block, we will deduce this formula from the " principle of 
 work " by the same kind of reasoning as we adopted in the case 
 of the wheel and compound axle. We leave the student, however 
 to apply the "principle of moments," whereby he should get the 
 same results, 
 
WESTON'S DIFFERENTIAL PULLEY BLOCK. 
 
 75 
 
 As will be gathered from an inspection of the accompanying out- 
 side view and the small diagram showing the directions of the forces 
 and their arms, it w r ill be seen that the 
 apparatus consists of three parts (i) 
 an upper block ; (2) an endless chain ; 
 (3) a movable lower block or snatch- 
 block. The upper block has a hook 
 with swivel joint, from which the iron 
 frame is suspended. In the centre of 
 this frame is a turned steel axle on 
 which rotates a 
 couple of pul- 
 leys cast in one 
 piece, and there- 
 fore rigidly con- 
 nected together. 
 The one pulley 
 is slightly larger 
 than the other, 
 and both pulleys 
 have V-grooved 
 peripheries with 
 side ridges or 
 teeth cast on 
 the inner sides 
 of the grooves, 
 so as to fit the 
 
 pitch of the links 
 
 * c ,1 , . SKELETON FIGURE OF 
 the chain, WESTON > S DIFFERENTIAL 
 which passes PULLEY BLOCK. 
 
 over them and 
 
 thereby prevent it slipping over the 
 surface of the pulleys. The lower or 
 movable pulley is simply an ordinary 
 smooth V-grooved pulley with swivel 
 and hook like that already described 
 under the heading "Snatch Block." 
 The endless chain is an ordinary open- 
 linked chain of uniform pitch and size of link. It passes from 
 the position where the hand or pull, P, is applied, over the larger 
 pulley of the upper block, underneath the lower pulley, over the 
 smaller of the upper block pulleys, and back to the starting-point. 
 (See also the small figure.) When a pull, P, is applied at this part 
 of the chain (if there were no friction), it would be transmitted with 
 undimimshed value throughout its whole length where the tension 
 
 WESTON'S DIFFERENTIAL 
 
 PULLEY BLOCK. 
 (BY HOLT & WILLETT.) 
 
76 LECTURE VII. 
 
 can act ; but, as we shall see afterwards, a large proportion of this 
 force is absorbed in overcoming friction. The stress due to the 
 load W is divided equally between the two vertical parte of the 
 chain connected to the lower block, and if W is moved through any 
 
 distance, the stress must act through double that distance. 
 
 The Principle of Work applied to Weston's Differential 
 Pulley Block and Tackle. Theoretically (i.e., leaving friction 
 out of account, the weight of the hanging part of the chain and 
 tlie weight of the lower block), we have by the principle of wwk, 
 in one revolution of the upper pulleys 
 
 P x its distance = W x its distance. 
 
 P x circumference of\_W (difference of the circumferences of 
 the larger pulley J ~ 2 1 the larger and smaller pulleys, 
 
 Px27rR = (27rR-27rr) 
 (Dividing each side of the equation by 2r) 
 PxR 
 
 2lv 
 
 (i) The Theoretical Mechanical Advantage or ratio of W to P 
 is found directly from the above equation by simple transposition. 
 W 2R 
 
 (2) The Velocity Ratio (or ratio of the distance passed through 
 by P to the distance passed through by W in the same time) is 
 also found in the same way. 
 
 P'S distal 27rR 2R* 
 
 " Ws distance i(2n-R - 2irr) R r 
 
 Or, the velocity ratio has the same numerical value as the 
 theoretical advantage. 
 
 EXPERIMENT I. With a Weston's differential pulley block, 
 having in the upper block one pulley with an effective radius of 
 4" (i.e., from the centre of the pulley to the centre of the chain 
 which passes round it), and a smaller pulley with an effective 
 radius of 3^", you can just lift a total load of 100 Ibs. (including 
 the dead weight, the lower block, and the hanging parts of the 
 chain) by a pull of 20 Ibs. on the chain. 
 
 * Dividing numerator and denominator by r does not alter the fraction 
 
WESTON'S PULLEY BLOCK AND TACKLE, 77 
 
 In this case the theoretical advantage and the velocity ratio are 
 ach equal to 
 
 2R 2 x4* 8 16 
 
 Or, the pull on the forward side of the chain must act through 
 1 6 ft. for every foot the load is raised. 
 
 (3) The Actual or Working Advantage of the machine is, how- 
 ever, only as 
 
 W IPO Ibs. 6 
 P = 20 Ibs. = f 
 
 (4) The Work put in in lifting W i ft. is 
 
 Px 16 = 20 Ibs. x i6' = 320 ft.-lbs. 
 
 (5) The Work got out is = W x i = 100 Ibs. x i' = 100 ft.-lbs. 
 
 //:\ m w - Work got out 100 ft. Ibs. 
 
 (6) The Efficiency \* =^ r- 5 ir=~ = --- * n = -3125. 
 
 Work put in 320 ft. Ibs. > 1 *->' 
 
 (7) The Percentage Efficiency is 
 
 = .3125x100 = 31.25%. 
 
 This is a very low efficiency for a machine, but it accounts for 
 one of the useful properties of the Weston's differential pulley 
 block viz., that you can lift a weight by it, then let go your 
 hold of the chain, and the weight will remain hanging in the 
 exact position you left it, without overhauling the chain in the 
 slightest degree. It is therefore an extremely useful appliance 
 in engineering workshops where, for example, a slide valve and 
 its valve casing port face have to be scraped so as to fit each other. 
 After rubbing the valve on the port face, you can lift the valve 
 by aid of a Weston's block, and leave it hanging, without any 
 fear of its overhauling the chain which supports it, until you have 
 scraped off the high or hard parts from the port face, when you 
 can lower it for another rub. Or, in the case of having to adjust 
 the centres of a heavy job to be turned in a lathe, you can lif t the 
 job from the lathe by a Weston's block, and leave it hanging 
 quite free at the most convenient height to be acted upon, until 
 you are ready to lower it again into position. Of course, with 
 such apparatus, although the theoretical advantage is great, the 
 actual or working advantage is small ; yet this property of not 
 overhauling is of such importance that appliances possessing it are 
 constantly being used in every engineering workshop. 
 
 Cause of the Load not Overhauling the Chain. In the 
 first place, the chain cannot slip round the pulleys of the upper 
 block, because the links of the chain fit into the notches or 
 
78 LECTURE m 
 
 between the outstanding teeth or ridges cast in their grooves. 
 In the second place, the friction between the pulleys of the upper 
 block and their axle is so great, that more than 50 % of the " work 
 put in " is expended in overcoming it. 
 
 To prove this, take the case of the above experiment. When 
 the pull P and the load W are both in action, the downward 
 pressure (due to these two forces alone) between the pulleys and 
 their axle is 120 Ibs. (TOO Ibs. for W 4- 20 Ibs. for P). Now, 
 remove the 20 Ibs. pull, and you only relieve the pressure causing 
 friction by ; for TOO Ibs. (or of 120 Ibs.) is still there. But 
 friction is practically proportional to the pressure in such a case, 
 and therefore, although the pull required to lift the load be 
 removed, of the total friction will remain at the upper block, 
 and the friction at the lower block is unaltered. In lifting the 
 load of TOO Ibs. i ft., there was put into the machine 320 ft.-lbs., 
 or 220 ft.-lbs. was lost work, required solely to overcome friction. 
 Consequently, to lower the load of 100 Ibs. i ft. there would have 
 to be expended at least of 220 ft.-lbs., or not less than 183 ft.-lbs. 
 But the load can of itself only give out 100 ft.-lbs. in descending 
 i ft. ; therefore it must be assisted by at least 83 ft.-lbs. 
 (183 ft.-lbs. 100 ft.-lbs.) put into the chain on the slack side, or 
 where it comes down from the smaller pulley. 
 
 This principle of the weight not running down (or overhauling, 
 as it is technically termed) is common to all machines wherein 
 more than 50 % of the force applied is spent in merely overcoming 
 frictional resistance.* 
 
 * The student should be most earnestly warned against such expres- 
 sions as the following, which are only too common in books dealing with 
 Applied Mechanics : " By increasing the number of sheaves in a pair of 
 pulley blocks, the power may be increased." Now, power, or the rate of 
 doing work, can never be increased by any continuously acting mechanical 
 device, so long as the work given out depends directly on the work put in. 
 It is simply the force which can be augmented, whilst the distance 
 through which it acts is diminished. Of course, in the case of a pile- 
 driver where the weight is lifted slowly and let go suddenly, so that the 
 rate of giving out work is greater than the rate of putting it in, it is true 
 that the power is increased. But this is not a continuous acting mechani- 
 cal device in the sense referred to above. The fundamental principle to 
 be observed is, that no more work can be got out than has been put in. 
 The term " mechanical powers " should also be avoided, and the expres- 
 sions " simple machines " or " mechanical elements " used instead. 
 
LECTURE VII. QUESTIONS^ 79 
 
 LECTUBB VII. QUESTIONS. 
 
 1. Sketch and describe the wheel and compound axle, or Chinese wind- 
 lass. Apply the " principle of moments " and the " principle of work " to 
 find the formula for the relation between the force applied and the weight 
 raised by this machine. 
 
 2. In a compound wheel and axle the diameter of the two parts of the 
 axle are 5 and 6 inches respectively. The weight raised, viz., W y hangs 
 from a single movable pulley in the usual manner, and is supported by a 
 pressure, P, applied perpendicularly to a lever handle 15 inches in length. 
 Find the ratio of P to W. Sketch and describe the compound wheel and 
 axle, and state its inconveniences. Ans. i : 60. 
 
 3. A force of 20 Ibs. draws up W Ibs. by means of a wheel and compound 
 axle. The diameter of the wheel is 5 feet, and the diameters of the parts 
 of the compound axle are 9 and 1 1 inches respectively ; find W. Ans. 
 1 200 Ibs. 
 
 4. In a compound wheel and axle, let the diameter of the larger axle be 
 8 inches, and the radios of the smaller one 2 inches, while the force applied 
 to the handle passes through 47*12 inches in one revolution. Find the 
 ratio of the velocity of the handle to that of the weight being raised. 
 Ans. 7-5:1. 
 
 5. Explain the mechanical principle upon which Weston's pulley block 
 is constructed, and give a skeleton diagram showing the direction of all 
 the forces at work. If the weight which is being raised is left hanging, 
 and the pull removed, why does the weight not descend T 
 
 6. Sketch and describe Weston's differential pulley block. If the diame- 
 ters of the pulleys are 4 and 4} inches, what weight can be raised by a 
 force of 20 Ibs. ? If the weight to be raised is half a ton, what force must 
 be applied to the leading side of the chain ? (Neglect friction.) Ans. 
 360 Ibs. ; 62-2 Ibs. 
 
 7. Determine the relation between P and W in Weston's differential 
 pulley block (i) by the "principle of moments" ; (2) by the "principle 
 of work. " 
 
 8. If a weight is raised by a Weston's differential pulley block at the 
 rate of 5 ft. per minute, and the diameters of the pulley of the compound 
 sheaves are 7 and 8 inches respectively, at what rate must the chain be 
 hauled ? Work out answer in full from the principle of work. Ans. 80 ft. 
 per minute. 
 
 9. By experiment with a Weston's differential pulley block it was found 
 that a pull of 15 Ibs. on the leading side of the chain was required to lift 
 a weight of 60 Ibs. (including the weight of the lower pulley and hook). 
 The dimensions of the apparatus were radius of larger pulley, 2 inches ; 
 radios of smaller pulley, 175 inches. Find (i) the theoretical advan- 
 tage ; (2) the actual or working advantage ; (3) the efficiency or modulus ; 
 (4) the percentage efficiency of the apparatus. Why does the weight 
 remain suspended when there is no pull on the chain T Ans. (i) 16 : i ; 
 (2) 4 : i ; (3) -25 ; (4) 25. 
 
 10. In a Weston policy block, the diameters of the two pulleys are 8 ms. 
 and 7 ins. respectively, and it is found that a pull of 25 Ib. is sufficient 
 to raise a weight of 240 Ib. Find the efficiency of the tackle. Ans. 30%. 
 
 (C. & G., 1903, O., Sec. A-) 
 
 11. Describe how you would proceed to determine experimentally (i) the 
 velocity ratio, (2) the mechanical efficiency of a Differential Pulley Block. 
 
 (B. of E., 1904.) 
 I 
 
( 8o ) 
 
 LECTURE 
 
 CONTENTS. Graphic Demonstration of Three Forces in Equilibrium- 
 Parallelogram of Forces Triangle of Forces Three Equal Forces in 
 Equilibrium Two Forces acting at Right Angles Resolution of a 
 force into Two Components at Right Angles Resultant of Two Forces 
 acting at any Angle on a Point Resultant of any number of Forces 
 acting at a Point Example I. Stresses in Jib Cranes Examples 
 II. III. Stresses on a Simple Roof Example IV. Questions. 
 
 IN Lecture I. we explained and illustrated how a force may be 
 represented by a straight line both in direction and magnitude, 
 and we defined the terms components, equilibrant, resultant, re- 
 solution, and composition of forces. We will now discuss briefly 
 the case of three forces in equilibrium when acting towards or 
 from a point, as well as the parallelogram and the triangle of 
 forces with examples, before taking up the inclined plane and 
 friction. 
 
 Graphic Demonstration Of Three Forces in Equilibrium. 
 EXPERIMENT I. Take a black board which (for convenience of 
 handling and demonstration before a class) may be of the form 
 shown by the accompanying figure. Select two movable clamps, 
 each fitted with a small V-grooved pulley about 2 inches in dia- 
 meter, with a minimum of friction at their bearings, and fix them 
 to the outside of the board as indicated. Pass a very fine flexible 
 cord over the pulleys, and attach to the ends of this cord S hooks. 
 Hang from these hooks weights of say 24 oz. and 32 oz., and from 
 the cord (anywhere between the pulleys) another cord with an 
 S hook and a weight of, say, 40 oz. After a few up-and-down 
 oscillations these three weights will come tp rest in the definite 
 position shown by the figure, and if you disturb them from this 
 position they will invariably return to it again. Consequently, 
 you conclude that the three forces acting from their common 
 point of attachment are in equilibrium, and that the force 40 oz 
 is the equilibrant of the two forces 24 oz. and 32 oz. 
 
 * This Lecture may require two meetings of a class when the students 
 have had no previous training in Theoretical Mechanics. In any case, it 
 will be well to spend at least one with a revisal hour before the written 
 examination, which should now take place upon the work gone over since 
 the beginning of the session, prior to the Christmas holidays. 
 
FORCES IN EQUILIBRIUM. 
 
 81 
 
 With a piece of finely pointed white chalk, draw lines (from the 
 point where the three forces act) on the black board parallel to the 
 cords, and plot off from this point to any convenient scale (say by 
 aid of a two-foot rule) distances along them to represent their 
 respective magnitudes. Extend from the same point in an upward 
 vertical direction another line, and mark it off to represent 40 oz. 
 This line evidently corresponds, in point of application, direction, 
 and magnitude, to the resultant of the components (24 oz. and 
 $2 oz.), for it is equal and opposite in direction to their equili- 
 
 GRAPHIC REPRESENTATION OF FORCES IN EQUILIBRIUM. 
 
 brant. From the extremity of this resultant draw lines joining 
 tlie outer ends of the components (24 oz. and 32 oz.). Then you 
 a parallelogram whoso adjacent sides from the point of 
 
 have 
 
 application, represent, both in direction and magnitude, the com- 
 ponent forces, and whose diagonal represents, also both in direc- 
 tion and magnitude, their resultant. 
 
 If any other pair of convenient weights be selected and applied 
 in the same way, you can find an equilibrant and resultant for 
 them. From these experiments you conclude that a general prin- 
 ciple, termed the " parallelogram of forces," is true without having 
 recourse fa any special mathematical reasoning. 
 
LECTUKE VIII. 
 
 Parallelogram of Forces. If two forces, acting simultane- 
 ously towards or from a point, be represented in direction and 
 magnitude by the adjacent sides of a parallelogram, then the re- 
 sultant of these forces will be represented in direction and magnitude 
 by the diagonal of the parallelogram which passes through their point 
 of intersection. 
 
 For example, let any two forces, P and Q, act from the point 
 at any convenient angle, say 60, then, if OA and OB be 
 
 plotted to scale to represent these 
 forces in direction and magnitude, 
 the diagonal OD of the parallelo- 
 gram OADB will represent in direc- 
 tion and to the same scale their 
 resultant R. But the resultant E, 
 is equal and opposite in direction 
 to a force E, which would exactly 
 balance the effect of P and Q, or to 
 a force represented in direction and 
 in magnitude by the line DO. Further, since the side AD is equal 
 and parallel to the side OB, it may be taken to represent Q in 
 direction and magnitude. Hence we have the three sides of the 
 triangle OAD taken in the order OA, AD, DO, representing in 
 direction and magnitude three forces, P, Q, E, in equilibrium, 
 acting from the point O. Hence we have a general proposition 
 termed the " triangle of forces," or a deduction from the "paral- 
 lelogram of forces." 
 
 Component 
 
 PARALLELOGRAM OF FORCES. 
 
 TRIANGLE OF FORCES. 
 
 Triangle of Forces. If three forces acting towards or from a 
 point are in equilibrium, and a triangle be drawn with its sides 
 
TRIANGLE OF FORCES. 83 
 
 respectively parallel to those forces taken in due order, then the 
 forces will be represented to scale by the sides of the triangle. 
 
 CONVERSELY : If three forces acting towards or from a point 
 are represented in direction and to scale by the sides of a triangle 
 taken in due order, these three forces are in equilibrium. 
 
 For example, let the three forces P, Q and E act from the 
 point 0, and be in equilibrium. Draw a triangle with its sides, 
 P, Q, E, respectively parallel to these forces ; then the sides of 
 this triangle, taken in that order, represent to the same scale these 
 forces. Or, if the triangle, whose sides are respectively P, Q and 
 E, represent in direction and to scale the three forces P, Q and 
 E, as they act from a point 0, these forces are in equilibrium. 
 We have shown by a dotted line the resultant R, and its direc- 
 tion as opposed to E, by the same side of the triangle. 
 
 It is quite evident that if the forces P, Q and E acted 
 towards the point 0, instead of from it, the triangle P, Q, E 
 would still represent these forces in magnitude, but the direction 
 of all the arrows would have to be pointed the opposite way. 
 
 SPECIAL CASES. Three Equal Forces in Equilibrium. It can 
 easily be proved by the apparatus used for Experiment I., or by construe 
 tion, that if you have three equal forces in equilibrium they must act at 
 120 from each other, and that the triangle representing their directions 
 and magnitudes will be an equilateral triangle, or a triangle whose angles 
 are each equal to 60. 
 
 Two Forces acting at Hight Angles. In this case it can be 
 
 E roved by the same apparatus, or by Euclid, Book I. Prop. 47, that any two 
 Drees P and Q, acting at right angles to each other, have a resultant K, or 
 arc balanced by a third force E, of such magnitude that 
 
 Consequently, if you hare any two forces in the proportion of 3 to 4 
 acting at right angles to each other, their resultant will proportionate!/ 
 be 5. 
 
 For, suppose P = 3a, Q=4<z, where a is any number of units of force. 
 Then, R 
 
 .. K= 
 .*. P: 
 
 Or, P:Q:R = 3 : 4 : 5 . 
 
 Conversely, if any two forces in the proportion of 3 to 4 units are 
 balanced by a third force proportionately of 5 units, the forces 3 and 4 
 must be acting at right angles to each other. 
 
8 4 
 
 LECTURE VIII. 
 
 Resolution of a Force into Two Components at Right Angles 
 to each other.* Let R be the force to be resolved, P and Q the com- 
 ponents, and let K make an angle, 0, with the force Q. 
 
 Then . 
 
 And . 
 
 Also . 
 
 . R . Cos 9 = Q ; f or Cos = 
 
 R 
 
 . E . Sin = P ; for Sin d = ? 
 
 B - =i= Tane 
 
 Resultant of Two Forces acting at any Angle on a Point. 
 
 The proof of this general case must be left to the Author's Advanced 
 Treatise on Applied Mechanics, but the formula may be given here, viz. : 
 
 R 2 = P 2 + Q 2 + 2 P x Q Cos a 
 
 where P and Q are any two forces, R their resultant, and a the angle 
 between the directions of the forces P and Q. 
 
 If P=Q, then 
 
 R 2 = P 2 + P 2 + 2P 2 Cos a = 2P 2 + 2P 2 Cos a 
 
 or, R 2 = 2 P 2 (i + Cos a) 
 
 /. R = 2P Cos 2. 
 
 2 
 
 = 4P 2 Cos - 
 (Since Cos a = 2 Cos 2 --!) 
 
 Resultant of any Number of Forces Acting at a Point. Let 
 3 j, P, 2 , P a , &c., be any number of forces acting at a point ; then, by the 
 parallelogram of forces find a resultant, R,, 
 for P, and P a ; and a resultant, E y for R, and 
 P s ; and so on. The last resultant will be the 
 resultant of all the forces. 
 
 Example I. Forces 3, 5 and 7 units act 
 from a central point O at equal angles. Find 
 the resultant. 
 
 AKSWEB. Let OA, OB and OC represent 
 the forces in direction and magnitude. Then 
 you can follow out the above rule and find a 
 resultant for, say, 3 and 5 call this R, ; and 
 finally find a resultant for R, and 7. But it is 
 obvious that you may subtract 3 units from 
 each OA them without affecting the result, 
 since the forces are acting at equal angles 
 from each other. This will destroy one of 
 them, and leave OBj to represent 2 units, 
 and OC, to represent 4 units. Then, by the 
 parallelogram of forces you find the resultant 
 R = 3'5 units. 
 
 * The reverse of this may be applied to the composition of two or more 
 forces acting at a point in one plane, but we will leave the demonstration 
 of such problems, as well as that of the polygon of forces, to our Advanced 
 Book on Applied Mechanics, 
 
STRESSES IN JIB CRAVES. 
 
 Stresses in Jib Cranes. As a practical example of the appli- 
 cation of the " triangle of forces," take the case of an ordinary 
 hand-worked jib crane. The load is suspended from the hook 
 H of the snatch-block SB ; or, in the case of a crane for lifting 
 light loads quickly, to a simple hook with a swivel attached directly 
 
 Note. See under page heading " Notes 
 and Questions " at the end of this 
 Lecture, a Fig. and description of Butters 
 Brothers and Co.'s PATENT NEW JIB 
 ARRESTER. 
 
 HAND-WORKED JIB CRANE. 
 INDEX TO PARTS. 
 
 J represents Jib. 
 JP Jib pulley. 
 SB Snatch-block. 
 
 H Hook. 
 
 CP represents Central post. 
 
 F Framing. 
 WG Wheel gear. 
 
 C Chain. 
 TR Tie-rods. 
 
 to the end shackle of the chain C, as it comes down from the jib 
 pulley JP, instead of the chain passing round a snatch-block 
 pulley, and up to an eyebolt near the point of the jib. 
 
86 
 
 LECTURE VIII. 
 
 1 i) The load produces a tension on the chain C. 
 
 (2) A thrust along the jib J, from the jib pulley to the 
 eye-bolt connecting the shoe of the jib to the bottom of the 
 framing F. 
 
 (3) A tension in the tie-rods from the top of the framing to 
 their connection with the top of the jib. 
 
 (4) This tension on the tie-rods produces a horizontal pull, 
 tending to bend and break the crane-post CP, where it leaves the 
 upper foundation plate-bearing and joins the framing. Cranes 
 for heavy lifts require a back balance weight to counteract this 
 force. (See the third figure in Lecture XIII.) 
 
 (5) It also causes an upward pull, tending to unship or lift the 
 crane-poet from its bearings in the upper and lower foundation 
 plates. 
 
 The directions and values of these stresses will be better under- 
 stood by the student after considering a particular example. 
 
 EXAMPLE II. In a hand-worked jib crane of the form shown 
 by the above figure, the length of the jib is 30 feet, the lengths of 
 the tie rods are 25 feet each, and the vertical distance between 
 the attachments of the tie-rods and of the jib to the framing, is 
 12 feet. Find the stresses produced on these parts of the crane 
 
 FRAME ANP STRESS PIAGRAV FOR A JIB GRANS. 
 
v STRESSES IN JIB CRANES. 87 
 
 by a load of 10 tons hung from the hook, neglecting all other 
 stresses produced by the weight of the several parts of the 
 crane. 
 
 ANSWER. First, draw a "frame diagram," or figure to scale, 
 representing the directions and the lengths of centre lines of the 
 various parts under stress. As shown by the frame diagram of 
 the accompanying upper figure, AB represents the 1 2 feet vertical 
 distance between the foot of the jib and inner ends of the tie-rods 
 marked post, BC represents the 30 feet jib, AC the 25 feet tie- 
 rods, and CW the lo-ton load all to the same scale. 
 
 Now, it is evident from an inspection of this figure that the load 
 W causes 
 
 (1) A vertical downward tension on the chain from C to W. 
 
 (2) A thrust or compression along the jib from C to B, which 
 produces an equal and opposite reaction from the framing at B 
 along the jib from B to C. 
 
 (3) A tension on the tie-rods from A to C. 
 
 (4) This tension on the tie-rods may be resolved into a hori- 
 zontal pull or force from A towards the direction of W, tending 
 to bend or break the poet about B. 
 
 (5) Also, a vertical upward pull or force in the post from B 
 toward^ A. 
 
 The student should mark the directions of these various forces 
 by arrowheads on his frame diagrams. 
 
 Second, draw a " stress diagram," viz., ab, vertical and to a con- 
 venient scale, to represent the downward force of the jo-ton load 
 on the point C ; be, parallel to the reaction along the jib from B to 
 C ; and ca, parallel to the tension in the tie-rods from G to A. 
 
 Then, by " the triangle of forces," since we have three forces 
 acting from the point C (viz., load, reaction along jib, and tension 
 in tie-rods) in equilibrium, and since we have drawn a triangle with 
 its sides respectively parallel to these forces taken in due order ; 
 the forces will be represented to one scale by the sides of this 
 triangle. Consequently, ab represents the load to scale ; be the 
 reaction along the jib; and ca the tension in the tie-rods. Now, 
 this tension on the tie-rods may be resolved into vertical and 
 horizontal components by the method already described in this 
 Lecture ; therefore, a vertical line, ad, represents the vertical 
 component or upward pull on the post, and dc the horizontal pull 
 on the same, both in direction and to the same scale as ab repre- 
 sents the load. By applying the scale to which ab has been 
 drawn to represent 10 tons (viz., yV'to i ton), be shows 25 tons; 
 ca, 20-7 tons (which would be 10-35 tons on eac * 1 tie-rod if they 
 were parallel to each other, but more if inclined from the jib-head 
 to the outside of the framing) ; cd, 20 tons; and ad t 4 tons. 
 
88 
 
 LECTURE VIII. 
 
 Force 
 Diagram 
 
 VC 
 
 TENSION IN THE TIE-ROD 
 AND THRUST IN THE JIB 
 OF A CRANE. 
 
 EXAMPLE III. In a common crane the jib is 15 feet long, and 
 the tie-rod 1 2 feet. The tie-rod is attached to the crane post at a 
 point 5 feet above the foot of the jib. If a weight of 6 tons be 
 hung from the point of the jib, find the tension in the tie-rod and 
 the thrust in the jib. 
 
 ANSWER. First draw the frame diagram as explained in Ex- 
 ample II., marking the lengths of the parts by dotted lines and 
 
 arrow-heads. (The student in his 
 diagrams should also mark the direc- 
 tions of the stresses.) 
 
 Second, on the line of action of 
 the weight W draw ca to scale to 
 represent the direction and magni- 
 tude of the weight, 6 tons. Then 
 draw cb parallel to the jib, and ab 
 parallel to the tie-rod. The triangle, 
 cab, represents by its sides to one 
 scale the magnitudes of the forces 
 viz., 14.4 tons tension in the tie- 
 rods and 1 8 tons thrust or reaction 
 in the jib.* 
 
 Stresses on a Simple Roof. 
 
 EXAMPLE IV. The weight on each principal of a simple 
 triangular roof is i ton. Find the stresses on the points of 
 support and in the Ssveral members of the principal. 
 
 ANSWER. First, draw a frame diagram of the principal, where 
 AB and AC represent the direction and length of the rafters, 
 and BO represents the tie-beam. 
 
 Then, the whole weight may be supposed to act in a vertically 
 downward direction, AW, from the junction of the rafters through 
 the middle of the tie-beam. This weight naturally produces a 
 
 W 
 
 pressure at B and at of . It also produces at these points 
 
 reactions K x and K 2 , each equal to , since the whole is sym- 
 metrically balanced about the central vertical line AW. Further, 
 there is a stress of compression along the rafters in the directions 
 AB and AC, and consequently an equal and opposite reaction 
 
 * We have purposely used the letters ABC and abc differently placed 
 from the previous figures in Example II., and have drawn the stress dia- 
 gram in a different position, in order to teach the student that he must not 
 depend upon his memory with regard to letters, but upon a clear under- 
 standing of the "triangle of forces." Students should draw their frame 
 and stress diagrams to as large a scale as their exercise books will admit. 
 
STRESSES IN JIB CRANES. 
 
 8 9 
 
 along those members from B to A, and from C to A. Also, there 
 is an equal tension on the tie-beam from its centre towards B and 
 towards C. 
 
 Second, draw the stress diagram for the three forces that are in 
 equilibrium at the bearing C (viz., the vertical downward pressure 
 
 W 
 
 , the horizontal tension along the tie-beam and the reaction along 
 
 the rafter from C to A) by plotting DC as a vertical line to scale to 
 
 W 
 
 represent , or 10 cwt., and drawing DE parallel to AC, and 
 
 producing the tension on the tie-beam until it meets this line DE. 
 Therefore, the other sides of the triangle DCE represent in 
 
 R 
 
 cwt. 
 
 FEAME AND STEESS DIAGBAMS OP A SIMPLE ROOF. 
 
 direction and to the same scale as DC represents 10 cwt. ; the ten- 
 sion on the tie-beam by CE, equal to loj cwt. ; and the reaction 
 along the rafter by ED, equal to 1 5 cwt. 
 
 In a precisely similar manner the stresses at the bearing B may 
 be found by the " triangle of forces." 
 
 We will leave the more complicated questions in graphic statics 
 to our book on the Advanced Stage of Applied Mechanics, since 
 we believe the elementary or tirst year's student will find that 
 what has been included in this Lecture is sufficient to enable him 
 to understand what will be brought before him in the future 
 Lectures of this book, as well as prepare him for answering the 
 various problems which are likely to be asked of him. 
 
90 LECTURE VIII. QUESTIONS. 
 
 LECTURE vm. QUESTIONS. 
 
 1. State the principle of the parallelogram of forces, and explain how you 
 would prove the truth thereof by experiment. A vertical force of 50 Ibs. 
 is balanced by two forces of 30 Ibs. and 40 Ibs. Find their directions and 
 the angle between them. 
 
 2. Represent the point of application, the direction and the magnitude 
 (to a scale of ^ inch to a pound) of the following forces : 10 Ibs. acting 
 northwards, 15 Ibs. acting eastwards, 20 Ibs. acting southwards, and 25 Ibs. 
 acting westwards, all from one point. Find their resultant, and its direc- 
 tion. Ans. 14*14 Ibs. acting south-west. 
 
 3. State the principle of the triangle of forces. Three forces, P, Q and 
 R, act from or towards a point, and are in equilibrium. Show graphically 
 how you would represent their magnitude and direction by the three sides 
 of a triangle taken in order. Explain the converse of this question. 
 
 4. Two ends of a piece of cord are fastened to two nails in a wall 8 ft. 
 apart in a horizontal line. The cord is 10 feet in length, and has a knot 
 4 ft. from one end, from which point a weight of 25 Ibs. is suspended. 
 Find by construction the stresses on the nails, and indicate their direction 
 by arrows. A ns. 22'$ Ibs. ; 17-5 Ibs. 
 
 5. Show how to resolve a force into two components at right angles to 
 each other. A force of 100 Ibs. acts at (ist) 30, (2nd) 45, (3rd) 60, (4th) 
 75 to the horizontal. Find by construction the vertical and horizontal 
 components for each case, and prove your results by calculation. 
 Ans. Vertical components 150 Ibs. ; 707 Ibs. ; 86'6 Ibs. ; 96*6 Ibs. 
 Horizontal components : 86 '6 Ibs. ; 70*7 Ibs. ; 50 Ibs. ; 25-8 Ibs. 
 
 6. Sketch an ordinary hand-worked jib-crane. Explain its action by an 
 index to parts, and show how the various stresses due to a load on the 
 chain act, by aid of a frame and a stress diagram. Nine tons is hung 
 from the end of the jib of a crane, which is inclined to the horizontal at 
 an angle of 60. If the compression on the jib is 16 tons, find by frame 
 and stress diagrams the tension on the tie-rod. Ans. 9-4 tons. 
 
 7. In a crane, show the method of estimating the tension of the tie-rod 
 and thrust on the jib when a given weight is hung from the end of the jib 
 If the load = 6 tons, and the tension of the tie-rod (which makes an angle 
 of 60 with the vertical) = 18 tons, find by a diagram drawn to scale the 
 thrust on the jib. Ans. 21 '6 tons. 
 
 8. In a common crane the jib is 30 ft. long and the tie-rod 24 ft. The 
 tie-rod is attached to the crane-post at a point 10 ft. above the foot of the 
 jib. If a weight of 10 tons be hung from the point of the jib, find by con- 
 structing a frame and a stress diagram (a) the tension on the tie-rod ; 
 (b) the thrust on the jib; (c) the horizontal pull on the post ; (d) the up- 
 ward pull on the same. Ans. (a) 24 tons ; (b) 30 tons ; (c) 2i - 2 ; (d) ii'2 
 tons. 
 
 9. A symmetrical pair of steps, hinged together at the top and con- 
 nected together by a string at the bottom, stands on a smooth horizontal 
 plane. If the length of each side be 3 feet 3 inches, and the string be 
 3 feet in length, find the tension o the string when a person of 140 Ibs. in 
 weight stands on the steps at a height of 2 feet from the ground ? How 
 is the tension of the string affected as the person ascends the steps ? 
 Ans. 24-8 Ibs. When the person is at the very foot of the steps the 
 
LECTURE VIII. QUESTIONS. Ql 
 
 tension =o; when he is at the top, the tension is a maximum of nearly 
 36-25 Ibs. 
 
 10. A rectangular trap-door measuring 4 feet square and weighing 
 75 Ibs. is hinged with one edge horizontal, and is supported in the 
 horizontal position by a chain which is connected with the middle point 
 of the outer edge of the trap-door, and with a point vertically over the 
 middle point of the edge in which the hinges are fixed, but 7 feet 
 above it. Sketch the arrangement, and determine the tension upcn the 
 chain and the reaction on the hinges. Ana. T = K = 43 Ibs. 
 
 (I.C.E. Feb. 1902.) 
 
 11. A load, W, of 2000 Ibs. is hung from a pin, P., at which pieces AP 
 and BP, meet like the tie and jib of a crane. The angles WPB and WPA 
 are 30 and 60 respectively. Show by a sketch how to find the forces in 
 AP and BP. Distinguish as to a piece being a strut or a tie. Ant, Strut, 
 3464 Ibs, ; tie, 2000 Ibs. 
 
 12. Two pieces in a hinged structure meet at a pin, and a load is 
 applied at the pin. Show how we find the pushing or pulling forces in the 
 pieces. Describe an apparatus which enables your method to be illus- 
 trated. 
 
 13. The weight of a chain hanging from two points of support is 
 500 Ibs. Its inclinations to the horizontal at the points of support are 30* 
 and 50 respectively ; what are the tensions at the points of support f 
 Ans. 326-4 Ibs. ; 440 Ibs. 
 
 14. The weight of a chain hanging from two points of support is 
 220 Ibs. ; its inclinations to the horizontal at the points of support are 25 
 and 42 respectively ; what are the tensions in the chain at the points of 
 support? Ana. ^=177 '6 Ibs. ; T^=2i6'6 Ibs. 
 
 15. A chain weighing 800 Ib. is hung from its two ends, which are 
 inclined to the horizontal at 40 and 60 respectively. What are the 
 forces in the chain at the points of suspension ? Ans. 400 and 600 Ibs. 
 
 (B. of E., 1903.) 
 
 16. In a lo-ton crane the jib is 35 ft. long, the tie 30 ft. long, and the 
 crane post 15 ft. high. Neglecting the effect of the tension in the chain, 
 obtain the longitudinal forces in the tie and jib Ans. Stress in tie = 
 20 tons ; stress in jib = 23 tons. (C. & G., 1903, O., Sec. B.) 
 
 17. A machine 5 tons in weight is supported by two chains ; one of these 
 goes up to an eyebolt in a wall and is inclined 20 to the horizontal ; the 
 other goes up to a roof principal and is inclined 73 to the horizontal ; find 
 the pulling forces in the chains. You may use a graphical or any other 
 method of calculation you please. Ans. 1*47 and 47 tons. 
 
 (B. of E. 1904.) 
 
 1 8. A cord is wrapped six times round four parallel rods, which pass 
 through the four corners of a rectangle, and are perpendicular to the plane 
 of the rectangle ; the long sides of the rectangle are 2^ times the length 
 of the short sides. The cord, while being wrapped round the rods, is kept 
 taut with a uniform tension of 7^ pounds. Find graphically, or in any 
 other way, the force exerted by the cord on any one of the rods. 
 Ans. Force on any one rod = 64 Ibs. (B. of E., 1905.) 
 
 19. A roof trnss consists of two rafters, equally inclined, connected by a 
 horizontal tie at a distance of a feet below the apex. If the total span be 
 I feet and the load carried at the apex be w tons, show that the tension in 
 the tie rod is wL^a tons. (C. & G M 1904, 0., Sec. B.) 
 
92 NOTES AND QUESTIONS. 
 
 Many fatal accidents have taken place due to the falling of 
 the jibs of hand and steam worked cranes. The following 
 simple device has been recently applied by a Glasgow firm with 
 the object of avoiding such accidents. 
 
 1909 PATENT JIB ARRESTER, BY BUTTERS BROTHERS & Co. 
 GLASGOW. 
 
 In the event of any of the jib gear or of the jib rope of the 
 crane breaking, the spring A contracts, thus relieving the rope 
 B, which holds the clutch C in position. The two springs at the 
 point of the jib marked D then contract, and pull the clutch C 
 inwards. This catches the lift rope E, on the pulley at the point 
 of the jib F, thus holding up the jib and preventing it from 
 falling, as well as keeping the load "W in position. 
 
( 93 ) 
 
 LECTURE IX. 
 
 CONTENTS. Inclined Planes The Inclined Plane without Friction 
 When the Force acts Parallel to the Plane Example I. When the 
 Force acts Parallel to the Base Example II. When the Force acts 
 at any Angle to the Inclined Plane Example III. The Principle of 
 Work applied to the Inclined Plane Example IV. Questions. 
 
 Inclined Planes, An inclined plane is a plane surface inclined 
 to the horizontal, whereby a certain force may be used to raise a 
 greater weight to a desired height than could be done by applying 
 it directly to elevate the weight vertically. Inclined planes are 
 also used for easing down weights with less retarding force than 
 would be necessary to lower them vertically. In another form, 
 called the wedge, inclined planes are employed for splitting bodies, 
 or different parts of the same body, asunder, as in the case of the 
 steel wedge used by the woodman to split up logs for firewood and 
 other purposes. Wedges are also used for forcing bodies together, 
 and for fixing them tightly in a desired position ; or for elevating 
 them through a small distance, as in the case of the levelling of 
 the heavy cast-iron sole-plate of an engine. And further, as we 
 shall have occasion to prove, the well-known screw, in whatever 
 form it may be ap^jted, is simply an inclined plane of a particular 
 shape. 
 
 The Inclined Plane without Friction. In the first place, 
 we will consider the inclined plane with a body placed thereon and 
 kept in position by a force applied to the body, when all friction 
 between the plane and the body is supposed to be absent or negligible 
 i.e., both the plane and the body are assumed to be perfectly 
 smooth. There are three cases of this statical problem. 
 
 (i) When the force supporting the body acts parallel to the 
 inclined plane. 
 
 (2) When the force acts horizontally. 
 3) When the force acts at any angle to the plane. 
 Case I. Let the force P act parallel to the plane, and let the 
 accompanying figure represent a vertical section through the plane 
 and the e.g. of the body. Let a be the e.g. of the body; W its weight, 
 acting vertically downwards along the line aW ; P the necessary 
 pull (to keep the body in position) applied along the line P, 
 
94 
 
 LECTURE IX. 
 
 parallel to the plane AB ; and R the reaction from the plane (due 
 to the weight of the body resting thereon), acting along the line 
 aR, at right angles to the plane. 
 
 Also, let the length of the plane AB be indicated by I ; its 
 height, BC, by h ; its base, AC, by b ; and the angle of the plane 
 to the horizontal by a. 
 
 Now, by the " triangle of forces," since we have three forces, 
 W, P and R, acting at a, the eg. of the body, and since these 
 forces are in equilibrium, if we construct a triangle whose sides 
 
 INCLINED PLANE, CASE 1. 
 WHEN P ACTS PARALLEL TO PLANE AB. 
 
 are parallel to these forces, they will represent them in direction 
 and in magnitude. 
 
 Therefore, plot off along the line aW a distance 6, to repre- 
 sent the weight of the body W, to any convenient scale. From, b, 
 draw a line be parallel to P, and from, a, extend the direction of 
 R to, c, by the line ac. 
 
 Then, W : P : R : : ab : be : ca 
 
 But by Euclid the triangle abc is similar to the triangle ABC. 
 .-. ab: be : ca : : AB : BC :CA 
 AB:BC:CA:: I : h : b 
 R :: I : h : b 
 
 W : P 
 
 And, 
 Consequently, 
 
 Or, 
 
 Or, 
 
 Precisely the same results will be arrived at if (as shown by the 
 right-hand side of the figure) we considered the vertical side BC 
 
 P 
 
 W 
 
 h R b , P h 
 
 = 7' w = 7 ; and n =; 6 
 
 P 
 
 - Sin a ; 5 = Cos a 
 
 p 
 
 ; and = Tan a 
 
 W 
 
 W 
 
 R 
 
INCLINED PLANES. - 95 
 
 of the triangle ABC as representing W, and then have drawn a 
 line from C on the direction of AB, parallel to R. It will form 
 a useful exercise for the student if in every case he will plot down 
 both methods, and mark along the sides of the triangle of forces 
 the respective forces which they respectively represent. 
 
 EXAMPLE I. A weight of 100 Ibs. is supported on a smooth 
 inclined plane by a force P, acting parallel to the plane. If the 
 incline be i in 10, find P, and give the reasoning by which you 
 establish the result. 
 
 ANSWER. Draw a figure exactly the same as tnat accom- 
 panying Case 1. and mark W= 100 Ibs., 1= 10, and h = i. Then 
 by the " triangle of forces " : 
 
 P_ BC * * 
 
 W = AB I = To 
 
 P = W -L - .12? = 10 Ibs. 
 
 10 10 
 
 Case 2. Let the force P act parallel to the base, with the same 
 
 INCLINED PLANE, CASE 2. 
 WHEN P ACTS PARALLEL TO BASE AC. 
 
 signification for each of the forces and parts of the inclined plane, 
 and the same assumptions. Then plot off ere, along aW, to represent 
 W ; draw cb parallel to P, and extend the direction of K back- 
 wards along 06, until it meets cb at the point 6 
 Then W : P : R : : ac : cb : ba. 
 
 But by Euclid the triangle acb is similar to the triangle ACB. 
 
 .-. ac : cb : ba ::AC:CB : BA 
 And, AC :CB: BA:: b : h : I 
 
 G 
 
96 
 
 LECTURE IX. 
 
 Consequently, 
 
 W : P ! R 
 
 
 I 
 
 Or, | 
 
 _ *. 
 
 R 
 
 W 
 
 = - ; and 
 
 P 
 
 R ~ 
 
 Or, 
 
 P 
 
 <Sxr 
 
 = Tan 
 
 a '> w 
 
 = Seoo; and- = i 
 
 Precisely the same results will be arrived at if (as shown by 
 the right-hand side of the figure) we considered the vertical side 
 BC of the triangle ABC as representing W, and then draw a line 
 from C parallel to R, and a line BD, parallel to P, to meet the 
 line CD. 
 
 EXAMPLE II. A force of 100 Ibs. is supported on a smooth 
 inclined plane by a force P acting parallel to the base. If the 
 incline be i in 10, find P. 
 
 ANSWER. Draw a figure exactly the same as that accom- 
 panying Case 2, and mark W = 100 Ibs., 1= 10, and h = i. 
 
 Then, by the " triangle of forces " : 
 
 JP __ CB h i i ri 
 
 W 
 
 A<J 
 
 100 
 
 ^99 9-95 
 = 10-05 Ibs. 
 
 9'95 9*95 
 
 Case 3. Let the force P act at any angle 6 to the inclined plane 
 AB. With the same signification for each of the forces and parts 
 of the inclined plane, and the same assumptions, plot off from a, 
 
 INCLINED PLANE, CASE a.' 
 
 "WHEN P ACTS AT ANY ANGLE TO PLANE 
 
INCLINED PLANKS. 97 
 
 along the line aW, a distance am, to any convenient scale to 
 represent the weight of the body W. From this point, m, draw 
 a line mn parallel to P, and .extend the direction of R backwards 
 to meet this line. This small triangle, amn, will be a " triangle 
 of forces," for W, P and R, which are in equilibrium about the 
 e.g. of the body at a. 
 
 But in this case the student will probably realise the proof of 
 the problem more easily if he considers BC as representing to 
 scale the weight W, and then draws CD parallel to R, and DB 
 parallel to P, 
 
 When W : R : P : : BC : CD : DB, 
 
 or the triangle BCD is the " triangle of forces," representing the 
 forces W, R and P in direction and magnitude by the sides BC, 
 CD and DB respectively. 
 
 P DB R = CD. d P_55 
 Or ' W = BC'W BC' R~CD 
 
 If we resolve the force P (which acts at the angle 6 to the in- 
 clined plane) parallel to the plane, then we can treat the com- 
 ponents of P exactly in the same way as we did the simple force 
 P in Case 1. 
 
 If we resolve P into the direction of R, then this component 
 acts with R, and is evidently balanced by the resolved part of W 
 in the same direction i.e., along the line, an. 
 
 EXAMPLE III. A weight of 100 Ibs. is supported on a smooth 
 inclined plane by a force P, acting at 60" in an upward direction 
 from the inclined plane. If the incline be i in 10, find P. 
 
 ANSWER. Draw a figure exactly the same as that accompanying 
 Case 3, and mark W= 100 Ibs., 1= 10, A= i, and 6 = 60*. 
 
 Then by the " triangle of forces," BC represents W, and DB 
 represents P to scale. Measuring their respective lengths we get 
 
 P = W?? =100 H =20 Ibs. 
 
 BC ioo 
 
 Principle of "Work applied to tLe Inclined Plane. 
 
 Referring to the figure for Case 1, let the body, whilst under 
 
 the action of the three forces W, P and R, be moved the whole 
 
 length of the incline. Therefore P acts from A to B, and at the 
 
 same time W acts through a vertical height CB. Consequently, 
 
 neglecting friction as before, we have by the " principle of work " 
 
 The work put in = The work got out 
 
 P x its distance = W x its distance 
 
 PxAB = WxCB 
 
98 LECTURE IX. 
 
 But this is precisely the same result as we got by applying the 
 principle of the "triangle of forces." Hence, the "principle of 
 work" agrees with the "triangle of forces" in respect to the 
 inclined plane. 
 
 Cases 2 and 3 may be treated by the student in exactly the 
 same way, and the correct results will be the same as those found 
 by the " triangle of forces." 
 
 - EXAMPLE IV. An inclined plane is used for withdrawing 
 barrels from a celJar by securing two ropes to the top of the 
 incline at B, then passing them down the incline, half round 
 the barrel, and up to the horizontal 
 platform at the top of the incline, where 
 two men pull on the ropes in a direction 
 parallel to the plane. If the weight, W, 
 of the barrel is 200 Ibs., the length, Z, of 
 the incline 20 ft., and the height 10 ft., 
 find, by the principle of work, the least 
 force which must be exerted by the two 
 RAISING BARRELS BY men? and the work expended, neglecting 
 
 -------- 5 
 
 ?NCL U INED PLA^E. AN f*** in drawing the barrel from the 
 
 cellar. 
 
 Let the accompanying figure represent a vertical cross section 
 through the middle of the barrel and the inclined plane. Then a 
 statical force, P, applied at the e.g. of the barrel, would just balance 
 its weight, W, and the reaction from the plane (not shown). 
 By the principle of work, neglecting friction 
 The work put in = The work got out. 
 P x its distance = W x its distance. 
 
 ?xl = Wxh. 
 
 P=W 7 = 200^' -i oo Ibs. 
 
 I 20 
 
 But by passing the rope round the barrel, as explained in the 
 question, this force P is halved on the ropes (see Lecture VI. on 
 the pulley and snatch-block). Therefore the least force which 
 the two men must exert in order just to move the barrel wiH be 
 
 .122.60 bs. 
 
 2 2 
 
 But this force acts through a distance 2^ = 40' ft. ; therefore the. 
 work expended will be 
 
 p 
 
 x 2 2 = 50 Ibs. x 40' = 2000 ft.-lbs. 
 
 Or, work got out =? W * /? =F 299 Ibs. x 10' = 20OO ft.-lbs. 
 
INCLINED PLANES. _; ; 99 
 
 In this question we have a combination of the pnlley and the 
 inclined plane. The inner ends of the two ropes being fixed at 
 the top of the inclined plane, the force with which the men act 
 on the free ends is communicated throughout the ropes, so that 
 the stress in the ropes on each side of the barrel balances the 
 force P, that would be required to move the barrel up the incline 
 if applied at its centre of gravity. 
 
 Or, the theoretical advantage due to the pulley part of the 
 
 . . P 2P 2 
 
 system is, . . . p = ^r = - 
 
 Then for the inclined plane part we have by the " triangle of 
 forces," or by the " principle of work," a theoretical advantage of 
 
 Therefore, the total theoretical advantage is the product of the 
 two separate advantages, viz. 
 
 2P W_2 2 4 
 
 P X P" = 7 X I = I 
 
 Consequently, a force of i Ib. applied at the free end of the rope 
 would balance a weight of 4 Ibs. on the incline. Or, as in the 
 question, and, neglecting friction, a barrel weighing 200 Ibs. requires 
 a pull of 50 Ibs. to move it up the inclined plane. 
 
 We have simply split up the total advantage in this way to 
 show the student that the combined advantages of the several 
 parts of a compound machine must equal the advantage of the 
 whole. We might have said at once, as we have done before in 
 other cases 
 
 The Theoretical Advantage = p- = - = y 
 
 NOTE. I have this day (Sept. 9, 1892) witnessed the interesting opera- 
 tion of lowering four very large 25-ton steam boilers of the marine type, 
 down an incline of about 100 feet in length by the method described in the 
 foregoing question. One man, by aid of an ordinary block and tackle, 
 supplied the requisite restraining force on the free end of the rope. 
 
100 LECTURE IX. QUESTIONS. 
 
 LEOTUBE IX. QUESTIONS. 
 
 1. Prove by the triangle of forces (drawn to scale) the relation between 
 the weight W of a body resting on a smooth inclined plane, the reaction, 
 R, from the plane, and the force, P, necessary to just balance the weight 
 (i) when the force, P, acts parallel to the plane ; (2) when it acts parallel 
 to the base ; (3) when it acts at an angle, 6, to the plane. 
 
 2. A ball, weighing 100 Ibs., rests on an inclined plane, being held in 
 position by a string which is fastened to a bracket so as to be parallel to 
 the plane. The height of the plane being of the length, find the tension 
 of the string and the pressure perpendicular to the plane. Establish your 
 results by reasoning on known principles, such as the principle of work or 
 that of the parallelogram of forces. Am. P = 33'3 Ibs., and B=94*3 Ibs. 
 
 3. Prove the relation between W, P, and R, acting on a body resting on 
 a smooth inclined plane by the "principle of work" for cases 1, 2 and 3 
 in this Lecture. An incline is i ft. in vertical height for 15 in length. A 
 weight of 100 Ibs. rests on the plane and is held up by friction ; make a 
 diagram for estimating the pressure on the plane, and find its amount. 
 Ans. 99'7 Ibs. 
 
 4. Friction being neglected, find the force, acting parallel to the plane, 
 which will support i ton on an incline of i ft. vertical and 10 ft. along the 
 incline. Prove the formula which you employ. If the incline were i ft. 
 vertical and 280 ft. along the incline, find the force in pounds which would 
 support i ton. Ans. 224 Ibs., and 8 Ibs. 
 
 5. A smooth incline plane has a vertical side of i ft., and a length of 
 10 ft. ; what work is done in pulling 10 Ibs. up 8 ft. of the incline 1 Am. 
 8 ft. -Ibs. 
 
 6. When a body is raised through a given height, how is the work done 
 estimated ? A body weighing 8 cwt. is drawn along 100 ft. up an incline, 
 which rises 2 ft. in height for every 5 ft. along the incline ; the resistance 
 of friction being neglected, find the work done. Ans. 35,840 f t.-lbs. 
 
 7. A smooth incline is 8 ft. long, and the total vertical rise from the 
 bottom to the top thereof is 2 ft. What amount of work is performed in 
 drawing a weight of 100 Ibs. up 4 ft. of the incline, and what is the least 
 force which will do this work ? Ans. 100 ft.-lbs. ; 25 Ibs. 
 
 8. Friction is neglected, and it is found that a force acting horizontally 
 will move 10 Ibs. up 5 ft. of an incline rising i in 4. Find the work done, 
 and find also the force parallel to the plane which will just support the 
 weight of 10 Ibs. Ans. 12*5 ft.-lbs. ; 2*5 Ibs. 
 
 9. A car laden with 20 passengers is drawn up an incline, one end of 
 which is 1 60 ft. above the other ; the car, when empty, weighs 3 tons, and 
 the average weight of each passenger is 140 Ibs. Find the number of 
 fU-lbs. of work done in ascending the incline, neglecting friction. 
 Ans. 1,523,200 ft.-lbs., or 680 ft. -tons. 
 
 10. It will be observed that draymen sometimes lower heavy casks into 
 cellars by means of an inclined plane and a rope. One end of the rope is 
 secured to the upper end of the inclined plane, and is then passed under 
 and over the cask, the men holding back by means of the loose end. Now, 
 supposing the incline to be at an angle of 45 degrees, explain the mechanical 
 principles that are here applied, and find the advantage. Ans. 21/2:1. 
 
 1 1. A barrel weighing 5 cwt. is lowered into a cellar down a smooth slide 
 inclined at an angle of 45 degrees with the vertical. It is lowered by means 
 of two ropes passing under the barrel, one end of each rope being fixed, 
 while two men pay out the other ends of the ropes. What pull in Ibs. 
 must each man exert in order that the barrel may be supported at anj 
 point ? -<* Ant. 99 Ibs. nearly. 
 
LECTURE X. 
 
 CONTENTS. Friction Heat is Developed when Force overcomes Friction 
 Laws of Friction Apparatus for Demonstrating First and Second 
 Laws of Friction Experiment I. Example I. Angle of Repose or 
 Angle of Friction Experiment II. Diagram of Angles of Repose 
 Limiting Angle of Resistance Experiment III. Apparatus for 
 Demonstration of the Third Law of Friction Experiment IV. 
 Lubrication Anti-Friction Wheels Ball Bearings Work done on 
 Inclines, including Friction Example II. Questions. 
 
 Friction. Whenever a body is caused to slide over another 
 body, an opposing resistance is at once experienced. This natural 
 resistance is termed friction.* The true cause of friction is the 
 roughness of the surfaces in contact. The smoother the sliding 
 surfaces are made the less will be the friction. Friction cannot, 
 however, be entirely eliminated by any known means, for even 
 the most microscopical protuberances on the smoothest of sur- 
 faces seem to fit into corresponding hollows on other equally 
 smooth places, so that some force is required to make the one 
 body slide over the other. 
 
 Friction has its advantages as well as its disadvantages. For 
 example, if it were not for friction we could not walk, neither 
 could a locomotive start from a railway station, nor could it be 
 brought to rest in the usual speedy manner. Friction is also 
 essential to the utility of nails, screws, wedges, driving belts, <fec. 
 On the other hand, power is often expended in overcoming 
 friction with the result of much wear and tear in machinery. 
 For example, in the case of working the slide valves of locomotive 
 engines as much as twenty horse-power is required in moving 
 these essential parts when running at full speed, f 
 
 It is the duty of the engineer to reduce friction to a minimum 
 in the case of the bearings of engines, shafting, and machines 
 generally, in order that a minimum of work may be expended in 
 moving them. He has, however, also to devise means of pro- 
 ducing a maximum of friction in the case of certain pulleys, grips, 
 clutches, brakes, and such like appliances, where motion has to 
 be transmitted by aid of friction, or bodies in motion (such as a 
 
 * French writers call friction a passive resistance, because it Is only 
 apparent when one body tends to move or pass over another. 
 
 f See the Author's "Elementary Manual on Steam and the Steam 
 Engine," page 182, for an arithmetical example. 
 
102 LECTUKE X. 
 
 moving train) have to be brought to rest quickly when nearing a 
 station. 
 
 Heat is Developed when Force overcomes Friction. 
 When a body is kept moving by a force, part (or in certain 
 cases it may be the whole) of the mechanical force is expended in 
 overcoming fractional resistance. This lost work is directly trans- 
 formed into heat in the act of overcoming the frictional resistance 
 through a distance. For example : A person slips down a vertical 
 rope by holding it between his hands and his legs. The force of 
 gravity impels him downwards, overcoming the frictional resist- 
 ance between his hands and limbs and the rope, with the conse- 
 quence that they become severely heated, especially if he happens 
 to slip down quickly. A boy takes a run, and then slides along a 
 level piece of ice. The foot-pounds of work stored up in him just 
 before he begins to slide are expended partly in overcoming the 
 frictional resistance between the soles of his boots and the ice, 
 and partly in the frictional resistance between his clothes and the 
 air. As a consequence, he will find that by the time he gets to 
 the end of the slide his soles are considerably warmed. If the ice 
 were perfectly level, infinitely long, and if there were absolutely 
 no friction between it and his boots, and if there were no fric- 
 tional resistance between him and the air, then he would slide on 
 for ever 1 If we could diminish the frictional resistance between 
 the skin of a ship and the water, and between the exposed parts 
 of the ship and the air, to nothing, then all that would be required 
 to transport her across the Atlantic would be a strong force 
 applied at the start until she attained the desired speed, when she 
 would proceed forward, and arrive at her destination with undi- 
 minished velocity ! In reality, however, we find it necessary to 
 employ steam engines of 10,000 horse-power continuously in order 
 to propel an Atlantic " greyhound " of 5000 tons at twenty 
 knots in the calmest of weather. About one-half of this power 
 is absorbed in overcoming the frictional resistance of the ship 
 through the water and air, and the other half in the frictional 
 and other losses due to the working of the propelling machinery. 
 Examples of the conversion of mechanical work into heat are 
 so familiar to you all, being in fact brought prominently before 
 your notice every day of your existence, that we need not further 
 enlarge upon this question except to remind the student of 
 Dr. Joule's discovery of the rate of exchange between heat and 
 work. He found by experiment that if work is transformed into 
 heat, every 772 ft.-lbs. of work will produce i heat unit, or that 
 quantity of heat which would raise i Ib. of water i Fahr.* 
 
 * For further examples and an explanation of Dr. Joule's experiments 
 ?ee the Author's Treatise on Steam and the Steam Engine. 
 
FRICTION. 
 
 103 
 
 Laws of Friction. From 1831 to 1 8 34 General Morin carried 
 out an extensive series of experiments at Metz on friction for 
 plane surfaces, with different areas, pressures and velocities, from 
 which he arrived at certain conclusions. These conclusions were 
 for a long time regarded as constituting the fundamental laws of 
 friction. They have been since proved to beonly true within the 
 limits of his experiments, for, they do not hold good for great 
 pressures and high velocities, neither are they true for fluid, 
 rolling, or axle friction. For the latest and most reliable experi- 
 ments we must refer to the Proc. of the Inst. of Mechanical 
 Engineers, 1883, lS8 5> 1888, and 1891. 
 
 ist Law. Friction ia directly proportional to the pressure between 
 two surfaces, if they remain in the same condition. 
 
 2nd Law. Friction is independent of the areas in contact. 
 
 3rd Law. Friction is independent of velocity. 
 
 Apparatus for Demonstrating First and Second Laws of 
 Friction. Nevertheless, it will be both interesting and instruc- 
 tive to students to have these three laws demonstrated by the 
 following simple apparatus : 
 
 G.P 
 
 APPARATUS FOR DEMONSTRATING THE FIRST AND SECOND LAWS 
 OP FRICTION. 
 
 INDEX TO PARTS. 
 
 IP represents Inclined plane. I B represents Box for planes. 
 
 GP Guide pulley. PP Pinching pin. 
 
 Q Quadrant. | LL Legs. 
 
IO4 LECTURE X. 
 
 The inclined plane IP is fitted with a joint at its left-hand 
 end, and after slackening the pinching pin PP, it may be raised 
 to any desired angle or fixed in a level position by tightening the 
 pin. The desired position is found by reading off the angle oppo- 
 site the plumb-ball line on the graduated degree scale of the 
 quadrant Q. In the box B may be kept planes of glass, brass, 
 iron, steel, &c., as well as the different kinds of wood to be experi- 
 mented upon. These planes are fixed on IP, in a central position, 
 by means of a catch, and the bodies to be laid upon them should 
 be fitted with a small hook opposite their e.g., to which a fine 
 flexible silk cord can be attached and passed over the guide- 
 pulley GP, which should turn very freely on its bearings.* The 
 pull P is best effected by attaching to this cord a small tin pail 
 into which shot may be dropped one by one until the body moves 
 freely on the plane. The pail and shot may then be unhooked 
 and weighed in a balance. 
 
 Demonstrations of the First Law of Friction. EXPERI- 
 MENT I. Fix the inclined plane IP in a truly horizontal posi- 
 tion. Take from the box B, 
 say, a long strip of planed 
 yellow pine and a small block 
 of the same kind of wood, 
 and let its weight be W. 
 Adjust the strip along the 
 middle line of IP by means 
 PEOOF OF FIRST LAW OF FRICTION. of the sneck or catch, and 
 
 place the block therein. At- 
 tach the silk thread to the hook on the forward side of the block, 
 and pass the same over the practically f rictionless pulley. Hang a 
 little tin pail from the free end of the silk thread, and drop small 
 shot one by one into the pail until the block moves freely over the 
 yellow pine strip when aided by a little tapping on the table. 
 Unhook the pail containing the shot, and weigh it as carefully as 
 you weighed the block of yellow pine. Let it equal P units. 
 
 Then P is the force which just overcomes the directly opposing 
 passive resistance, called friction, between the surface of the yellow 
 
 * The guide-pulley bracket should be fitted with a stiff joint and with a 
 telescope arm, so that the pulley may be raised or lowered in order to 
 bring the direction of the pull P on the cord parallel to the plane, or 
 parallel to its base, or adjusted to any desired angle with respect to the 
 plane, in order to demonstrate Cases i, 2 and 3 of the inclined plane in 
 Lecture IX. By having, say, a " slot along the middle of the plane, and 
 by lowering the pulley, Case 2, wherein the pull on the body is parallel to 
 the base, may be readily demonstrated; and by pulling out the telescope 
 arm of the bracket, and turning up the bracket, Case 3, wherein the pull 
 makes an angle, 0, with the plane, may be verified.- 
 
FRICTION. 105 
 
 p 
 
 pine block and strip ; and the ratio ^ is termed the co-efficient of 
 
 friction. Now put another block of weight W on the top of the one 
 just tested, so as to double the pressure on the sliding surface, and 
 put in shot until the block moves when aided by a little vibration, 
 so as to overcome the greater resistance to starting the body in 
 motion than to keep it moving.* You find on weighing the pail 
 and shot that it is now 2?. Consequently the co-efficient of 
 
 2 p P 
 
 friction has not altered, for =^- is the same fraction as ^. 
 
 EXAMPLE I. Suppose you take a very small block of woocf 
 (say i/io" thick, 2" long and i" broad; in fact, so light that its 
 weight is negligible), and place a i-lb. weight on the top of it ; you 
 will find that 575 oz. are required to cause motion of this piece 
 of wood over the surface of the yellow pine strip. You therefore 
 conclude that the co -efficient of friction is 
 
 P c-7C oz. 
 ft= Wi6oz, = * 353 ' r friction = '353 W. 
 
 Now, place a 2-lb. weight on the upper piece of wood, and you 
 find that it requires more shot in the pail to move it. Weigh the 
 pail and the shot again just after you have obtained free move- 
 ment of the one bodf over thi other, and you will fi*4 that it 
 amounts to 11.5 oz. 
 
 P 11.5 
 
 Consequently, ^ = - -353 as before. 
 
 If, however, you put a lo-lb. weight on the upper piece of 
 wood, you will obtain a different result, thus proving the first law 
 and the variation therefrom; because in this latter case the 
 pressure is so great, compared with the first and second experi- 
 ments, that the grains of the upper piece of wood enter those of 
 the lower, and bring into play another condition of affairs viz., 
 the gripping action of the one set of grains on the other set. If 
 you had taken a large plank of yellow pine, weighing, say, 100 Ibs., 
 and had placed it on another similar plank, the co-efficient of 
 friction would have a certain value. If you had even put a loo-lb. 
 weight on the upper plank, the co-efficient of friction might not 
 have varied perceptibly. But if you placed a weight of 1000 Ibs. 
 
 * Statical friction, or the friction of repose, is that resistance which 
 opposes the commencing of the motion. If a body be allowed to rest on 
 another for some time, it requires more force to move it than if it had only 
 been stationary for a few seconds. 
 
106 LBCTtJKB X. 
 
 on the upper plank, the co-efficient of friction would be con- 
 siderably altered. Hence you observe that this first law only 
 holds good between narrow limits.* 
 
 Angle of Repose, or Angle of Friction. EXPERIMENT II. 
 Another way of proving the first law of friction is to disconnect 
 the silk thread and the shot-pail from the upper body, and tilt up 
 the inclined plane to such an angle, a, with the horizontal that, 
 (with the aid of a little tapping) the weighted block of yellow 
 pine just slides slowly down the incline. Here we have simply 
 the force of gravity acting on the body and overcoming friction. 
 At the moment the body just begins to slide we have the weight, W, 
 of the body acting vertically downwards, R the reaction from the 
 plane at right angles to the surface, and F, the passive resistance 
 of friction, acting parallel to the plane in the direction of aP 
 in the first figure in this Lecture. Now, these three forces act 
 from the e.g. of the body, and they are in equilibrium. R is equal 
 to the resolved part of W at right angles to the plane (or R = 
 W Cos a), and it represents the pressure between the surfaces. 
 F is the resolved part of W, parallel to the plane (or F = W Sin a). 
 
 TT 
 
 and ^ is the co-efficient of friction. 
 
 XV 
 
 F _ W Sin a _ m _ h _ height of plane 
 
 " R W Cos a b base of plane 
 
 The angle a, to which the plane must be inclined before the 
 free body will slip over the fixed one, has been termed the " anyle 
 of repose" or " angle of friction" 
 
 Therefore, ike tangent of the angle of repose is equal to the co- 
 efficient of friction. 
 p 
 
 But was proved by the previous experiment to be also equal 
 
 to the co-efficient of friction, 
 
 P F h 
 
 * ' W - R - 1 
 
 Or, P *= /iW and F 
 
 * Sir Robert Stawell Ball, when Professor of Mechanism at the Royal 
 College of Science, Ireland, tried a careful experiment in the above way 
 with a smooth horizontal surface of pine 72" x 1 1", and a slide, also of pine, 
 9"x9" grain crosswise. He loaded and started the slide, and applied a 
 force sufficient to maintain it in uniform motion, and he found that on 
 increasing the load from 14 to 112 Ibs., by increments of 14 Ibs., the co- 
 efficient of friction diminished from -336 to -262. From these experiments 
 he constructed the empirical formula for this case that F = '9 + -266 R, 
 where F is the factional resistance and R the reaction from the surface or 
 net load- 
 
FRICTION. 
 
 107 
 
 where fi is the Greek letter universally adopted to represent co- 
 efficients of friction. 
 
 The accompanying figure is a diagram of the " angles of repose ' 
 for various common materials, together with the numerical 
 values of or their co-efficients of friction. 
 
 Jbrlcat e 
 ro rtO~ 
 
 A.NGLES OP REPOSE. 
 
 Limiting Angle of Resistance, or Sliding Angle. A 
 
 third way of proving the first law of friction is to place the bodies 
 so that the sliding surface is perfectly level. Then begin by 
 pressing the upper body through the intervention of a compres- 
 sion spring-balance fitted with a sharp point, so that it will not 
 slip off, and with a clinometer to indicate the angle through which 
 it is tilted away from the perpendicular. Now gradually incline 
 your pressure to the perpendicular, until you arrive at such an 
 angle as will just cause the upper body to slide over the under 
 one. This angle is termed the "sliding angle" or " limiting angle 
 of resistance" because it is the limit, or maximum angle which 
 the reaction from the surface can make with the perpendicular to 
 the surfaces, for the reaction must act in the directly opposite 
 direction to the pressing force.* Again, apply the spring- balance, 
 but with double the registered pressure, and you can just incline 
 this force to the same angle as before. If, however, you pre& f . 
 with ten times the former force, you would probably be able to 
 act at a greater angle than before. It will be seen from this 
 experiment that 
 
 The Limiting Angle of Res-istance = The Angle of Repose. 
 
 * Here the weight of the tipper body is supposed to be negligible in 
 Comparison with the inclined pressure upon jt t 
 
io8 
 
 LECTURE X. 
 
 PKOOF OF SECOND LAW OF FRICTION. 
 
 Demonstration of the Second Law of Friction. Ex* 
 PERIMENT III. Take a block of planed yellow pine, and cut it into 
 two equal pieces at right angles to the planed surface. Place one 
 piece on the horizontal strip of yellow pine (used in previously 
 demonstrating the first law), with the planed side next to it, 
 
 and put the other piece on 
 the top of it, as shown by 
 the second figure in this 
 lecture. Now ascertain the 
 horizontal pull, P, required 
 to overcome friction. Then 
 attach the top piece to the 
 bottom one, as shown by the 
 accompanying figure, so that 
 
 the area of the surface in contact is doubled, and you will find 
 that the same horizontal force, P, will cause it to move. If 
 you take a long planed block and cut it into ten equal pieces, 
 each of the same size as one of the above pieces, and try the 
 experiment in a similar manner, you will be able to increase 
 the area of contact tenfold, and you will then find that the 
 
 p 
 ratio ^ is not exactly the same with the surface of one block 
 
 in contact with the strip, as when the surface of the whole ten 
 came into action at once. The result of increasing the area in 
 contact may also be tried by placing the blocks on the inclined 
 plane, and observing the angle to which the plane is tilted when 
 they begin to slide down the plane. 
 
 APPARATUS FOR DEMONSTRATING THE FIRST AND 
 THIRD LAWS OF FRICTION. 
 
 INDEX TO PARTS. 
 
 L represents Lever. 
 
 F 
 SR 
 SP 
 TP 
 
 q 
 
 Fulcrum. 
 Small roller. 
 Scale-pan. 
 Test-piece. 
 Cylinder. 
 
 H represents Handle. 
 
 R 
 
 S 
 
 P 
 
 GS 
 
 Rack. 
 
 Spiral spring. 
 
 Pointer. 
 
 Graduated scaled 
 
 Base of apparatus. 
 
IKICnON LUBRICATION. 109 
 
 Demonstration of the Third Law of Friction. The pre- 
 ceding figure represents the apparatus belonging to the Applied 
 Mechanics Department of the Royal College of Science, South 
 Kensington (as described by Prof. Goodeve in his " Manual of 
 Applied Mechanics "), for demonstrating the first and third laws 
 of friction. 
 
 If the weights, W, be removed from the scale-pan SP, then 
 there will be but a slight pressure between the lower surface of 
 the test-piece TP, and the roller cylinder C. Consequently, on 
 turning the handle H in the direction of the arrow, there will be 
 a slight pull on the cord, causing the pointer P to move a degree 
 or two over the graduated scale GS. The pointer should there- 
 fore be set back to zero. 
 
 EXPERIMENT IY. Put a weight, W, of say 5 Ibs., into the 
 scale-pan, and turn the cylinder slowly by the handle as before. 
 The pointer deflects so many degrees. Increase the weight W 
 to 10 Ibs., and the pointer instantly indicates twice the amount 
 of friction ; put in 15 Ibs., and it shows treble the friction ; thus 
 demonstrating the first law. Then turn the handle faster and 
 faster, and the pointer remains fairly stationary, thus proving 
 within certain limits that friction is independent of the velocity.* 
 
 Lubrication. Lubricants, such as tallow, grease, soft soap, 
 and many kinds of oils, are used to reduce friction. Both skill 
 and knowledge are required to decide upon the best kind of lubri- 
 cant and the proper amount for different cases. Lubrication and 
 lubricants should receive greater attention from the engineer, for 
 the satisfactory working and length of life of most machines de- 
 pend so largely upon effective lubrication. Where very heavy 
 pressures and high speeds are experienced as in some cases of 
 electrical machinery, it pays to use the very best kind of oil, and 
 to distribute it to all the bearings from one common centre under 
 pressure by means of a force-pump. It thereby flows in a con- 
 tinuous stream through the bearings to a filtering tank, from 
 which it is again and again pumped on its soothing mission for 
 months on end, without change or great loss in quantity. This is 
 a very different state of matters from the " travelling oil-can " 
 system, where the amount applied may vary, and the times of appli- 
 cation may be erratic, according to the opinion of the attendant. 
 
 Anti-Friction Wheels. In the case of delicate machinery, 
 such as in Atwood's machine for ascertaining by experiment the 
 acceleration of gravity, and in Lord Kelvin's mouse-mill for 
 driving the paper rollers of his Syphon Recorder, when receiving 
 
 * Sea Molesworth'i Pocketbook of Engineering Formulae, and the Trans- 
 actions of the Institution of Mechanical Engineers, for results of friction 
 experiment! with shafts ran at different speeds. 
 
I 10 
 
 LECTURE X. 
 
 ANTI-FRICTION 
 WHEELS. 
 
 telegraphic signals from long submarine cables, anti-friction 
 wheels are used for the purpose of reducing the friction to a 
 minimum. The accompanying figures illustrate one pair of 
 anti-friction wheels. The spindle S, which carries the driving- 
 wheel, instead of resting on two ordinary 
 bearings, is supported by two wheels at 
 each end, so that a rolling contact is pro- 
 duced between it and the wheels. This 
 form of contact implies far less friction to 
 begin with, than a sliding or scraping con- 
 tact. Besides, the small amount of force 
 required to overcome the friction between 
 the spindle and the rims of these wheels, 
 has a great advantage or leverage given 
 to it, in as far as, it acts with an arm 
 equal to the radius of the wheels FW\ and 
 FW 2 . This enables it to turn them with 
 great ease at a slow rate in the very small 
 bearings Bj and B 2 . 
 In merely overcoming friction at a bearing, there is a con- 
 siderable advantage in using large pulleys ; for, the force necessary 
 at the periphery of the pulley to overcome the friction at the bear- 
 ing, is inversely proportional to the radius or diameter of the 
 pulley. (See Lecture XI. fig. i). 
 
 Ball Bearings. Another example of the effect of rolling 
 contact reducing friction is found in the use of ball bearings, 
 which are now so common in all kinds of cycles and in high-class 
 foot-driven lathes.* 
 
 When it is necessary to move heavy beams, guns, &c., a common 
 practice is to place them on rollers or on two channel iron 
 girders o with round cannon-shot between them, when a com- 
 paratively small force, properly applied, will have the desired 
 effect. 
 
 We will have to return to this subject in the Advanced Course 
 when dealing with the friction between shafts and their bearings, 
 and the various means that have been adopted for minimising 
 the same. In the meantime, we will complete this Lecture with 
 an example of work done on an incline when friction is included. 
 Work done on Inclines, including Friction. The method 
 of calculating the work expended in moving a body along a smooth 
 inclined plane was fully dealt with in Lecture IX. ; consequently, 
 the student is prepared, after what has been said about friction in 
 
 * Refer to Lecture XVI., p. 183. 
 
WORK DONE ON INCLINES INCLUDING FRICTION. I I I 
 
 this Lecture, to consider the case of pulling a body up or down 
 a plane when the co-efficient of friction between the body and the 
 plane is known. 
 
 The total work expended is evidently divisible into two distinct 
 portions 
 
 1 i ) The work done with or against tJie action of gravity, accord- 
 ing as the body is moved down or up the inclined plane = W x h 
 (where h is the height of the plane). 
 
 (2) The work done against friction = F x I (where I is the length 
 of plane passed over). 
 
 The work to be done against friction is the same whether the 
 body is urged up or down tho incline; for it is equal to the 
 r^-effiomnt of friction x the reaction of the plane x the distance 
 through which it is moved. 
 Or, FxZ = /ixRxJ 
 
 But by Lecture IX. B x 1 = W x b; .. F x Z = /z x W x b 
 Or, the work done against friction in moving a body along the 
 inclined distance I, is equal to the work done in moving the same 
 body along a horizontal distance b, equal to the base of the incline. 
 
 If the work to be done in overcoming friction, is equal to the 
 work capable of being done on the body by gravity, the body will 
 be in equilibrium, and the inclination of the plane is equal to the 
 angle of repose. 
 
 If the work to be done in overcoming friction is less than the 
 work which gravity can do on the body, the body will slide down 
 the incline, or, in technical language, the machine will overhaul. 
 
 EXAMPLE II. What is the co-efficient of friction, and how is it 
 ascertained ? There is an inclined plane of i foot vertical to 
 10 feet horizontal; what work is done in moving 700 Ibs. 5 feet 
 along the plane, the co-efficient of friction being -08 ? (S. and A. 
 Exam. 1892.) 
 
 w 
 FIGURE FOB EXAMPLE II. 
 
 ANSWER. The co-efficient of friction for two bodies in contact 
 is the passive resistance (opposing the motion of the one over the 
 ofher) divided by the reaction or normal vressure between the surfaces 
 in contact 
 
 Friction F 
 
 i.e. , Co-emcient or fnction = -^ . = ^ = u. 
 
 Reaction R 
 
 For methods of ascertaining co-efficients of friction, see the text in this 
 Lecture, -a 
 
112 LECTURE X. 
 
 Total work done = work done against gravity + work done against frwtion. 
 
 Referring to the accompanying figure, we see that 
 
 Si) Work done against gravity = W x DE 
 2) Work done against friction = F x AD 
 
 Total work done = W x DE + F x AD 
 
 We have therefore only to substitute the numerical values corre- 
 sponding to these letters in order to arrive at the result. From the ques- 
 tion W = 7oo Ibs. From the figure we see that DE is parallel to BC ; con- 
 sequently by Euclid the A", ADE, and ABC are similar in every respect ; 
 and therefore 
 
 DE : BC : : AD : AB ; or, DE= B * AP . 
 
 Ai> 
 
 But, also by Euclid, AB= ^ AC 2 + BC 2 = V io 2 +i 2 = 10-05 ft. (nearly) 
 Consequently, DB-?55. =JJL = - 497 ft. 
 
 And, F * AtB 
 
 From the question we are told that |i = -08, and we learn from Lecture IX. 
 that 
 
 B : W : : AC : AB ; or, R= WAC = 7io =696 . s lbs . 
 
 .'. F=/iR=-o8x 696-5 = 55-72 Ibs. 
 
 Hence Total Work = W x DE + FxAD 
 
 = 7oo * -497' + 55-72x5' 
 
 = 347-9 ft.-lbs. + 278-6 ft.-lbs. 
 
 = 626-5 ft. -Ibs. 
 
 NOTE. For the work done against friction quite a simple way would have been to hard 
 taken the formula deduced on the previous page 
 
 Vix. : FxZ=MXWx&=/mXWxAE= o8X7cx>X4-97 = 278'6 ft.-lbs. 
 
 * IB-BO A 8 -*p-22*=- w 
 
 APPROXIMATE ANSWER. Since the inclination of the plane is so 
 very small in this case, we might have assumed that 
 
 R = W; AB = AC, andDE = JBC 
 Then, 
 
 (1) Work done against gravity = W x DE = 700 x % = 350 ft.-lbs. 
 
 (2) Work done against friction = F x AD=-o8 x 700 x 5 = 280 ft.-lbs. 
 
 % TQtal work = W x DE + F x AD = 350 + 280= 630 ft.-lbs. 
 
LECTUEE X. QUESTIONS. - 113 
 
 LECTURE X. QUESTIONS. 
 
 1. What is friction, and how does it act ? What is developed ^ hen force 
 overcomes friction ? How do you measure the result ? 
 
 2. Explain by sketches and concise description how the laws of friction 
 may be tested experimentally. What is meant by the " co-efficient of fric- 
 tion," "angle of repose," "angle of friction," and " sliding angle " or 
 11 limiting angle of resistance " ? 
 
 3. How is the co-efficient of friction between two surfaces ascertained 
 approximately by experiment? When two rough surfaces are pressed 
 together, how much may the line of pressure be inclined to the common 
 perpendicular to the surfaces in contact before motion ensues ? 
 
 4- What is the co-efficient of friction when the angle of repose is 
 (a) 5 42' ; (6) 1 1 18' ; (c) 16 4* ; (d) 21 48' ; (e) 26 36' : (/) 30 ; (g) 45 T 
 Draw the angles to scale. Ans. (a) -i ; (b) -2; (c)3; (d) -4; (e) -5 ; 
 
 (/)'5774: (9) i. 
 
 5. An inclined plane is 100 feet long and 20 feet high. A body weighing 
 100 Ibs. is pulled up from the bottom to the top, and then down again. If 
 the co-efficient of friction between the body and the plane is -5, what 
 work was expended in each case ? What would require to be the co- 
 efficient of friction in order that the body might just slide down of its 
 
 own accord? Ans. 6,900 ft.-lbs. ; 2,900 ft.-lbs. ; /i= _= ^.=-204. 
 
 b 12 
 
 6. What is the co-efficient of friction, and how is it ascertained ? There 
 is an inclined plane of i foot vertical to 5 feet horizontal ; what work is 
 done in moving 100 Ibs. through 100 feet along the plane, the co- efficient 
 of friction being ! 1 Ans. 2940 ft.-lbs. 
 
 7. An incline is 80 feet long, with a rise of 20 feet. A body weighing 
 100 Ibs. is drawn 40 feet along the incline ; what work is expended if the 
 co-efficient of friction is -6 ? Ans. 3,323 ft. -Ibs. 
 
 8. A weight of 5 cwts. resting on a Horizontal plane requires a horizontal 
 force of loo Ibs. to move it against friction. What is the co-efficient of 
 friction ? Ans. -18. 
 
 9. A plank of oak lies on a floor with a rope attached to it. When the 
 rope is pulled horizontally with a force of 70 Ibs. it just moves, but when 
 pulled at an angle of 30 to the floor a force of 60 Ibs. moves it. What is 
 the weight of the plank and the co-efficient of friction between it and the 
 floor ? Ans. 116-6 Ibs. ; -6. 
 
 10. Suppose a locomotive weighs 30 tons, and that the share of this 
 weight borne by the driving wheel is 10 tons. Then, if the co-efficient of 
 friction between the wheels and the rails be -2, what load will the engine 
 draw on the level if the required co-efficient of traction be 10 Ibs. per ton 
 of train load ? What load will this engine draw at the same rate up an 
 incline of i in 20 ? Ans. 448 tons (including engine) ; 36- 72 tons (in- 
 cluding engine). 
 
 11. State the laws of friction, and explain the contrivance known as 
 friction wheels. What is the advantage of ball bearings for bicycles? 
 Sketch in section such a bearing. 
 
 12. What are lubricants, and for what purposes are they used in machin- 
 ery ? What kind of lubricant would you use for the moving parts of a 
 very high-speed engine and direct-driven dynamo, and how would you 
 apply it so as to be able to use it over and over again ? 
 
114 LECTURE X. QUESTIONS. 
 
 13. What is friction ? What is meant by limiting friction, by sliding 
 friction, and by the co-efficient of friction ? A weight of 5 cwts. resting 
 on a horizontal plane, requires a horizontal force of 108 Ibs. to move it 
 against friction. What in that case is the value of the co-efficient of 
 friction 1 Ans. "192. 
 
 14. How would you experimentally determine the nature of the friction 
 between clean, smooth surfaces, say of oak, and what sort of law would 
 you expect to find ? 
 
 15. Describe any experiment which you have made or seen for finding 
 the laws of solid friction. What are the laws so found ? Are they quite 
 true ? How do they differ from the laws of fluid friction ? 
 
 16. Sketch and describe an apparatus for determining the co-efficient 
 of sliding friction between two planed surfaces of oak. 
 
 If you have made this or a similar experiment describe the behaviour 
 of the sliding piece, and any troubles you may have had. State how you 
 would conduct the experiment so as to establish the principal facts 
 concerning such friction. (B. of E., 1902.) 
 
 17. The tractive resistance of a train weighing 335 tons is 1 1 Ibs. per ton. 
 If the effective horse-pov er of the lo 'emotive is 600, estimate the uniform 
 speed obtainable when ascending an incline of I in 200. Ans. 2662 ft. 
 per min. or 30 miles per hour. (C. & G., 1904,0., Sec. A.) 
 
NOTES AND QUESTIONS. 
 
LECTURE XL 
 
 CONTENTS. Difference of Tension in the Leading and Following Parts of 
 a Driving Belt Brake Horse-Power transmitted by Belts Examples 
 I. II. Velocity Ratios in Belt Gearing Examples III. IV. Open and 
 Crossed Belts Fast and Loose Pulleys Belt Gearing Reversing 
 Motions Stepped Speed Cones with Starting and Stopping Gear 
 Driving and Following Pulleys in Different Planes Shape of Pulley 
 Face Questions. 
 
 WE shall devote this Lecture to the transmission of power by 
 belting and to belt-gearing. 
 
 Difference of Tension in the Leading and Following 
 Parts of a Driving-Beit. In Lecture VI., when discussing the 
 case of the simple pulley, we assumed that the belt or rope passing 
 over the pulley was perfectly flexible, and that there was no fric- 
 tion at the axle of the pulley. Conse- 
 quently, we found that equal weights 
 would balance each other, or that the 
 tension of the two sides of the belt 
 were equal. A little consideration of 
 the subject will show that when one 
 pulley is driven from another one by 
 an endless belt or rope, the tension on 
 the driving side must be greater than 
 that on the following side. 
 
 i. Take the case of an ordinary 
 vertical pulley with its axle or shaft 
 resting in two bearings (one on each 
 side of the pulley), with a belt or rope 
 passed over it, and with weights at- 
 tached to the free ends of the same. 
 Here we must have a certain amount 
 of friction between the axle and its 
 bearings, which can only be overcome 
 by a force applied to the circumference 
 of the pulley. 
 
 T , T? _ ( Force required to overcome friction at the circum- 
 
 .Lieu . J? ! 1 fsrence of the axle or shaft. 
 
 DIFFERENCE OF TENSION 
 DUE TO FRICTION. 
 
DIFFERENCE OF TENSION. 117 
 
 Let . . r l = Radius of the axle. 
 
 p _ /Force required to overcome the friction of the axle 
 t \ when acting at the circumference of the pulley. 
 
 . . r 2 = Radius of the pulley to centre of belt. 
 Then, F t xr 1= F 2 xr 2 . 
 
 [Weight attached to the left-hand side of the belt, 
 Let . . W = j and which therefore produces a tension on the 
 I slack side = T,. 
 
 (Least weight on the right-hand side of the belt 
 that will produce motion, and which therefore 
 produces a tension on the driving side=Tj. 
 
 Then taking moments about the centre of the axle, we have 
 
 VV l x r 2 + F 2 x r 2 = W 2 x r a 
 Or, . . T.xr f + F f xr, = T d x r a 
 
 Dividing both sides of the equation by r, we get 
 
 T.+F,=T d 
 
 .-. F 2 = T d -T, 
 
 Or, expressed in words, the force F 2 , acting at the circumference 
 of the pulley (which is required to overcome the friction of the 
 axle) is equal to the tension T d on the driving or forward side of 
 the pulley, minus the tension T, on the slack or following side. 
 
 In order that the periphery of the pulley may move at the same 
 rate as the under face of the belt, we must have sufficient tension 
 on each part of the latter, and the co-efficient of friction between 
 them must not have less than a certain value. Too great adhe- 
 sion between them would result in a loss of work, for in that case 
 an extra force would have to be applied solely for the purpose of 
 pulling the belt from the pulley. 
 
 2. Take the case of one vertical pulley of diameter D, driving 
 another vertical pulley of diameter d by means of an endless belt, 
 rope, or chain in the direction of 
 the arrows shown on the accom- 
 panying figure. Whenever the 
 pulley T> is moved, the tension on 
 the driving side T d tends to 
 stretch the belt on that side, and 
 this tension increases until the 
 
 pulley d begins to move ; whereas BELT DRIVIKG . 
 
 the tension on the following or 
 slack side, T,, is gradually diminished until the difference of the 
 tensions (T d - T a ) produces a uniform velocity of the belt. Of 
 course the tension on the slack side must be sufficient to pro- 
 
ITS 
 
 LECTURE XI. 
 
 vent the slipping of the belt on either of the pulleys if the 
 periphery of the driven pulley is to keep pace with the peri' 
 phery of the driving one. In order that there may be a mini- 
 mum chance of the belt slipping, its slack side should always 
 run from the top side of the driving pulley. By so arranging the 
 drive, the sag of the belt on the slack side will cause it to encom- 
 pass a greater length of the circumferences of both pulleys. The 
 motion of the belt will be easier, and the wear and tear of the 
 bearings will be less, because there will be less total stress 
 (T d + T,) tending to draw the pulleys together for the transmis- 
 sion of a certain horse-power, than if the slack side left the under 
 side of the driving-pulley. Referring to the previous figure, if the 
 slack side leaves the top side of the pulley D, it grips the same 
 from position 4, round the back of the pulley to 5, and the pulley 
 d from 6 round to 3 ; whereas, if D were rotated in the opposite 
 direction, we should have the slack side entering on it at i, and 
 only gripping it as far as position 8 ; entering on d at 7, and only 
 gripping it to position 2, thus having far less grip on the pulleys 
 and thereby encouraging the natural tendency of the belt to slip 
 on the pulleys.* 
 
 Brake Horse-power transmitted by Belts. 
 
 Let. . V = Velocity of belt in feet per minute. 
 . . P = (T d T 4 ) the net pull causing motion in Ibs. 
 
 Then, B.H.P.= YP Y ( T ^~ T ) 
 
 33> 33>- 
 
 Let . . D Diameter of driving pulley in feet = 2r. 
 Then . vD *= Circumference of driving pulley in feet = 2wr. 
 Let . n = Number of revolutions of pulley per minute. 
 
 Then . Y = nDn = 2irrn = velocity of belt (with no slip). 
 
 And, the B.H.P. = 7rPnP = 27rmP 
 
 33><> 33> 
 
 EXAMPLE I. A pulley 6' in diameter is driven at 100 revolu- 
 tions per minute and transmits motion to another pulley by means 
 of a belt without slip. If the tension on the driving side of the 
 belt is 1 20 Ibs. and on the slack side 20 Ibs., what is the brake 
 horse-power being transmitted ? 
 
 * The previous figure should have been drawn with the full and dotted 
 .Ines at T,, reversed, but the student will easily follow the explanation. 
 
BRAKE HORSE- POWER TRANSMITTED BY BELTS.. 1 1 9 
 ANSWER. Here r = 3'; n = 100; P = (T d - T.) = (i 20 - 20) = loolbs. 
 
 B H P = 27rmP = 2 x y x 3 x IPO x IPO _ 5 . 71 
 33,000 33> 
 
 EXAMPLE II. What must be the number of revolutions per 
 minute of a driving pulley 6' in diameter, in order that it may 
 transmit 5-71 B.H.P. by a belt to another pulley, if the net pull 
 on the belt Ls 100 Ibs. ? 
 
 ANSWER. Here we have the same data to go upon as in 
 Example I., except that we are given the B.H.P. instead of the 
 revolutions per minute. Then, transposing every quantity 
 except n (the revolutions per minute) to one side of the above 
 equation, we have 
 
 n = (B.H.P.) x 33,000 = 5-71x33,000 = 100 r 
 27irP 2 x Hf x 3 x 100 
 
 In precisely the same way, if you were given the power to be trans- 
 mitted, the revolutions per minute, the difference of tension on the two 
 sides of the belt, and you were asked for the diameter of the pulley, the 
 formula would appear thus 
 
 n _(B.H. P.) x 33,000 
 
 7TP 
 
 If it was the difference of tension in the belt that was asked for, then 
 
 You would (after arranging the formula in this way, so as to keep the 
 unknown quantity on one side of the equation) simply have to substitute 
 the numerical values corresponding to the different symbols, and then 
 cancel out the figures in numerator and denominator, in order to reduce 
 the long multiplication and division to a minimum, and thereby arrive at 
 the result as quickly as possible. 
 
 Velocity Ratios in Belt Gearing. Let two or more pulleys 
 be connected by belting in the manner shown by the accompany- 
 ing figure. Then, if there is no slipping of the belts, the circum- 
 
 VELOCITY RATIOS IN BELT GEARING. 
 
 f erential speeds of the pulleys will be the same as the velocity of 
 the belts passing round them. 
 
120 LECTURE XI. 
 
 Let Dj, D, * Diameters of the drivers* 
 F 1? F, = Diameters of the followers. 
 N Di , N" Dj = Number of revolutions per minute of the drivers. 
 N p , N p = Number of revolutions per minute of followers. 
 
 Then, taking the first pair of pulleys, Dj and F, we have 
 Circumferential speed of driven = Circumferential speed of follower i. 
 
 (Divide both sides by *) 
 
 Or, The product of the diameter of -\ 
 the driver and its number of 
 revolutions per minute. 
 
 i The product of the dia- 
 meter of the follower 
 and its number of re- 
 
 ( volutions per minute. 
 
 D "N" 
 Or ...... ^-jg . . . (1) 
 
 i.e., The ratio of the diameters of the pulleys is in the inverse 
 ratio of their speeds or revolutions per minute. 
 
 Treating the motion of the second set of pulleys in exactly the 
 same way, we have 
 Circumferential speed, of D f = circumferential speed, of F, 
 
 (Divide both sides by TT) 
 
 D,N Dj = F,N,, 
 
 <* ..... f = ?' ' (2) 
 
 J a -^ 3 
 
 (But the revolutions of F, and of D 2 are the same) .'. N ri = N D 
 Or, . . . D, N, - F,N, .-. N F> - ?i/?' 
 
 X 
 
 _F,NF 
 
 Consequently . . ^ -^ 
 
 -rr 1 
 
 i 
 
 / (Dividing both numerator) ND = _D!_ = F t x F, 
 I and denominator by N Fi )J ]^ p i D 2 = D X x D, 
 
 ^a 
 
 Or, we might have arrived at the same result by multiplying equation* 
 (1) and (2) together. Thus 
 
VELOCITY RATIOS IN BELT GEARING. 121 
 
 Or, Speed of first driver m Product of diameters of followers 
 
 Speed of last follower Product of diameters of drivers 
 
 Or, . N Di x Dj x D, = N Fj x F t x F, 
 t.., Speed of first driver x dia-} _ (Speed of last follower x diame- 
 . meters o/JAe drwers J ~ ( ters of the followers. 
 
 In the same way we may treat any number of drivers and 
 followers by this general formula viz., 
 Speed or number of revolutions} 
 
 per minute of the first driver [ 
 x the successive diameters oft 
 the drivers ) 
 
 Speed of the last follower x the 
 successive diameters of the 
 followers. 
 
 Precisely the same rule holds good for discs driven by contact 
 friction and for wheel gearing, as you will find from the next 
 lecture ; but in friction gearing and wheel gearing the driver and 
 the follower move in different directions, whereas in belt gearing 
 they move in the same or in the opposite direction, according as 
 the driving belts are " open " or " crossed." 
 
 EXAMPLE III. Referring to the previous figure, suppose that a 
 driving pulley, D p is connected by a belt to a follower, F p 
 whilst it moves at 100 revolutions per minute. If the diameter 
 of the driver is 6' and of the follower 3', what will be the number 
 of revolutions per minute of the follower 1 
 
 By the previous formula for two pulleys, 
 D t x N Di = F t x N ?i 
 
 XT D, x N D 6' x 100 
 .. -N Pi -t_ * = 200r.p.m. 
 
 *! 3 
 
 EXAMPLE IT. Referring to the previous figure, suppose that 
 a driving pulley D a (4' diameter), is geared to a follower, F t 
 (2' in diameter), and that a second driver D, (4/ diameter), fixed 
 to the same shaft as F p is geared to a second follower F, (i' dia- 
 meter). If D makes 60 revolutions per minute, what is the speed 
 ofF,? 
 
 By the previous formula for four pulleys, 
 N Di x D t x D, = N p> x F t x F f 
 
 ** =480 
 
 r.p.m. 
 
 Open and Crossed Belts. By referring to the next figure, 
 the student will observe that the left-hand end view shows what 
 
122 LECTURE XI. 
 
 is termed an open belt, OB, and that the right hand end view 
 shows a crossed belt, 013. In the case of open belts, the driver 
 and the follower rotate in the same direction (as may be seen from 
 the second and third figures in this Lecture) ; whereas, with 
 crossed belt driving, the follower revolves in the opposite direction 
 to that of the driver, just as it does when direct friction or wheel- 
 gearing is used. 
 
 Fast and Loose Pulleys. As will be seen from the two front 
 views in the next set of illustrations, the open and the crossed 
 belts are shown passing from the broad driving-pulleys DP, to 
 the broad loose pulleys LP. Loose pulleys are generally bushed 
 with gun-metal, and then bored out so as to fit their shafts easily. 
 This permits them to rotate without turning the shaft upon which 
 they bear. The pulleys, FP, are keyed hard on to the shafts, so 
 that when the belt is forced over upon them by means of the 
 shifting forks, SF, the machines connected with the same are set 
 agoing. This simple combination of fast and loose pulleys there- 
 fore enables a machine to be stopped or started at pleasure, without 
 interfering with the motion of the driving pulley and the belt. 
 In ordinary cases where there is only one driving belt required, 
 the loose pulley is of the same breadth as the fixed pulley.* 
 
 Belt- Gearing Reversing Motions. In many kinds of 
 machine tools it is desirable to be able to drive the tool first in 
 one direction and then in the opposite direction, as well as to 
 start or stop it. This is frequently effected by a combination of 
 open and crossed belts with fast and loose pulleys, as illustrated 
 by the accompanying figure. 
 
 From what has just been said about open and crossed belts, as 
 well as fast and loose pulleys, the student will have no difficulty 
 in understanding this arrangement of reversing gear. If applied 
 to a machine for planing metals, the shaft which is keyed to the 
 fixed pulley FP would be connected either through wheel gearing 
 and a rack, or through a central screw, to the travelling table of 
 machine upon which the job to be acted upon is secured. When- 
 ever the table had been moved backwards to the end of the 
 required stroke by the crossed belt, the shifting fork SF would 
 be pushed forward by an outstanding arm or kicker attached to 
 
 * See the set of figures after the next, where E l is the driving belt 
 engaging the fixed pulley, FP ; and where LP is the loose pulley, to which 
 the belt may be shifted by means of the shifting-fork, SF, whenever it is 
 desirable to stop the speed cones, SCj, SC^ and the machine to which they 
 are connected. In the first front view of the next set of figures, the driving 
 pulley, DP, and both of the loose pulleys, LP, are drawn too narrow. They 
 should have been represented half as wide again, in order to prevent the 
 belts slipping over the outside edge, when the other belt is shifted on to 
 the fixed pulley situated between them. 
 
BELT-GEARING REVERSING MOTIONS. 
 
 123 
 
 the side of the table at such a position as would cause the crossed 
 belt CB to be shifted from the central fixed pulley to its loose one, 
 and at the same time bring over the open belt from its loose pulley 
 to the central fixed one. Whenever the planing tool had finished 
 its cut on the metal, the shifting fork would be pulled backward 
 by another similar outstanding arm or kicker (also attached to 
 the travelling table of the planing machine, at a position just 
 beyond the end of the required stroke for the particular job under 
 operation), thereby shifting the open belt OB from FP, to its loose 
 pulley, LP, and at the same time pulling over the crossed belt, CB, 
 
 L^FP 
 
 FRONT VIEW, 
 Forward and Return at same Speed. 
 
 FRONT VIEW. 
 Quick Return. 
 
 BELT GEARING REVERSING MOTIONS. 
 
 INDEX TO PARTS. 
 
 DP represents Driving pulleys. 
 FP Fixed pulleys". 
 LP Loose pulleys. 
 
 OB represents Open belts. 
 CB Crossed belts. 
 
 SF Shifting forks. 
 
 from its loose pulley to the central fixed pulley, thus causing the 
 table to make the return stroke. 
 
 '.The left-hand front view, with its accompanying end views, shov 
 the necessary arrangements when the forward and backward 
 velocities of the table are equal. The right-hand front view illus- 
 trates the case wherein the backward or non-planing motion is 
 intentionally made quicker than the forward or cutting stroke, so 
 as to save time, by having the back motion fixed pulley, FP, and its 
 corresponding loose one, LP, made smaller than the forward set. 
 The end views for this latter case would be similar to the former 
 onp. with the exception that the crossed belt would engage P smaller 
 
124 
 
 LECTUKE XI. 
 
 pulley of the same size as shown by the front view. This latter 
 arrangement can evidently be employed to obtain a fast or a slow 
 motion in the same direction, by simply having both belts open or 
 both crossed. 
 
 Stepped Speed Cones with Starting and Stopping Gear. 
 
 In many machines, such as lathes, planers and other machine 
 
 tools, it is very desirable not only to be able to start and stop 
 them, but also to alter their speed so as to suit different classes of 
 
 END VIEW. SIDE VIEW. 
 
 STEPPED SPEED CONES WITH STARTING AND STOPPING GEAB. 
 
 INDEX TO PARTS. 
 
 HB t ,HB 2 represent Hangingbrackets 
 for supporting 
 shaft, &c. 
 
 SCj,SC 2 Speed cones. 
 FP Fast pulley. 
 LP Loose pulley. 
 
 B,,B 2 represent Belts. 
 
 H Handle. 
 SB Sliding bar. 
 SF Shifting fork. 
 W Weight to fix SB in 
 positions ^ ^.. 
 
 work, without affecting the motion of the prime motor or that of 
 the shop driving-shaft. These objects are generally attained by a 
 combination of fast and loose pulleys with what are termed 
 " stepped speed cones." The accompanying side and end views 
 illustrate the arrangement as usually carried out in engineering 
 works. When the starting-handle, H, is turned to the right 
 hand, it pulls over the sliding-bar, SB, with its shifting-fork, SF, 
 which moves the belt, B p from the loose pulley, LP, to the fixed 
 one, FP ; thus setting the speed cones, SG p SC 2 , and thereby the 
 
DRIVING AND FOLLOWING PULLETS. 12$ 
 
 machine in motion. When the handle is turned to the left, it 
 pushes the sliding-bar and shifting-fork also in that direction, 
 thus moving the belt from the fixed to the loose pulley, which 
 allow the cones and machine to come to rest. In each case the 
 weight, W, causes a notch in the sliding-bar to engage with its 
 left-hand supporting bracket, thereby preventing the shifting fork 
 from pushing the driving belt too far, or off either pulley, and at 
 the same time ensuring that it remains in the desired position. 
 Both supporting brackets for the sliding-bar, SB, are merely 
 right-angle extensions from the hanging brackets, HBj, HB,, 
 which carry the upper shaft with its cone and pulleys. 
 
 The upper and lower speed cones, SCj, SO,, are generally made 
 of the same size and shape, but they are always keyed to their 
 respective shafts in opposite directions. Consequently, if it 
 should be desirable to run the machine fast for light work, the 
 belt, B r is shifted on to the largest pulley of the upper cone and 
 the smallest one of the lower cone. If the machine is required 
 to move slowly for heavy cuts, then the belt is placed on the 
 smallest upper pulley and the largest lower one. Any desired 
 intermediate speed between these extremes is obtained by adjusting 
 the belt on one or other of the remaining sets of pulleys of the 
 upper and lower cones. 
 
 The student can easily prove to himself (by drawing down the 
 arrangement to scale) that such stepped speed cones, if connected 
 by a crossed belt on one pair of its pulleys, will produce the same 
 tension in the belt with any other pair.* With open belt-driving 
 the tightness of the belt will not be the same when on one pair of 
 the pulleys as when on another ; but the difference is so small 
 that it can generally be disregarded in practice without having 
 recourse to tightening or slackening the same. 
 
 Driving and Following Pulleys in Different Planes. It 
 is often necessary to drive a follower placed in a different plane 
 from the driver. The accompanying set of illustrations show 
 very clearly how this is effected. The important precaution to be 
 observed is, that the leading or on-going part of the belt must 
 enter upon the follower in a fair or direct line with its plane o/ 
 rotation. If this rule be attended to, then power may be trans- 
 mitted between two non-parallel shafts, as shown by the first 
 tigure, even if their centi-e lines are in planes at right angles to 
 each other i.e., when the belt is working with quarter-twist. 
 When two shafts are in planes at right angles to each other, and 
 
 * The algebraical proof of this will be considered in our "Advanced 
 Book on Applied Mechanics." The student should refer to the general 
 view and to the detail drawings of the stepped speed cones in the foot- 
 driven screw-eutting lathe illustrated in Lecture XVI. 
 
126 
 
 LECTURE XI. 
 
 Tullis's Thick-sided Leather 
 Chain Belt, Working Quarter- 
 twist, and Transmitting l^ower 
 between two shafts which are 
 not parallel. No Guide Pul- 
 leys are required for this drive. 
 
 Flat Belt Working Quarter-twist and 
 Transmitting Power between two right- 
 angled shafts, with leading Guide Pulley 
 (GP) to remove the twist from the Belt 
 before it enters upon the Follower, and to 
 give the belt more grip on the pulleys. 
 
 Flat Belt Transmitting Power 
 over Guide Pulleys between two 
 non -parallel shafts in the same 
 plane. 
 
 Flat Belt Transmitting Power between 
 two parallel shafts not in the same plane 
 by aid of guide pulleys (GP). 
 
SHAPE OF PULLEY FACE. I2/ 
 
 it is found desirable to remove the twist from the belt before it 
 enters upon the follower, then a guide-pulley, GP, must be used 
 & shown by the second figure. When the shafts are parallel, but 
 not in the same plane, then the power must be transmitted by aid 
 of two guide-pulleys, as seen from an inspection of the third 
 figure. Or, should the shafts not be parallel, but in the same 
 plane, two guide-pulleys are necessary, as in the fourth figure. 
 Guide-pulleys, if supported by spindles running in adjustable 
 bearings or brackets, may be made serviceable as tightening- 
 pulleys for the purpose of taking up the slack of the belt, and 
 thus giving the necessary grip for transmitting more power with 
 a steadier drive than can be obtained without them. 
 
 Shape of Pulley Face. The student will have observed that 
 the faces or rims of the fast and loose pulleys, as well as those of 
 the stepped cones in the previous set of figures, are slightly 
 curved. This convex curvature, or double coning, is purposely 
 done in order to ensure that the belt may ride easily and fairly 
 in the centre line of the pulley face without inclining to either 
 side. A flat band, if placed on the smaller end of a revolving 
 straight conical pulley, will naturally tend to rise to the larger 
 end of the cone. Consequently, if each half of the face of a 
 pulley is coned (or, which amounts to the same thing, if the rim 
 of the pulley be curved so as to have its largest diameter in the 
 middle of its face), each half of the breadth of the belt will have 
 an equal tendency towards the middle of the pulley's rim. When 
 very fast driving and sudden severe stresses are brought to bear 
 upon a machine, as in the case of circular saws, morticing 
 machines, and emery-wheel grinders, it is found necessary to fit 
 the pulleys with side flanges, in addition to curving their rims, in 
 order to prevent the belts from sliding off the pulley's face to one 
 "ide or to the other. 
 
 N.B. The student may be referred to the Author's Text-book on 
 "Applied Mechanics and Mechanical Engineering," Chapters XVII. and 
 XVIII., for further information on belt, rope, and chain gearing. 
 
128 LECTUKE XL QUESTIONS 
 
 LECTURE XI. QUESTIONS. 
 
 1. In machinery, where one pulley drives another by means of an end- 
 Jess belt, there is a difference of tension in the two parts of the belt. Why 
 is this ? The pulley on an engine shaft is 5 feet in diameter, and it makes 
 100 revolutions per minute. The motion is transmitted from this pulley 
 to the main shaft by a belt running on a pulley, and the difference in tension 
 between the tight and slack sides of the belt is 1 15 Ibs. What is the work 
 done per minute in overcoming the resistance to motion of the main shaft ? 
 
 Ant. 180,642 ft.-lbs. 
 
 2. Deduce from the "principle of work" a formula for the brake 
 horse-power transmitted by a belt. The pull on the driving side of a belt 
 is 200 Ibs. and on the following side 100 Ibs., whilst the belt has a velocity 
 of 990 ft. per minute. Find the number of units of work performed in two 
 minutes and the B.H.P. transmitted. Ans. 198,000 ft.-lbs., 3 B.H.P. 
 
 3. State and prove the rule for estimating the relative speeds of two 
 pulleys connected by a belt. Also, the velocity ratio between the first 
 driver and the last follower in belt gearing, where there are two or more 
 drivers and a corresponding number of followers. [A main shaft carrying 
 a pulley of 12 inches diameter and running at 60 revolutions per minute, 
 communicates motion by a belt to a pulley of 12 inches diameter, fixed to 
 a countershaft. A second pulley on the countershaft, of 8| inches dia. 
 meter, carries on the motion to a revolving spindle which is keyed to a 
 pulley of 4^ inches diameter. Sketch the arrangement and find the 
 number of revolutions per minute made by this last pulley. Ana. 123-5 
 
 4. Two pulleys are connected by a driving belt, and the sum of their 
 diameters is 30 inches ; one pulley makes 2 revolutions while the other 
 makes 3 revolutions ; find their respective diametere, Ans. 18", 12". 
 
 5. An engine works normally at 106 revolutions per minute. At that 
 speed it was found that it drove by belting a dynamo at 420 revolutions 
 per minute, but to show off the electric lights at their normal candle 
 power the dynamo had to be run at 460 revolutions per minute. At what 
 speed was the engine being driven 1 Ans. 116 revolutions per minute. 
 
 6. A pulley of 3 feet radius rotates at 100 revolutions per minute and 
 transmits motion to another pulley of 36 inches diameter. If there is 
 10 per cent, slip on the belt what will be the speed of the follower ? 
 What will be the net driving pull on the belt if 5 B.H.P. is transmitted by 
 it ? Ans. 180 revolutions per minute ; 97-2 Ibs. 
 
 7. Sketch an arrangement of pulleys and bands for obtaining a reversing 
 motion from a shaft driven at a constant rate in one direction, and describe 
 the action of the combination. 
 
 8. Sketch a combination of fast and loose pulleys as used for setting in 
 motion, or stopping machinery. Explain the construction adopted for re- 
 taining a flat belt upon a pulley, pointing out where the fork is to be 
 applied, and why. 
 
 9. Sketch and describe a good form of slow forward and quick return 
 for a shaping machine. 
 
 10. Sketch and describe an arrangement for driving the table of a plan- 
 ing machine by means of a screw, so that the table may travel 50 per cent, 
 faster in the return than in the forward or cutting stroke. 
 
LECTURE XI. QUESTIONS^ I2Q 
 
 11. What is the object of using guide-pulleys in machinery ? Mention 
 Instances of their use, and show how the directions of their axes are 
 ascertained. 
 
 12. Describe, with a sketch, the mode of reversing the motion of the 
 table in a planing machine, when a screw is employed to drive the table. 
 
 13. A rope transmits 20 horse-power to a rope pulley of 8 feet diameter ; 
 draw a section of the rope in its groove. If the pulley makes 100 revolu- 
 tions per minute, what is the speed of the rope in feet per minute ? What 
 is the difference of the tensile forces in the rope on the two sides of the 
 pulley ? As it is the difference between the tensile forces in a belt or rope 
 that is important for power, why is it necessary to have any pull on the 
 slack side ? ^n*. 2513 ft. per min., and 262-5 lbs - 
 
 14. What are cone or speed pulleys ? Describe the use of such pulleys 
 in any machine with which you are acquainted. The spindle of a lathe 
 can, by moving the belt on its cone pulleys, be driven at four hundred 
 revolutions per minute when at its greatest and at 100 revolutions per 
 minute when running at its lowest speed. If the revolutions of the 
 driving shaft are kept constant throughout, and the largest diameter of 
 the speed cones is 20", what must be the diameter of the smallest steps 
 on the pulleys ; the speed pulleys on the two shafts being of the same 
 size ? Sketch the pulleys in position. An*. 10 inches. 
 
 15. Upon what does the limiting difference of tensions in the tight and 
 slack sides of a moving belt depend? If the working stress in a 
 belt of sectional area a square inches be /pounds per square inch, and the 
 ratio of the tensions in the tight and slack sides be m, find the horse-power 
 that can be safely transmitted when the speed of the belt is v feet per 
 second. (0. & G., 1903, 0., Sec. A.) 
 
 550 
 
 1 6. A belt transmits 60 H.P. to a pulley 16 inches in diameter running 
 at 263 revolutions per minute. What is the difference of the tensions on 
 the tight and slack sides 1 Ans. T d - T, = 1796 lbs. (B. of E., 1904-) 
 
 17. State the condition that has to be satisfied in order that the same 
 belt may work on two or more pairs of pulleys keyed to parallel shafts 
 (i) when the belt is crossed, (2) when open. Explain also why a belli 
 always climbs to the section of the pulley where the diameter is greatest. 
 
 (C. & G., 1905, O., Sec. A.) 
 
 18. Explain any means with which you are acquainted for determining 
 the brake horse-power of an engine provided with a fly-wheel. 
 
 A rope is wrapped once round a fly-wheel. One end of the rope carries 
 a weight of 500 lbs., and the other end is led upwards, and is attached to 
 a spring balance. When the revolutions are 105 per minute, the pull in 
 the spring balance varied between 10 and 20 lbs. If the diameter of the 
 wheel be 8 ft., and of the rope I in., find the average brake horse-power. 
 Ant. 39-6 B.H.P. (C. & G., 1005, 0., Se^. A.) 
 
130 
 
 LECTURE XII 
 
 CONTENTS. Velocity Ratio of Two Friction Circular Discs Pitch Surfaces 
 and Pitch Circles Pitch of Teeth in Wheel Gearing Rack and Pinion 
 Velocity Ration in Wheel Gearing Example I. Principle of Work 
 applied to Wheel Gearing Examples II. III. Questions. 
 
 Velocity Ratio of Two Circular Friction Discs. If two 
 
 truly centred circular discs or cylindrical rollers, having their 
 shafts parallel to each other and free to turn in fixed bearings, 
 be brought into firm contact ; then, if one of them be driven 
 round, and if there be no slipping, the other one will rotate in 
 the opposite direction with the same circumferential speed or 
 surface velocity (see the next figure). 
 
 Consequently, their velocity ratio will be in the inverse ratio to 
 their diameters. 
 
 This may be proved in exactly the same way as we found the 
 velocity ratio of two pulleys driven by a belt in Lecture XI. 
 
 Let D x = Diameter of the driving disc. 
 F x = Diameter of the following disc. 
 N D = Number of revolutions per minute of D r 
 N, = Number of revolutions per minute of F r 
 
 Then, The peripheral velocity of D l = Peripheral velocity of F t 
 i.e ..... irD^-irFjlSr^ 
 
 Or, .... D 1 N D1 = F 1 N F1 
 i.e. The Driver's diameter x its speed = Follower's diam? x its speed. 
 
 Speed of the Driver = Diameter of Follower 
 Speed of the Follower Diameter of Driver 
 
 This velocity ratio may also be proved in the following way : 
 
 Let the two circles centred at A and B represent a cross 
 
 flection of the two friction discs in contact at C ; and let them 
 
 move by rolling contact through the angles 6 and <j> respectively 
 
 in the same time. 
 
 Since the magnitude of an angle in circular measure is 
 
PITCH SURFACES AND PITCH CIRCLES. 13! 
 
 always = the length of the arc subtended by the angle at the centre of 
 \he circle -r the radius of the circle. 
 
 VELOCITY RATIO OF Two CIRCULAR Discs. 
 
 Then, 
 
 \ a 
 
 But, the arc DC = the arc EC since there is no slipping. 
 Consequently, 
 
 DC 
 
 The angular velocity of circle A _ 6 r, _ r 3 
 The angular velocity of circle B < EC r l 
 
 Or,* 
 
 The angular velocity or speed of driver, A _ Radius of follower B 
 The angular velocity or speed of follower B Radius of driver A 
 
 Pitch Surfaces and Pitch Circles. In the case of the two 
 discs or rollers just considered, their cylindrical surfaces are termed 
 the pitch surfaces; and the two circles in the previous figure 
 (which is simply a representation of their cross section, or section 
 in the plane of their rotation) are called the pitch circles. 
 
 * The angular velocity of a rotating disc is the angle described by its 
 radius in unit time. 
 
 The relation between angular velocity and linear velocity may be shown 
 thus: Let w=the angular velocity; whilst v = the linear velocity of a 
 point at radius r from the centre of motion when the disc makes * revo- 
 lutions in unit time ; 
 
 Then wxr=v ; or, w=-; but v=2vrn t 
 
132 LECTURE XII. 
 
 When the resistance to motion of the follower is great, the 
 discs have to be provided with teeth in order to prevent slipping. 
 
 Consequently, the pitch surfaces and the pitch circles of such 
 toothed rollers, toothed wheels, or spur wheel and pinion, are the 
 surfaces and the circles of their rolling contact.* 
 
 Pitch of Teeth in Wheel Gearing. The linear or the 
 circular distance from the centre of one tooth to the centre of the 
 next one, or the distance from one edge of a tooth to the corre- 
 sponding edge of its neighbouring one, as measured on the pitch 
 circle, is termed the pitch of the teeth of a wheel. 
 
 Let D = Diameter of a wheel or pinion at its pitch circle. 
 p = Pitch of the teeth in the wheel or pinion. 
 n = Number of teeth in the wheel or pinion. 
 ThenvrD = p x n 
 
 For the circumference of the pitch circle must bo equal to the 
 pitch between any two neighbouring teeth x the number of teeth 
 in the wheel or pinion ; since the pitch between each pair of teeth 
 must be the same all round the pitch circle, otherwise the wheel 
 would not gear properly with any other wheel or pinion of the 
 same pitch. 
 
 Back and Pinion. If a straight bar of iron be furnished with 
 teeth on one side it is called a rack. It may therefore be con- 
 sidered as a wheel of infinite radius. When a rack has a pinion 
 of the same pitch geared with it, the two form the useful combi- 
 nation termed the rack and pinion. It is employed for moving 
 to and fro the tables of planing machines and large saw benches, 
 as well as for elevating and lowering sluices in dams, &c. 
 
 Pinion 
 
 END VIEW. SIDE VIEW. 
 
 BACK AND PINION APPLIED TO A SAW-MILL TABLE. 
 
 The accompanying illustrations show the second of these appli- 
 cations, where two parallel racks are fitted to the under side of 
 
 * When a large toothed wheel gears with a small one, the larger is 
 termed a &pur- wheel and the smaller a pinion. It is not possible in the 
 space allotted to this elementary manual to enter into the best forms of 
 the teeth of different kinds of wheel gearing. This subject is taken up in 
 our "Advanced Text Book," Vol. I., Part II. 
 
VELOCITY RATIO IN WHEEL GEARING. 133 
 
 two movable tables or platforms. Upon the upper side of one 
 of the tables is laid a log of wood adjusted in the desrred position 
 by wedges. The tables are each carried and guided by four 
 rollers turning on fixed spindles. To the projecting end of the 
 pinion shaft there is fitted a lever handle, so that by merely 
 turning this handle in one direction, the racks, tables, and log of 
 wood are pushed forward upon the projecting circular saw which 
 revolves between the platforms, and if turned in the opposite 
 direction they are drawn backwards. The pinions with their shaft 
 and handle, have no linear motion, for the shaft is simply free 
 to rotate in fixed bearings. 
 
 The rack and pinion with their handle constitute a modification 
 of the wheel and axle, or lever and winch barrel, where the re- 
 sistance offered by the rack and its load is overcome by a force 
 applied to the handle. Every revolution of the handle turns the 
 pinion, and consequently moves the rack through a linear dis- 
 tance equal to the circumference of the pinion's pitch circle. 
 The principles of moments and of work can therefore be applied 
 to this machine in exactly the same way as we applied them to 
 the wheel and axle and the winch. 
 
 If P = Pull acting on the handles, 
 R = Radius of handle, 
 r = Radius of pinion's pitch circle, 
 W = Weight or resistance overcome ; 
 
 Then . . P x 2n-R = W x 2wr 
 P x R = W x r 
 
 Theoretical advantage . . * 
 
 . 
 
 P r W s velocity 
 
 Velocity Ratio in Wheel Gearing. From what has been 
 said about belt gearing, pitch surfaces, pitch circles, and pitch of 
 teeth, it must be at once apparent to the student that the same 
 rule which was worked out in Lecture XL, in connection with 
 belt gearing, will equally apply to the case of wheel gearing, 
 where there are an equal number of drivers and followers. In 
 the accompanying figure, where there are three drivers and three 
 followers, 
 
 Let D I} D,, D, = Diameters of the drivers. 
 F p F f , F, = Diameters of the followers. 
 N D , N Pt = Number of revolutions in the same time of th 
 first driver and the last follower. 
 
 Then, following the same reasoning as was expounded in Lecture 
 XL for the velocity ratio of belt gearing, we have 
 
134 
 
 LECTURE XII. 
 
 The speed of the first driver x ' 
 the successive diameters of 
 the drivers 
 
 x T> 2 x D 3 
 
 Or, . 
 
 :The speed of the last follower 
 x the successive diameters 
 of the followers. 
 
 N F3 x F L x F 2 x F 3 
 
 Fs/ T^ v 1R^ 
 i X -C X J? - 
 
 x D 2 X D 3 
 
 The speed of first driver _ Product of the diameters of the followers. 
 tThe speed of last follower Product of the diameters of the drivers. 
 In the above equation we may substitute the radii, or the cir- 
 
 H 
 
 SIDE VIEW. PLAN. 
 
 WHEEL GEAKING IN A TEIPLE PUKCHASE WINCH 
 
 cumferences, or the number of teeth in the drivers and in the 
 followers respectively, for their diameters ; consequently, 
 
 Let r D , r D3 , r D = Radii of the respective drivers. 
 C D[> C v C D ^ = Circumferences ,, 
 
 Ti 1 , n D , n 3 = Number of teeth in 
 
 3 = Radii of the respective roilowers. 
 
 /, C F 2 , C F 3 = Circumferences 
 S w_, , n p * = Number of teeth in 
 
 Then, 
 Or, 
 
 N 
 
 x r 
 
 J), 
 
 , XT* 
 1 Dl 
 
 L x C D x C D x CD = N F x C r x C F x 
 
 123 312 
 
PRINCIPLE OF WORK APPLIED TO WIIEEL-GEAKING. 135 
 
 Or, N D x nv x W D x n^ = N^ x n > a 
 
 EXAMPLE I. Three drivers of 10, 20, and 30 teeth each, gear 
 respectively with three followers of 40, 80, and 120 teeth each. 
 Ascertain the velocity ratio between the first driver and the tost 
 follower. 
 
 By the above formula 
 
 N D x TI DI x n D ^ x n 3 = N F;j X n ^ x n p ^n r ^ 
 N n Fl x n p x n F 
 
 Substituting the coire-^ 
 
 spending numerical I N D _ 40 x 80 x 120 4x4x4 64 
 values for the letters, j ^ ~~ i x 20 x 30 = i "1 
 we get } 
 
 Principle of Work applied to Wheel-gearing. Beferring 
 to the previous figure, it is perfectly evident from the former 
 applications to other machines of the " principle of work," that, 
 neglecting friction, the force applied (to the handles of the 
 machine) x the distance through which it acts, will be equal to the 
 weight raised x the distance through which it is elevated. 
 
 Let P = Push applied to the handles in Ibs. 
 
 R = Radius or leverage at which P acts. 
 
 w W = Weight raised by the rope on the barrel B. 
 
 r = Radius or leverage with which W acts. 
 
 Dj, D,, D s = Diameters of the driving wheels. 
 F,, F 8 , F,= Diameters of the following wheels. 
 N D = Number of revolutions of the first driver, "D v or of 
 
 the handles, H. 
 N P = Number of revolutions in the same time of the last 
 
 follower, F 3 , or of the barrel, B. 
 Then, by the principle of work and neglecting friction 
 
 P x its distance * = W x its distance. 
 t.., . P x 2?rR x N D = W x 2?rr x Np 
 (Divide both sides of the equation by 2ir) 
 .% PxRxN D =WxrxN P 
 ~ PxR N F P 8 N Ps xr 
 
 = or = 
 
 * It is evident that in order to 'obtain the distance through which P 
 acts, we must multiply the circumference of the circle described by the 
 handles by the number of revolutions they make ; and in the same way 
 the circumference of the fearrel must be multiplied by the revolutions 
 which it makes in the same time, in order to get W's distance. 
 
 Of THE 
 
 UNIVERSITY 
 
136 LECTURE XII. 
 
 But by the previous equation for velocity ratios, 
 N F D, x D 2 x P, 
 ND'-^XF.XF, 
 P x R D! x D 2 x P 8 
 "W^Tr = F 1 xF a xF, 
 
 Or, P x R x F! x F 2 x F 3 = W x r x D x x D 2 x D, 
 Hence the general rule for work done in wheel-gearing "Pxits 
 leverage x the diameters (or radii, or circumferences, or number of 
 teeth) of all the followers = W x its leverage x the diameters (or radii, 
 or circumferences, or number of teeth) of all the drivers. 
 
 EXAMPLE II. It four men exert a constant force of 15 Ibs. 
 each on the handles of a compound crab or winch (such as that 
 illustrated by the previous figure), and if the leverage of the 
 handles is 15", whilst the weight to be raised acts on the barrel 
 or drum at a leverage of 5'', what load will they lift if the 
 respective diameters of the drivers are 12", 20", and 20"; and 
 of the followers, 36'', 80'' and 100", neglecting friction? 
 
 ANSWER. In this case, P = 4x15 = 60 Ibs.; R=i5"; r=5"; 
 1^=12"; D 8 =2o"; D 3 =2o"; ^ = 36"; F, = So", and F 3 = 100". 
 
 By the above formula and by substituting the corresponding 
 numerical values \ve have 
 
 P x R x F t x F 8 x F 3 = W x r x D L x D, x D, 
 60 x 15'' x 36" x 80" x ioo"= W x 5" x 12" x 20" x 20" 
 
 3345 
 60 x 4ft xg0x $0x|00 
 
 $ x U x x ^0 
 Or, .... W = 6o x 3 x 3 x 4X 5 = 10,800 Ibs. 
 
 EXAMPLE III. If 40 % of the force applied to the handles be 
 absorbed in overcoming internal friction in the above example of 
 a winch, what weight can then be raised by the four men, each 
 acting, as before, with a constant force of 15 Ibs. ? 
 
 ANSWER. If 40 % of the applied force be lost in overcoming 
 friction, then only 60 % is left for effective work, or the efficiency 
 or modulus of the machine is said to be 0-6.* 
 
 Consequently, 100 : 60 : : 10,800 Ibs. : x Ibs. 
 
 60 x 10,800 
 .. x= -? = 6480 Ibs. 
 
 100 
 
 * The term modulus of a machine is only another expression for the more 
 ppropriate phrase, efficiency of a machine. 
 
LECTURE XII. QUESTIONS. 137 
 
 LECTURE XII. QUESTIONS. 
 
 f . When two circular discs with fixed centres are in firm contact and 
 roll uniformly together, state and prove the rule for estimating their 
 relative speeds of rotation. 
 
 2. Define the pitch circle of a toothed wheel. When two pitch circles, 
 A and B, of diameters 2 and 3 respectively, roll together, prove that the 
 angular velocity of A is to that of B as 3 to 2. Three spur wheels, A, B, C, 
 with parallel axes, are in gear. A has 8 teeth, B has 32 teeth, and C has 
 42 teeth. How many turns will A make upon its axis while C goes round 
 8 times ? Why is B termed an idle wheel J Ant. 42 turns. (Sec Aote to 
 Question 10 re'ldle Wheel.) 
 
 3. What is the pitch of a tooth in a spur wheel ? Two parallel shafts, 
 whose axes are to be as nearly as possible 2 feet 6 inches apart, are to be 
 connected by a pair of spur wheels, so that while the driver runs at 100 
 revolutions per minute, the follower is required to run at only 25 revolu- 
 tions per minute. Sketch the arrangement, and mark on each wheel its 
 diameter and the number of teeth, supposing the pitch of a tooth to be 
 ii inch. 
 
 Ans. The follower is 48 inches diameter with 120 teeth. 
 The driver is 12 30 
 
 4. Define the "pitch surface" and the "pitch circle" of a toothed 
 wheel. Two parallel axes are at a distance of 10 inches, and they are to 
 rotate with velocities as the numbers 2 and 3 respectively. What should 
 be the diameters of the pitch circles of a pair of wheels which would give 
 this motion. Find pitch of teeth on the smaller wheel if the larger has 
 24 teeth 1 Ans. 12 ins. and 8 ins. ; 1-57 inch. 
 
 5. Sketch and describe the "rack and pinion " and give instances from 
 personal observation of its application. A pinion of 3-2" diameter has 
 teeth of i" pitch, and gears with a straight rack applied to a sluice gate. 
 If the weight of the sluice and rack be 100 Ibs. and the lever handle 
 descnbes a circle of 40*2" in each turn, what force must be applied to the 
 handle to lift the gate ? How many feet will the sluice be lifted by six 
 turns of the handle ? Ans. 25 Ibs. ; 5 ft. 
 
 6. Sketch the arrangement known as the rack and pinion. Apply the 
 " principle of moments " and the " principle of work" to find the relation 
 between the force applied and the weight raised by aid of this machine. 
 A pinion has sixteen teeth of |-inch pitch in gear with a rack. If the 
 pinion makes 3^ turns, through what distance has the rack been moved ? 
 If the pinion is turned by a "handle 14 inches long, and with a force of 
 35 Ibs. applied to the handle, find the force with which the rack is urged 
 lorward. A ns. 49 inches ; 223 Ibs. 
 
 7. Deduce the formula for the velocity ratio in wheel gearing where there 
 are three drivers and three followers, and state the rule derived therefrom 
 in general terms. Three drivers of 20, 30, and 40 teeth respectively gear 
 with three followers of 40, 60, and 80 teeth. If the first driver makes 160 
 revolutions, how many revolutions will the last follower make 1 Ans. 20. 
 
 8. In the previous question, if the handles attached to the first driver 
 have each a radius of 15% and the drum connected to the last follower be 
 15" diameter, what force must be applied to the handles in order that 
 they may lift 1120 Ibs. supposing that the efficiency of the machine is 
 70 per cent. 1 Ans. 100 Ibs. 
 
 9. The hour and minute hands of a clock are on the same arbor or axis, 
 and the hour hand takes its motion from the minute hand. Devise some 
 train of wheels for connecting the two hands. 
 
138 LECTURE XII. QUESTIONS. 
 
 10. How would you determine the "pitch circles," and the proper 
 " pitch of the teeth " for a pair of spur-wheels ? What would be the 
 diameter of the pitch circle of a spur-wheel having 80 teeth of f-inch pitch 1 
 Ant. 19 inches. 
 
 Three spur-wheels A, B, C are on parallel axes, and are in gear. A has 
 so teeth, B has 35 teeth, and C has 55 teeth. How many revolutions upon 
 its axis will be made by A for every 4 revolutions of C ? Why is B called 
 an idle wheel and what is its use ? Ans. 22 revs. 
 
 Note re "Idle or Intermediate Wheel" When a wheel is carried on a 
 separate axle and is interposed between two other wheels (or is introduced 
 into a train of wheels), merely for the purpose of changing the relative 
 directions of rotation of the first and last wheel, then such intermediate 
 wheel is called an idle wheel, because it does not affect the numerical 
 value of the train, but only its sign. For examples, see Vol. I. of my 
 " Text Book of Applied Mechanics and Mechanical Engineering," Lectun* 
 
HOTES AND QUESTIONS* 139 
 
( 140 ) 
 
 LECTUKE XIII. 
 
 CONTENTS. Single-purchase Winch or Crab Example I. Double-pur- 
 chase Winch or Crab Example II. Wheel Gearing in Jib-Cranes 
 Questions. 
 
 IN this Lecture we will apply the principles and formulae dis- 
 cussed in the previous one to a few practical applications of 
 gearing in machines for lifting weights. 
 
 Single- pur chase Winch or Crab. The comparatively small 
 working advantage of the simple hand-driven wheel and axle or 
 
 SINGLE-PURCHASE WINCH OR CRAB. 
 
 By Messrs. Loudon Bros., Glasgow. . 
 
 handle and winch barrel (illustrated in Lecture V.) renders it unfit 
 for lifting greater weights than one or two hundredweight. Con- 
 sequently, whenever heavier loads have to he raised by manual 
 
SINGLE-PURCHASE WINCH OR CRAB. 141 
 
 labour, one of the most useful machines that can be employed is 
 the single-purchase crab. As will be seen from the accompanying 
 perspective view, this machine consists of a pair of lever handles 
 fitted to the squared ends of a round shaft carrying a pinion. 
 This pinion gears with a spur-wheel keyed to a lower shaft, upon 
 which is also fixed a drum or barrel. To a hook or eye on the 
 inside neck of the left-hand flange of this barrel the rope or chain 
 (to be connected to the load) is attached. Therefore, the turning 
 of the handles causes the barrel to rotate and wind the rope upon 
 it, thereby elevating the load. Both shafts turn in bearings 
 bored in the cast-iron end standards or A frames. These frames 
 are bound tightly together and kept at a fixed distance apart by 
 three wrought-iron collared stays, secured on the outside by screw 
 nuts. To the outside right-hand end of the barrel shaft there is 
 keyed a friction pulley acted on by a steel brake-strap, for the 
 purpose of enabling the labourers to lower a load gently or 
 quickly without enduring the stress and danger of pulling back 
 on the handles. In fact, after applying the brake-strap by its 
 outstanding handle, they can lift the claw pawl which is hinged on 
 the top stay (and which keeps the pinion in gear with the spur- 
 wheel when in the position shown on the figure) and by pulling 
 the upper shaft to the right, disengage the pinion from its wheel. 
 Then, by adjusting the pawl into the other groove of this shaft, 
 they are free to lower the load by the brake without having the 
 handles flying round. Between the right hand flange of the 
 barrel and its neighbouring A frame there is a ratchet-wheel (not 
 seen on the figure). This ratchet-wheel is generally cast along 
 with the barrel. Its pawl, which is hinged to the inner side of 
 the standard, can therefore be dropped down so as to engage 
 with a tooth of the stop-wheel, whenever it is necessary to cease 
 heaving up a heavy weight; thereby preventing the machine 
 overhauling, and giving the labourers freedom to leave the handles 
 and attend to other duties. 
 
 EXAMPLE I. In a single-purchase crab the lever handles are each 
 1 6" long, the diameter of the barrel is 8" ; the pinion or driver 
 has 12 teeth, and the wheel or follower 60 teeth. If two mei 
 apply a constant force of 20 Ibs. each to the handles, and are just 
 able to raise a weight of 600 Ibs. to a height of 20 feet in two 
 minutes, find (i) the theoretical advantage; (2) the working 
 advantage ; (3) the work put in for every foot the weight is 
 Hf ted ; (4) the work got out for every foot the weight is lifted ; 
 (5) the efficiency; (6) the percentage efficiency of the machine; 
 (7) the H.P. developed by the two men. 
 
 ANSWEB. Referring to the notation in last Lecture, we have 
 P = 2 x 20 ibs. 40 ltw. ; R=i6"; r = 4"; n D = 12 teeth; n, = 6o 
 
142 
 
 LECTUKE 
 
 teeth ; W T = the theoretical weight that would be raised if there 
 were no friction j W A = 600 Ibs. (the actual weight raised) ; h = 20 
 feet. 
 
 W 
 
 ( i ) Theoretical advantage = -pr 
 
 By the principle of work (neglecting friction.) 
 P x by its distance* = W T x its distance.* 
 
 P X 27rR X n y = W T X 27JT X 71 D 
 
 P x K, x n = WT x r 
 
 (Substituting the above numerical values we get) 
 
 i 
 
 ft x 
 
 _, W T 800 20 
 
 Consequently, . . -^~ = -- = -T 
 j- 40 * 
 
 W A 600 Ibs. 15 
 
 (2) Working advantage =^p~= 40 lbs . = Y 
 
 (3) Work put in for every foot W A is raised. From equation 
 (i) we see that for every foot W A is raised P must have gone 
 
 20 
 through 20 feet, since the velocity ratio is 
 
 .'. P x 20 = 40 Ibs. x 20 = 800 ft.-lbs. 
 
 (4) Work got out for every foot W A is raised 
 
 = W A x i' = 6oo Ibs. x i' = 600 ft.-lbs. 
 
 . _ . Work got out 600 ft.-lbs. 
 
 (<;) The efficiency *= ^TF r 2 . =o - FT-TI = !$ 
 
 Work put in 800 ft.-lbs. /;> 
 
 (6) Tlie percentage efficiency =-75 x 100 = 75% 
 
 / \ rr-L TT r> j 7 7 T J7 . , Work put in per minute 
 
 (7) The H.P. developed by the two men = 
 
 33,000 
 
 -u- v 8000 ft.-lbs. 1 i 
 
 ^.H.r. = - = bare, or of a horse-power per man. 
 
 * It is evident that 
 
 P x i turn of handles Number_of teeth in the driver 
 W T x i turn of barrel = NumbeTof teeth irTthelollower. 
 \)r, . . P x 27rK : W 
 
DOUBLE-PURCHASE WINCH OR CRAB. 
 
 143 
 
 Double-Purchase Winch or Crab. It will be observed, from 
 an inspection of the accompanying photographic view of a " Double- 
 purchase Crab," that the chief difference between it and the single- 
 purchase one is, that it has another pinion and wheel, with a view 
 of increasing the actual or the working advantage, and thus 
 enabling the same manual force to lift a greater load, although 
 by taking a longer time. It is also larger, heavier, and stronger. 
 
 DOUBLE-PURCHASE WINCH OB CRAB. 
 By Messrs. London Bros., Glasgow. 
 
 As will be seen from the figure, it may be used as a single-purchase 
 winch by simply lifting the claw-pawl hinged on the top stay, and 
 pushing the handle shaft forward until its left-hand pinion gears 
 with the large spur wheel, and then letting the pawl drop on to 
 bearing to the right hand of the two collars on this shaft. By so 
 doing, the right-hand pinion or first driver (when in double-pur- 
 chase gear) is freed from the first follower, and both are inactive 
 during the time it is used in single purchase, but the second 
 
144 
 
 LECTURE XIII. 
 
 driver is still in gear and is turned round by the spur wheel. The 
 brake strap pulley is keyed to the second shaft (carrying the first 
 follower and second driver), and can be used for lowering the load 
 without the handles coming into action (as described in the pre~ 
 vious case) by placing the claw-pawl between the two collars in the 
 first motion shaft. When the pawl is in this position, both of the 
 pinions on this shaft are out of gear. The machine may be locked 
 arid the load left suspended by dropping the ratchet into the 
 ratchet-wheel cast on the right-hand end of the barrel in the same 
 way as with the single-purchase crab. A triple-purchase winch 
 was illustrated in Lecture XII., and the student should again 
 refer to the plan and the side elevation of its gearing. 
 
 EXAMPLE II. Four men exert a force of 20 Ibs. each, on the 
 handles of a double-purchase crab, which are 15" long. The 
 driving pinions have 12 teeth each, the followers 24 and 48 teeth 
 respectively, and the diameter of the barrel is 10". Find the 
 weight that can be raised if 25 per cent, of the work put in be 
 absorbed in overcoming friction. 
 
 ANSWER. Here P = 4X2o = 8o Ibs. ; R = 1 5" ; n DL = 1 2 ; n Dt = 
 12; 7i Fi = 24; n F> = 48; r=s". 
 
 By the formula deduced in the previous lecture from the 
 principle of work (neglecting friction), 
 
 P x R x n f x n p = W T x r x n D xn D 
 
 324 
 After cancelling, we get 
 
 8ox3X 2 x 4 = W T = 1920 Ibs. 
 
 If 1920 Ibs. of work be expended by the men and 25 per cent. 
 of this be lost work, there remains 75 per cent, as useful work. 
 
 Or, . . 100 : 75 :: 1920 Ibs. : W A . 
 
 Weight actually raised = W A = 1440 Ibs. 
 
 Wheel Gearing in Jib Cranes. In Lecture VIII. the side 
 view of a jib crane was given for the purpose of exemplifying the 
 stresses on the jib, tie-rods, and central pillar. We now illustrate 
 A swing jib crane on a bogie and rails, to show that the frame- 
 work and lifting gear are simply those of an inverted double- 
 purchase crab with the toothed wheels placed outside the standards 
 instead of inside as in the ordinary winch. The snatch block 
 pulley (previously referred to in Lecture VII.), to the hook of 
 which the load is attached, doubles the theoretical purchase or 
 advantage of the winch gearing, and therefore one, two or more 
 men can lift nearly double the weight by aid of this simple 
 addition to the machine. Large crones of th'S description are 
 
WHEEL GEARING IN JIB CRANES. 
 
 145 
 
 fitted with slewing or horizontal turning gear, to enable the load 
 when lifted to be swung round before depositing it in a truck. 
 
 hold of a ship, or on a machine tool. This latter gear consists of 
 a horizontal wheel on the top of the vertical central cast-iron 
 
146 LECTURE 
 
 supporting boss, with which is geared a bevel pinion, actuated by 
 aid of a lever handle. 
 
 In order to prevent the whole machine being capsized by a 
 heavy load, there is a back balance weight, and further the bogie 
 wheels can be clamped to the rails. The back balance weight also 
 tends to cancel the severe right angle stress on the central pillar 
 which was specially taken notice of in Lecture VIII. We will 
 defer the description of heavy steam power cranes, tripods and 
 shear legs to our Advanced Course. 
 
 LBCTUEB XIII. QUESTIONS (continued) 
 
 10. What do you understand by the efficiency of a machine, and how is 
 it measured? In a single purchase crab, the pinion has 12 teeth and the 
 wheel has 78 teeth, the diameter of the barrel being 7 inches, and the 
 length of the lever handle 14 inches. It is found that the application of a 
 force of 15 Ibs. at the end of the handle suffices to raise a weight of 
 280 Ibs. Find the efficiency of the machine. Ans. 072 ; or 72 per cent. 
 
 jo. In a crane an effort of 122 Ibs. just raises a load of 3265 Ibs. What 
 is the mechanical advantage 1 If the efficiency be 60 per cent., what is 
 the velocity ratio 1 Ans. Mech. Adv. 2676 : i ; Vel. Ratio 44-6 : I. 
 
 (B. of E., 1903.) 
 
 11. In a crane, a force of 3 Ibs. applied at the handle is found to raise a 
 weight of 42 Ibs., and a force of 8 Ibs. a load of 120 Ibs. If the 
 relation between the force applied and the weight raised is represented by 
 the straight line law, obtain the equation expressing the relationship 
 between them ; and if the velocity ratio between the force applied and 
 the weight raised is 18, estimate the efficiency of the crane when lifting a 
 load of 200 Ibs. Am. F=5W/78 + 4/i3 ; 84*6%. 
 
 (0. & G., 1904, 0., Sec. A.) 
 
 12. In an electrically driven overhead crane a weight of 5 tons is 
 raised at the rate of 90 feet per minute. What is the horse-power? 
 Convert this into watts. The motor drives through gearing whose 
 efficiency is 70 per cent. How many amperes of current must be supplied 
 to the motor at a voltage of 220 if the efficiency of the motor is 87 per 
 cent. ? Ans. 30-5 H.P. 22,787 watts ; 170 amperes. 
 
 [Note that i horse power = 746 watts ; and i ampere multiplied by i volt 
 is i watt.J (B. of E. 1905.) 
 
LECTUEE Xm. QUESTIONS. 
 
 LECTUBE XIII. QUESTIONS. 
 
 1. Where wheelwork is employed to modify motion, as in a crane, or in 
 the double-geared headstock of a lathe, how is the change of motion 
 calculated ? Write down the formula employed. 
 
 2. Sketch a side elevation and end view of a single purchase crab, and 
 describe the same by aid of an "index to parts." Apply the principle of 
 work in solving the following question : The lever handle of a crab is 
 three times the diameter of the drum, and the wheelwork consists of a pinion 
 of 16 teeth driving a wheel of 80 teeth ; what weight will be lifted by a 
 force of 30 Ibs. acting at the end of the lever handle 1 An*. 900 Ibs. 
 
 3. Describe, with a freehand sketch, a single purchase lifting crab. The 
 leverage of the handle of the crab is 16 in., and there is a pinion of 20 teeth 
 driving a wheel of 100 teeth, the diameter of the barrel being 8 in. Assign 
 the relative proportions of the working parts, and estimate the theoretical 
 advantage. What weight would be raised by a man exerting a force of 
 
 15 Ibs. on the lever handle, neglecting friction ? Ans. 300 Ibs. 
 
 4. A weight of 4 cwt. is raised by a rope which passes round a drum 
 3 feet in diameter, having on its shaft a toothed wheel also 3 feet in 
 diameter. A pinion, 8 inches in diameter, and driven by a winch-handle 
 
 1 6 inches long, gears with the wheel. Find the force to be applied to the 
 winch-handle in order to raise the weight. Ans. 112 Ibs. 
 
 5. In a lifting crab the lever handle is 14 inches long, the diameter of the 
 drum is 6 inches, and the wheel and pinion have 57 and n teeth respect- 
 ively. Find the weight in pounds which could be raised by a force of 50 
 Ibs. applied to the lever handle, friction being neglected. Ans. 1209 Ibs. 
 
 6. In a crane there is a train of wheelwork, the first pinion being driven 
 by a lever handle ; and the last wheel being on the same axis as the chain 
 barrel of the crane. The wheelwork consists of a pinion of 1 1 gearing 
 with a wheel of 92, and of a pinion of 12 gearing with a wheel of 72, the 
 diameter of the barrel being 18 inches and that of the circle described by 
 the end of the lever handle being 36 inches ; find the ratio of the pull to 
 the weight raised, friction being neglected. Ans. 1 1 : 1 104. 
 
 7. In a 3O-ton crane the tension of the chain as it runs on the winding 
 barrel is 7$ tons, the barrel is 2 feet in effective diameter, and the spur 
 wheel connected with it is 4 feet in diameter on the pitch line ; what 
 pressure will come upon the teeth of the spur wheel, supposing such 
 pressure to act on the pitch line (friction is neglected) ? A n.. 375 tons. 
 
 8. The crank of an engine is 2' long, and the diameter of the fly-wheel 
 is icf ; also the fly-wheel has teeth on its rim, and drives a pinion 3' in 
 diameter. If the mean pressure on the crank pin be 7^ tons, what is 
 the mean driving pressure on the teeth of the pinion 1 Ans. 3 tons. 
 
 9. Draw to scale a side elevation, end view and plan of a double purchase 
 crab, and describe the same by aid of an " index to parts." If four men 
 each exert a constant force of 15 Ibs. on the handles of such a crab ; if the 
 handles have a leverage of 1 6 inches whilst the barrel is 16 inches diameter, 
 and if the drivers have 12 teeth each while the followers have 24 and 60 teeth 
 respectively ; find the weight which they could balance neglecting 
 friction. If 30 per cent, of the work put in, be taken up in overcoming 
 friction, what load can they lift? State (i) theoretical advantage; 
 (2) working advantage ; (3) work put in when lifting the load i foot ; 
 (4) the work got out ; (5) the percentage efficiency ; (6) the height 
 through which they would lift the load in i minute if each man de- 
 veloped i H.P. AM. 1200 Ibs. ; 840 Ibs.; (i) 20 : i ; (2) 14: 1 (3) 1200 
 ft. -Ibs. ; (4) 840ft.-lbs-, (5) 70 per cent.; (6) 13- 75 ft. 
 
( '48 ) 
 
 LECTURE XIV. 
 
 CONTENTS. Screws The Spiral, Helix, or Ideal Line of a Screw Thread 
 The Screw viewed as an Inclined Plane Characteristics and Con- 
 ditions to be Fulfilled by Screw Threads Different Forms of Screw 
 Threads Whitworth's V-Threads Whitworth's Tables of Standard 
 V-Threads, Nuts and Bolt Heads Seller's V-Thread The Square 
 Thread The Hounded Thread The Buttress Thread Right and 
 Left-hand Screws The Screw Coupling for Railway Carriages 
 Single, Double and Treble Threaded Screws Backlash in Wheel and 
 Screw Gearings Questions. 
 
 Screws. Every one is more or less familiar with the form and 
 uses of the screw nail for securing pieces of wood together, and of 
 the bolt with its nut for fixing metal plates m position ; but every 
 one is not so familiar with the principle upon which screws are 
 generated and act, or with the best shape to be given to a screw 
 under different circumstances. We shall therefore endeavour 
 in this Lecture to explain these points in an elementary manner, 
 instancing a few examples of the practical applications of screws, 
 but reserving for the following Lecture questions on the work 
 done by screws and their efficiency. 
 
 The Spiral, Helix, or Ideal Line of a Screw Thread. 
 A very good idea of the form of a screw is obtained from the accom- 
 panying figure, which represents one means of elevating or trans- 
 
 SPIRAL OK SCREW FOR MOVING GRAIN. 
 
 ferring grain, flour or other powdered substances from one part 
 of a milling works to another. It consists of a steel band twisted 
 around a cylindrical shaft in a continuous and uniformly pitched 
 spiral. This shaft and screw are placed in a trough, tube or 
 pipe. The grain or powdered substance is fed in at one end 
 of the pipe, and by rotating the screw with a wheel or lever fixed 
 
THE SCREW. 
 
 149 
 
 to one end of the shaft, the loose material is gradually pressed 
 forward until it reaches the other end, from which it may be 
 dropped into sacks or put through another process. It is evident 
 from an inspection of the figure that as the screw is turned round 
 by the lever, the particles of matter are forced along the face of 
 the continuous inclined plane formed by the spiral steel band. 
 
 The principle upon which the screw acts is, therefore, a combination 
 of tfw inclined plane and the lever. 
 
 To bring this view of the case still more forcibly before the 
 student, take a cylinder and fix along the side thereof parallel to 
 
 FORMING A SCREW THREAD ON A CYLINDER, 
 
 its axis (by gum or drawing pins) a rectangle, ACDE, of paper 
 or white cloth, having its sides, AC and DE exactly equal to the 
 circumference of the cylinder. Then, when the envelope is wound 
 round the cylinder by the turning of the handle, H (in the direction 
 shown by the arrow at P), it exactly covers its cylindrical surface. 
 On the outside of this rectangle when unfolded, draw any con- 
 venient number of parallel inclined black lines, AB, &c., equi- 
 distant from each other as shown by the figure, and again wrap 
 it round the cylinder. These lines will be found to form a con- 
 tinuous spiral, helix, or screw-thread Line from one end cf the 
 cylinder to the other. And the side AC of the right-angled 
 triangle ACB forms the circumference, BC the pitch, AB the 
 length of the thread (for one complete turn of the cylinder), and 
 the angle BAC is the inclination or angle of the screw. 
 
 The Screw Viewed as an Inclined Plane. Take another 
 cylinder having an evenly pitched screw-thread line drawn upon 
 it. Cut a sheet of flexible cardboard into the form of a right- 
 angled triangle with its height BC or h equal to the pitch (or dis- 
 
LECTURE XIV. 
 
 tance between two consecutive threads when measured parallel 
 to the axis of the cylinder) ; AC or b equal to the circumference 
 of the screw and wrap it round the cylinder, taking care to keep 
 BC parallel to the axis. Then the hypothenuse AB or length I 
 of the inclined plane will coincide with the contour of the screw- 
 thread for one complete turn, and BAG or, a, is the angle of the 
 thread to the plane at right angles to the axis of the cylinder. 
 
 Now conceive this screw-thread instead of being a mere line 
 to be an inclined plane of known breadth, as in the case of the 
 grain elevator.* Let the total weight of material being urged 
 
 FIGURE TO PROVE THAT A SCREW THREAD is AN INCLINED PLANE. 
 
 forward or upwards by the turning of the screw be W Ibs., and 
 let the resistance due to this load be uniformly distributed along 
 the screw thread or inclined plane. Then, comparing the first and 
 the third figures, it is evident that any small portion of the load 
 having a weight ~YV 2 Ibs. will have a corresponding reaction R 2 Ibs., 
 and will require a part P 2 Ibs. (of the total force, P, applied to 
 turn the screw at the radius at which this portion is situated) to 
 move it along the screw-plane against the frictional resistance F r 
 Imagine the work done to be transferred to the inclined plane, 
 AB, then any portion of the load having a weight W t Ibs. will 
 have a corresponding reaction R x Ibs., and will require a part 
 P x Ibs. (of the total force, P, applied parallel to the base to pull 
 the whole load up the inclined plane) to move it along the plane 
 against the frictional resistance F r Now, these forces act in 
 identically the same way as the second case of the inclined plane, 
 which was discussed in Lecture IX., consequently 
 
 W 
 
 R 
 
 AC 
 
 AC 
 
 b 
 
 CB 
 
 CB 
 
 h 
 
 AB 
 
 AB 
 
 I 
 
 * Or, that the screw-thread has a certain depth as measured radially 
 from the axis of cylinder. 
 
SCREW THREADS. I$I 
 
 Or P __ C B _ h eight _ h _ pitch of thread 
 
 W AC base b circumference of screw. 
 
 We therefore see that a screw may be treated as an inclined 
 plane where the force turning the screw i.e., overcoming the 
 resistance to motion acts parallel to the base of the incline The 
 same reasoning may be applied to any screw turning in a nut or 
 to a nut turning on a screw. 
 
 Characteristics of and Conditions to be Fulfilled by 
 Screw Threads. The essential characteristics of a screw-thread 
 are its pitch, depth, and form. 
 
 The principal conditions to be fulfilled by a screw-thread are : 
 (i) efficiency; (2) strength; (3) durability. 
 
 (1) The efficiency depends on the pitch and the friction, and 
 hence on the pitch and form of thread. 
 
 (2) The strength depends upon the form or the shearing thick- 
 ness and depth, or area of the cross section parallel to the axis. 
 
 (3) The durability depends chiefly on the depth that is, upon 
 the extent of bearing surface. 
 
 Different Forms of Screw Threads. Sir Joseph Whitworth, 
 the famous tool and gun manufacturer, was so impressed with 
 the great inconvenience and loss of money which arose from the 
 use of different pitches and forms of threads for screws and nuts, 
 that he published the following tables giving the dimensions of 
 what has now become known as the Whitworth standard. Prior 
 to 1841, the year in which Whitworth proposed the adoption of 
 standard sizes for screws, and for several years afterwards, differ- 
 ent engineering works in this country not only used different 
 pitches for screws of the same diameter, but it was no uncommon 
 thing to find a want of uniformity in the same shop. Now, 
 every one in Great Britain and her colonies uses the Whitworth 
 standard sizes for V-threaded bolts and nuts of J-inch and upwards, 
 and the British Association standard for smaller screws in electrical 
 and philosophical instruments. 
 
 Whitworth's V Thread.* The following figures of a Whit- 
 worth thread and nut, together with the tables, will serve to 
 give full information regarding the number of threads per inch 
 for different diameters of screw-bolts, nuts and bolt-heads, <fcc. 
 
 The angle between opposite sides of the threads and of the 
 intervening spaces is 55. One-sixth of the depth of the thread 
 is rounded off at both the top and the bottom for the purpose of 
 preventing a sharp nick at the bottom (which would weaken a 
 
 * For a description of Whitworth's screw-taps, plates, stocks, dies and 
 combs, see " Workshop Appliances " bj Professor Shelley. And for a table 
 of the B. A. Standard for Small Screws, see Munro and Jamieson's Electrical 
 Rules and Tables, i6th ed., p. 67. 
 
LECTURE XIV. 
 
 WHITWORTH'S STANDARD FOR SCREWS WITH ANGULAR THREADS. 
 
 -8 
 
 
 ITS 
 
 T3 
 
 
 ,-g 
 
 CO 
 
 
 .g 
 
 1| 
 
 Old Sizes, 
 Inches. 
 
 New Standa 
 Decimals of 
 Inch. 
 
 No. of Threa 
 per Inch. 
 
 Old Sizes, 
 Inches. 
 
 New Standai 
 Decimals of 
 Inch. 
 
 No. of Threa 
 per Inch. 
 
 Old Sizes, 
 Inches. 
 
 New Standar 
 Decimals of 
 Inch. 
 
 48 
 
 
 O'lOO 
 
 12 
 
 
 0'600 
 
 4 
 
 2f 
 
 2*375 
 
 40 
 
 
 
 0*125 
 
 1 1 
 
 1 
 
 0-625 
 
 4 
 
 2 2 
 
 2-500 
 
 32 
 
 
 0*150 
 
 II 
 
 
 0*650 
 
 4 
 
 2 i 
 
 2*625 
 
 24 
 
 
 0*175 
 
 II 
 
 
 0-675 
 
 3 
 
 2j 
 
 2*750 
 
 24 
 
 
 O'2OO 
 
 II 
 
 
 0-700 
 
 
 21 
 
 2-875 
 
 24 
 
 
 0-225 
 
 IO 
 
 1 
 
 0-750 
 
 3i 
 
 3 
 
 3'OOQ 
 
 20 
 
 4 
 
 0-250 
 
 10 
 
 
 0-800 
 
 34 
 
 
 3^5 
 
 2O 
 
 
 0-275 
 
 9 
 
 1 
 
 0-875 
 
 31 
 
 3^ 
 
 3-50 
 
 18 
 
 
 0-300 
 
 9 
 
 
 0-900 
 
 3 
 
 31 
 
 375 
 
 18 
 
 
 0-325 
 
 8 
 
 I 
 
 I'OOO 
 
 3 
 
 4 
 
 4*00 
 
 18 
 
 
 0-350 
 
 7 
 
 Is 1 
 
 1*125 
 
 2j 
 
 44 
 
 4-25 , 
 
 16 
 
 1 
 
 0-375 
 
 7 
 
 ii 
 
 1-250 
 
 2 
 
 
 4-50 
 
 16 
 
 
 0-400 
 
 6 
 
 '1 
 
 1-375 
 
 2| 
 
 4f 
 
 475 
 
 14 
 
 
 0*425 
 
 6 
 
 ii 
 
 1-500 
 
 2| 
 
 5 
 
 5-00 
 
 14 
 
 
 0-4^0 
 
 5 
 
 if 
 
 1*625 
 
 2| 
 
 54 
 
 
 14 
 
 
 0*475 
 
 5 
 
 if 
 
 1*750 
 
 2| 
 
 
 5*50 
 
 12 
 
 
 
 0-500 
 
 44 
 
 Ig 
 
 1*875 
 
 
 5S 
 
 575 
 
 12 
 
 
 0-525 
 
 4 
 
 2 
 
 2-000 
 
 2 3 
 
 6 
 
 6'oo 
 
 12 
 
 
 0-550 
 
 4i 
 
 2 
 
 2*I25 
 
 
 
 
 12 
 
 
 0-575 
 
 4 
 
 24 
 
 2-250 
 
 
 
 
 WHITWORTH VEE THREAD. 
 
 Angle of thread = 55. One -sixth 
 of depth is rounded off at top and 
 bottom. 
 
 Number of threads to the inch in 
 square threads = J number of those in 
 angular threads. 
 
 Depth of threads = 0*64 pitch for 
 angular = 0*475 pitch for square 
 threads. 
 
 WHITWORTH'S GAS THREADS. 
 
 Diameter in Inches. 
 
 I 
 
 i 
 
 t 
 
 i 
 
 1 
 
 i 
 
 ij 
 
 i* 
 
 ij 
 
 2 
 
 No. of threads per inch 
 
 28 
 
 19 
 
 19 
 
 M 
 
 14 
 
 II 
 
 ii 
 
 ii 
 
 ii 
 
 II 
 
WHITWORTH'S STANDARD NUTS AND- BOLTS. 153 
 
 WHITWORTH SCREW NUT. 
 
 WHITWORTH'S STANDARD NUTS AXD BOLT-HEADS. 
 
 _^ 
 
 
 1 
 
 ^ 
 
 ll 
 
 ^ 
 
 
 1 
 
 3 
 
 *i 
 
 i 
 
 5 
 
 fc 
 
 m 
 
 
 3 
 CQ 
 
 S 
 
 
 
 
 * 
 
 11 
 
 ll 
 
 1 
 
 H 
 
 i* 
 
 H 
 
 ll 
 
 i 
 
 || 
 
 1^ 
 
 "S ~ 
 
 ~Sn 
 
 3 
 
 
 S 
 
 ts5 
 
 "** 
 
 
 
 -S S 
 
 
 
 g 
 
 
 
 g 
 
 ES 
 
 1 
 
 g 
 
 ^ 
 
 1 
 
 |3 
 
 5 
 
 
 c- 
 
 H 
 
 5* 
 
 .S 
 
 a 
 
 
 g 
 
 g 
 
 l 
 
 1 
 
 0-338 
 
 I 
 
 0-1093 
 
 0-0929 
 
 ii 
 
 1*8605 
 
 il 
 
 0^43 
 
 0-942 
 
 T* 
 
 0-448 
 
 l'* 
 
 0*1640 
 
 0*1341 
 
 ii 
 
 2*0483 
 
 i| 
 
 1-0937 
 
 o6 7 
 
 i 
 
 0-525 
 
 i 
 
 0-2I87 
 
 0*1859 
 
 if 
 
 2*2146 
 
 if 
 
 I*203I 
 
 1615 
 
 A 
 
 0*6014 
 
 A 
 
 0-2734 
 
 0-2413 
 
 i-l 
 
 2-4134 
 
 i% 
 
 I-3I25 
 
 2865 
 
 1 
 
 0-7094 
 
 1 
 
 0-3281 
 
 0*2949 
 
 i| 
 
 2-5763 
 
 i| 
 
 I*42l8 
 
 3688 
 
 A 
 
 0*8204 
 
 TB 
 
 0-3828 
 
 0*346 
 
 if 
 
 27578 
 
 if 
 
 1*5312 
 
 4938 
 
 i 
 
 0-9191 
 
 i 
 
 0-4375 
 
 0*3932 
 
 is 
 
 3-0183 
 
 U 
 
 1*6406 
 
 5904 
 
 T^ 
 
 I 'OH 
 
 I 9 ? 
 
 0-4921 
 
 Q-4557 
 
 2 
 
 3' I 49 I 
 
 2 
 
 175 
 
 7154 
 
 1 
 
 101 
 
 1 
 
 0-5468 
 
 0*5085 
 
 -: 
 
 3-337 
 
 2g 
 
 i -8593 
 
 8404 
 
 H 
 
 2O I 
 
 H 
 
 0-6015 
 
 0-571 
 
 2i 
 
 3'546 
 
 2i 
 
 1-9687 
 
 1*9298 
 
 1 
 
 3012 
 
 I 
 
 0-6562 
 
 0*6219 
 
 2| 
 
 375 
 
 2| 
 
 2*0781 
 
 2*0548 
 
 H 
 
 '39 
 
 H 
 
 0-7109 
 
 0*6844 
 
 2j 
 
 3-894 
 
 2| 
 
 2-1875 
 
 2*1798 
 
 i 
 
 4788 
 
 | 
 
 0*7656 
 
 07327 
 
 2| 
 
 4-049 
 
 2| 
 
 2-2968 
 
 2*3048 
 
 H 
 
 '5745 
 
 II 
 
 0*8203 
 
 07952 
 
 2| 
 
 4*181 
 
 2| 
 
 2-4062 
 
 2-384 
 
 i 
 
 I -6701 
 
 i 
 
 0-875 
 
 0*8399 
 
 3 
 
 4-53I 
 
 3 
 
 2-625 
 
 2-634 
 
154 
 
 LECTURE XIV. 
 
 bolt or a nut), as well as for ease in manufacturing them, since it 
 would be practically impossible to maintain such perfectly sharp 
 edges in the stocks and dies or in the combing tools with which 
 such bolts and nuts are generally screwed. Besides, it would be 
 most inconvenient to handle such sharp-pointed screws if they had 
 edges tapering right off to 55, and, moreover, it would serve no 
 useful purpose, for such a thin edge cannot materially add to the 
 strength of a screw-thread. 
 
 The Whitworth thread is stronger than any other, except 
 the buttress one, since its thickness at the bottom of the thread 
 is nearly equal to the pitch of the screw. The compression or 
 grip is considerably greater than with the square thread, because 
 the pitch is only half as much for the same size of bolt. The 
 efficiency of the Whitworth V-thread as a means of transmitting 
 motion is, however, small, since the reaction being at right angles 
 to the face of the thread, a large part of the force employed in 
 turning the screw is expended in tending to burst the enveloping 
 nut. This very inefficiency, however, adds to its utility as a 
 binder for all kinds of machinery, since a properly fitted nut when 
 once screwed down, will not run back or overhaul, unless the pitch 
 be very great and the threads be well oiled. 
 
 Seller's V-Thread. In the United States of America, Seller's 
 V-thread is used. It differs from the 
 Whitworth V-thread in that the angle 
 between the opposite sides of the thread 
 and between the spaces is 60 instead 
 of 55, also the depth is reduced by a 
 sharp flat top and bottom, equal to one- 
 eighth of the total depth, instead of 
 being rounded. This is rather a curious 
 divergence from the usual American practice, where almost all 
 other parts of their excellent machine tools are beautifully 
 rounded off by symmetrical curves. 
 
 The Square Thread Since the bearing surface in this 
 thread is very nearly at right angles to the direction of pressure 
 and resistance it is much used for transmitting motion. Of the 
 
 P--*? 
 
 SELLER'S V THREAD. 
 
 SQUARE THREAD. 
 
SCREW THREADS. 
 
 155 
 
 force applied to turn this screw there is only a small percentage 
 dissipated in tending to burst the nut ; consequently, its efficiency 
 is greater than that of the V-thread. As will be seen from the 
 accompanying figure, the thickness of the thread and the width of 
 the space are made equal, in single- threaded screws, therefore the 
 shearing thickness is greatly reduced, and consequently its strength 
 is less than the V-thread. The durability is, however, greater 
 than in any other form of screw, for there is a larger bearing sur- 
 face presented in the best manner to resist pressure. 
 
 The Rounded Thread This form is simply a modification 
 of the square thread, in order to facilitate the quick engaging and 
 
 Fif-H 
 
 ROUNDED THREAD 
 
 disengaging of a leading motion screw by its nut in machine 
 tools, or where a screw has to be subjected to rough usage. Its 
 efficiency and durability are less than the square thread, but its 
 strength is much greater, since the shearing thickness is greatly 
 increased by the fillets at the bottoms of the thread. 
 
 The Buttress Thread. In such cases as the raising and 
 lowering of heavy guns for the purposes of sighting and loading 
 them, where the pressures are always 
 in one direction, then this form of 
 thread is adopted, because its strength 
 is a maximum, the loss due to friction 
 is a minimum, and there is very little 
 tendency to burst the nut. The 
 efficiency is quite equal to that of the 
 square thread, although the durability 
 is lessened by the fact that a certain 
 amount of wear would diminish the depth of the thread. The 
 strength is, however, nearly double, since the shearing thick- 
 ness is double. It therefore possesses the advantages of the V 
 and the square thread where pressures have to be applied in one 
 direction. 
 
 A slight toodificaiion of the buttress thread is used for wood 
 screws. These bolts take a very firm hold of any material into 
 which they can be screwed. Consequently, they are used for 
 screwing thick planks of wood together, and binding down plates or 
 
 BUTTRESS THREAD. 
 
156 
 
 LECTURE XIV. 
 
 other planks where vibration and stresses would start and lessen 
 the grip of the ordinary V-thread. They are much used by ship 
 
 COACH OB WOOD SCREW WITH SEMI- BUTTRESS THREAD. 
 
 carpenters and erectors of light scaffolding, and are sometimes 
 called holding-down bolts. 
 
 Right- and Left-hand Screws. A right-hand screw, when 
 being turned forward or into a nut, rotates in a right-handed 
 way or in the direction of motion of the hands of a watch, whereas 
 
 RIGHT HANDED SCREW 
 
 LEFT HANDED SCREW 
 
 a left-hand screw moves in the opposite or left-handed direction, 
 as shown by the direction of the circular arrows in the above 
 figure. 
 
 The Screw-coupling for Railway Carriages is a very good 
 
 SCREW COUPLING FOR RAILWAY CARRIAGES, 
 
SCREW-COUPLING FOR RAILWAY CARRIAGES. 157 
 
 example of the use of right- and left-hand screws. When two 
 carriages are brought together, the free link hanging from the 
 hook of one of them is placed on the hook of the other one. The 
 porter then turns the central lever by rotating the ball in a circle, 
 thereby screwing lx>th the right- and the left-hand screws into 
 their respective nuts, which consequently draws the hooks to ward 
 each other, and couples the carriages tightly together. 
 
 EXAMPLE. If the pitch of each screw is J", the length of the 
 lever arm or distance from the axis of the screw to the centre of 
 the ball is 14"; and if the railway porter pulls the ball with a 
 force of 40 Ibs. when the carriages are brought tightly together, 
 what will be the tension on the screw threads ? 
 
 ANSWER. Here ;?= J"; 6 = 27rll = 2 x x 14" = 88"; P = 4olbs. 
 
 The formula for the ratio of P to W in the case of a single 
 screw given in this Lecture is 
 
 But there are two screws, and for every complete turn made by 
 P, the stress W would be moved through twice tho pitch of one 
 screw or through 2 x J" = i". 
 
 P x b 40 x 88 
 .-. W = - = - - = 3520 Ibs. 
 
 NOTE. We may answer this question directly from the " Principle of 
 Work." Students 'should be trained to work out each question from first 
 principles rather than from formulae ; for, by a too free use of formulae thej 
 are apt to lose sight of principles. 
 
 Let the lever make one complete turn, then eacli nut will advance along 
 its own screw a distance equal to thepitch. Therefore the two nuts, and con- 
 sequently the two carriages, will be brought nearer by a distance equal to 
 twice, the" pitch, or, = 2 x p. 
 
 By the principle of work, and neglecting friction 
 
 Work got out = Work put in 
 Or, . . Wx2/> = Px2irR 
 _Px2irR 
 
 Or, 
 
 Single, Double, and Treble-threaded Screws. As has been 
 previously stated, both the efficiency and the forward distance 
 traversed in a single turn of a screw are directly as the pitch of 
 
158 LECTURE XIV. 
 
 the thread, but the strength is proportional to the area of its cross 
 section. Now, if for any purpose requiring a rapid movement of 
 the nut or of a screw, the pitch must be increased ; and if the 
 screw consisted of a single- threaded square one, where the depth, 
 thickness of the thread, and the width of the groove are each 
 equal to half the pitch, the strength of the shaft upon which the 
 screw is cut would be unnecessarily reduced. If the groove be 
 made shallower and narrower, then two threads with two spaces 
 having the same pitch as the single one, can be cut upon it so as 
 to present about the same area of bearing surface to the pressure 
 and at the same time afford quite as great a shearing thickness 
 without interfering with the velocity ratio.* If a very great 
 velocity ratio should be required, then three or more threads with 
 corresponding grooves may be cut in the shaft and nut. 
 
 Backlash in Wheel and Screw-Gearings. Backlash is the 
 slackness between the teeth of wheels in gear or between a screw 
 and its nut. Suppose that two wheels are in gear, and that you 
 move one of them in a certain direction until it turns the other, 
 and then reverse the motion ; if you can now move the pitch circle 
 through, say, | inch, before the second wheel responds, this distance 
 is the amount of backlash. In the same way, suppose you turn a 
 screw in one direction until its nut moves, and then reverse the 
 motion, the angle or proportion of a turn which you can now 
 make before the nut responds, is the backlash of the screw and its 
 nut. If a great amount of backlash be present in wheel-gearing, 
 it causes vibration and a disagreeable rattling noise ; and where 
 severe stresses and sudden stoppages are common, the teeth are 
 liable to be stripped. It can only be thoroughly prevented by 
 cutting the teeth most accurately of the best rolling contact form 
 by a tooth-cutting machine. All screws and nuts that are much 
 worked are liable to backlash as they become worn, although when 
 new they may have been very free from it, so that the best way 
 of taking up the slack is to form the nut in two parts with flanges 
 connected by screw-bolts, which may be tightened from time to 
 time so as to take up the wear, and thus keep one side of the 
 threads in one half of the nut, bearing hard against one side of the 
 threads of the screw, and those m the other half against the other 
 Bide. 
 
 * The screw of the fly-press, figured on p. 249, is a double-threaded one. 
 
LECTUBE XIV. QUESTION 
 
 LECTUEB XIV. QUESTIONS. 
 
 1. Explain how a screw is a combination of the lever and inclined 
 plane, and illustrate your remarks. Find the theoretical advantage or 
 ratio of W to P in the case of a screw of I inch pitch and 3-2 inches 
 diameter ; if the lever or spanner key be 7 feet long. Ans. 528 : 1. 
 
 2. Given a cylinder and a sheet of paper of sufficient size to cover the 
 cylindrical surface, show how you would trace an evenly pitched spiral or 
 screw line on the cylinder. Mark on your sketch the pitch, circumference, 
 \nd angle of the screw-thread. 
 
 3. Trace a screw-thread line on a cylinder. Draw a triangle to repre- 
 sent the pitch, circumference and angle of the thread, and show the 
 direction of all the forces on the supposition that there is a total pressure, 
 Wlbs., on the end of the cylinder acting parallel to its axis and balanced by 
 a force, P Ibs., acting at its circumference in a plane at right angles to the 
 axis, with a total friction of F Ibs. on the screw-thread. 
 
 4. What are the essential characteristics of a screw-thread ? Upon 
 which of these do (i) the efficiency, (2) the strength, (3) the durability of a 
 screw depend? 
 
 5. Sketch and describe all the forms of screw-threads which you have 
 seen in practice. State their representative advantages and disadvantages, 
 and for which kind of work each kind is most suitable. 
 
 6. Define the pitch of a screw. In the Whitworth angular screw-thread, 
 what is the angle made by opposite sides of the thread ? To what extent 
 is the thread rounded off at the top and bottom ? Distinguish between a 
 tingle and a double-threaded screw ; in what cases should the latter be 
 used ? Why are holding down bolts made with angular threads ? 
 
 7. Distinguish between a right-handed and a left-handed screw. Sketch 
 the screw-coupling which is commonly used to connect two railway 
 carriages, and explain the action of the combined screws. If the pitch of 
 each acrew is f inch and the lever-arm from the axis of the screw to the 
 centre of the ball is 12 inches, with what force will the carriages be pulled 
 together by a force of 50 Ibs. applied to the ball on the end of the arm ? 
 Ans. 5028 Ibs. 
 
 8. Draw a single, double, and treble square-threaded screw to a iVth 
 scale, where the outside diameter of the screw-thread is 10 inches and the 
 pitch 6 inches. Explain the advantages of using a double or treble thread 
 instead of a single one for transmitting rapid motion against a considerable 
 resistance. 
 
 9. Why is the angular-threaded Whitworth or Seller's screw better 
 adapted than the square, rounded, or buttress thread for the bolts which 
 are used to bind pieces of machines, &c., together ? 
 
 10. What is meant by backlash? How may backlash be prevented in a 
 crew, and in wheel gearing 2 
 
LECTUEE XV. 
 
 CONTENTS. Efficiency, &c., of a Combined Lever, Screw, andPulleyGear 
 Example I. Bottle Screw- Jack Example II. Traversing Screw- 
 Jack Screw Press for Bales Screw Bench Vice Example III. 
 Endless Screw and Worm- Wheel Combined Pulley, Worm, Worm- 
 Wheel and Winch Drum Worm- Wheel Lifting Gear Example IV. 
 Questions. 
 
 Efficiency, &c., of a Combined Lever, Screw, and Pulley 
 Gear. Construct an apparatus of the following description, 
 having a horizontal Whitworth V-screw of, say, p" pitch, with 
 cylindrical ends and flanges supported by bearings, so that the 
 screw cannot move longitudinally, but with a nut free to travel 
 from one end of the screw to the other, along a slide or guide 
 
 
 APPARATUS FOR DEMONSTRATING THE ACTION AND 
 EFFICIENCY OF SCREW GEAR, 
 
 INDEX TO PARTS. 
 
 P represents Pull on pulley rope 
 
 W represents Weight to be lifted. 
 GP Guide Pulley. 
 N Nut. 
 
 fci 
 
 Screw. 
 
 Radius of pulley. 
 Base or support 
 
SCBZT, LEV23, ASL PULLEY GEAR. l6l 
 
 which prevents it from turning round. Apply a force, P, to a 
 rope passed over the V-grooved pulley of radius, R, keyed to the 
 end of the screw shaft, until it moves the nut with the hook, 
 rope, and weight, W, attached thereto, ao shown by the accom- 
 panying side elevation, plan and end view of the apparatus.* 
 
 EXAMPLE I. If the radius, R, of tho turning-pulley be 12", the 
 pitch, p, of the screw i", and the gross pull, P, required to lift a 
 weight of TOO Ibs. be 4 Ibs. : find (i) the velocity ratio; (2) the 
 theoretical advantage ; (3) the working advantage ; (4) the work 
 put in to lift W i foot ; (5) the work got out ; (6) the percentage 
 efficiency. 
 
 ANSWER. We have got in this question all the necessary data required 
 to find the various answers except n, the number of turns which the screw 
 will have to make in order to lift W i foot. Since the pitch of the screw 
 is i", each turn thereof will elevate or lower the weight i", according as it 
 is turned the one way or the other ; consequently, if the screw makes 
 12 turns, the nut and the weight will move through 12", therefore n=i2 
 turns. 
 
 * It is evident that, in addition to the friction between the screw and 
 the nut, there is friction at the several bearings, at the nut slide, and in 
 the bending of the ropes. Consequently, if the student were to place in 
 fucceftsion weights at W of, say, 10, 20, 30, 40 Ibs., &c., and ascertain by aid of 
 a Sailer's spring balance (hooked into the rope which passes round the turn- 
 ing-pulley), the corresponding pulls required to lift these several weights, 
 and to plot down the results on squared paper with the weights as abscissa 
 and the pulls as ordinates, and then to draw a line through the inter- 
 sections of the vertical and horizontal lines drawn through the correspond- 
 ing values, he would obtain a characteristic curve for the friction of the 
 machine as a whole. If he took the precaution to balance the initial 
 friction of the machine (when there was no weight attached at W) by 
 hanging such a small weight at P as would just move the nut towards 
 the turning-pulley, he would find upon repeating the above experiments 
 (keeping the small additional weight on all the time) and replotting the 
 results as now recorded by the spring balance, that the second frictional 
 curve would approach much nearer to a straight line than the former one. 
 In fact, its deviation therefrom would simply prove that the friction of the 
 movable bearing surfaces was not directly proportional to the load. To 
 arrive at the characteristic friction curve for the screw alone, he would 
 have to find out by trial the proportion of the several pulls applied, which 
 were spent in overcoming friction at all other points except between the 
 screw and the nut. To those students who have the time and opportunity 
 for carrying out experiments in applied mechanics, the apparatus illustrated 
 above will prove interesting and instructive. The figures are drawn from 
 the machine constructed in the author's engineering workshop for the 
 purpose of enabling his students to make similar tests to those suggested 
 above. A square, or a rounded, or a buttress-thread may be substituted 
 for the V- Whit worth one, and sound information may thus be obtained 
 about different forms of screws, which will make a stronger and more 
 lasting impression on some students than merely studying books. 
 
1 62 LECTURE XY. 
 
 By the principle of work : 
 
 / \ m-i rr j * T> * - P s distance in i turn of driving pulley 
 
 (1) The Velocity Ratio _, ,. . . . ... jfP^- * 
 
 v ' W s distance in the same tune 
 
 ~ _ 0* of pulley _ 2?rB. _ 75 '4 
 
 pitch of screw p 1 
 
 (2) The Theoretical Ad-\ _ Weight lifted if there were no friction 
 
 vantage . . I Pull applied 
 
 ^ p 1 
 
 (3) The Working Advan- \ _ W = 100 Ibs. = 25 
 
 tage . . J P 4 Ibs. 1 
 
 (4) The Work Put in to] 
 
 y X I 2 
 
 (5) The Wart 0* va | = w ,_ lbs ,. fl . 
 
 raising W i /ooi j 
 
 (6) 31 - _ Effid 
 
 cieTicy . J J 
 
 Work got out 
 w " Work put in X 
 
 IPO ft.-lbs. # 
 
 301-56 ft,ibs: = 
 
 Bottle Screw-Jack. The importance of the screw as a 
 simple machine for exerting great pressures, is very well ex 
 emplified by the screw-jack. This tool is used for replacing 
 locomotives and railway carriages upon their rails, for elevating 
 heavy girders into position, or for overcoming any great resistance 
 through a small space which cannot be effected by a labourer and 
 a lever. As will be seen from the accompanying figure it consists 
 of a strong hollow bottle-shaped casting, with a projecting handle 
 for facilitating the carrying of the tool from one place to another. 
 In the upper end of the casting a square-threaded screw is cut 
 
 * It is evident that with such a low percentage efficiency the weight 
 when hanging from the rope will not be able to overhsrul the machine. 
 The student can calculate what pitch of screw would be required with the 
 same co-efficient of friction before overhauling could take place. 
 
BOTTLE SCREW-JACK, x 163 
 
 parallel with the axis, and into this nut there is fitted a steel 
 
 screw terminating in a spherical head, having two holes bored 
 
 through it at right angles to each other. Into 
 
 one or other of these holes an iron lever bar 
 
 is fixed, so that by pulling or pushing on the 
 
 outer end of the bar the screw is turned, and 
 
 thus the head is gradually raised from the 
 
 base. To avoid the tearing, grinding action 
 
 that would ensue between the head and the 
 
 object acted upon, the former is provided with 
 
 a loose crown fitted on a central pin projecting 
 
 from the round head. 
 
 Let L = Length of the lever arm in inches 
 from centre of jack to where the 
 force is applied. 
 
 p = Pitch of screw in inches. 
 
 P=^Pull or push applied at radius L. 
 
 ft W = Weight lifted or resistance over- 
 come. 
 
 Then, by the Principle of Work, and neglect- 
 ing friction, we have in one turn of lever 
 
 P x its distance = W x its distance 
 !* P ~"Wxp 
 
 -C ' 
 
 BOTTLE SCKEW- 
 JACK. 
 
 EXAMPLE II. A weight of 10 tons has to be lifted by a screw- 
 jack, in which the pitch of the screw is J". What length of lever 
 will be required if a force of 70 Ibs. be applied at the end of it ? 
 (i) Neglecting friction ; (2) if the modulus or efficiency of the 
 tool is only '4. 
 
 ANSWER. (i) By the previous formula (neglecting friction) 
 
 1120 
 
 L 
 
 X27T 
 
 /?0X#X22 22 
 
 (2) Taking friction into account we see from the question that 
 the efficiency is = -4, therefore the percentage efficiency is 40, or 
 60 per cent, of the work put in is lost work required to overcome 
 friction between the screw and its nut. But as the length of the 
 lever is directly proportional to the work put in, the theoretical 
 length of the lever found above is only 40 per cent, of the actual 
 or working length required. 
 
164 
 
 LECTURE XV. 
 
 .-. 40: 1001:25-45 :L 3 
 L 100x25-45 = 63 . 6 ,, 
 
 40 
 
 Traversing Screw-Jack. It is very often convenient, when 
 using a strong heavy screw-jack, to be able to move the head a 
 short distance to one side or the other, when near the object to 
 which it is to be applied ; or, after having raised a load with one 
 or more jacks, to be able to traverse the jacks forward or back- 
 ward through a short distance until the load is brought into 
 
 TRAVERSING SCREW-JACK WITH KATCHET-LEVERS. 
 (By P. & W. MacLellan, Glasgow. ) 
 
 the desired position. These movements may be effected with a 
 jack of the form shown by the accompanying figure. Further, 
 this jack is provided with a side foot-step attached to and pro- 
 jecting from the lower end of the vertical screw. This foot- step 
 can be placed under the flange of a low beam or rail, where it would 
 be inconvenient or perhaps impossible to get the top head under- 
 neath the same. The nut of the horizontal traversing screw is 
 
SCREW PRESS FOR BALES. 165 
 
 formed in, or fitted to the bottom of the vertical casting, and this 
 screw is turned by a ratchet-lever which may be slipped on to one 
 or other of the squared ends of its shaft. The upward and down- 
 ward movement of the vertical screw is also affected by a ratchet- 
 lever, and in this case without turning the screw, for the ratchet- 
 wheel is fixed to the nut of its screw. The pawl of the ratchet 
 may be locked on one side or the other, so as to enable the ratchet- 
 wheel and the vertical screw-nut to be turned round in either 
 direction for elevating or lowering the load. 
 
 Screw Press for Bales. When soft goods or hay have to be 
 transported they may be squeezed into small bulk by means of a 
 
 SCREW PEESS FOB BALES. 
 (By London Bros., Glasgow.) 
 
 screw press, and bound firmly when under the press, by strips of 
 hoop-iron passed round them and then riveted before the pressure 
 is relieved. The bound bundle is then termed a bale. The 
 operation will be understood by an inspection of the accompany- 
 ing figure. The loose material is placed in the space between the 
 rigid base and the movable plate of the press, the doors are closed 
 and locked, the pressman applies himself to the end of the lever 
 with a force, P, thereby turning the nut of the screw and forcing 
 the movable plate downwards with a pressure, W, until the 
 
i66 
 
 LECTURE XV. 
 
 desired compression of the goods has been attained. The doors 
 are then opened and the strips of hoop iron (which were previously 
 or are now placed in the grooves of the base and upper plate) are 
 brought together and riveted. The lever is then turned in the 
 opposite direction to relieve the pressure, and the bale is removed 
 to the store or ship to make room for another quantity of goods 
 being subjected to a similar action. 
 
 The same formula as we used for the screw-jack and for 
 Example I. in connection with the combined lever, screw and 
 pulley gear, naturally applies to this press, and to any similar 
 appliance, such as a letter-copying press.* 
 
 Screw Bench Vice. A bench vice is essentially an instru- 
 ment for seizing and holding firmly any small object whilst it is 
 being acted upon by a chisel, file, drill, saw, or emery cloth, &c. 
 Looking at the figure which illustrates the following example, it 
 will be seen that the vice is a combination of two levers, a square- 
 threaded screw, and a nut. The object to be gripped is placed 
 
 END VIEW. SIDE VIEW. 
 
 SCEEW BENCH VICE. 
 
 between the serrated jaws JJ. The lever handle H, on being 
 turned, forces the screw S into its long nut, and thereby presses 
 forward the outer jaw upon the object, by aid of the flange on the 
 screw-head. This jaw is a lever, having a fulcrum at F, and there- 
 fore the pressure on the object is less than that on the screw- 
 collar in the proportion of SF to OF. The bent flat spring 
 between the limbs of the fixed and movable jaws serves to force 
 the movable jaw away from the fixed one when the screw is turned 
 backwards, and thus relieves the object without having to pull 
 
 * Kefer to index for page where the illastration of the Fly-press occurs.. 
 The statical pressures produced by this machine when used for punching 
 holes, &c., may be treated in the same way. 
 
BENCH YICE. * 167 
 
 this jaw back by the hand. It will be observed that the fixed 
 jaw should have been continued to the floor level by a vertical 
 supporting leg, in the case of such a big vice intended for rough 
 heavy engineering work. 
 
 EXAMPLE III. Sketch an ordinary bench vice. Apply the 
 principle of work to find the gripping force obtained when a man 
 exerts a pressure of 20 Ibs. at the end of a lever 18 inches long, 
 the screw having four threads per inch, the length from the 
 hinge to the screw being 18 inches, and the length from the 
 hinge to the jaws being 24 inches. (S. <fe A. Exam. 1892.) 
 ANSWER. Let P represent Pull on end of handle H = 20 Ibs. 
 Q ,, Resistance offered by screw at S. 
 B Reaction, or gripping force, exerted 
 
 on object at O. 
 
 L Length of handle H= 18 inches. 
 p Pitch of screw S = J inch. 
 
 Suppose the handle, H, to make one complete turn under the 
 action of a constant force, P, at the extremity thereof, against a 
 constant resistance, Q, acting along the axis of the screw. 
 
 [The student will observe that we suppose the forces P and Q to be con- 
 stant, which is not correct for such a large movement as a complete turn 
 of the handle, but which may be assumed here for the sake of simplicity. 
 The reason for this is, that the resistance, R, will vary with the com- 
 pression produced on the object at O. However, the ratio between P 
 and R will remain a constant quantity.] 
 
 The work done by P during one turn of handle = P x 2irL. 
 And on Q, during the same time =Q xp. 
 
 But, by the Principle of Work 
 
 Work done by P = Work done on Q 
 
 .*. Px 2?rL = Q Xp 
 
 Substituting the 
 numerical values 
 22 
 
 20X2X XlS" = QxJ' K* 
 
 .-. Q= 20X2X2 7 2Xl8x4 ~ 9 S'-43 ". 
 But by the Principle of Moments 
 
 24 4 
 
 t.., B = x 9051-43 = 6788-67 Ibs. 
 
1 68 LECTURE XV. 
 
 Endless Screw and Worm-Wheel.* When a screw is 
 rotated between fixed bearings so that it cannot move longi- 
 tudinally, it is called an endless screw, because the threads of the 
 screw seem to travel onwards without ending.! When such a 
 screw gears with a toothed wheel, having its teeth set obliquely 
 at the same angle as the threads of the screw so as to bear evenly 
 thereon, the wheel is termed a worm-wheel. The endless screw is 
 sometimes called the worm, no doubt from its resemblance to that 
 well-known humble animal which, when coiled up for rest, would 
 not turn upon any one unless trod upon. 
 
 By this arrangement, motion may be transmitted from one shaft 
 to another at right angles to each other, without any possibility 
 of the machine overhauling ; for although the velocity ratio is very 
 great, the efficiency is comparatively small considerably under 
 50 per cent, with single-threaded screws owing to the friction 
 between the worm and the wheel. J 
 
 It is most important for the student to comprehend that if the 
 screw be a single-threaded one, it must make as many turns as there 
 are teeth on the wheel, for every revolution of the latter. If the 
 screw is a double-threaded one, then for each revolution thereof it 
 drives the wheel through a distance equal to the distance between 
 tu-o teeth on the pitch circle, and if treble-threaded through the 
 pitches of three teeth. Thus, if N equal the number of teeth in 
 the worm-wheel, then, with a single-threaded screw, for every 
 
 turn of the same, the wheel will move a distance of ^ ; with a 
 
 double-threaded worm ^, and with a treble-threaded one j* 
 
 and so on. 
 
 The endless screw and worm-wheel is used in a very great 
 variety of circumstances, from the turning of a big marine engine 
 when in port, to the delicate movements in a telescope or a micro- 
 scope. 
 
 Combined Pulley, Worm, Worm-wheel and Winch 
 Drum. This combination is shown by the accompanying end 
 and side views drawn from an experimental piece of apparatus it 
 
 * Refer to the next figure. 
 
 t The term perpetual screw would express more exactly its action, foj 
 when in motion, it continually screws the worm-wheel round. 
 
 + The greater the diameter of the screw and the smaller its pitch is, the 
 better will be its bearing on the teeth of the wheel, but then the efficiency 
 will be so small that there will be no chance of overhauling. This is the 
 condition to be observed when the screw is intended to drive the wheel. 
 If, however, it should be required to drive the screw by the wheel, or 
 necessary that overhauling should take place, then the screw must be small 
 In diameter, its pitch very great, and either double or treble threaded. 
 
COMBINED PULLEY, WOKM, WORM-WHEEL, ETC. 
 
 169 
 
 the Author's Laboratory, which is used by the students for ascer- 
 taining the efficiency of the machine, and for finding the co-efficient 
 of friction between the endless screw and worm-wheel. 
 
 END VIEW. SIDE VIEW. 
 
 PULLET, WORM, WORM-WHEEL AND WINCH DRUM. * 
 INDEX TO PARTS. 
 
 P represents Pull applied to 
 
 R 
 
 Wm 
 
 WW 
 
 Radius of pulley. 
 Worm or endless 
 
 screw. 
 Worm wheel. 
 
 N repre?ents Number of teeth in 
 
 WW. 
 D Drum, or diameter of 
 
 winch barrel. 
 
 r Radius of drum, D. 
 
 W Weight to be lifted. 
 
 By the Principle oj Work (neglecting friction), if the drum, 
 makes one turn, and if the worm be a single-threaded sciew, 
 
 P x its distance = "W x its distance 
 Or, P x 2?rRN = W x 2?rr 
 
 (divide both sides by zv) 
 
 P x RN = W x r 
 
 _ 
 * W~~RN 
 
 * Find out what is wrong with the above drawing. 
 
 t It will be evident to the student that, given any four of these five 
 values, he can change this formula so as to find the fifth one ; and, that 
 he can experiment with this machine in precisely the same way as has 
 been already explained in the case of the wheel and axle, block and tackle, 
 Weston's pulley block and screw, &c., to ascertain its working advantage, 
 co-efficient of friction and efficiency. 
 
LECTURE XV. 
 
 Worm-wheel Lifting Gear. The accompanying figure 
 shows a practical application of the endless screw and worm- 
 wheel for the same purpose as the Weston's differential block is 
 used viz., the lifting of weights without fear of the tackle over- 
 hauling. A light-driving endless chain passes 
 over a V-grooved pulley having ridges or teeth 
 on the inner sides of the grooves, so as to fit 
 the pitch of the links of the chain. This 
 pulley is keyed to the outer end of a worm 
 spindle, whose screw gears with a worm- 
 wheel fixed to or cast along with a second 
 V-grooved ridged pulley or drum, over which 
 is passed the movable end of a heavier lifting 
 chain after it has been reeved under a snatch- 
 block pulley. In fact, it is simply the previous 
 experimental apparatus in a handy and com- 
 pact form. 
 
 EXAMPLE IY. If in lifting tackle of the 
 above description the driving pulley has a 
 radius B = 5", the number of teeth in the 
 worm-wheel N = 20, and the driven pulley a 
 radius r = 5" ; what weight suspended from 
 the snatch-block hook could be lifted by a 
 force of 10 Ibs. applied to the forward side of 
 the light chain (i) Neglecting friction, (2) if 
 the modulus or efficiency of the whole appa- 
 ratus were only "25 ? 
 
 ANSWER. (i) Applying the previous formula, and taking 
 account of the fact that the lifting chain is combined with a 
 snatch-block, we have 
 
 PxRxN 2X10X0X20 
 
 (2) Owing to friction, weight of chain and snatch-block, the 
 actual result obtainable is only '25, or 25 per cent, of this theo- 
 retical value ; consequently 
 
 100 : 25 : 400 : x 
 x= 5x400 = 1Q01bs 
 
 IOO 
 
LECTURE XV. QTTESTIONS. I? 1 
 
 LECTURE XV. QUESTIOITS. 
 
 1. A horizontal screw, of I inch pitch, is fitted to a sliding nnt which is 
 pulled horizontally by a cord passing over a fixed pulley, and having a 
 weight, W, attached to it. To the free end of the screw there is fixed a 
 pulley of 20 inches diameter, from the circumference of which a weight, 
 ", hangs by a cord. Find the ratio of P to W. Ans. i : 62-8. 
 
 2. In a set of combined lever, screw, and pulley gear, like that illustrated 
 before Example I. in this Lecture, R = 6", P = 2 Ibs., W = 50 Ibs., and 
 the pitch of the screw is such that there are 2 threads to the inch ; find (i) 
 velocity ratio, (2) theoretical advantage, (3) working advantage, (4) work put 
 in to lift W i ft., (5) work got out, (6) percentage efficiency. Ans. (i) 75-4 : i ; 
 (2) 75-4 : I ; (3) 25 : I ; (4) 150-8 ft. -Ibs. ; (5) 50 ft.-lbs. ; (6) 33-1 per cent. 
 
 3. Describe, with sketches, the construction of an ordinary lifting jack 
 in which the weight is lifted by means of a screw and nut. If the screw 
 be i inch pitch, the lever 20 inches long, and the pressure applied at the 
 end of the lever be 30 Ibs. ; what weight can be lifted (neglecting friction) t 
 (Take IT 3'i4i6.) Ans. 3770 Ibs. 
 
 4. In a screw-jack, where a worm-wneel is used, the pitch of the screw is 
 | inch, the number of teeth on the worm-wheel is 16, and the length of 
 the lever is 10 inches; find the gain in pressure. Ans. P : W : : i : 1609. 
 
 5. What practical objection is .there to the use of screw gear of any 
 description for obtaining great pressure ? Take for example the case of 
 the screw-lif ting jack. Sketch in vertical section and plan, and describe, a 
 traversing one to lift say 20 tons. Explain how the screw of the jack is 
 raised and lowered without being turned round. 
 
 6. Sketch and describe the construction and action of a screw press for 
 pressing goods so as to make them into bales for transport. What force 
 must be applied at the end of a screw press lever 8' 4" in length, in order 
 to exert on the goods a total pressure of 22,000 Ibs. when the pitch of the 
 screw is i" ? If 60 per cent, be lost in friction, what pressure would result 
 from the application of this force on the lever? Ans. 35 Ibs. ; 8800 Ibs. 
 
 7. Sketch an ordinary bench vice. Apply the principle of work to find 
 the gripping force obtained when a man exerts a pressure of 15 Ibs. at the 
 end of a lever 15 inches long, the screw having 5 threads per inch, the 
 length from the hinge to the screw being 12 inches, and the length from 
 the hinge to the jaws being 16 inches. Ans. 5303-6 Ibs. 
 
 8. Explain, with a sketch, the manner in which the principle of work is 
 applied in determining the relation of P to W in the case of the endless 
 screw and worm-wheeL The lever handle which works the screw being 
 14" long, the number of teeth in the worm-wheel 20, and the load being a 
 weight of icoo Ibs. hanging upon a drum 12" diameter on the worm-wheel 
 shaft, find the force to be applied at the end of the lever handle in order 
 to support the weight. Ans. 21-43 ^ DS - 
 
 9. Explain the mechanical advantage resulting from the employment 
 of an endless screw and worm-wheel. The lever handle which turns an 
 endless screw is 14" long, the worm has 32 teeth, and a weight W, 
 hangs by a rope from a drum 6" diameter, whose axis coincides with that 
 of the worm-wheel. If a pressure P be applied to the lever handle, find 
 the ratio of P to W. Ans. P : W : : 3 : 448. If in this question the 
 worm oe changed to (i) a double, and (2) a treble-threaded screw, what 
 Trill be the respective ratios of P to W? Ans. (i) i : 747 ; (2) i : 49-8. 
 
172 LECTURE XV. QUESTIONS. 
 
 10. Describe, with the aid of a sketch, how the pressure upon the book 
 is obtained in an ordinary copying-press. What should be the length of 
 the double-ended lever, supposing that the force be always applied simul- 
 taneously to both ends of the lever, in order that with a screw having 
 6 threads to the inch, the combination may have a mechanical advantage 
 of 216? Ans. Length of lever 6 inches. 
 
 11. Sketch in vertical section the common screw or bottle lifting jack. 
 The lever in such a jack is single ended, and measures 24 inches in length, 
 the pitch of the screw is f inch. What force applied at the end of the 
 lever would be required to raise a load of 22 cwt., the effect of friction 
 being neglected 1 Ans. 6 '12 Ibs. 
 
 12. Describe, with a sketch, the construction of an ordinary screw-jack 
 with a lever handle and screw. If the pitch of the screw be f inch, the 
 length of the lever handle 29 inches ; what load could be lifted, neglecting 
 friction, by a force of 19 Ibs. applied to the end of the lever handle ? 
 Ans. 2 tons. 
 
 13. Describe either a screw-jack (pitch of screw ^", handle 19" long) or a 
 simple winch for lifting weights up to I ton by one man. What is the 
 mechanical advantage neglecting friction? Describe what sort of trial 
 you would make to find its real mechanical advantage under various loads, 
 and what sort of result would you expect to find ? 
 
 14. The diameter of the safety valve of a steam boiler is 3 inches. The 
 weight on the end of the lever is 55 Ibs., and the distance from the centre 
 of the valve to the fulcrum is 4-5 inches. What must be the length of the 
 lever from the centre of the valve to the point of suspension of the weight, 
 in order that the valve will just lift when the pressure of steam in the 
 boiler is 80 Ibs. per square inch ? Neglect the weight of lever and valve. 
 Ans. 41 '3 inches. 
 
 15. Make a correct draughtsmanlike drawing of the next to the last 
 figure in this lecture. Point out distinctly what is wrong with the end 
 view if the side view be taken as correct. 
 
 16. Describe how you would proceed to determine experimentally for a 
 screw-jack (i) the velocity ratio, (2) the actual effort required to lift a 
 given load. 43. of E., 1905.) 
 
HOTES AND QUESTIONS. 
 
( '74 ) 
 
 LECTURE XYI. 
 
 CONTENTS. General Idea of the Mechanism in a Screw-cutting Lathe 
 Motions of the Saddle and Slide Rest Velocity Ratio of the Change 
 Wheels Rules for Calculating the Required Number of Teeth in 
 Change Wheels Examples I. II. Movable Headstock for a Common 
 Lathe Descriptions of a Screw-cutting Lathe and of an Electrically 
 Driven Hexagon Turret Lathe, with Frontis-Plates and complete sets 
 of Detail Drawings Questions. 
 
 General Idea of the Mechanism in a Screw-cutting 
 Lathe. We will devote this Lecture to giving a general idea of 
 the mechanism by which screws are cut in lathes, and the velocity 
 ratio of the screw to be cut to the leading screw, together with a 
 description of a complete set of illustrations prepared from work- 
 ing drawings of a new self-acting screw-cutting lathe. 
 
 Referring to the following figure, and to the general view of the 
 6-inch centre screw-cutting lathe (further on), it will be seen that 
 the round metal bar on which the screw is to be cut is placed 
 between the steel centres of the fixed and movable headstocks of 
 the lathe. This bar has an eye-catch on its end next to the fixed 
 headstock, which engages with a driving-stud connected to the 
 face-plate. In order to obtain the necessary force to cut the 
 screw, and to reduce the speed of the workshop motion shafts (in 
 the case of a power lathe, or of the treadle shaft in the case of a 
 foot lathe) to the required velocity, the fixed headstock is sup- 
 plied with back motion gearing. The back wheels may be put 
 into or out of gear with the lathe spindle wheels at pleasure, by a 
 simple eccentric motion (in the case of the lathes herein illus- 
 trated), or, as is sometimes effected, by sliding the back shaft 
 forward, so that its wheels clear those on the lathe spindle, and 
 then fixing it there, by a tapered pin fitting through a hole in the 
 framing and a groove cut in the shaft. But, when the back 
 motion is required for the purpose of making a slow heavy cut, 
 the follower F, is thrown out of gear with the stepped cone 
 pulley, so that the driver D L (which is keyed to the cone) may 
 turn the follower Fj ; and the driver D s (which is keyed to the 
 same spindle as F t ) rotate the follower F, (which is keyed to the 
 lathe spindle), and hence revolve the face-plate and the bar, out of 
 which the screw is to be formed. 
 
MECHANISM IN A SCREW-CUTTINa LA.THK. 
 
 175 
 
 On the back extension of the lathe spindle there is fixed a 
 change wheel or small driver, d, which gears with a follower, f 
 (keyed to the left-hand end of the leading or parent screw), either 
 direct in the case of the cutting of a very finely pitched left- 
 handed screw; or, through the intervention of a transmitting, or 
 
 Isc 
 
 
 JjQS 
 
 CHILD 
 
 \\\\\\\\ 
 
 i\\U\U\\\\\\\v\\ 
 
 13 
 
 PLAN SHOWING CUTTING A SCREW. 
 
 END VIEWS SHOWING CHANGE WHEELS. 
 FOE RIGHT-HANDED SCREWS. FOE LEFT-HANDED SCREWS. 
 
 GENERAL IDEA OF MECHANISM IN A SCREW-CUTTING LATHE. 
 
 SC represents Screw fro be cut 
 or child. 
 
 LS 
 SR 
 
 Leading screw, 
 
 or parent. 
 Slide rest. 
 Drivers of fixed 
 
 headstock. 
 
 INDEX TO PARTS. 
 
 F, f F 2 represents Followers of fixed 
 
 headstock. 
 
 d, f Driver and fol- 
 lower of change 
 wheels. 
 
 IP Idle pinion of 
 change wheels. 
 
 what is technically termed an idle, pinion, IP, in the case of a 
 medium-pitched right-hand screw. (See also the end views of the 
 change wheels above.) 
 
 It will therefore be seen that there are two independent motions 
 to be considered (i) the reducing gear from the speed of the 
 
176 LECTURE XVL 
 
 driving cone to that of the lathe-spindle or bar to be operated 
 upon ; and (2) the multiplying or reducing gear between the 
 lathe-spindle and the leading screw. The former of these will be 
 at once understood from the figures, and from what was said in 
 regard to wheel-gearing in Lecture XII. 
 
 We shall now consider the second motion. Remembering that 
 the pitch of the parent or leading screw is fixed and unalterable, 
 and that on its truth depends to a large extent the accuracy with 
 which the child, or screw to be cut, can be formed, it will be clear 
 that we have only to connect these two parallel shafts with suit- 
 able gearing in order to transmit, by aid of the " copying prin- 
 ciple " the characteristics of the parent to the child.* This may 
 be done in an equal or magnified or diminished degree, according 
 as the pitch of the screw to be cut is equal or greater or less than 
 that of the leading screw. 
 
 Motions of the Saddle and Slide Best. The base of the 
 slide rest, or the saddle as it is technically termed, bears upon and 
 is guided by the truly-planed shears (or upper framing of the lathe) 
 parallel to the line joining the centres of the fixed and movable 
 heads. In turning a right-handed screw the saddle is moved from 
 the movable headstock towards the fixed one, or from right to left, by 
 clasping it to the leading or guiding screw with a split nut attached 
 to the under side of the saddle. In cutting a left-handed screw 
 the saddle is moved by the same means, but in the opposite direc- 
 tion, i.e., from left to right. In other words, it travels in the direc- 
 tion towards which the threads of the screw to be cut are inclined 
 forward. 
 
 To the upper side of the saddle is bolted the slide-rest sur- 
 mounted by the tool-holder. The rest is provided with two in- 
 dependent sliding motions, each actuated by a hand-turned screw, 
 and guided by a true plane surface with dovetailed sides. These 
 motions (for the purposes of turning parallel work and screws) 
 are fixed at right angles to each other, the lower one being 
 parallel to the centre line of the lathe, and the upper one at right 
 angles thereto. Both motions are therefore independent of each 
 other and of the sliding motion of the saddle. The turner is 
 thereby enabled to adjust the cutting tool with great delicacy and 
 accuracy with reference to the job to be operated upon, irrespective 
 of the automatic travel of the supporting saddle. 
 
 Velocity Ratio of the Change Wheels. As has been men- 
 tioned already, the change wheels are interposed between the 
 
 * It is reported that Sir Joseph Whitworth, feeling the importance of a 
 thoroughly true leading screw, spent an immense deal of money upon the 
 scraping and finishing of a parent screw for a first-class lathe, from which 
 many of the best screws in this country have been copied. 
 
CHANGE WHEELS. 177 
 
 back end of the lathe spindle and the leading screw, for the pur- 
 pose of transferring motion to the saddle, and determining, that 
 the cutting tool shall be moved through a definite pitch for each 
 rotation of the cylinder to be turned or screwed. Every turn of 
 the leading screw moves the saddle and cutting tool through a 
 distance equal to its pitch, and consequently if the bar to be 
 screwed, turns at the same rate as the leading screw, the pitch ot 
 the screw cut upon it, will be the same as that of the leading 
 screw. If it moves faster than the leading screw, the pitch will 
 be less; and if slower, the pitch will be correspondingly greater. 
 It therefore follows as a matter of course, that if we fit wheels on 
 the lathe spindle and on the leading screw of the same diameter, 
 or having the same number of teeth, the screw being cut will 
 have the same pitch as the leading screw. If we fix a small 
 
 Einion, or one with few teeth, on the lathe spindle ai^d a wheel of 
 irge diameter, or many teeth on the leading screw, the pitch of 
 the screw to be cut will be small, compared with that of the leading 
 screw. Or, if the number of turns per minute of the leading 
 screw be greater than that of the screw being cut, the pitch of 
 the latter will be greater than that of the former, and vice versa* 
 Rules for Calculating the Required Number of Teeth 
 in Change Wheels. The following rules simply express the 
 previous reasoning in the form of proportion. In applying them, 
 the student should again refer to the end views of the change 
 wheels in the first figure of this Lecture. 
 
 Pitch of ncrrw to be cut_ No. of teeth in ist driver x No. in 2nd driver. 
 1'itch of guiding screw No. of teeth in ist follower x No. in 2nd follower. 
 Let p e = Pitch of screw to be cut in inches, or fraction of 
 
 inch, between two threads. 
 
 p a = Pitch of guiding screw .. 
 
 d r d, = Diameters or number of teeth in drivers. 
 > Juft = Diameters or number of teeth in followers. 
 
 Then, 
 Or, 
 
 * What was said in Lectures XII. XIII. and XIV. enables the student to 
 pee clearly the velocity ratio between the cut screw and the leading screw. 
 We need scarcely remind the student that the above statements refer to 
 the pitch of a screw as the distance between two consecutive threads, and not to 
 the number of threads per inch. If the number of threads per inch of its 
 length are taken as the pitch, instead of the distance between two threads, 
 the reverse ratio will hold good. Since a pitch of " means 4 threads to 
 the inch, a pitch of $" means 3 threads to the inch, and a pitch of " means 
 2 threads to the inch. Or, the number of threads per inch is inversely 
 proportional to the distance between two consecutive threads of the screw. 
 
LECTURE XVL 
 
 When the train of wheels is a compound one, as in this case, the 
 two intermediate multiplying or reducing wheels, / and d v are 
 fixed to any outstanding movable arm or quadrant at the left-hand 
 end of the lathe, so as to bring them into gear with d l and /,. (See 
 second view of the previous figure.) 
 
 If the train of wheels is a simple one, as in the first, third, and 
 fourth views referred to above, where there is only one driver, d, 
 and one follower,/, with, when necessary, one or more idle pulleys, 
 IP, simply for the purpose of connecting d and /and of giving J 
 the desired direction of rotation, then 
 
 **-7, 
 JV / 
 
 Should the pitch of a screw be expressed by the number of 
 threads per inch of its length as is usually the case in tables of 
 screws and change wheels then you can either convert this 
 number into the pitch proper, by taking its reciprocal (i.e., by 
 making the number of threads per inch the denominator of a 
 fraction, with i for the numerator) or you may say 
 
 Let t c = Threads per inch of screw to be cut. 
 ^ = Threads per inch of guiding screw. 
 
 Then, since the number of threads per inch are inversely 
 proportional to the distance between any two consecutive threads, 
 
 * PC 
 
 Or, . . t x ^ x d a = t x /, x f t 
 
 If the train is a simple one, then 
 
 i-=/;or,^ x d-i ff x/ 
 la a 
 
 EXAMPLE* I. The lathe illustration further on, has a guiding 
 screw of J" pitch, or 4 threads to the inch. Calculate the 
 number of teeth in the change wheel to be fixed to the end of the 
 guiding or leading screw in order to cut a screw of 8 threads td 
 the inch when the driver on the lathe-spindle has 40 teeth. 
 
 Compare the answer with the change-wheel table printed above the 
 general view of the screw-cutting lathe, further on in this Lecture. 
 
 ANSWER. Here t e = 8 ; t 4 ; d = 40 ; and you are required 
 to find/. 
 
NUMBER OF TEETH IN CHANGE -WHEELS. 179 
 
 By above formula, 
 
 ! =/ ; or, * - .*./~ 8 x * = 80 teeth. 
 
 t g d ' 4 40 4 
 
 By using the previous formula, we have p c = J" and p g = J" 
 ... & * s r,|=^... /= i44o = 8_X 
 
 P<7 / 4 / 8 4 
 
 It is at once evident from this example that you avoid having to 
 multiply and divide by sometimes awkward fractions if you 
 consider the number of threads per inch as the measure of the 
 pitch of the screw, instead of the distance between two threads. 
 
 EXAMPLE II. The guiding screw of a lathe is %' pitch, and 
 you are required to cut screws of $" and ^V' pitch respectively. 
 Determine the number of teeth in the follower, given the use 
 of a driver having 20 teeth. 
 
 ANSWER. For a screw of ^" pitch, or 10 threads per inch, and 
 using a driver of 20 teeth, we get by the above formula for a 
 simple train, 
 
 i_ = / ; or IO - IO X I0 - 100 -/ 
 t g d' '22 10 20 d 
 
 For a screw of " pitch the number of threads per inch will be 
 20, and using a driver of 20 teeth, we find from the formula 
 for a compound train 
 
 i. '_x/ t 
 
 < d, 
 
 20 x 40 
 
 Here we multiplied numerator and denominator by 20, in order to 
 obtain suitable wheels, of which d v will have 20 teeth. (See in 
 the previous figure the second of the end views showing change 
 wheels.) 
 
 Movable Headstock for a Common Lathe. Before 
 describing a complete screw-cutting lathe we will explain the use 
 and construction of this part of a common small lathe for ordinary 
 work. As will be seen from the accompanying rough sketch, it 
 consists of a cast-iron poppet-head planed on its under side, so as 
 to engage the breadth of the top of the shears. It may be bolted 
 thereto in any desired position (along the length of the bed) by 
 an underneath iron plate placed across the shears, and a single 
 vertical bolt. The upper portion of the head is cylindrical, and 
 is bored for about seven-eighths of its length to receive a round 
 
i8o 
 
 LECTURE XVI. 
 
 hollow steel mandril, M, and for the remaining one-eighth to 
 receive the spindle S. The mandril is fitted in front with a 
 tapered centre, 0, and behind with a screw nut, N". The centre 
 is for carrying one end of the job to be operated upon by the 
 turning tool, and the nut is for engaging the screwed part of the 
 spindle S. On the back end of the spindle there is a collar, c 
 (kept in position by a larger collar or guard, G, with small screws), 
 and a hand-wheel, H\V.* Consequently, by turning this wheel in 
 
 MOVABLE HEADSTOCK FOR A COMMON LATHE. 
 
 one direction the mandril and its centre are forced forward, and 
 when moved in the opposite direction they are screwed backwards. 
 To prevent the mandril turning round, it is fitted with a longi- 
 tudinal slot on its underside, into which fits the flattened or 
 rounded end of a small screw, s. A fixing stud, FS, with a 
 handle, enables the mandril to be clamped to the head when it 
 has been adjusted by the hand wheel and screwed spindle. 
 
 Description of a Screw-cutting Lathe. By the favour of 
 Messrs. John Lang & Sons we are enabled to give a general view, 
 with a complete set of reduced working drawings, carefully 
 indexed to every detail, of the very strong and superior 6-inch 
 centre screw-cutting lathe, lately presented to the Author's Elec- 
 trical Engineering Laboratory and Engineering Workshop by 
 Mr. Andrew Stewart, of Messrs. A. & J. Stewart, and Clydesdale. 
 This lathe weighs, with all its chucks and supernumerary parts, 
 over 1 5 cwt. It has a bed 6 feet long, and admits a bar 3 feet 
 
 * This arrangement of collar and guard is neither good nor strong, 
 although frequently adopted in the case of small foot-lathes. The collar 
 should be inside the bored head, behind the nut N. 
 
MOVABLE HEADSTOCK OF A SCBEW-CUTTING LATHE. l8l 
 
 2 inches between its centres. The bed is 9^ inches broad and 
 6 J inches deep. The gap is 9 inches wide and 6 inches deep ; 
 consequently the lathe can swing a job of 20 inches diameter clear 
 of the leading screw, and one of 24 inches diameter when this 
 screw is withdrawn from its bearings. The speed-cone has three 
 pulleys, each 2 J inches broad, the diameter of the largest being 
 8 inches and that of the smallest 4 inches. 
 
 The makers have planed and scraped the bed to a true bearing 
 surface, and have so fixed the gap piece that it cannot wear loose 
 or spring the bed. 
 
 k 
 
 _EH_ 
 
 N\\\\\\\Vy\\\\\\\\\\\V ' , , ._!. 
 
 *-y - ' -j---^ ^ , 
 
 END VIEW. LONGITUDINAL SECTION. 
 
 MOVABLE HEADSTOCK OP SCREW-CUTTING LATHE. 
 INDEX TO PARTS. 
 
 JBP represents Bottom part. 
 TP Top part. 
 
 S Spindle, or mandril. 
 SC Steel centre. 
 SS Steel screw. 
 HW Hand wheel. 
 
 BH represents Binding handle. 
 AS Adjusting screw. 
 
 ES 
 
 EH 
 
 B 
 
 P 
 
 Eccentric spindle. 
 Eccentric handle. 
 Bolt for clamping. 
 Plate under B. 
 
 The movable or loose headstock is gripped to the bed by an 
 eccentric motion worked by a handle, so that it may be instantly 
 clamped in position without the trouble of finding a key to fit 
 the usual nut, and then screwing it gradually home. The upper 
 part of this head, which carries the mandril or spindle, has a side 
 adjustment by means of a side screw, whereby the steel centre 
 may be truly aligned with the corresponding centre of the fast 
 headstock, or it may be moved to the one side or to the other in 
 the 'case of taper turning. A small oil-holder is cast on the back 
 side of the head to facilitate the oiling of the steel centre without 
 having to look for an oil-can. 
 
182 
 
 LECTURE XVI. 
 
FRAMING, ETC., FOE A SCREW -CUTTETO LATHE. 183 
 
184 
 
 FAST HEADSTOCK OF SCREW-CUTTING LATHE. 
 
 
 H- 
 
 8 
 
 1 
 
 g^l. 
 
 o 1S&I 
 
 a j M 
 
 su cc >- J5 
 
 .3 3 ?, | > 
 ^ 
 
 &mi * 
 
 ilil:- * 
 
 3 I so 
 
 ipr^i==i!i t^ii 
 M liiidi^^K-|Uil] 
 
 rrTr^ r^ C3 H &- Ld h-J WWW 
 
DESCRIPTION OF A SCREW-CUTTING ..LATHE. 
 
 I8 S 
 
 The spindle of the fast headstock is made of hard crucible steel 
 ground accurately cylindrical, where it fits into parallel gun- 
 metal bearings. These bearings are of extra diameter and length. 
 This spindle is bored hollow for 1 2 inches of its length, in order 
 to admit small rods for making terminals and screws in electrical 
 engineering work. The speed-cone is turned inside and outside, 
 and properly balanced. A specially strong and simple reversing 
 gear has been fitted to the back end of this headstock, whereby 
 the machine-cut steel pinions for turning right and left hand 
 screws may be put into or out of gear by simply depressing or 
 elevating a reversing handle. The back-motion gear is actuated 
 by means of a handle and eccentrics on each end of the back- 
 motion shaft ; whilst the front wheel (or last follower, F 2 , as we 
 have symbolled it in the formula) is locked to the cone or thrown 
 out of gear therewith in the usual way viz., by a bolt fitting 
 into a sliding slot in the cone and a projecting nut on the side of 
 the toothed wheel. 
 
 The saddle has T slots on its upper side for the purpose of 
 bolting work to it that requires boring out, and which necessitates 
 the removal of the slide rest. A quick hand traverse motion is 
 provided for the saddle by means of a rack and pinion motion, 
 quite independent of the sliding motion of the leading screw. 
 The leading screw is turned to the standard pitch of J inch, or 
 four threads to the inch. The engaging nut is made in halves, 
 so that it may grip the leading screw fairly at the top and bottom 
 of the threads.* 
 
 * In order to make the construction and action of the split nut which 
 engages the leading screw- 
 clearer, we show here an en- 
 larged view with the halves of 
 
 the nut, N< *N, slightly 
 
 apart, and the disc handle re- 
 moved, so as to bring into 
 full view the two eccentric 
 slots, ES, which guide the two 
 steel pins, P and P, fixed on 
 N and N. By comparing this 
 view with the others under 
 heading " Saddle and Slide," 
 the student will see how, by 
 merely turning the disc handle 
 DH the disc D is moved round 
 through nearly a quarter of a 
 circle, and the eccentric slots 
 ES cause the pins, P, P, to 
 move closer to or further away ENLAEGED VIEW OF SPLIT NUT FOB 
 from the centre of the disc D, LEADING SCREW, &c. 
 
 and consequently move the 
 two parts of the nut, N, N, in or out of gear with the leading screw. 
 
I 86 SADDLE AND SLIDE FOR A SCREW-CUTTING LATHE. 
 
 1". 
 
 . oo -H ^3 be far, 
 
 02 -g be 3 fl 
 
 H-IJM .'! 
 
 ^ S#!3A.S| | I 
 
 iieisi J 5d 
 
 PH 
 
 g 3 fr o . 
 
 2 ^s,||l||- 
 
 g ^^-a-g S).s|.S 
 
 g 0^*1*3^ 
 
 p EH tt EH H h t-5 ^5 ^ 
 
 S 
 
 , a_r 
 
CHANGE-WHEEL GEAR. 
 
 i8 7 
 
 *" r-3 02 ^^ 
 
 ~" A~* rC ^ 
 
 be ticS t>o fe i 
 
 flC S3 i 60 
 
 .2 i-i ^-** CD zr 
 
 ^ W E? QQ vU C 
 
 
 
 ^iliii 
 
 fe ft 
 
1 88 LECTURE XVI. 
 
 A compound slide rest is fitted to the top of the saddle, having 
 large bearing surfaces with adjustments for taking up the wear, 
 and a swivel arrangement for conical boring. 
 
 All the toothed wheels, including the change-wheels, have had 
 their teeth cut directly from the solid casting, by the makers' 
 special tool for that purpose, so that back-lash, and consequently 
 noise and vibration arising from fast-speed driving may be mini- 
 mised as far as possible. 
 
 The driving shaft has anti-friction steel roller-bearings. It is 
 connected to the foot-treadle at each end by a pulley, chain, and 
 crank. The driving-cone is so stepped that the belt has equal 
 tension on any corresponding pair of driving and driven pulleys. 
 It is sufficiently heavy to act as a fly-wheel. It is balanced along 
 with the treadle to secure an easy, steady drive. A power-drive 
 may be applied if desired, but the author believes that, as students 
 should work in pairs or in sets of three in a laboratory, they will 
 take a deeper interest in their experiments if they have turned 
 out everything by their own skill and labour, than if motive power 
 were freely supplied to them. 
 
 Of heavy chucks there are a very complete set, including a four- 
 jaw expanding chuck, clement driver, drill chucks for both the 
 fast and loose head spindles, <fec. 
 
 The student should now go over each drawing most carefully 
 by aid of the corresponding index to parts, and compare the 
 drawings with an actual screw-cutting lathe. 
 
 Hexagon Turret Lathe. Before completing this section it 
 will be necessary to illustrate one of the latest and most important 
 labour-saving appliances. The hexagon turret lathe is principally 
 used for repetition work, and it is generally found in large en- 
 gineering shops (where this class of work is carried on to a large 
 extent) to pay the firm to spend some time in designing suitable 
 cutting tools. The lathe with hexagon turret shown in frontis- 
 plate, and the several detail views of turret head and saddle as 
 designed by Messrs. Alfred Herbert, Ltd., Coventry, who have 
 kindly given me assistance in supplying the necessary drawings 
 and photos from which these figures were reproduced. The 
 special features of this lathe are (i) the ease with which the 
 speeds and feeds can be changed, also the feeds reversed 5(2) the 
 greatly improved set of tool holders on the turret head ; (3) the 
 improved lever operated chuck for gripping the bar ; and (4) the 
 new stop motion, by which each tool has its own longitudinal and 
 transverse stops. 
 
 Description of Lathe. Fig. i on page 189 shows a direct- 
 geared motor-driven headstock for this lathe. The motor is 
 
HEXAGON TURRET LATHE. 
 
 189 
 
 Front Elevation and Plan of 
 Back Friction Clutch 
 
 FIG. i. SECTIONAL PLAN OP AN ELECTRICALLY-DRIVEN HEXAGON TURRET 
 LATHE, SHOWING ARRANGEMENT OF GEARS AND FRICTION CLUTCHES. 
 
 M represents Electric motor. 
 
 MS Motor spindle. 
 
 \ P 6 ,, Pinions. 
 
 \-F 6 Followers or wheels. 
 
 IS ,, Intermediate gear stud. 
 
 Lj-Ko Keys fixed in shafts. 
 
 JSj, CS 2 Friction clutch sleeves. 
 
 l v R.J Rollers. 
 
 P Plunger or wedge. 
 
 FS Friction screw. 
 
 FP >, Friction plate. 
 
 AP Adjusting pin. 
 
 FC Friction clutch centre. 
 
 FBS Friction J^ack shaft. 
 
 C f Collar or ruff on shaf tFBS 
 
 SS Sliding shaft. 
 
 ^2, L 3 Levers for working SS. 
 
 INDEX TO PARTS. 
 
 PP represents Pulley for working force 
 
 pump. 
 LS .. Lathe spindle. 
 
 RCj, RC 2 ,, Friction ring carriers. 
 
 G lf G 2 Gluts. 
 
 TO Toggle carrier. 
 
 T Toggle. 
 
 S 15 S 2 Studs. 
 
 FRj, FRo Friction rings. 
 
 S 3 , S 4 Shoes for friction r 
 and toggle carrier. 
 
 Lj Lever for actuating R 
 
 and RC 2 
 
 BS Back gear shaft. 
 
 B!, B 2 Bearings for LS. 
 
 B 3 , B 4 Bearings for BS. 
 
LECTURE XVI. 
 
 J 
 2 . 
 
 (So202WOo26o^HOWcc 
 
 isli: 
 
 tn 
 
 c5 . ^pq 
 
 * 
 
 HOwl 
 
 ^S N , ** 
 
 **#%*##? * 
 
 nti \ / if Jj&J <(n, / 
 
HEXAGON TUKRET LATHE. 
 
LECTURE XVI 
 
 ^ J c/2 ^ 
 
 02 PQ PQ m 02 PH O 02 W 
 
OF THE 
 
 UNIVERSITY 
 
 OF 
 
HEXAGON TURRET LATHE. 
 
 STAD SL/DE 
 ADJUST/NO SCREW 
 
 FUNNEL FOR 1MB ft J GATING 
 3TADy LIQ CLAMP 
 
 ffACK P/N/ON SMAfT 
 
 STOP CLAMP 
 
 \ 
 
 WORM StiA, 
 
 CUTTER CLAMP/MG SCPW* 
 CUTTER SL/DE 
 TOP iNCLIME 
 
 BODV 
 
 ELEVATING 
 
 SCREWS *- ItJ- 
 
 SUPPORT 
 FACE 
 
 CUTTER SL/DE G/B SCffHS& 
 CUTTER SL/DE CLAMP 
 
 THE PATENT ROLLER STEADY TURNER. 
 
 The following improvements have been made in this lathe : 
 
 1. The motor is mounted on a hinged base plate, by which it can be 
 elevated and depressed, and it drives the headstock by means of a belt. The 
 swinging movement of the motor serves to adjust the tension of the belt. 
 
 2. The geared headstock provides sixteen speeds within itself, thus 
 enabling a constant speed motor to be used, and renders the lathe equally 
 applicable to direct or alternating-current motors. 
 
 3. The driving pulley of the lathe contains an epicyclic reverse gear for 
 giving the spindle speeds in every direction. 
 
 4. The feed motion now gives nine feeds instead of three as 
 formerly. 
 
 5. The turning tool holders are fitted with roller steadies, enabling 
 much higher cutting speeds to be used than formerly. 
 
 6. A scale and adjustable pointer are fitted to the bed to enable direct 
 measurements to be made without using a rule. 
 
CHUCKS TURRET-HEAD AND SADDLE. 1 93 
 
 of 8 B.H.P. It is a semi-enclosed motor with speed variation 
 obtained by inserting resistance in the field coils from a 60- 
 point shunt controller. Consequently any desired speed is 
 easily got between 1150 and 1850 revolutions per minute. The 
 motor M and spindle MS is attached to the rawhide pinion P,. 
 This pinion drives by means of the compound gear train F v P,, 
 F 2 with P 3 , the first shaft FBS. As will be seen from the sec- 
 tional plan the shaft FBS carries two spur pinions P 3 and P 5 , 
 either of which can be put in or out of gear with follower wheels 
 F 3 and F 5 by means of the friction clutches CS^ CS 2 . The levers 
 L S , L 3 with central spindle SS actuate the friction clutches 
 CS 1? CS 2 . Hence, for every speed of the motor this shaft FBS 
 can give two different speeds to the sleeve or followers F 3 and F 4 
 on the lathe spindle LS which is driven from shaft FBS. These 
 follower wheels correspond to the cone pulley on an ordinary 
 lathe, and drive the spindle LS direct through the friction ring 
 carrier RC 2 , which slides along the key K, fixed in the lathe 
 spindle LS ; or, through double gearing P 4 , F 4 , BS, P 6 , F 6 and 
 friction ring carrier RCj connected to lathe spindle LS by key 
 K 2 , the change from the one to the other is made by means 
 of the lever L r The friction clutches on the shaft FBS are of 
 the expanding ring type operated by means of wedge surfaces P 
 and rollers R 1? R 2 , whilst the clutches RC 1? R0 3 on the spindle 
 LS are also of the expanding ring type, but actuated by toggles T. 
 (See Index for toggle joints.) 
 
 The lathe shown by the frontispiece is of a later type. 
 It is driven by a constant speed motor mounted on a hinged 
 base plate. This base plate provides for the necessary tightening 
 of an endless belt. 
 
 Chucks. The lathe may be fitted either with an automatic 
 or universal chuck. The former chuck is recommended where the 
 bars to be worked are practically straight and cylindrical, whilst 
 the latter is used on bars which are badly out of round or not 
 straight. 
 
 Looking at the front outside view of lathe, it will be seen how 
 that the bar to be turned is gripped by an automatic chuck 
 worked by the longer handle parallel to and beside L r 
 
 The automatic chuck may be opened and closed whilst the 
 machine is running, and it has the advantage of holding finished 
 work without bruising. 
 
 Turret-Head and Saddle. It will be seen from the frontis- 
 plates and the accompanying views, Figs. 2 and 3, that the saddle 
 can be moved either by hand or power, as in the ordinary lathe. 
 In the former method of working, the saddle is moved by nvans 
 
194 
 
 LECTURE XVI. 
 
 of the pilot handle PH acting through wheels SPj, SW : , RP to the 
 rack R; whilst, by the second method, motion is taken from the 
 
 FIG. 5. THE "COVENTRY" SELF-OPENING Dix HEAD WITH 
 ROUGHING AND FINISHING ATTACHMENT. 
 
 F IGt 6. TURNING TOOLS VIEW SHOWING RIGHT AND LEFT-HAND 
 
 CUTTEES. 
 
 feed or traverse shaft FS through the spur wheels SP 8 and worm 
 W, worm wheel WW, and wheels SP p SW 1? RP X to the rack R. 
 
CHANGE WHEEL GEAR, ETtt 195 
 
 The worm wheel is carried in a cradle CH, hinged round the 
 shaft WS, and held in position by the stop or trip lever TL and 
 spring SS 2 . Whenever the hand traverse is to be used, the 
 traverse worm W is released by the cradle dropping due to its 
 own weight, after releasing the catch or trip lever TL. 
 
 When the handle TL is pressed to the right, the bar connected 
 to it turns on a fulcrum, and causes the worm cradle handle CH 
 to be dropped, thus stopping any further traverse of the saddle 
 along the bed of the lathe. This traverse may also be stopped 
 automatically by means of the trip lever TL coming into contact 
 with the stop S, which is set at the desired position on the stop 
 rod SR, and actuating the feed trip lever FTL. The hexagon 
 head can be rotated into any position by turning the handle H, 
 plunger pinion PP, and plunger P 2 , so that P, is freed from the 
 cast steel bush B. 
 
 Bolted to the faces of the hexagon head are the specially 
 shaped tools, while boring bars may be passed through the 
 circular holes shown in the hexagon head and held firmly by 
 bolts. 
 
 Change Wheel Gear. It will be seen from the illustrations 
 in Fig. 4 that the traverse or clutch shaft CS is rotated from the 
 mandril through the feed pulley FP, and the pairs of wheels SW^ 
 SW 6 . Any pair of wheels may be put into gear by means of the 
 hand lever HL, which acts upon the internal sliding rod SR, and 
 slides a cotter key SO so as to fix SW 4 , SW 5 or SW 6 to the shaft 
 SR. The handle or reversing lever RL puts the claw clutch RC 
 in gear with bevel pinions BP X or BP, to give either a forward or 
 backward rotation or stoppage of the shaft CS. Consequently, 
 three variations of speed for cutting purposes may be obtained 
 by altering the position of the vertical hand lever HL, while the 
 direction of rotation of the feed shaft may be changed by turning 
 the reversing lever RL. 
 
 Turning Tool Holders as supplied with the lathe are one of 
 its most important features. A general idea of the different 
 tools on the turret or hexagon head is obtained from the frontis- 
 plate. When another tool is required to be fetched up to the 
 work, this is effected by turning the short handle H seen near top 
 of saddle downwards, thus releasing the turret, which can now be 
 turned as desired, afterwards bringing the handle back to its 
 normal, horizontal position, and firmly fixing the hexagon head. 
 Separate views of the turning tool holder, cut-off tool rest and 
 self-opening die head for the hexagon head are also given, see 
 Fig. 6 on page 194 and plate facing page 193. 
 
 Rapid Turning with High Speed Steel. The student- 
 will be interested to learn something regarding the recent im- 
 
196 
 
 LECTURE XVI. 
 
 provements which have been made in rapidly turning out work 
 with the most modern electric-driven lathes, and the new kinds 
 of high speed steel. Consequently, we have made a short 
 abstract from three papers read recently before Engineering 
 Societies where these matters were specially dealt with and 
 discussed.* 
 
 FIG. 7. CUT-OFF TOOL REST SHOWING STEPPED CUT-OFF TOOL. 
 
 Description of Lathe. The lathe on which all the tests 
 were made was supplied by Armstrong, Whitworth & Co., and was 
 on.e of their 15" centre screw-cutting lathes, taking in 9' 6" 
 between the centres, but for these experiments 18" headstocks 
 were fitted. The fast headstock had both double and treble back 
 
 * See Proc. Inst. Eng. and Ships, in Scotland, vol. xlvii. 1904, for Mr. 
 Charles Day's Paper on " Experiments with Rapid Cutting Steel Tools." 
 Also see Mr. J. M. Gledhill's paper on "High Speed Tool Steel," read 
 before the Coventry Engineering Society in March 1904, and Mr. P. V. 
 Vernon's paper on " Speeds of Machine Tools," read before the Manchester 
 Association of Engineers on November 14, 1903. 
 
EESULTS I9/ 
 
 gears, the ratios being 14.9 to i and 42.5 to i. The headstock 
 was specially fitted with a 3-step cone suitable for a 6" belt. The 
 lathe was driven by a direct current shunt-wound motor of 120 
 E.H.P., with a large air-cooled rheostat. The speed of the motor 
 could be varied between 150 and 300 revolutions per minute at 
 no load on the lathe, and from 60 to 300 revolutions with heavy 
 cuts by means of the rheostat. The lathe was driven by two 
 intermediate countershafts having 10" belts. 
 
 Results. The diagrams show the maximum cutting speeds 
 successfully used in each experiment made with rapid cutting 
 steel tools, and the curves show the average speeds during each set 
 of trials. Figs. 8 and 9 show the average results obtained from 
 those tools which finished in such condition as to warrant atten- 
 tion, whilst the dotted lines in the figures show the maximum 
 results obtained with any tool which finished in a satisfactory 
 condition. 
 
 It was found that no single make of steel proved to be superior 
 to all others in every respect, but it would appear that the 
 average curves are those which may be taken as standards of all 
 round comparison for use in general engineering shops. The 
 following empirical formulae give approximately the cutting speeds 
 which may be adopted for different areas of cut upon different 
 materials, and the curves show the results obtained therefrom : 
 
 For soft steel S * g6 + 1 2 
 
 A x 0.013 
 
 medium steel S = *' 23 + 5 
 Ax 0.015 
 
 , hard steel S= A T ' 77 +5 
 Ax 0.027 
 
 soft cast iron S = + 20 
 
 A x 0.02 
 
 i-5 
 
 Ax 3 
 
 hard S= 
 
 Where: S = Cutting speed in ft. per minute. 
 
 A = Area of section of cut in sq. inches, i.e., traverse 
 in inches multiplied by the depth of cut in inches, 
 
198 
 
 LECTUKE XVI. 
 
 FIG. 8. VARIATION OF CUTTING SPEED WITH AREA OF CUT STEEL. 
 
 3 1 
 
 FIG. 9. VARIATION OF CUTTING SPEED WITH AREA OF CUT CAST IRON. 
 
POWEB BECORDS CUTTING SPEED RESULTS, ETC. 199 
 
 Power Records. Very careful records were taken of the 
 power used for the various cuts and spaces. Also, data were 
 obtained of the forces brought to bear on the cutting tools. 
 
 Cutting Speed Results. Table I. gives the average results 
 obtained from the tools which finished in good condition. The 
 horse-powers given are the gross horse-powers as calculated from 
 the readings of electrical instruments attached to the motor, and 
 include the motor losses with any countershaft friction. The nei 
 horse-powers required to overcome the resistance to cutting are 
 not given in the Table. These are only required for determining 
 the cutting force on the tool point. 
 
 Endurance Results. Table II. gives the average result* 
 obtained from soft forg< d steel and medium cast iron, which 
 maintained their average cutting edge in fair condition foi 
 60 minutes or longer. 
 
 Comparison of Results. For the purposes of comparing 
 results which may be obtained with the new steels against those 
 obtainable with ordinary Mushet steel and ordinary water- 
 hardened steel, tools made of the.se materials were tested, and 
 the average results are also given in Table II. It will be noted 
 that the new steels give decidedly improved results, and that 
 with them the cutting speed can be about twice as fast as with 
 ordinary Mushet steel, and three or four times as fast as with 
 ordinary water-hardened steel. 
 
 An item of interest which may be mentioned here is that the 
 ordinary Mushet steel can be very greatly improved by treating 
 it in the same manner as the new steels when tempering. This 
 is a point of value, as it enables greatly improved results to be 
 obtained from existing tools. 
 
 It will be seen from an inspection of Table I. that, when much 
 metal has to be removed, this may be done not only more quickly, 
 but also with a less expenditure of energy per Ib. of material 
 removed, if a heavy cut is taken at a comparatively low speed in 
 preference to a lighter one at a high speed. 
 
 Cutting Forces and H.P. for Lathes. The figures showing 
 the cutting force on tool points should prove of great service to 
 machine-tool designers. 
 
 The information regarding the horse-powers is worthy of 
 special attention, for it is this element which perhaps will form 
 the greatest difficulty in the way of using existing lathes 
 efficiently with the new high-speed cutting tools. A lathe on 
 which a cut of f " x J" on soft steel can be taken is by no means 
 
2OO 
 
 LECTURE XVI. 
 
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202 LECTURE XVI. 
 
 an abnormal one, and this duty can be carried out on most good 
 lathes of, say, 12" centres, but the driving cones, the countershaft, 
 and the belts connected with few such lathes would be suitable 
 for 24 h.p. Further than this, the line shafts in most engineering 
 shops are too light to drive many lathes using 20 h.p. each, or 
 anything approaching that figure. 
 
 Example. As a small example of what the Herbert Turret 
 
 Lathe (see previous figs.) has done with this new high-speed steel, 
 
 we quote the following from the Proceedings of the Coventry 
 
 Enginering Society, as found in the paper read by Mr.J. M. 
 
 Gledhill on March 4, 1904 : 
 
 Sample No. i was a i" bolt 6" long in shank, with i" round 
 head, i^" deep, and with the point screwed for 2". 
 
 This was finished complete in 5 minutes 28 seconds, using 
 " A. W. " tools. The following are the details : 
 Reducing i-J- in. bar to i in., 6in. 
 
 long, 96 cuts per inch . . 4 min. o sees. 
 Screwing ..... 40 
 
 Cutting- off .... 28 
 
 Idle movements, etc. . . . 20 
 
 Total . . 5 min. 28 sees. 
 
 The above time does not include the facing of the back of the 
 head, which would require J minute. 
 
 This bolt, made in the ordinary way with ordinary tool-steel, 
 with a good operator, requires 15 minutes. 
 
 Forging and Hardening the Tools. In forging, annealing, 
 and hardening crucible steels it is essential that the most suitable 
 temperatures should be found for all of these processes, and then 
 accurate means be taken to ensure such temperatures being 
 actually obtained as near as practically possible. This can only 
 be effected by the skilful use of pyrometers or other scientific 
 heat-recorders, for to work on the old-fashioned lines of judging 
 by the eye is no criterion of actual temperature, and is no longer 
 advisable. It is now known that every composition of steel has 
 its own definite temperature that is best suited for obtaining 
 from it the most satisfactory results, and the nearer this can be 
 worked to the better, any deviation from the correct temperature, 
 up or down, involving a corresponding difference in the efficiency 
 of the steel.* 
 
 * Students may refer here to the illustrated descriptions of these Pyro- 
 meters in the Author's Elementary and Advanced Steam Books. 
 
FORGING AND HARDENING THE TOOLS 203 
 
 Having obtained a bar of, say, the Armstrong- Whitworth ot 
 " A. W." brand of tool steel, it is necessary to cut off the required 
 lengths ; and this must be done at a forging heat. The lengths 
 must not be broken off cold, as this tends to cause cracks in the 
 bars. For forging, the steel should be placed in the fire, and 
 slowly but thoroughly heated, taking care that the heating has 
 penetrated to the centre of the bar, and then forged at a bright 
 red heat. Whilst forging, the bar should not be allowed to get 
 lower than a good red. After the tool is forged it should be laid 
 down in a dry place and allowed to cool slowly. To harden the 
 tool, the nose only should be raised to a white melting heat and 
 then cooled with an air blast. 
 
 To obtain the maximum efficiency from this " A. W. " steel it 
 is essential that the nose of the tool shall be raised to a white 
 melting heat as described, for if during this heating the point 
 of the nose becomes fused or melted, no harm whatever has been 
 done. The tool is then ready for use after grinding on a wet 
 stone. 
 
 Another method which may be described of preparing the tools 
 is as follows : 
 
 Forge the tool as before, and when cold roughly grind to shape 
 on a dry stone or dry emery wheel. The tool then requires 
 heating to a white heat, just short of melting, and cooling in the 
 air blast. This method also lends itself for cooling the tools in 
 oil; before which the temperature has to be lowered from the 
 white heat to a good red heat (about 1600 F.) either by the air 
 blast or in the open, and the tool then quenched in oil. Tools 
 hardened by the latter method are specially good where the 
 retention of a sharp edge is a desideratum, as in finishing tools, 
 capstan and automatic lathe tools, brass workers' tools, <fec. 
 Nothing has yet been found to be so good for air hardening 
 steels as the wet sandstone. 
 
 Specific Gravity Test. Where engineering firms buy large 
 quantities of various steels, it is advisable to have samples cut off, 
 to ascertain the various specific gravities and tabulate them. The 
 specific gravity of the steel of any one particular brand and 
 maker is found to be fairly constant. Consequently the firm will 
 be able to identify from amongst any number the manufacturer of 
 the steel which they find by trial to be most suitable for their 
 purpose by this method. 
 
204 LECTURE XVI. QUESTIONS. 
 
 LECTURE XVI. QUESTIONS. 
 
 1. Sketch the fast headstock of a double-geared lathe, and explain the 
 contrivance for increasing or diminishing the speed of the mandril. In 
 the headstock of a lathe a pinion of 20 teeth drives a wheel of 60, and a 
 second pinion of 20 drives another wheel of 60; compare the rates of 
 rotation of the first driving pinion and of the mandril of the lathe. 
 Ans. 9 : i. 
 
 2. Why is a lathe often back-geared ? Sketch a section through the 
 headstock showing the arrangement. If the two wheels have 63 and 63 
 teeth respectively, and each pinion has 25 teeth, find the reduction in the 
 velocity ratio of the lathe epindle due to the back-gear. Ans. 6*35 : i. 
 
 3. Make a vertical longitudinal section through the movable or loose 
 headstock of a lathe, showing precisely the manner in which a screw anc 
 nut are applied to produce the necessary movement of the centre which 
 supports the work. Name the materials of which the several parts are 
 made. 
 
 4. What is the use of the guide-screw in a lathe ? Where is ifc usually 
 placed ? Show by sketches the precise manner in which the slide rest is 
 connected with or disengaged from the guide-screw. 
 
 5. Describe and show by sketches the means by which the slide rest of 
 a lathe may be connected with the leading screw. If the slide rest 
 traverses the bed at the rate of i^ feet when the leading screw makes 56 
 revolutions, what is the pitch of the screw thread 1 Ans. % inch. 
 
 6. Sketch and describe the mechanism by which the saddle of a screw- 
 cutting lathe can be made to travel automatically in either direction along 
 the lathe bed while the speed pulleys run always in the same direction. 
 
 7. How is the copying principle applied in a screw-cutting lathe ? 
 Describe a method of throwing a self-acting screw-cutting lathe in and 
 out of gear, and of reversing it by means of a belt and overhead pulleys. 
 (See Fig. 5 in Lecture XI.) 
 
 8. Explain the use of the quadrant for change wheels in a screw- cutting 
 lathe by making a sketch showing it in its position on a lathe with the 
 wheels in gear. (See the general and the end views of the 6" screw 
 cutting-lathe bed, and " Index to Parts " for the part marked CP.) 
 
 9. Explain the mode in which change wheels are employed in a screw- 
 cutting lathe. The leading screw being of J-inch pitch, give a sketch 
 of the arrangement of change wheels required for cutting a screw of 15 
 threads to the inch, marking the numbers oE the teeth on each wheel. 
 
 10. Sketch and describe the mechanism for cutting a screw with five 
 threads to the inch in a lathe where the guide screw has three threads 
 to the inch. Assign suitable numbers to the wheels which you would 
 employ. 
 
 11. The leading screw of a lathe is |-inch pitch and right-handed. 
 Sketch and describe the arrangement whereby you would employ the lathe 
 for cutting a screw of -inch pitch, and left-handed^. 
 
 12. Describe the operation of cutting a screw in a lathe, showing the 
 wheels required, and how they are placed to cut a right-handed screw 
 with eight threads to the inch in a lathe whose leading screw is of J-inch 
 pitch. 
 
LECTURE XVI. QUESTIONS. 
 
 13. Explain the use of change wheels in a screw-cutting lathe. It is 
 desired to cut a screw of f-inch pitch in a lathe with a leading screw of 
 four threads to the inch, using four wheels. If both screws be right- 
 handed, what wheels would you employ ? 
 
 14. The leading screw in a self-acting lathe has a pitch of inch ; show 
 an arrangement of change wheels for cutting. a screw of f-inch pitch. 
 
 15. You are required to cat a left-handed screw of five threads to the 
 inch in a lathe fitted with a right-handed guide screw of $-inch pitch. 
 Show clearly by the aid of sketches the change wheels which you would 
 employ for the purpose, indicating how they would be respectively carried, 
 and the number of teeth in each wheel. 
 
 1 6 What do you understand by a single-geared, a double-geared, and a 
 treble-geared lathe ? Give such sketches as will show clearly the arrange- 
 ment of the headstock in each of these cases. 
 
 17. Given a screw-cutting lathe with aright-handed leading screw with 
 four threads per inch sketchan arrangement for cuttinga left-hand thread 
 of eleven threads per inch. What gear wheels would be required ? 
 
 1 8. A driving shaft runs at 100 revolutions per minute, and carries a 
 pulley 22 inches in diameter from which a belt communicates motion to a 
 pulley 12 inches in diameter carried upon a counter-shaft. On the 
 counter-shaft is also a cone pulley having steps, 8, 6, and 4 inches in 
 diameter respectively, which gives motien to another cone pulley with 
 corresponding steps on a lathe spindle. Sketch the arrangement in front 
 and end elevation, and find the greatest and least speeds at which the 
 lathe spindle can revolve. Ans. 366'6 revs, per minute, and 91 '6 revs, per 
 minute. 
 
 19. Describe and sketch the arrangement of the mechanism by which 
 the saddle of a lathe is traversed by hand along the bed. 
 
 If the slide rest of a screw-cutting lathe when in gear with the leading 
 screw moves along the bed fora distance of 14", while the leading screw 
 makes 56 revolutions, what must be the pitch of the thread on the leading 
 screw 1 Ans. inch. 
 
 20. Give free-hand sketches of the front and back outside views of the 
 Hexagon Turret Lathe described in this Lecture with index to parts. 
 
 21. Give a sectional plan with index to parts of the electrically-driven 
 Hexagon Turret Lathe, showing the detailed arrangement of motor and 
 gears with friction clutches for the fixed headstock. 
 
 22. Give a plan with front and end elevations, together with an index to 
 parts, showing the general arrangement of Saddle and Turret for the 
 Hexagon Lathe. 
 
 23. Give a plan with front and end elevations of the general arrange- 
 ment of the apron for the Hexagon Turret Lathe. Indicate each part by 
 letters and an index. 
 
 24. Give a front and end elevation with index to parts of the general 
 arrangement of the feed motion of the Hexagon Turret Lathe. 
 
 25. What is meant by " rapid turning high speed tool steel " 1 Indicate 
 the variations of cutting speed with area of cut-steel and cast-iron by 
 diagrams plotted on squared paper with an example of the h.-p. required 
 under certain conditions. 
 
 26. Describe the process of forging and hardening rapid-cutting tool 
 steel, and state how it differs from ordinary tool steeL 
 
2O6 NOTES AND QUESTIONS. 
 
 27. Describe, with sketches, the mechanism for giving an automatic 
 feed to the cutting tool of a lathe or shaping machine, and how it is put 
 in or out of action, and the amount of feed varied. (B. of E., 1902.) 
 
 28. On the headstock spindle of a lathe is keyed a speed cone, the 
 greatest and least diameters of which are 10 ins. and 5^ ins. respectively. 
 It is driven from a similar speed cone keyed to a counter-shaft which makes 
 350 turns a minute. The back gearing is of the usual type, the spur-wheels 
 concentric with the headstock spindle having 62 and 30 teeth, gearing with 
 wheels having 18 and 50 teeth respectively on the back spindle. Sketch 
 and describe the arrangement, and find the greatest and least speeds at 
 which the headstock spindle can run. (C. & G., 1903, 0., Sec. A.) 
 
 Ans. Without gearing: 6j6'$6 revs, per minute, and I92'5 revs, per 
 
 minute. 
 With gearing : 1107 revs, per minute, and 33*5 revs, per minute. 
 
 29. The gearing of a capstan engine is arranged as follows : Fixed on the 
 crank shaft is a double-threaded worm, which gears with a worm-wheel of 
 50 teeth ; keyed to the worm-wheel shaft is a spur-wheel (A) of 22^ ins. 
 diameter pitch circle, and very approximately 3^-in. pitch, driving another 
 wheel of 40 teeth, which is keyed to the same shaft as the holder round 
 which the cable passes. Find the number of teeth in the \\ heel A ; and if 
 the effective diameter of the cable-holder is 24 in^., find the number of 
 revolutions the engine must make to heave in 90 ft. of cable. 
 
 (C. & G., 1904, O., Sec. A.) 
 
 Ans, Nnmber of teeth on wheel A = 20. Revs, of engine required to 
 heave in 90 ft. of cable = 715. 
 
 30. It is required to transmit a velocity ratio of 80 to I by a train of 
 toothed wheels ; no pinion of this train of wheels is to have less than 
 16 teeth, and no wheel is to have more than 90 teeth. Determine a 
 suitable train of wheels, and the number of teeth in each wheel. 
 
 (B of E., 1905,) 
 . _64x8ox64_8o 
 
 ! ~l6xi6xi6~T* 
 
 31. In the gear for an electrically-driven turret, the motor shaft is 
 provided with a single-threaded worm gearing with a worm- wheel of about 
 23^ ins. diameter, and i^-in. pitch, keyed to a spindle, which carries a 
 wheel of 12 teeth, which gears with a wheel of .about 30^ ins. diameter 
 and 4-in. pitch. Keyed to the spindle of the latter is a spur-wheel of 
 15 ins. diameter, gearing with a circular rack on the turret, and which is 
 15 ft. diameter. It is found that the shortest and longest times to turn 
 the turret through 270 are 52 seconds and 21 minutes respectively. Find 
 the corresponding revolutions per minute of the motor. 
 
 (C. & G., 1905, 0., Sec, A.) 
 Ans. Number of teeth on worm-wheel = 49 ; 
 ,, ,, ,, spur-wheel =24; 
 
 Revs, of motor per minute. 1018 and 42. 
 
( 207 ) 
 
 LECTURE XYII 
 
 CONTENTS. Hydraulics Definition of a Liquid Axioms relating to a 
 Liquid at Rest Transmission of Pressure by Liquids Pascal's Law 
 "Head" or Pressure of a Liquid at Different Depths Total Pres- 
 sure on a Horizontal Plane immersed in a Liquid Lord Kelvin's 
 Wire-testing Machine Total Pressure on any Surface immersed in a 
 Liquid Examples I. II. Questions. 
 
 Hydraulics. Hitherto the student's attention has been con- 
 fined to solid bodies, which were supposed to remain perfectly 
 rigid and unchanged when acted upon by forces. We shall now 
 direct his consideration to the properties and applications of 
 another great division of matter viz., liquids which possess the 
 marked opposite character of mobility under the action of forces. 
 In nature we do not meet with either perfectly solid or perfectly 
 liquid bodies; and consequently the practical engineer, when 
 applying the formulae of the physicist to his machines and 
 hydraulic works, has to make certain allowances according to 
 circumstances, with the aid of constants predetermined by experi- 
 ence and experiment. 
 
 The most common and the most useful liquid with which the 
 engineer has to deal is that of water. Hence the term " hydraulic 
 engineer," as applied to persons who direct and guide the action 
 of waters, as in the case of the water supply for a town, or for 
 navigation purposes, or for the transmission of force and power. 
 The term hydraulics, therefore, comprehends hydro- statics y which 
 is the science of liquids in equilibrium^ and hydro-kinetics, the 
 science of liquids in motion. We shall only have space in this 
 manual for an elementary inquiry into the former of these two 
 divisions of hydraulics.* 
 
 Definition of a Liquid. A liquid is a collection of particles 
 which are perfectly movable about each other. In consequence of 
 this property, a liquid requires some external force or resistance 
 to keep its particles together, such as the sides of a vessel ; for its 
 molecules can be displaced by the smallest force, and are readily 
 divided from each other in any direction.f 
 
 * For Viscous Fluids, see p. 224. 
 
 t The late Prof. Clerk Maxwell distinguished solids from liquid* in the 
 following manner: "Bodies which can sustain longitudinal pressure, 
 
 o 
 
208 
 
 LECTURE XVII. 
 
 a v 2 ss 
 
 Axioms relating to a Liquid at Rest. It follows directly 
 from the above definition, that when equilibrium exists 
 
 (1) l\efree surface of a liquid at rest is horizontal ; 
 
 (2) Any surface of a liquid at rest is everywhere perpendicular to 
 the force which acts upon it ; 
 
 (3) A liquid at rest acted on by a force presents a surface whijh 
 is everywhere perpendicular to the direction of the force ; 
 
 (4) A surface supporting a liquid at rest reacts everywhere per- 
 pendicularly to the pressure of the liquid ; 
 
 (5) In all cases of pressure on or from liquids at rest, action and 
 reaction are equal and opposite. 
 
 If such were not the ..case, equilibrium could not exist, and 
 motion of the liquid would take place. 
 
 Transmission of Pressure by Liquids. Take a tight vessel 
 
 filled with a liquid and fitted 
 with four f rictionless piston- valves, 
 Y 15 Y 2 , V 3 , Y 4 , of the same area. 
 Let the outward pressure on these 
 valves be balanced by spiral 
 springs, arranged so that they 
 indicate the forces applied to 
 them. Now apply an inward 
 force of, say, i or 5 or 10 Ibs. to 
 the spiral spring of valve Y x , then 
 instantly the other three springs 
 TEANSMISSION OF PEESSUEE of the other valves will register 
 BY LIQUIDS. an outward pressure of the same 
 
 (Horizontal Section.) amount as that applied. If the 
 
 other valves had been of different areas from valve Y p their 
 springs would have registered pressures corresponding with the 
 ratio of their areas to the area of valve Y r Or the pressure per 
 square inch on valve Y x is communicated throughout the liquid 
 to the other valves, and to every square inch of the internal sur- 
 face of the vessel, with undiminished effect. 
 
 Pascal's Law. Fluids transmit pressure equally and in all 
 directions* In the case of solids pressure is only transmitted 
 
 "however small that pressure may be, without being supported by lateral 
 pressure, are called solids, and those which cannot are termed liquids." A 
 perfect liquid is therefore one in which there is absolutely no resistance to a 
 change of shape, although there may be practically an infinite resistance to 
 change of volume. We say practically because, although liquids are more 
 or less compressible to a very small extent, yet the amount is so small as 
 to be negligible in the case of most engineering problems. 
 
 * Here the word fluid hag been used instead of liquid, as being more 
 general, since the term fluid includes both liquids and gases. Refer to 
 p. 2, Lecture I., for the distinction between a liquid and a gas, 
 
PRESSURES DUE TO LIQUIDS. 2 09 
 
 along the line of its action, and therefore we have in this law an 
 exemplification of the fundamental distinction between solids an<S 
 fluids. In Lecture J1X. we will explain several machines that 
 depend upon the principle enunciated by Pascal's law for their 
 action. 
 
 Head or Pressure of a Liquid at Different Depths. 
 Imagine a very small horizontal area, a (for instance, a square 
 inch), situated at a depth or height, h, inches from the free surface 
 of a liquid, and that the vertical column from, a, to the surface 
 becomes solidified -without in any way disturbing equilibrium; It 
 is evident that the horizontal and the vertical forces on the solid 
 column must be separately in equilibrium, otherwise motion would 
 ensue. But the only vertical forces are the weight of the column 
 downward and the pressure of the surrounding liquid upwards 
 on the base, a. Therefore, 
 
 The pressure upwards = weight of the prism. 
 
 Or, . . */* p = haw. 
 
 Where, ?, is the weight of every inch of its height or the weight 
 of a cubic inch of the column. But the area, a, and the weight, 10, 
 are constant quantities for any particular unit of area and kind 
 of liquid. Hence 
 
 Pressure varies directly as the depth from the free surface. 
 
 Or, p ec h. 
 
 The technical term " head " expresses the above fact in a single 
 word. For, when speaking of the working pressure per square 
 inch due to a supply of water for a mill wheel or turbine, we say 
 it has i o or 20 or 30 feet of head, meaning thereby the pressure 
 due to a difference of level of so many feet, from the free surface 
 of the water as it enters the supply pipe to the free surface of 
 the tail race or discharge pipe. Since every foot of " head " of 
 water gives in round numbers a pressure of i Ib. per square inch, 
 we might have said that the pressure was 5 or 10 or 15 Ibs 
 respectively per square inch. Consequently, 
 
 Pressure varies directly as the head. 
 
 Total Pressure on a Horizontal Plane immersed in a 
 Liquid. Take a vessel of any shape having a horizontal base, 
 and fill it with a liquid to any known height. Then from the 
 above rule it follows that, 
 
 height in inches from base to sur- 
 
 The Total Pressure on the lose = . 
 
 face x area of base in square 
 
 inches x weight of a cubic 
 inch of the liquid. 
 For, pressure per sq. in., p = haw t when, a= i square inch. 
 
210 
 
 LECTUEE XVII. 
 
 Consequently, if the total area of the horizontal plane be equal 
 to, a, square inches, instead of i square inch. 
 
 The Total Pressure = haw. 
 
 This shows that the shape of the vessel containing the liquid, and 
 the total weight of water in the vessel, do not in any way affect the 
 total pressure on the base. For, it depends solely on the difference of 
 level between the base (or immersed plane) and the free surface, on 
 the area immersed, and on the weight per unit volume or specific 
 gravity of the liquid. 
 
 This property results in what used to be termed the hydrostatic 
 paradox, which is very well illustrated by Lord Kelvin's apparatus 
 for testing the tensile strength and percentage elongation of the 
 sheathing wires used for covering and protecting the insulated 
 conductors of submarine cables. 
 
 Lord Kelvin's Wire-testing Machine, or Hydrostatic 
 Paradox. W represents the wire to be tested, which is 
 fixed to the clips 4 C r HB is a circular hydrostatic bellows, 
 
 3' diameter, with india-rubber 
 sides. WD 2 is the bottom wooden 
 disc attached by bolts to an iron 
 tripod T, which is connected at its 
 centre to the clip C 2 ; while WD X is 
 an upper wooden disc rigidly fixed 
 to the wooden framing WF. H is 
 a handle keyed to the screwed 
 spindle S. HS is a hydrostatic 
 scale, fixed behind the vertical glass 
 tube which is fitted into a short 
 brass cylinder passing through 
 WD t and into HB. ES is the scale 
 for measuring the percentage elon- 
 gation. The upper end of this scale 
 is fixed to the wire W, and the 
 lower end is free. There is a clip 
 pointer P which is affixed to 
 
 each wire before testing it, and moved up or down until it is 
 opposite to the zero of the scale ES. 
 
 Method of Testing Wire by this Machine. (i) Turn the handle 
 H backwards until C t is as far down as it can get. (2) Fix wire 
 in clips, and attach the pointer P so as to be opposite the zero of 
 scale ES. (3) Turn the handle H forward, thus lifting WD 3 , and 
 stretching the wire, by forcing water up the glass tube in front of 
 HS. This gives the necessary " head," A, or pressure due to the 
 difference in level between the free surface in the glass tube and 
 the bottom of the wooden base WB 2 . The area in square inches 
 
 THOMSON'S HYDROSTATIC 
 WiEE-TESTING MACHINE. 
 
TOTAL PRESSURE DUE TO LIQUIDS. 211 
 
 of this base gives, #, and hence the total pull on the wire is = haw. 
 (4) Note the elongation by the scale ES, and the total tensile stress 
 by the scale HS, at the moment the wire breaks. WD a falls upon 
 stops, so as not to injure the india-rubber hydrostatic bellows 
 HB. 
 
 This machine was used in 1872-73 by the Author and others 
 in testing all the sheathing wire for the Western and Brazilian 
 Company's cables. The homogeneous wire gave an average of 
 55 tons per square inch. 
 
 In this machine we see that, owing to the quaqua versus principle 
 enunciated above a few pounds weight of water can produce a 
 stress of many hundreds or even thousands of pounds by simply 
 giving it "head" through a small tube in connection with an 
 enlarged area. 
 
 When the sides of a vessel taper towards the top, as in the case 
 of a wine bottle, the liquid pressing vertically upwards upon them 
 produces a reaction on the base, which makes up for the want of 
 weight of liquid which would be naturally due to direct vertical 
 pressure in the case of a cylindrical vessel. 
 
 Total Pressure on any Surface immersed in a Liquid. 
 Let a surface of any shape be immersed in a liquid of any kind 
 to any depth, as illustrated by the following figures. Then, by 
 applying the previous proofs, and a property of the "centre of 
 
 END VIEW. SIDE VIEW. 
 
 PEESSURE ON ANY SURFACE IMMERSED IN A LIQUID. 
 gravity " (which affirms that the mean perpendicular distance from 
 any plane, is equal to the distance from the e.g. of the surface to that 
 plane), we find, that the total pressure on the immersed surface is 
 represented by the following equation : 
 P = HAW.* 
 Where P = Pressure (total) in Ibs. 
 
 H = Height from e.g. to free surface in feet.* 
 ,, A = Area in square feet.* 
 W = Weight of a cubic foot of the liquid.* 
 * The student will observe that we have suddenly jumped from height* 
 
212 
 
 LECTURE XVII. 
 
 ==^^ 3ft. 
 
 EXAMPLE I. Find the total pressure on the bottom of a 
 cubical tank having a bottom 4' x 4' and filled with water to a 
 depth of 4'. 
 
 ANSWER. By the above formula 
 P = HAW. 
 P = 4 ' x (4' x 4') x 62-5 Ibs. = 4000 Ibs. = 1-8 tons. 
 
 . We may here remark that 62*3 Ibs. is the weight of a cubic foot 
 of fresh water at 65 F (see Appendix for Useful 
 Constants), whereas 62-5 Ibs. is the value at the 
 maximum density of water, or 39 F. 
 
 EXAMPLE II. A rectangular tank for 
 holding water has a vertical side whose 
 dimensions are 3 feet vertical by 4 feet 
 ^P3j_^ horizontal. An open pipe is inserted into 
 
 I , | the cover of the tank, and water is poured 
 
 JL'<> iiF in until the level in the pipe is 7 feet 
 
 ^ ' ' above the base of the tank. Find the 
 
 pressure on the vertical side and the reduc- 
 tion of pressure when the water in the 
 pipe is allowed to sink ij feet. (The 
 weight of a cubic foot of water = 62.5 Ibs.) 
 (S. and A. Exam. 1890.) 
 
 ANSWER. In the first case, 
 Height from e.g. of side to free surface = H t = 5.5'. 
 Area of this vertical side in sq. ft. = A = 3' x 4"= 12 sq. ft. 
 Weight of a cubic foot of water = W =62.5 Ibs. 
 
 By the above formula, 
 The total pressure P^H^AW. 
 
 .-.?! = 5.5' x 12x62.5 = 4125 Ibs. 
 
 In the second case, when the free surface is lowered by i J ft., 
 everything remains the same except the H, which is now reduced 
 from Hj to H 2 = 4'. 
 By the formula, 
 
 P 3 = H 2 AW. 
 
 .'. P 2 = 4X 12 x 62.5 = 3000 Ibs. 
 
 Consequently, the reduction in pressure is the difference between 
 these pressures. 
 
 Or (P t - P 2 ) - 4125 Ibs. - 3000 Ibs. -1125 Ibs. 
 
 in inches to those in feet, areas in square inches to those in square feet, 
 and weights of cubic inches to those of cubic feet. This is because the 
 usual units of measurement in hydraulics are feet, square feet, and cubic 
 feet. Before attempting the more difficult questions on page 213, he 
 ihould study a few pages of the next Lecture. 
 
LECTURE AVJLL. QUESTIONS, 2 1 3 
 
 LECTURE XVII. QUESTIONS. 
 
 1. Define the terms liquid, hydro-statics, hydro-dynamics, and hydraulics. 
 
 2. Give the chief properties of a liquid, stating wherein it differs from a 
 solid and a gas. 
 
 3. Describe and illustrate any experiment, other than the one referred to 
 in this Lecture, to prove the law of transmission of pressure by liquids. 
 State Pascal's law. 
 
 4. Describe the nature of fluid pressure. A mass of stone when in water 
 appears to be lighter than when it is situated in the open air. Will you 
 explain the cause of this fact, and state the difference of weight per cubic 
 foot of water displaced ? 
 
 5. What is meant by " head" in relation to water supplies for developing 
 power t Give an example. 
 
 6. Explain how the pressure on the base of a vessel depends solely upon 
 the area of the base and its depth from the free surface. Illustrate your 
 remarks by showing a series of connected vessels of very different shapes, 
 but with each of their bases of the same size and on the same level, and 
 filled with water to the same height. 
 
 7. Sketch and describe Sir Wm. Thomson's wire-testing machine, and 
 explain how such a great force is obtained thereby from such a small 
 quantity of water. 
 
 8. How is the pressure of water on a given area ascertained ? A tank, 
 in the form of a cubical box, whose sides are vertical, holds 4 tons of water 
 when quite full ; what is the pressure on its base, and what is the pressure 
 on one of its sides ? Ans. 4 tons ; 2 tons. 
 
 9. A water tank is 13 feet square and 4 feet 6 inches deep ; find the pres- 
 sure upon one of the sides when the tank is full. Ans. 8226-56 Ibs. 
 
 10. State approximately the increase of pressure to which a diver would 
 be exposed when working at a depth of 50 feet below the surface of fresh 
 water. Ans. About 22 Ibs. per square inch. 
 
 11. In the vertical plane side of a tank holding water, there is a rectan- 
 gular plate whose depth is I foot and breadth 2 feet, the upper edge being 
 horizontal, and 8 feet below the surface of the water ; find the pressure on 
 the plate. Ans. 1062-5 Ibs. 
 
 12. The base of a rectangular tank for holding water is a square, 16 
 square feet in area. The sides of the tank are vertical, and it holds 250 
 gallons of water when quite full. Find the depth of the tank and th 
 pressures on each side and on the base when quite filled with water. 
 
 Am. 2-5 feet ; 781-25 Ibs. ; 2500 Ibs. 
 
 13. A rectangular tank for holding water has a vertical side whose 
 dimensions are 4 feet vertical by 5 feet horizontal. An open pipe is in- 
 serted into the cover of the tank, and water is poured in until the level in 
 the pipe is 10 feet above the base of the tank. Find the pressure on the 
 vertical side and the reduction of pressure when the water in the pipe is 
 allowed to sink 2 feet. Ans. 10,000 Ibs. ; 2500 Ibs. 
 
 14. A gauge in a water pipe indicates a pressure of water equal to 40 Ibs. 
 on the square inch. What is the depth of the point below the free 
 surface ? Sketch and explain the action of some form of gauge suitable 
 for the above purpose. Ant. 92*16 ft. 
 
FRESH 
 
 LECTURE XVIII. 
 
 CONTENTS. Useful Data regarding Fresh and Salt Water Examples 
 I. II. III. IV. Centre of Pressure Immersion of Solids Law of 
 Archimedes Floating Bodies Example V. Atmospheric Pressure 
 The Mercurial Barometer Example VI. Low Pressure and Vacuum 
 Water Gauges Example VII. The Siphon Distinction between 
 Solids, Liquids and Gases Definitions of perfect, viscous, and elastic 
 Fluids Cohesion Questions. 
 
 tTseful Data regarding Fresh, and Salt Water. We will 
 commence this Lecture by giving some useful data regarding the 
 weights, <fcc., of fresh and salt water, and then work out a few 
 more examples for the pressures on immersed surfaces, finishing 
 with the immersion of solids in fluids, &c. 
 
 Specific gravity * = i . 
 
 cubic foot weighs 62-5 Ibs., or 1000 oz. 
 
 gallon weighs 10 Ibs., or 160 oz. 
 
 ton occupies 35-84 cubic feet. 
 
 atmosphere = 1 4' 7 Ibs. per sq. in. = 29*92 in. 
 
 mercury = 33*9 (say 34) ft. head of water. 
 
 foot of head = -43 Ib. on sq. in. 
 
 Ib. on the sq. in. = 2-308 ft. head. 
 H.P. in a waterfall = cubic ft. per minute x 
 
 head x 62-5 -f- 33,000. 
 
 Specific gravity * = 1-026. 
 
 i cubic foot weighs 64 Ibs. 
 
 i gallon weighs joj Ibs. 
 
 I ton occupies 35 cubic ft., or 2iSj gallons. 
 
 EXAMPLE I. A cubical box or tank 
 with a closed lid, the length of a side of 
 which is 4 feet, rests with its base hori- 
 zontal, and an open vertical pipe enters 
 one of its sides by an elbow. The tank 
 is full of water, and the pipe contains 
 water to the height of i foot above the 
 top of the tank. What are the pres 
 sures of water on the top, bottom, and 
 sides of the tank ? (Given the weight of 
 a cubic foot of water = 62 J Ibs.) (S. and 
 A. Exam. 1887.) 
 
 * Specific gravity is the ratio of the weight of a given bulk of a sub- 
 8tar.ce, to the weight of the same bulk of pure water. ~ 
 
 SALT WATER 
 
1 
 
 EXAMPLES OF PRESSURE OX IMMERSED SURFACES. 215 
 
 ANSWER. (i) For the pressure on the top 
 
 The depth of e.g. of the top from free surface = H = i'. 
 
 . . Total pressure on top = HAW = i' x (4' x 4') x 62-5 Ibs. = 
 1000 Ibs. 
 
 (2) For the pressure on the bottom 
 
 The depth of e.g. of the bottom from the free surface = H = 5*. 
 
 .. Total pressure on bottom = HAW = 5' x (4' x 4') x 62-5 Ibs. = 
 5000 Ibs. 
 
 For the pressure on e ich of the sides 
 
 depth of e.g. of each side from the free surface = H = 3'. 
 
 . . Total pressure on each side = HAW = 3' x (4' x 4') x 62 -5 Ibs. 
 = 3000 Ibs. 
 
 EXAMPLE II. A cylindrical vessel, 30 
 inches long and 6 inches in diameter, is 
 sunk vertically in water, so that the base, 
 which is horizontal, is at a depth of 25 
 inches below the surface of the water. 
 Find the upward pressure in pounds on 
 the base of the vessel The weight of a 
 cubic foot of water is 62 1 Ibs., and TT = 
 3-1416. (S. and A. Exam. 1889.) 
 
 ANSWER. The depth of e.g. of the base from the free surface 
 is = H = -^ = 2j^-'= 2*083 feet. 
 
 7T - 2 2 , 
 
 Ti.e area of the base = A = - = ('5 x -5 ) = T90 sq. tt. 
 
 4 7x4 
 
 The weight of a cubic foot = W = 62*5 Ibs. 
 
 .. The total pressure on base = HAW = 2-083 x -196 x 62-5 = 
 25-5 Ibs. 
 
 EXAMPLE III. A water tank, 8 feet long and 8 feet wide, with 
 an inclined base, is 12 feet deep at the front and 6 feet deep at 
 the back, and is filled with water. Find the pressure in Ibs. on 
 each of the four sides and on the base ; water weighing 62 J Ibs. 
 per cubic foot. 
 
 ANSWER. In answering a question of this kind the student 
 will find it best to draw a figure representing the water tank and 
 the positions of the centres of gravity of each side and of the base 
 in the manner shown by the accompanying illustration. The only 
 point that presents any difficulty is the e.g. of the side DEFC 
 and of the correspondingly opposite one. This might be done by 
 first finding the e.g. of the D DEFH, viz., G 2 ; second, of the A 
 HFC, viz., G 3 ; third, by joining these two points with a line GG 3 , 
 
2l6 
 
 LECTUEE XVIII. 
 
 and taking a distance along it from G 2 towards G 3 inversely pro- 
 portional to the areas of the n DEFH and the A HFC ; this 
 would give a point G 6 the e.g. of the whole side = 4*6' from surface. 
 But it will evidently be easier to treat the pressures on the Q 
 and A separately, and then to add them together in order to 
 obtain the total pressure on the whole side DEFC. 
 
 / 
 
 
 
 4 
 
 &. 
 
 1 i 
 
 
 SHOWING POSITIONS OF THE CENTEBS OF GRAVITY. 
 
 G, represents centre of gravity of area ABCD 
 G 2 DEFH 
 
 G 3 HFC 
 
 G ENMF 
 
 Let Hj, H 2 , &c., represent depths of G 1? G 2 , &c. 
 Then H! = JDC = 6' ; H 2 = JEF = 3'. 
 
 G 3 is J of HC below the line HF (see Lecture III., re position of 
 e.g. of certain areas). 
 
 G 5 is at a d^pth below the surface = the mean between the edges 
 BC and FM of the base BCFM. 
 
 Total pressure on area 
 
 ABCD = H 1 A 1 W= 6' x (i 2' x 8') x 62-5 = 36,000 Ibs. 
 3 / x(6 / x 8') x 62-5 = 9,000 Ibs. 
 
CENTRE OF PRESSURE. 
 
 2I 7 
 
 H 3 A 3 W = 8'x(-, x8')x6 2 -5 = 12000 Ibs. 
 
 DEFC = DEFH + HFC - 9000 + 1 2000 = 21000 
 H 4 A 4 W = 3'x(6' x 8') x 62-5 = 9000 
 H 5 A 5 W = 9'x(io'x8')x62-5 =45000 
 
 EXAMPLE IY. A sluice gate 
 is 4 feet broad and 6 feet deep, 
 and the water rises to a height 
 of 5 feet on one side and 2 
 feet on the other side. Find 
 the pressure in pounds on the 
 gate. 
 
 ANSWER. The net pressure 
 on the sluice gate is evidently 
 equal to the difference of the 
 pressures on the two sides. 
 
 Total pressure on 
 
 NET PRESSURE ON SLUICE GATE. 
 
 Back side = H,A,W = 2*5' x (4' x 5') x 62.5 = 3125 I DS - 
 Front side = H 2 A 3 W = i' x (4' x 2') x 62.5 = 500 
 
 Subtracting the front from the back) _oftOK Ib 
 pressure we fret the net pressure / 
 
 Centre of Pressure. In the case of a plane area immersed 
 in a liquid, the " centre of pressure " is the point at which the re- 
 sultant of all the pressures of the fluid acts. If the plane be 
 horizontal, the resultant naturally acts at the centre of the figure, 
 and therefore the centre of pressure agrees with the centre of 
 gravity of the figure. In the case of a vertical rectangle, having 
 one of its edges in the surface of liquid, like a dock-gate or a 
 sluice, the centre of pressure will be at a point f of the depth from 
 the free surface and at the middle of the breadth of the immersed 
 portion. We will have to prove this in our Advanced Course, 
 and perhaps refer to the position of the centre of pressure in 
 other cases. 
 
 Immersion of Solids. Archimedes' Discovery. If a solid 
 be immersed in any fluid (whether liquid or gas), it displaces a 
 quantity of that fluid equal to its own volume. This is evident 
 from the principle of impenetrability viz., " two bodies cannot 
 occupy the same space at the same time." 
 
 Hence we have a simple method of determining the volume of 
 
2l8 LECTURE XVIII. 
 
 any irregular body by plunging it into a liquid, and noting the 
 cubic contents of the liquid displaced, by letting it run into a 
 measure of known volume, such as a graduated jar. This prin- 
 ciple was first discovered by Archimedes, a philosopher of Syra- 
 cuse, in the year 250 B.C. The story of this discovery is related 
 by Yitruvius, who states that Hero, a king, sent a certain weight 
 of gold to a goldsmith to be made into a crown. Suspecting that 
 the workman had kept back part of the gold, he weighed the 
 crown, but found that it was the same as the weight of the gold 
 previously sent by him to the goldsmith. He was, however, not 
 satisfied with this test, so he consulted Archimedes, and asked 
 him whether he could find out if the crown was adulterated. 
 Not long afterwards the philosopher, on going into his bath 
 (which happened to be full of water), observed that a quantity of 
 the water was displaced. He immediately conjectured that the 
 water which ran over must be equal to the volume of the immersed 
 part of his body. He was so overjoyed at the discovery that he 
 jumped out of the bath and ran naked to the king, exclaiming, 
 EvprjKa I fvpTjKa I (I have discovered ! I have found out !) He then 
 began to experiment with the crown by taking a quantity of pure 
 gold of the same weight, and observed its displacement in water. 
 Next he ascertained by the same process the volume of the same 
 weight of silver, and finally the volume of the crown, which 
 actually displaced more water than its equivalent weight of pure 
 gold. In this interesting manner the fraud of the artificer was 
 detected, to his great astonishment and chagrin, and a Law of 
 Nature was discovered. 
 
 Floating Bodies. A body is said to float in a fluid when it is 
 entirely supported by the fluid. In order that a body may float, 
 
 the forces acting upon it must 
 be in equilibrium. Now, as 
 may be seen from the case 
 illustrated by the accom- 
 panying figure, there are 
 only two forces to be con- 
 sidered viz., the weight of 
 the body acting vertically 
 downwards through its centre 
 of gravity G lf and the pres- 
 
 CONDITIONS OF EQUILIBRIUM IN THE 8Ure f the fluid acting VOT- 
 
 CASE OF A FLOATIN-G BODY. tically upwards through the 
 
 centre of gravity G, of the 
 
 displaced fluid. The horizontal pressures of the fluid on the 
 body are in equilibrium by themselves, and simply tend to com- 
 press it so that they do not aflect the question. The upward 
 
FLOATING BODIES. 
 
 2I 9 
 
 pressure of the fluid must be equal to the weight of the body, 
 otherwise the body would rise or sink 
 
 Also, the weight of tne body must be equal to the weight of 
 the fluid displaced. This will be evident when we remember that 
 the total upward pressure of the fluid on the surface ACB is 
 equal to the weight of the fluid which formerly filled that space. 
 But, since equilibrium still exists when the body is floating, it is 
 clear that the weight 01 the body must also be equal to the 
 weight of the fluid displaced. If the body be wholly immersed it 
 will be pressed upwards with a force equal to the weight of the 
 fluid which it displaces. Hence the statement known as the 
 
 Principle of Archimedes. When a body is wholly or partially 
 immersed in a fluid it is pressed vertically upwards by the fluid 
 with a force equal to the weight of the fluid which it displaces and 
 this force may be taken to act at the e.g. of the displaced fluid. 
 
 As a natural deduction from the above proof we eonclude that 
 a body cannot float in a liquid of less specific gravity than itself. 
 A solid glass or metal ball will float in mercury, but not in water. 
 If the specific gravity of a body be the same as that of a liquid, it 
 will float totally submerged. If the body and the liquid are each 
 incompressible, the body will float indifferently at any depth. If 
 the body be incompressible, but be placed in a compressible fluid, 
 such as air, the body will rise or fall until it finds a place where 
 its mean specific gravity is the same as that of the displaced gas. 
 This is exemplified by the case of a balloon filled with a gas lighter 
 than air. It rises until it arrives at a height from the earth where 
 the combined weight of the machine and the gas contained therein 
 are equal to the weight of the 
 same volume of air.* 
 
 EXAMPLE V. A rectangular 
 tank, 4 feet square, is filled 
 with water to a height of 3 feet. 
 A rectangular block of wood, 
 weighing 125 Ibs., and having 
 a sectional area of 4 square feet, 
 is placed in the tank, and floats 
 with its sides vertical and with 
 this section horizontal How 
 much does the water rise in the 
 tank, and what is now the 
 pressure on one vertical side 
 of the tank? (S. and A. Exam. 1892.) 
 
 ANSWEB. Let AjB x be the original surface of the water in the 
 tank before the block was immersed, A,B, the surface after 
 immersion of the block. 
 
 
 * We 
 
 Cours*. 
 
 must leave the subject of mtacentres, fee., to our Advanoad 
 
220 LECTURE XVIII. 
 
 Let Vj = volume represented by A^CD. 
 V 2 = . A 2 B 2 CD. 
 
 z> = of water displaced by block, 
 
 represented by abed. 
 x = amount by which the water rises in the tank 
 
 when the block is immersed. 
 Then clearly, V 2 = "V\ + v. 
 Or V.-V^v. 
 
 N ow V a - V x = volume represented by A^BjBjA,, 
 
 = cross sectional area of tank x x t 
 = 4 2 xa;=i6aj cub. ft. 
 
 i6x = v or x= ^- ft. 
 
 16 
 
 But, by the principle of Archimedes we know that 
 
 The, weight of water displaced by block = The weight of the block. 
 
 s. = i25 Ibs. 
 
 v = = 2 cub. ft. 
 
 x = -^ = - ft. = 1 J inches. 
 
 Next, we have to find ^e pressure on one of the vertical sides 
 of the tank. Here the depth of the centre of gravity of the area 
 of the side subjected to pressure below the free surface of the 
 water is 
 
 Total pressure on side = P = HAW 
 
 Or, ... P = 1220-7 Ibs. 
 
 Atmospheric Pressure. Surrounding the earth's surface 
 there is a deep belt of air, which gets rarer and lighter the 
 higher we rise from the earth. If we consider the case of a com- 
 plete vertical column of this air, we find that it produces an 
 average pressure on the earth's surface of about 15 Ibs. ; or, in- 
 other words, we say that the atmosphere produces an average 
 pressure of 1 5 Ibs. on the square inch, for we find that it will 
 balance a vertical column of mercury of about 30 inches, or a 
 vertical column of water of 34 feet. We do not experience any 
 inconvenience from this normal pressure of the atmosphere,. 
 because we are so constituted as to be able to resist it. Should we r 
 however, enter the closed compressed air-chamber of the under- 
 ground workings of a railway tunnel (such as those in operation- 
 
THE MERCURIAL BAROMETER. 221 
 
 at the present time for the construction of the London Tube 
 Railway), or the caissons of a great bridge while they are being 
 sunk (as in the case of the Forth Bridge), or go down into the sea 
 in a diving-dress or diving-bell, then we do feel a most uncom- 
 fortable sensation in our ears, eyes, <fcc. Or, if we climb a very 
 high mountain, or rise far into the air in a balloon, we have a some- 
 what similar sensation, but due to an opposite effect viz., a 
 diminution from the normal pressure to which we are accustomed. 
 
 The Mercurial Barometer. The pressure of the atmosphere 
 is usually measured by a mercurial barometer, which consists of 
 a vertical tube of glass about 33 inches long, of uniform calibre, 
 hermetically sealed at the top end, into which has been carefully 
 introduced mercury freed from air. The lower end dips into an 
 open dish containing a quantity of that liquid metal. Conse- 
 quently the pressure of the atmosphere acting on the mercury in 
 the open dish forces it up inside the tube to a height directly 
 proportional to its pressure, since there is supposed to be a 
 perfect vacuum between the upper surface of the mercury and 
 the closed end of the glass tube. 
 
 EXAMPLE VI. Suppose the height of mercury as registered by 
 a mercurial barometer is 30 inches, and that the specific gravity 
 of mercury be taken as 13-6, what would be the height in feet of 
 a water column which would support the same atmospheric 
 pressure ? 
 
 ANSWER. i : 13-6 : : 30 inches : x 
 
 .*. x = 30 x 13-6 = 408" = 34 feet. 
 
 Low Pressure and Vacuum Water Gauges.* It is often 
 necessary for the engineer to measure low pressures or vacuums of 
 gases. For example, in the supply of illuminating gas to a 
 town, or in the pressure of air feeding a boiler furnace by natural 
 or forced draught, or the vacuum produced by a chimney- 
 stalk; or, in the case of the vacuum in a coal mine produced 
 by a furnace below the earth, or by a guibal fan situated 
 near the upcast shaft, <fcc. In such cases, as well as in many 
 others where low pressures have to be observed, the force is not 
 reckoned by pounds per square inch, or by inches of mercury 
 sustained in a vertical column, but by the number of inches of 
 water which the pressure will support or which the vacuum will 
 detract from the atmospheric pressure. 
 
 The accompanying figure illustrates the apparatus usually 
 employed in determining such low pressures. It consists of a 
 
 * For a description of mercurial pressure and vacuum gauges, as well aa 
 Bourdon's high-pressure and vacuum gauges, refer to the Author's Elemen- 
 tary and Advanced Books on " Steam and Steam Engines." 
 
222 
 
 LECTUKE XVIII. 
 
 W 
 
 simple bent U glass tube with a scale between the vertical legs of 
 the U, divided into inches and tenths of an inch, so that either 
 the pressure or the vacuum may be read off in inches of water 
 pressure, according as the forward pressure from the point of 
 supply is positive or negative in respect to the pressure of the 
 atmosphere. For example, let the leg of 
 the U tube next the cock be connected 
 to the gas pipe of a house, then the 
 pressure of the gas supply acts on the 
 water in the right-hand leg of the tube, 
 and forces it downwards, whilst the 
 water in the other leg rises correspond- 
 ingly. The reading observed on the 
 scale S, below or above the zero or 
 equilibrium line, has of course to be 
 doubled in order to ascertain the exact 
 total pressure in inches of water. If 
 the U tube be connected to a vacuum 
 or negative pressure, then the water 
 rises in the inner leg of the U tube, 
 owing to the greater pressure of the 
 atmosphere on the outer limb, and the 
 inches of water representing the amount 
 of the vacuum are accordingly read off 
 in the same way. For example, if the 
 apparatus be connected to the base of 
 a steam boiler chimney, or to the inlet of a guibal fan creating 
 a draught in a coal mine, then the suction produced forms a 
 vacuum which requires the supply of atmospheric air, and con- 
 sequently the air presses on the open water of the outer limb of 
 the U tube, and forces it downwards. The vacuum is therefore 
 observed and recorded by adding the inches of water below and 
 above the zero line. 
 
 EXAMPLE VII. A difference of level is observed of 4 inches 
 between the outer and inner limbs of a U tube water-gauge. 
 What is the pressure of the gas supply in Ibs. per square inch ? 
 
 ANSWER. A vertical column of 34 feet of water corresponds to 
 15 Ibs. pressure on the square inch. Consequently, 
 
 (34' x 12"): 4":: 15 Iks.:* 
 
 GAS PRESSURE GAUGE. 
 
 INDEX TO PARTS. 
 
 SC represents Steam or 
 
 gas cock. 
 
 GT Glass tube. 
 S Scale. 
 
 W Water. 
 
 34 
 
 12 
 
 = > or nearly -{- of a Ib. per sq. in. 
 
 The Siphon is simply a bent tube for withdrawing liquids 
 from a higher to a lower level by aid of the atmospheric pressure. 
 It is used in chemical laboratories and works for emptying acids 
 
THE SIPHON. , 223 
 
 from carboys, in breweries and distilleries for extracting beer 
 from vats and spirits from casks, in the crinal glass tube of 
 Lord Kelvin's recorder for conveying ink from the ink-pot to 
 the telegraph message-paper ; 
 and on a large scale for draining 
 low-lying districts, such as the 
 fens of Lincolnshire. 
 
 The conditions for the success- 
 ful working of a siphon are, 
 that 
 
 1. The liquid shall be carried THE SIPHON. 
 by the outer limb of the tube to 
 
 a lower level than the surface of the supply. 
 
 2. The vertical height from the free surface of the liquid being 
 drained to the top of the bend of the siphon shall not be greater 
 than the height of the water barometer at the time say only 30 feet 
 
 on account of the necessary deduction of 3 or 4 feet to be made 
 
 from the full height of 34 feet, due to having to overcome the 
 friction of the pipe. 
 
 3. The end of the siphon dipping into the liquid to be drained, 
 shall not become uncovered. 
 
 To start the siphon, either the tube must be filled with liquid, 
 the ends closed, and the siphon inverted, with the shorter limb 
 under the fluid to be drained, before uncovering the ends ; or, 
 whilst the end of the shorter limb is in the liquid a partial 
 vacuum must be formed in the siphon tube by extracting the air 
 from the end of the longer leg. 
 
 The principle upon which the siphon acts is as follows : 
 A vacuum having been formed in the tube, the pressure of the 
 atmosphere acting on the free surface of the liquid to be drained, 
 forces it up the shorter limb, and having turned the highest 
 point of the n the liquid descends the longer limb by the action 
 of gravity with a velocity proportional to the J 'difference of levels 
 between the outlet and the free surface of the source of 
 supply. The outflowing liquid is always acting as a water-tight 
 piston at the bend of the f|, and in this way keeping up the vacuum 
 there, until either the inlet and the outlet free surfaces come to a 
 level (when the siphon stops for want of " head "), or, when the 
 difference of level between the free surface of the supply and the 
 top of the bend exceeds the height supportable by the atmosphere, 
 when it stops for want of breath or atmospheric pressure. 
 
 p 
 
224 LECTURE XVIII. 
 
 .Distinction between Solids, Liquids, and Gases. At the 
 
 very commencement of this book we referred to the fact that Matter 
 exists under three conditions. 
 
 (i) Solids ; (2) Liquids ; (3) Gases. We shall now define and distin- 
 guish concisely between the three states of matter. 
 
 (1) A Solid is matter in such a condition, that the molecules cannot 
 move freely amongst themselves, and consequently it retains its shape 
 and volume unless acted upon by a force. 
 
 (2) A Liquid is a collection of inter-mobile particles of matter, which 
 offer great resistance to change of volume, but little to change of shape. 
 
 (3) A Gas is matter in its most subdivided state, and which readily 
 yields to the slightest force tending to change its shape or its volume. 
 
 We thus see that the chief characteristic distinctions between these 
 three states of matter are, that 
 
 (1) A Solid resists both change of shape and of volume. 
 
 (2) A Liquid only resists change of volume. 
 
 (3) A Gas resists neither change of shape nor of volume. 
 
 (4) A Fluid may be either a liquid or a gas. 
 
 (5) A Viscous Fluid is a liquid which offers more or less resistance to 
 motion amongst its particles, e.g., treacle, tar, and heavy oils, &c. 
 
 (6) An Elastic Fluid is a gas whose volume will increase indefinitely. 
 
 (7) Cohesion is a property of matter common to both solids and liquids. 
 It causes more or less resistance to the separation of the molecules of 
 matter. 
 
LECTURE XVIII. QUESTIONS. 225 
 
 LECTURE XVIII. QUESTIONS. 
 
 1. What are the respective specific gravities and the weights per cubic 
 foot and per gallon of fresh and of salt water ? 
 
 2. A cylindrical vessel, 120 inches long and 10 inches in diameter, is 
 sunk vertically in water, so that the base, which is horizontal, is at a depth 
 of 100 inches below the surface of the water. Find the upward pressure 
 in pounds on the base of the vessel. Ans. 284-2 Ibs. 
 
 3. A cubical box or tank with a closed lid, the length of a side of which 
 is 5 feet, rests with its base horizontal, and an open vertical pipe enters 
 one of its sides by an elbow. The tank is full of fresh water, and the pipe 
 contains water to the height of 10 feet above the top of the tank. What 
 are the pressures of water on the top, bottom, and sides of the tank 1 
 Ans. 15,625 Ibs. ; 23,437.5 Ibs. ; 19,531-25 Ibs. 
 
 4. A water tank 10' long, 10' wide, with an inclined base 10' deep at one 
 end and 5* at the other end, is filled with fresh water. Find the pressure 
 in pounds on each of the four sides and on the base. Ans. 31,250 Ibs. ; 
 7,812-5 Ibs. ; 18,229-16 Ibs. ; 52,500 Ibs. ; 10,421 *S Ibs.* 
 
 5. A lock gate is 12 feet wide, and the water rises to a height of 8 feet 
 from the bottom of the gate. What pressure in pounds does it sustain ? 
 The weight of a cubic foot of water is 62^ Ibs. Ans. 24,000 Ibs. 
 
 6. A vertical rectangular sluice gate, measuring 2 feet horizontal by 3 feet 
 vertical, is immersed so that its upper side is 4 feet below the surface of the 
 water pressing on it. Find the pressure on tbe gate : you are required to 
 explain the reasoning on which your calculation is founded., 
 
 Ant. 2062*5 Ibs. 
 
 7. What is meant by the " centre of pressure " in the case of a plane 
 surface immersed in & liquid ? If the plane be a horizontal circle, where 
 does the centre of pressure act ? If it be a vertical rectangle 10 feet wide 
 and 6 feet deep, immersed in water so that the upper edge of the rectangle 
 Is flush with the surface of the water, where does the " centre of pressure " 
 act ? Ans. at the centre of the circle ; 4 feet below surface of water. 
 
 8. State the law discovered by Archimedes, and the conditions for a bodj 
 In equilibrium floating in a liquid. A cy Under 10 feet long and 2 feet in 
 diameter floats in fresh water, with 2 feet prelecting from the surface ; find 
 tbe weight of the cylinder. Ans. 1571 Ibs. 
 
 9. A rectangular tank, 5 feet square, is filled with water to a height of 
 7 f feet. A rectangular block of wood, weighing 312-5 Ibs., and having a sec- 
 tional area of 5 square feet, is placed in the tank, and floats with its sH*s 
 vertical and with its section horizontal. How much does the water rise 
 in the tank, and what is now the pressure on one vertical side of the tank I 
 Ans. 2-4 inches ; 9875-4.^)8. 
 
 10. The mercurial barometer registers 31"; calculate the height of columns 
 Df fresh and of salt water that will balance the corresponding pressure. 
 Ans. 35-13 ft., 34-24 ft. 
 
 n. Sketch and describe a mercurial barometer. State how it is made, 
 and how it acts as a register of the pressure of the atmosphere. 
 
 12. Describe some simple form of gauge which would enable you to 
 measure the pressure at which gas is supplied, and explain the principle on 
 which it is constructed. 
 
 * In the answers given, ^12$ is assumed to be 11*3. 
 
226 LECTURE XVIII. QUESTIONS. 
 
 13. Sketch and explain the action of the siphon, and give a few practical 
 examples of its use. Also state under what circumstances it fails to work. 
 
 14. The bottom of a water-tank measures f in length and 3' 4" in width. 
 When the tank contains 900 gallons of water, what will be the depth of 
 the water, and what would be the pressure on the bottom, on each side 
 and. end of the tank respectively ? One gallon of water weighs 10 Ibs. 
 One cubic foot weighs 62-3 Ibs. Ans. Pressure on bottom of tank = 
 9000 Ibs. Pressure on each side = 8382 Ibs. Pressure on one end of tank - 
 3991 Ibs. Depth of centre of pressure is 6*2 feet. 
 
 15. Draw the diagram of water-pressure on the side of a tank wi h 
 vertical sides, 12 feet high, and filled with water. Deduce the vertical 
 depth of the centre of pressure below the top edge of the tank. 
 
 16. Name the chief physical properties of a liquid, and show in what 
 respect a liquid differs from a gas and from a solid. How is the pressure 
 of water on the vertical sides of a tank calculated ? 
 
 A water-tank is 10' long, 10' wide, and 10' deep. When it is filled with 
 water, what will be the force with which the water acts on one side of the 
 tank 1 Ans. 31,250 Ibs. 
 
 17. Describe how you would carry out an experiment to determine the 
 discharge of water through a sharp edged circular orifice on the sid of a 
 water tank. (B. of B., 1904.) 
 
LECTURE XIX. 
 
 CONTENTS. Hydraulic Machines The^Common Suction Pump Example 
 I. The Plunger, or Single-acting Force Pump Example II. Force 
 Pump with Air Vessel Continuous-delivery Single-acting Force 
 Pump without Air Vessel Combined Plnnger and Bucket Pump 
 Double-acting Force-Pump Example III. Centrifugal Pumps 
 Example IV. Questions. 
 
 Hydraulic Machines. The Common Suction Pump consists 
 of a bored cast-iron barrel PB, terminating in a suction pipe, SP, 
 fitted with a perforated end or rose R, which dips into the well 
 from which the water is to be drawn. The object of the rose is 
 to prevent leaves or other matter getting into the pump, that 
 might clog and spoil the action of the valves. At the junction 
 between the barrel and suction pipe there is fitted a suction valve 
 SY, of the hinged clack type faced with leather. The piston or 
 bucket B is worked up and down in the barrel of the pump by 
 the force P, applied to the end of the handle H, being commu- 
 nicated to it through the connecting link of the hinged piston- 
 rod PR. In the centre and at the top of the bucket is fixed the 
 clack delivery valve DY, which is also faced with leather in order 
 to make it water-tight. The bucket is sometimes packed with 
 leather ; but, as shown by the figure, a coil of tightly woven flax 
 rope wrapped round the packing groove would be more suitable 
 in the present instance. 
 
 Action of the Suction Pump. (i) Let the barrel and the suc- 
 tion pipe be filled with air down to the water-line, and let the 
 bucket be at the end of the down stroke. Now raise the bucket 
 to the end of the up-stroke by depressing the pump handle. This 
 tends to create a vacuum below DY ; therefore the air which filled 
 the suction pipe opens SY, expands, and fills the additional volume 
 of the barrel. Consequently, according to Boyle's law, its pres 
 sure must be diminished in the inverse ratio to the enlargement 
 of its volume.* This enables the pressure of the atmosphere 
 
 * The student may refer to Lecture XII. of the Author's Elementary 
 Manual on " Steam and the Steam Engine," for an explanation and demon- 
 stration of Boyle's law ; where it is shown that if p=the pressure of a 
 gas and t>=its volume, then at a uniform temperature pv=o constant, or 
 p varies as. 
 
228 
 
 LECTURE XIX. 
 
 (which acts constantly on the surface of the water in the well) to 
 force a certain quantity of water up^the suction pipe, until the 
 weight of this column of water and the pressure of the air 
 (between it and the delivery valves) balance the pressure of the 
 outside atmosphere. 
 
 COMMON SUCTION PUMP. 
 INDEX TO PARTS. 
 
 H represents Handle. 
 
 P Push or pull at A. 
 
 F Fulcrum of H. 
 PR Plunger rod. 
 IB Pump barrel. 
 
 S Spout. 
 
 SP represents Suction pipe. 
 R Rose. 
 
 SV 
 
 B 
 
 DV 
 
 Suction valve. 
 Bucket or piston. 
 Delivery valve. 
 
 (2) In pressing the bucket to the bottom of the barrel by 
 elevating the handle, the suction valve closes and the delivery 
 valve opens, thereby permitting the compressed air in the barrel 
 to escape through the delivery valve into the atmosphere. 
 
 (3) Raise and depress the piston several times so as to produce 
 the above actions over again, and thus gradually diminish the 
 volume of the air in the pump to a minimum. Then water will 
 have been forced by the pressure of the atmosphere up the suction 
 pipe and into the pump, if the bucket and the valves are tight, 
 
SINGLE ACTING POECE PUMP. 2 29 
 
 and if the delivery valve when at the top of its stroke be not 
 more than the height of the hydro-barometric column above the 
 water line of the well.* 
 
 (4) The bucket now works in water instead of in air. In fact, 
 the machine passes from being an air-pump to be a water one. 
 During the down-stroke water is forced through the delivery 
 valve. During the up-stroke this water is ejected through the 
 spout ; at the same time more water is forced up through suction 
 pipe and valve to supply the place of the vacuum created by the 
 receding piston. The water is therefore discharged only during 
 the up-stroke in the case of the pump illustrated by the figure. 
 Should it, however, be fitted with an air-tight piston-rod and pump 
 cover, and should the pump handle be moved rapidly, more water 
 will be taken into the barrel than can escape from the spout 
 during the up-stroke. Consequently, the compression of the pent- 
 up air between the surface of the water in the barrel and the 
 cover, will cause the water to flow out in a more or less continuous 
 stream during the down-stroke. In other words, the top cover 
 and the portion of the pump above the spout may be converted 
 into an air vessel, the precise action of which will be explained 
 later on. 
 
 EXAMPLE I. If the cross area of the bucket of a suction pump 
 be 20 sq. in. and if water be raised 24 ft. from its surface in the 
 well, what is the pull on the pump rod ? 
 
 ANSWER. The pull P on the pump rod is evidently equal to 
 the weight of a column of water of height H=24 ft., and the 
 area of the bucket in sq. ft. = A= 204- 144. Therefore, by the 
 formula employed for the pressure of a liquid on a surface in 
 Lectures XVII. and XVIII. 
 
 P = 24 'x x 62-5 = 208 Ibs. 
 
 144 
 
 The Plunger, or Single-acting Force Pump. The upper 
 or outer end of the barrel of this pump is provided with a stuffing- 
 box and gland, through the air-tight packing of which the solid 
 pump plunger works. 
 
 During the up or outward stroke of the plunger a vacuum is 
 
 * Theoretically, such a pump should be able to lift water from a depth 
 of 34 feet below the highest part of the stroke of the delivery valve, but 
 practically, owing to the imperfectly air-tight fitting of the piston and the 
 valves, it is not used for withdrawing water from wells more than 20 to 25 
 feet below this position of the delivery valve. In fact, such a pump fre- 
 quently requires a bucket or two of water to be poured into it above the 
 delivery valve in order to make it work at all, if it should have been left 
 for some time without being worked. 
 
230 
 
 LECTURE XIX. 
 
 created in the pump barrel, and consequently air is expanded 
 into it from the suction pipe. This pipe is attached to the flange 
 of the suction valve-box. During the down or inward stroke 
 the suction valve closes, and the pent-up air in the barrel is forced 
 through the delivery valve. This action goes on precisely in the 
 manner just explained in the case of the suction pump, until the 
 water rises into the barrel. Then the inward stroke of the plunger 
 drives water through the delivery valve to any desired height (or 
 against any reasonable back pressure, as in the case of a feed 
 
 THE PLUNGER FORCE PUMP. 
 
 SV represents Suction valve. 
 DV Delivery valve. 
 
 INDEX TO PARTS. 
 
 PB 
 PP 
 
 Ch 
 
 Checks for valves. 
 
 SB and G 
 
 represents Pump barrel. 
 , Pump plunger. 
 , Stuffing box 
 and gland. 
 
 pump for a steam boiler) consistent with the strength of the pump 
 and the power applied. 
 
 The eye of the plunger may be attached to a connecting-rod 
 actuated by a hand lever, as in the case of the suction pump, or it 
 may be worked from one eccentric or crank revolved by a steam 
 engine or other motor. 
 
 By whichever way it is worked, the force applied to the plunger 
 must be sufficient to overcome the friction between the plunger 
 
FOECE PUMP WITH AIR VESSEL. 
 
 231 
 
 and the packing, the resistance due to sucking the water from the 
 source of supply, and of driving the same up to the place where 
 it is delivered. 
 
 With this pump (as in the case of the suction pump), the water 
 is only delivered during one out of every two strokes of the 
 plunger, and consequently, in an intermittent or pulsating fashion. 
 In order to make the supply continuous we have to use one or 
 other of the devices about to be described. 
 
 EXAMPLE II. In a single-acting plunger force pump the cross 
 area of the plunger is 10 sq. in., and its distance from the surface 
 of the water in the well, when at the end of its outv* a~d or suction 
 stroke, is 20 ft. During the inward stroke the water is pumped 
 up to a height of 100 ft. above the end of the plunger. What 
 forces are required to move the pump plunger during (i) an " out," 
 and (2) an in-stroke (neglecting the forces to overcome friction). 
 
 ANSWER. (i) P x = H X AW =20' 
 
 10 
 
 x x 62*5 = 86-8 Ibs. pull. 
 144 
 
 10 
 
 (2) P 8 = H,AW = ioc/ x x 62-5 = 434 Ibs. pressure. 
 
 Force Pump with Air Vessel. In the following figure of 
 a force pump the only points of difference worth noticing between 
 
 FOECE PUMP WITH AIR VESSEL. 
 INDEX TO PARTS. 
 
 SP represents Suction pipe. 
 
 SV 
 B 
 
 PB 
 G 
 
 Suction valve. 
 Barrel of pump. 
 Plunger barrel. 
 Packing gland. 
 Plunger rod. 
 
 DV represents Delivery valve. 
 
 S 
 AV 
 AC 
 DP 
 
 Stop for DV. 
 Air vessel. 
 Air cock. 
 Delivery pipe. 
 
232 LECTURE XIX. 
 
 it and the previous one are : (i) The plunger, instead of being 
 solid, is a hollow trunk or barrel, with the connecting rod fixed 
 to an eye-bolt at its lower end. 
 
 (2) The suction and the delivery valves are both at one side, 
 instead of being fixed on opposite sides of the pump. 
 
 (3) There is an air vessel. 
 
 Action of the Air Vessel. During the inward or delivery stroke 
 of the plunger, part of the water forced from the barrel goes up 
 the delivery pipe, and the remainder enters the air vessel, and 
 consequently compresses the air in AV. During the out- 
 ward or non-delivery stroke of the plunger the compressed 
 air in the air vessel presses the rest of the water into the delivery 
 pipe. In this simple way a continuous flow of water is main- 
 tained in the delivery pipe, and with far less shock, jar, and noise 
 than in the previous case. Where very smooth working is re- 
 quired, an air vessel is also put on to the suction side of the 
 pump. Should the air in the air vessel become entirely absorbed 
 by the water, the fact will be noticed at once, by the noise and 
 the intermittent delivery. Then the pump should be stopped, 
 the air cock AC opened, and the water run out. When the air 
 vessel is full of air, the air cock should be shut and the pump 
 started again. 
 
 Continuous-delivery Pump without Air Vessel. A 
 fairly continuous delivery may be obtained by making the plunger 
 of the piston form, and the pump rod exactly half its area, as 
 shown by the accompanying figure. During the down stroke, 
 half the water expelled by the piston from the under side of 
 the pump barrel goes up the delivery pipe, and the other half 
 is lodged above the piston, to be in turn sent up the delivery pipe 
 during the up-stroke. Where very high pressures are required, 
 such as in the filling of an accumulator ram, pumps working on 
 this principle, but of the following form, are frequently used. 
 The *>- ' ,</n is precisely the same as in the one just described, and 
 th.8 ime index letters have been used, so that the student will 
 l;ave no difficulty in understanding the figure ; more especially 
 as the directions of motion of the piston and of the ingoing and 
 outflowing water have been marked by straight and feathered 
 arrows respectively. Where sea or acid water is used it may be 
 necessary to fit the pump barrel, PB, with a brass liner, L, to 
 prevent corrosion. 
 
 In accumulator and other kinds of high-pressure work it is not 
 advisable to use air vessels, because you cannot prevent the water 
 which enters the vessel absorbing air and carrying the same with 
 it to the hydraulic machines, where its presence would be most 
 objectionable, and because with, say, 750 to 1000 or more Ibs. 
 
CONTIGUOUS DELIVERY FORCE PUMP. 
 
 ^33 
 
 CONTINUOUS-DELIVERY FORCE PUMP WITHOUT AIR VESSEL. 
 INDEX TO PARTS. 
 
 IP represents Inlet pipe. 
 
 SV Suction valve. 
 
 CC Cover and check to 
 
 SV. 
 
 P Piston. 
 PR Pump-rod. 
 
 DV represents Delivery valve. 
 CO Cover and check to 
 
 DV. 
 
 DP Discharge pipe. 
 SB Stuffing-box. 
 
 G Gland. 
 
 CONTINUOUS-DELIVERY FORCE PUMP. 
 used ID Connection with the Armstrong Accumulator. 
 (See Indexes to previous Figures.) 
 
234 LECTURE XIX. 
 
 pressure per square inch, you would require a very large and very 
 strong air vessel before it could be of any service. If a pressure 
 of only 750 Ibs. per square inch were used, then, since the normal 
 pressure of the atmosphere is 15 Ibs. per square inch, the air in 
 the air vessel would be compressed to -f^, or -pyth of its original 
 volume, in accordance with Boyle's law. Consequently, with an 
 %ir vessel of 50 cubic feet internal volume, there would be only 
 i cubic foot of air in it, when the pump was in full action. 
 
 Combined Plunger and Bucket Pump. We have already 
 seen that a suction pump discharges water during the outward 
 stroke, and that a plunger pump discharges water during the 
 inward stroke ; consequently, by combining these two kinds, we 
 get a double-acting pump. By making the cross area of the 
 plunger half that of the barrel, half the water raised by the 
 bucket during the up-stroke goes into the delivery pipe, whilst 
 the other half fills the space left by the receding plunger. During 
 the down-stroke the plunger forces the latter half up the delivery 
 pipe. We do not happen to have a figure with which to illus- 
 trate these remarks, but if the student will first of all sketch a 
 complete vertical section of a suction pump like that shown by 
 the first figure in this lecture, and then draw a solid plunger, with 
 stuffing-box and gland, like that in the second figure, in place of 
 the pump rod and open cover in the suction pump, it will form a 
 useful exercise in the designing of such a pump. 
 
 Double-acting Force Pump. The pumps which we have 
 hitherto considered are all single-acting in this sense, that they 
 do not both suck and discharge water during every stroke. This 
 can, however, be accomplished by having two sets of suction and 
 delivery valves placed at each end of the pump barrel, as shown 
 by the accompanying figure. Then, during the outward stroke of 
 the piston the pump draws water from the source of supply 
 through the inlet pipe and suction valve SVj. At the same time 
 the piston forces the water in front of it through the delivery 
 valve DV, and outlet pipe. During the inward stroke, suction 
 takes place through SV, and discharge through DV X , all as 
 clearly shown by arrows in the drawing. The valves are pro- 
 vided with india-rubber cushions, IR, to ease the shock and mini- 
 mise the jarring noise due to their reaction and natural reverbe- 
 ration when they are suddenly opened and closed. 
 
 EXAMPLE III. In a double-acting force pump the vertical 
 height from the surface of the well to the point of delivery is 
 TOO feet. If the area of the piston equal i square foot, what is 
 the stress on the piston-rod during each stroke ? 
 
 ANSWER. Here we need not distinguish between the force 
 required during suction and delivery, for both actions take place 
 
DOUBLE ACTING FORCE PUMP. 
 
 235 
 
 during each stroke. We have only to deal with the net force 
 required to elevate a column of water to a height of 100 feet. 
 
 DOUBLE ACTING FORCE PUMP.* 
 INDEX TO PARTS. 
 
 SV, SV 2 represent Suction valves. 
 DV, DV 2 Delivery valves. 
 IK India-rubber 
 
 cushions. 
 
 B represents Barrel (liner). 
 P Piston (solid). 
 PR Piston-rod. 
 
 Neglecting friction, the stress on the piston rod will therefore 
 be the weight of a column of water of height 100' and cross 
 area = i sq. ft. 
 
 .'. P- HAW= 100' x i' x 62-5 = 6250 Ibs. pull and push. 
 If 30 per cent, of the force applied be spent in overcoming 
 friction, what will then be the stress on the pump- rod. Here 
 6250 is only 70 per cent, of the whole stress, for 30 per cent, of 
 the whole is lost force. 
 
 .'. 70 : TOO : : 6250 :x 
 x = 6250 x 100 = 
 
 * We are indebted for the above figure to Professor H. Robinson's book 
 on "Hydraulic Machinery," published by Messrs. Charles Griffin & Co. 
 Students should refer to Lecture XXIV. of the Author's Elementary 
 Manual on " Steam and the Steam Engine " for detailed illustrations and 
 description of the air and circulating pumps of the SS. " St. Rognvald." 
 
LECTURE XIX. 
 
 Centrifugal Pumps. The following illustration shows one ok 
 these pumps or reversed water turbines. They are often used in 
 preference to the reciprocating pumps previously described, when 
 large quantities of water have to be quickly elevated through a 
 small height, such as emptying graving docks and holds of vessels 
 
 sc 
 
 THE BON-ACCORD CENTRIFUGAL PUMP. 
 Designed and made by Drysdale & Company, Glasgow. 
 
 INDEX TO PARTS. 
 
 SF for Supporting flanges. 
 PC Pump casing. 
 CP door. 
 SP Suction pipe. 
 V Suction water guide. 
 
 IW for Impeller wheel. 
 
 S shaft. 
 
 N nut. 
 DP Delivery pipe. 
 -> Direction of flow. 
 
 or circulating the cooling water through the condenser tubes of 
 steam-engines, as well as for dredging soft- bottomed rivers. 
 
 The original type of centrif ugafpump had straight radial blades, 
 but it has been found, that if these are curved in the direction 
 and manner shown by the accompanying figure, there is less 
 shock due to the quick flowing water and greater efficiency. They 
 may be driven by belts, direct coupled steam-engfnes, turbines, 
 or electric motors as preferred. 
 
V 
 
 BON-ACCORD CENTRIFUGAL PUMP. 237 
 
 Details. As will be seen from the vertical cross-section and 
 side view, the chamber consists of a snail-like outer pump casing 
 P C, supported upon two flanged feet S F, connected to a suction 
 pipe S P, and a delivery pipe D P. In the centre is fitted the 
 shaft S, which carries an impeller wheel I W, that rotates 
 between the tapered inside faces of the pump casing P C, and 
 a removable side cover S C. 
 
 Action when circulating cold water through a steam condenser. 
 Should this pump be situated below the level of the supply water, 
 the air is driven out of the pump and its pipes by this head of 
 water. In such a case, the pump can be started straight away 
 by its motor. But, where the pump is situated above the suction 
 supply, then the mere rotation of the impeller wheel I W does 
 not produce a sufficient vacuum to make the water rise into it ; 
 and consequently, the pump casing has either to be filled with 
 water through the nipple hole (beside the lifting eye-bolt) or a 
 steam ejector with a sluice valve are added in certain cases. 
 Supposing that the pump is fairly started, then the mere rota- 
 tion of I W inside the water-tight casing not only gives the kinetic 
 energy and pressure to the water contained therein, to force the 
 same right through the delivery pipe D P and the condenser tubes, 
 butalso tokeepup the necessary vacuum in the purnp,so as to ensure 
 a continuous feed of water through the suction pipe. It will be 
 observed, thf.t the incoming water is divided by the sharp, knife- 
 edged portion of cast iron, at the volute V, and that it flows 
 equally up each side to the centre of the wheel, whereby the 
 same is subjected to balanced side pressures. 
 
 Should the interior of the pump require to be inspected, the 
 attendant may first open the cleaning door C D, but if he finds 
 that any adjustment is required, then he can take off the side 
 cover S C. When this is removed, he will obtain a clear view 
 of the whole of I W, and he may remove the same from the 
 tapered end of the shaft S, by unscrewing the nut N. 
 
238 LECTURE XIX. 
 
 EXAMPLE IV. A centrifugal pump is to lift 6'2 cubic feet of 
 water per second to a height of 7 feet ; how much horse-power must 
 be supplied to it if its efficiency is 45 per cent ? 
 
 It is direct-driven by a continuous current electro-motor whic v 
 works at 200 volts. How many amperes of current must be supplied 
 to the motor, if its efficiency is 85 per cent ? (B. of E., 1904.) 
 
 Useful work done) TTT i , * 
 
 by pump in lifting L = Wei g ht of w ^r ln , lbs ' x dlstance m ft - throu S V 
 the water J which it is raised. 
 
 =6-2 x 62-5 x 7 = 2712-5 ft.-lbs. 
 
 But, Efficiency of \ _ Useful work done by centrifugal pump in ft.-lbs. 
 Pump, ij f J Total work done in driving the centrifugal pump. 
 45 ) 2 7 I2 '5 ft.-lbs. 
 
 100 J ~~ Total work done. 
 
 .-.Total work done j 2?I2 . 5X 
 
 in driving Cen- j- = - ft.-lbs. 
 
 trifugal Pump J 45 
 
 And this work is done per second. 
 
 .*. H.P. required J 2 7i2' r x 100 
 
 to drive the Cen- ]-= =10'96H.P. 
 
 trifugal Pump J 
 
 Also i H.P. = 746 watts, and i watt = i volt x i ampere. 
 Hence. Watts) , 
 
 givenoutby Motor) = Io '9 6 * 746- watts. 
 
 And, Efficiency of) Watts given out. 
 Motor J ~~ Watts taken in. 
 
 85 ^ 10-96 x 746 
 
 100 J ^ Watts taken in. 
 
 .*. Watts taken in) 10-96 x 746 x 100 
 by Motor /" "~8j~ 
 
 But, Watts taken) n -n -rr ^ 
 
 in by Motor ) = Currenfc m Amperes x Pressure in Yolte. 
 
 .% Current in am-) _ Watts taken in by Motor, 
 peres J ~~ Pressure in Yolts, 
 
 }_ io - 96 x 746 x 100 
 85 x 200 
 
 4088 
 
 -5 =48-1 Amperes. 
 
LECTUEE XIX. QUESTION*. 239 
 
 LECTURE XIX. QUESTIONS. 
 
 1. Explain the manner in which the pressure of the atmosphere Is mada 
 serviceable in the case of the common suction pump. Sketch and explaip 
 by an index the details of this pump. 
 
 2. Describe, with a sketch, an ordinary suction or lifting pump, and ex 
 plain its action. If the diameter of the bucket is 4", and the spout is 20? 
 above the free surface of the well, what is the tension on the pump-rod in 
 the tsp-stroke ? Arts. 109 Ibs. 
 
 3. Sketch and describe a force pump, drawing a section so as to show 
 the packing of the plunger and the construction of the valves. How is an 
 air-vessel applied to such a pump ? Why is the air-vessel dispensed with 
 when pumping water into an accumulator ? 
 
 4. Explain the use of an air-vessel in connection with a force pump. 
 Sketch a section through a double-acting force pump, showing the valves 
 and the connection of the pump with the air-vessel, and explain the action 
 of the pump. Water is forced p to 100 feet above the air-vessel ; what 
 proportion of the volume of the air-vessel is occupied with water, and 
 what is the pressure of the air therein? Ans. 74*6 per cent. ; 43*35 Ibs. 
 per sq. in. above the atmospheric pressure. 
 
 5. The leverage to the end of the handle of a common force pump is five 
 times that to the plunger, ana the area of the plunger is 5 square inches ; 
 what pressure at the end of the lever handle will produce a pressure of 
 45 Ibs. per square inch on the water within the barrel 1 Ans. 45 Ibs. 
 
 6. A force pump is used to raise water from a well to a tank. The 
 piston has a diameter of r6", and is 20' above the free surface of the water 
 in the well, and 40' below the mouth of the delivery pipe leading into the 
 tank. Find the force required to work the pump (i) Neglecting friction; 
 (2) when 30% is spent in overcoming friction ; (a) when sucking, (6) when 
 forcing, (c) what is the work put in and got out per double stroke of 6" t 
 Ans. (a) (i) 17-45 Ibs- 5 (2) 24-93 Ibs. ; (&) (i) 34'9 Ibs. ; (2) 49-86 Ibs. ; (c) 
 37-39 ft.-lbs. ; 26-17 ft. -Ibs. 
 
 7. What is the difference between a double-acting and a single-acting 
 pump ? The area of the plunger of a force pump being 3 square inches, 
 find the pressure upon it when water is forced up to a height of 2of. 
 Ans. 26-04 Ibs. 
 
 8. Describe, with a sketch, some form of pump which will deliver half 
 the contents of the barrel at each respective up-stroke and down-stroke of 
 the pump-rod. Name the valves. 
 
 9. Sketch and describe a "double-acting force pump." If the diameter 
 of the piston be 12", the stroke 3', the distance from pump to well 2d, 
 from pump to position for delivering the water 40', and if the number of 
 strokes per minute be 40, what is (i) the theoretical horse-power required 
 to work the pump, (2) the actual, if 30 per cent, of the power be spent 
 against friction. Ans. (i) 1071 ; (2) 15-3. 
 
 10. What is the difference between a single and a double acting pressure 
 pump ? Sketch in section a double-acting force pump for working at high 
 pressure, showing the arrangement of valves, and indicate of what material 
 the several parts should be constructed. 
 
 11. Sketch and describe the construction and action of some form of 
 pump by which you could raise water from a well where the level of the 
 
 Q 
 
240 LECTURE XIX. QUESTIONS. 
 
 water is 45 feet below the surface of the ground. Explain fully where you 
 would fix the pump, and give reasons for the arrangement which you 
 propose to adopt. 
 
 12. Sketch in section and describe the action of the ordinary lifting 
 pump. In such a pump the pump -rod is f inch in diameter, and the pump 
 barrel is 5 inches in diameter, while the spout at which the water is 
 delivered is 20 feet above the surface of the pump bucket when the latter 
 is at its lowest point ; what", would be the maximum tension on the pump 
 rod in the upstroke of the pump, neglecting the weight of the pump rod 
 and the pump bucket, also the weight of water below bucket in suction 
 pipe (the weight of a cubic foot of water is 62*5 Ibs.) ? -4ns. 166*6 Ibs. 
 
 13. Describe the construction and action of an ordinary suction-pump 
 for raising water from a well. If 200 gallons of water are raised per hour 
 from a depth of 20 feet, and if the efficiency of the pump is 60 per cenk, 
 wfcat horse- power is being given to the pump ? Ans. -034. 
 
LECTURE XX 
 
 CONTENTS. Bramah's Hydraulic Press Bramah's Leather Collar Pack 
 ing Examples I. II. Large Hydraulic Press for Flanging Boiler 
 Plates The Hydraulic Jack Weem's Compound Screw and Hydraulic 
 Jacfc Example III. The Hydraulic Bear or Portable Punching 
 Machine The Hydraulic Accumulator Example IV. Questions. 
 
 Bramah's Hydraulic Press. This useful machine was in- 
 vented by Pascal, but he could not make the moving parts water- 
 tight. Bramah, about the year 1 796, discovered a means by which 
 this difficulty was effectually overcome ; and thus the instrument 
 has been handed down to us under his name. As may be seen 
 from the following figure, it consists of a single-acting force pump 
 in connection with a strong cylinder containing a plunger or ram, 
 which is forced outwards from the cylinder through a tight collar 
 by the pressure of the water delivered into the cylinder from the 
 force pump. 
 
 From what was said in Lecture XIX. about force pumps, we 
 need not particularise about this part of the machine, except to 
 gay that the suction and delivery valve boxes can be disconnected 
 from the pump, and the valve cover-checks removed at any time 
 for the purpose of examining the parts, or of legrinding the 
 valves into their seats. The plunger extends through a stuffing- 
 box and gland filled with hemp packing, and is guided by a cen- 
 trally bored bracket bolted to the top flange of the pump. The 
 lever fits through a slot in this guide-bar, whereby it has an easy 
 free motion, when communicating the force applied through it to 
 the pump plunger. The relief- valve RV has a loaded lever, ad- 
 justed like the lever safety valve in Lecture IV., so as to rise and 
 let the water escape when the pressure exceeds a certain amount. 
 It may also be used for taking the pressure of the object under 
 compression, or for lowering the ram R by simply lifting the 
 little lever and pressing down the table T, when the water flows 
 easily from the cylinder, and out of DP by the relief valve. The 
 delivery pipe DP is made of solid drawn brass, and the ram 
 cylinder is carefully rounded at the bottom end, instead of being 
 flat, in order that it may be naturally of the strongest shape.* 
 
 * Lo the case of largp cylinders for very great pressures, the lowtr oz 
 
LECTUEE XX. 
 
 2 4 2 
 
 The guide pillars are securely bolted to the base B by nuts and 
 iron washrs, not shown. The cup leather packing CL deserves 
 special attention, because it formed the chief improvement by 
 
 VERTICAL SECTION OF A BRAMAH HYDRAULIC PRESS, 
 Made in the Engineering Workshop of The Glasgow Technical College. 
 
 INDEX TO PARTS. 
 
 L represents Lever. 
 
 DV represents Delivery valve. 
 
 p 
 
 Pressure on L at A. 
 
 EV 
 
 , Eelief valve. 
 
 F 
 
 Fulcrum of L. 
 
 DP 
 
 , Delivery pipe. 
 
 B 
 
 L's connection with 
 
 EC 
 
 , Earn cylinder. 
 
 
 plunger's guide- 
 
 E 
 
 , Earn or plunger. 
 
 
 rod. 
 
 CL 
 
 , Cup leather packing. 
 
 PP 
 
 ,, Pump plunger. 
 
 T 
 
 , Top, table, or T piece. 
 
 Q 
 
 ,, Keaction or stress 
 
 W 
 
 , Weight lifted, or 
 
 
 on plunger PP. 
 
 
 total pressure on E. 
 
 PB 
 
 ,, Pump barrel. 
 
 CO 
 
 , Cross girder. 
 
 IP 
 
 ,, Inlet pipe. 
 
 GP 
 
 , Guide pillars. 
 
 SV 
 
 ,, Suction valve. 
 
 BB 
 
 , Base block. 
 
 inner end of the cylinder should be carefully rounded off, both inside and 
 outside. For, if left square, or nearly square, the crystals formed in 
 the casting of the metal naturally arrange themselves whilst cooling in 
 such a manner as to leave sn initial stress, and consequent weakness, invit- 
 ing fracture along the lines joining the inside to the outside corners 
 of the cylinder end. The severe shocks and stresses to which this weak 
 line of division is subjected during the working of the press would sooner 
 or later force out the end of the cylinder, in the shape of the frustum of 
 a cone, unless the cylinder had been made unnecessarily thick and strong 
 at the bottom end. 
 
BRAMAH'S LEATHER COLLAR PACKING. 243 
 
 Bramah on Pascal's press. It consists of a leather collar of fl 
 section, placed into a cavity turned out of the neck of the cylinder, 
 and kept there by the gland of the cylinder cover. The following 
 figure shows an enlarged section of Bramah's packing suitable for 
 a huge press, where the desired shape of the leather collar LC 
 is maintained by an internal brass ring, BR, and an outside metal 
 guard ring GE., resting on a bedding of hemp H. It will be 
 observed at once, from an inspection of this figure, that the water 
 which leaks past the easy fit between the plunger or ram R, and 
 the cylinder C, presses one of the sharp edges of the leather collar 
 against the ram, and the other edge against the side of the bored 
 cavity in the neck of the cylinder, with a force directly propor- 
 tional to the pressure of the water in the cylinder. By this simple 
 automatic action, the greater the pressure in the cylinder the 
 tighter does the leather collar grip the ram and bear on the 
 cylinder's neck. 
 
 Bramah's Leather Collar Packing. This collar is made 
 from a flat piece of new strong well-tanned leather, thoroughly 
 soaked in water, and forced into a metal mould of the requisite 
 
 ENLARGED VIEW OP BRAMAH'S LEATHER COLLAR FOB A 
 BIG HYDRAULIC PRESS. 
 
 INDEX TO PARTS. 
 
 B represents Ram. 
 C Cylinder. 
 G Gland of C. 
 LC Leather collar. 
 
 BR represents Brass ring. 
 GR Guard ring. 
 H Hemp bedding. 
 
 size and shape until it has assumed the form of a \J collar. The 
 central or disc portion of the leather is then cut out, and the cir- 
 cular edges are trimmed up sharp in the bevelled manner shown 
 by the above figure. 
 
U5CTTJRE XX. 
 
 Formula for the Pressure on the Ram of a Bramah Press. 
 
 Referring again to the first figure in this Lecture, it will be 
 found that by taking moments about the fulcrum at F, we obtain 
 the pressure or reaction Q on the plunger of the force pump 
 Therefore, neglecting weight of lever and friction, we get 
 
 PxAF-QxBF. .-. Q-^ 
 
 Further, by Pascal's law for the transmission of pressure by liquids, 
 enunciated in Lecture XVII., we know that the statical pressure 
 Q is transmitted with undiminished force to every corresponding 
 area of the cross section of the ram. 
 
 Or, , . Q : W : : area of plunger : area of ram. 
 
 .*. W x area of plunger = Q x area of ram. 
 Wx7rr a = Qx7rR a 
 
 Where r = radius of plunger, and R = radius of ram, both in the same unit^ 
 Substituting the previous value for Q, and dividing each side of the equa- 
 tion by TT, we get- 
 
 Since the radius of a circle is directly proportional to its diameter, we 
 may write the formula thus, where D is the diameter of the ram and d the 
 diameter of the plunger, both in the same unit 
 
 PxAF I? 
 
 EXAMPLE I. In a small Bramah press, P= 50 Ibs., AF= 20 in., 
 BF= 2 in., area of plunger = i sq. in., whilst area of rani = 14 sq. 
 in. Find W, neglecting friction and weight of lever. 
 
 ANSWER. By the above formula 
 
 w _PxAF 7rR a 
 
 BF X ^r7* 
 Substituting co x 20 IA 
 
 values, we get W = ^ x = 7000 Ibs. 
 
 2 T 
 
 EXAMPLE II. In Bramah's original press at South Kensington 
 the plunger is 3" in diameter, and it acts at a distance of 
 6 inches from the fulcrum, which is at one end of a lever 10 feet 
 3 inches long, carrying a loaded scale-pan at the other end. 
 What should be the pressure of the water in the press in order to 
 lift a weight of 3 cwt. in the scale-pan, neglecting the weight of 
 the lever ? Make a diagram of the arrangement. (S, and A 
 Exam. 1892.) 
 
HYDRAULIC PRESS. 245 
 
 ANSWER. Here d = 3 in., consequently the area of the plunger 
 / 'x" = s. in. BF = 6"; AF=io' 3" = i2"; 
 
 P = 3 cwt. = 3x112 = 336 Ibs. ; and we have to find the pressure 
 per sq. in. on the ram that will balance P, acting with the stated 
 advantage, since the area of the ram is not given. By the formula 
 
 w _PxAF x areaof 1 sq. in._ 33 6x 123" x I sq. i".- 9811bg 
 BF area of plunger 6" 7 sq. in. 
 
 .'. Pressure per sq. inch on ram of press = 984 Ibs. 
 
 Large Hydraulic Press for Flanging Boiler Plates, &c. 
 As an example of the practical application of the Bramah press 
 to modern boiler-making, the accompanying illustration shows 
 the form which it takes when worked by a high -pressure water 
 supply derived from a central accumulator, which may at the same 
 time be used to work cranes, punching, riveting, and other similar 
 machine tools, in the same works. 
 
 The operation of flanging, say the end tube-plates of the cylin- 
 drical barrel of a locomotive boiler, is carried out in the following 
 manner : The ram B, is lowered to near the bottom of the 
 hydraulic cylinder HC, thus leaving room to place the boiler 
 plate (which has been heated all round the outside edge) on the 
 movable table T r High-pressure water is then admitted from 
 the central accumulator to the auxiliary cylinders ACj thus forc- 
 ing the side rams SB, SB,, with their table T r and the plate P, 
 vertically upwards, until the upper surface of the plate bears hard 
 against the bearers B, B, or internal part of the dies. Water 
 from the same source is now admitted into the hydraulic cylinder 
 HC, which forces up the ram B-^ with its table T lf supporting 
 columns SO, SO, and the external part of the dies D, D, until 
 the latter has quietly and smoothly bent the hot edge of the 
 plate round the curved corner of the internal bearer B, B. The 
 ram B, is now lowered, carrying with it the table T x and dies 
 D, by letting out water from HC, and then the table T,, with 
 the flanged plate, are lowered by letting out water from AC. The 
 plate is removed from its table, allowed to cool, placed in position 
 in the barrel of the boiler, marked off' for the rivet holes, drilled 
 and riveted in the usual manner. The student will now under- 
 stand what a useful and powerful servant a hydraulic press is to 
 the engineer in the hands of a skilful workman, for it can be 
 made to do work in the manner indicated above in far less time, 
 and with far greater certainty of uniformity and exactitude, than 
 the boiler-smith could turn out, with any number of hammermen 
 to help him. It is fast replacing, the steam-hammer for crossing. 
 work, and the steam or belt-driven punching and riveting 
 
246 
 
 LECTURE XX. 
 
 LARGE HYDRAULIC PRESS FOR FLANGING BOILER PLATES.* 
 
 * The above figure is a reduced copy of one from Prof. Henry Kobinson's 
 book on " Hydraulic Machinery," published by Messrs. Charles Griffin & 
 Co., bnt it has been indexed according to the Author's style of symbols, 
 and described in an elementary manner. 
 
THE HYDRAULIC JACK. 247 
 
 INDEX TO PABTS. 
 
 HO represents Hydraulic cylinder. 
 
 R Ram of HC. 
 
 0, Columns supporting Y. 
 
 Y Yoke or cross- head. 
 
 BB Bearers of the internal die ring. 
 
 P Plate to be flanged. 
 
 DD 
 80 
 
 8R 
 AO 
 
 Dished die or external die ring. 
 Supporting columns for DD. 
 T-pu-ce or movable table for DD. 
 T-piece or movable table for P. 
 Side rams for T 
 
 e rams or r 
 Auxiliary cylind 
 
 ers. 
 
 machines, the steam screw and wheel-gear worked cranes, screw 
 and wheel-gear hoists, as well as the screw press for making up 
 bales of goods mencioned in Lecture XV. For with it, you can 
 bring to bear a force of a few pounds on the square inch or as 
 many tons, by merely turning the handle of a small cock, and 
 with a certainty of action unattainable by any other means. 
 
 The Hydraulic Jack is a combined hydraulic press and force 
 pump, arranged in such a compact form as to be readily portable, 
 and applied to lifting heavy weights through short distances. 
 It therefore effects the same objects as the screw-jack described 
 in Lecture XV., but with less manual effort or greater mechanical 
 advantage. 
 
 The base on which the jack rests is continued upwards in the 
 form of a cylindrical plunger, so as to constitute the ram of 
 the hydraulic cylinder HC. Along one side of this ram there is 
 cut a grooved parallel guide slot GS, into which fits a steel set 
 pin, screwed through the centre of a nipple cast on the side of the 
 cylinder (not shown in the drawings) for the purpose of guiding 
 the latter up and down without allowing it to turn round. The 
 top of the ram is then bolted with a water-tight cup leather CL, 
 by means of a large washer and screw-bolt. 
 
 The action of this cup leather is precisely the same as the 
 leather collar in the cylinder of the Bramah press already de- 
 scribed ; but it has only to be pressed by the water in one direction 
 viz., against the sides of the truly-bored cast-steel cylinder, 
 instead of against both the ram and the cylinder neck, as in the 
 previous case. The head 11 and upper portion of the machine is 
 of square section, and is screwed on to the hydraulic cylinder in 
 the manner shown by the figure. It contains a water reservoir 
 WR, which may be filled or emptied through a small hole by 
 taking out the screw-plug SP.* In the centre line of the head- 
 
 This screw plug SP is slackened back a little to let the air in or out 
 
248 
 
 LECTURE XX. 
 
 THE HYDRAULIC JACK. 
 
THE HYDRAULIC JACK. 
 
 249 
 
 piece there is placed a small force pump, the lower end of which 
 is screwed into the centre of the upper end of the hydraulic cylin- 
 der. This pump is worked by the up-and-down movement of a 
 handle placed on the squared outstanding end of the turned crank 
 shaft CS. To the centre of the crank shaft there is fixed a crank 
 0, which gears with a slot in the force-pump plunger P, and thus 
 the motion of the handle is communicated to the pump plunger 
 in a reduced amount, corresponding to the inverse ratio of the 
 lengths of the handle and the crank from the fulcrum or 
 centre of the crank shaft. By comparing the right-hand section 
 of the water reservoir, and the section on the line AB, with the 
 vertical left-hand section of the jack, it will be seen where the 
 inlet and delivery valves IV and DV are situated. On raising 
 the pump plunger P, water is drawn from WE, into the lower 
 end of the pump barrel through IV, and on depressing the 
 plunger this water is forced through the delivery valve DV into 
 the hydraulic cylinder, thus causing a pressure between the upper 
 ends of the cylinder and the ram, and thereby forcing the cylinder, 
 with its grooved head H, and footstep S, upwards, and elevating 
 whatever load may have been placed thereon. Both the inlet and 
 outlet valves>are of the kind known as " mitre valves." They 
 have a chamfer cut on one or more parts of their turned spindles, 
 so as to let the water in and out along these channels. The yalves 
 are assisted in their closing action by small spiral springs SS, 
 bearing in small cups or hollow centres, as shown more clearly in 
 the case of DV by the enlarged section on AB. 
 
 Weems' Compound Screw and Hydraulic Jack. This is a 
 jack combining some of the advantages of the ordinary screw-jack 
 with those of the hydraulic one. It is often desirable to be able 
 to bring the head or footstep into trial contact with the load 
 before applying the water pressure. This can easily be done by 
 turning the nut at the foot of the screw, cut on the ram of the 
 jack. The arrangement will at once be understood from the 
 figure. It will be observed that the load may also be lowered by 
 turning this nut, or by the screw-tap which permits water to flow 
 from the cylinder back into the cistern, as in the previous case. 
 The bottom nut may be screwed hard up to the foot of the 
 hydraulic cylinder, so as to sustain the whole load, and thus 
 prevent overhauling through leakage of the water. 
 
 When it is necessary to lower the load or the head of the jack, 
 
 of the top of the water reservoir when working the jack. There is gene- 
 rally another and separate screw plug opening (as will be seen by the 
 following figure of Weems' patent jack) for filling or emptying the water 
 reservoir, quite independent of the above-mentioned one, which is used in 
 tfci caje lor both purposes. 
 
250 
 
 LECTURE XX. 
 
 the relief valve or lowering screw, is 
 turned so as to permit the water to 
 escape from the hydraulic cylinder 
 
 back into the water reservoir, as 
 clearly shown by the drawing. 
 This may be done very gently by 
 simply giving this screw a very 
 small part of a complete turn ; 
 in other words, by throttling the 
 passage between the hydraulic 
 cylinder and the water reservoir. 
 Or it may be done quickly by 
 turning it through one or more 
 revolutions. This passage can 
 then be closed by screwing the 
 plug home on its seat. 
 
 Mr. Croydon Marks, in his 
 book on " Hydraulic Machinery," 
 illustrates and describes another 
 method of lowering the jack-head 
 (first introduced by Mr. Butters, 
 of the Royal Arsenal, Woolwich), 
 where, by a particular arrange- 
 ment, the inlet and delivery valves 
 are acted upon by an extra de- 
 pression of the handle, and conse- 
 quent movement of the pump 
 plunger. He also gives the main 
 dimensions, with a drawing, of 
 the standard 4- ton pattern as used 
 by the British Government, where 
 the ram has a diameter D = 2", 
 the pump plunger a diameter 
 d = i" ; and the ratio of the lever- 
 age of the handle to the crank is 
 1 6 to i. Therefore from the 
 previous formula we find that, 
 
 The Theoretical Advantage = 
 
 WBEMS' COMPOUND SCREW AND 
 HYDRAULIC JACK. 
 
 WAF 
 
 _ 
 
 ~ i X i 2 
 
THE HYDEAUUC BKAB. 251 
 
 And he instances two trials by Mr. W. Anderson, the Inspector- 
 general of Ordnance Factories, to determine the efficiency of these 
 jacks, where, with a pressure on the end of the working handle of 
 76 Ibs., the theoretical load should have been 76 Ibs. x theoretical 
 ad vantage = 76x64 = 4864 Ibs., instead of which it was only 
 3738 Ibs.; 
 
 . . 4864 Ibs. : 3738 Ibs. : 100 : x 
 
 Or, . . x= = 77 % efficiency 
 
 In a second trial, a load of 1064 ^ DS - required a pressure of 
 22 Ibs. on the handle, and consequently the efficiency at this 
 lighter load, as might be expected, was less, or only 74 % . 
 
 EXAMPLE III. With a hydraulic jack of the dimensions given 
 above, and of 77 % efficiency, it is desired to lift a load of 
 4 tons ; what force must be applied to the lever handle ? 
 
 ANSWER. By the previous theoretical formula, 
 
 w _ P x AF D> 
 
 -IF" '"? 
 
 . p _ W x BF # 
 
 ~sr~ "E 5 
 
 p_4X 224 x I X ^ 
 16 2 
 
 But the efficiency of the machine is only 77%: consequently 
 140 Ibs. is 77 per cent, of the force required 
 
 .. 77 : 100 :: 140 Ibs. ixlba. 
 
 x= I40XIOO = 181-81 Ibs. 
 
 77 
 
 The Hydraulic Bear, or Portable Punching Machine. 
 This is another very useful application of the hydraulic press and 
 force pump. It is used in every iron or steel shipbuilding-yard 
 and bridge-building works. By comparing the drawing with the 
 index to parts, and taking into consideration the fact that its 
 construction and action are so very similar to the hydraulic jack 
 already described in full detail, we need say nothing more than 
 direct the student's attention to the action of the raising cam, 
 and to the means by which the apparatus is lifted and suspended. 
 In order to raise the punch for the admittance of a plate between 
 it and the die D, the relief valve B-Y must first be turned back- 
 wards, and the lever L depressed. This causes the corner of the 
 raising cam RC to force the hydraulic ram HB upwards, and the 
 water from the hydraulic cylinder HC back into the water 
 
LECTURE XX. 
 
 reservoir \VR. The relief valve may now be closed and the plate 
 adjusted in position. Then the pump lever can be worked up 
 and down until the punch P is forced through the plate, and the 
 punching drops through the die D and the hole in the metal 
 frame MF, on to the ground, or into a pail placed beneath to 
 receive it. 
 
 SIDE VIEW AND SECTION. END VIEW AND SECTION. 
 
 THE HYDEAULIC BEAK, OK PORTABLE PUNCHING MACHINE. 
 
 INDEX TO PARTS. 
 
 PL represents Pump lever. 
 CS Crank shaft. 
 
 C 
 
 PP 
 WR 
 IV 
 DV 
 BV 
 
 Crank. 
 
 Pump plunger. 
 Water reservoir. 
 Inlet valve. 
 Delivery valve. 
 Belief valve. 
 
 HC represents Hydraulic cylinder. 
 CL Cup leather. 
 
 HB 
 
 BC 
 
 L 
 
 P 
 
 D 
 
 MF 
 
 Hydraulic ram. 
 Baising cam. 
 Lever for BC. 
 Punch. 
 Die ring. 
 Metal frame. 
 
 The whole bear is suspended by a chain (worked by a crane or 
 other form of lifting tackle) attached to a shackle, whose bolt 
 passes through a cross hole in the back of the metal frame MF, 
 just above, but a little to the front of the centre of gravity of the 
 machine. This hole and shackle are not shown in the drawing, 
 but the student can easily understand that the hole would be 
 bored a little above where the letters RC appear on the side view, 
 
TL8 HYDRAULIC ACCUMULATOR. 253 
 
 and that the chain would pass clear of the pump Jever, since it 
 works well to the right-hand side of the bear. 
 
 The Hydraulic Accumulator. The demand for hydraulic 
 power to work elevators, cranes, swing bridges, dock gates, presses, 
 punching and riveting machines, <fcc., being of an intermittent 
 nature at one moment requiring a full water supply at the 
 maximum pressure, and at another a medium quantity, whilst in 
 many cases all the machines may be idle it is evident that if 
 an engine with pumps were devoted to supplying this demand in 
 a direct manner, the power thereof would have to be equal to the 
 greatest requirements of the plant, and would have to instantly 
 answer any and every call from the same. In the case of a low- 
 pressure supply, as for lif ts, this difficulty is best overcome by 
 placing one tank in an elevated position at the top of the hotel or 
 building where the lift is required, and another tank below the 
 level of the lowest flat. Then a small gas engine working a two- 
 or three-throw pump, or a Worthington duplex steam pump, may 
 be used to elevate the water more or less continuously from the 
 lower to the higher tank. The " head " of water in the elevated 
 tank will, if sufficient, work the lift at the required speed, and 
 the discharged water from the hydraulic cylinder will enter the 
 lower tank, to be again sent round on the same cycle of operations. 
 Should the lift be stopped for any considerable time, then a float 
 in the upper tank, connected by a rope or chain with the shifting 
 fork for the belt-driven pumps (in the case of the gas engine) 
 will force the belt over on to the loose pulley, or shut off the steam 
 from the Worthington pump. And when the water falls in the 
 upper tank, the float will cause a reverse movement of the rope 
 and shift the belt to the tight pulley, or open the steam valve, and 
 so start the pumps. When the pressures required are great, such 
 as for cranes, <kc., where 700 Ibs. on the square inch is considered 
 a very medium pressure, an elevated tank would be out of the 
 question, for it would have to be fully 1600 feet high in order to 
 exert this force and to overcome friction. Under these circum- 
 stances recourse is had to a very simple and compact arrangement 
 called an accumulator, of which a lecture diagram is herewith 
 illustrated, without any details of cocks or valves, and automatic 
 stopping and starting gear. A steam engine or other motor 
 works a continuous delivery pump, of the combined piston and 
 plunger type, without the aid of an air vessel, as illustrated by 
 the fourth and fifth figures in Lecture XIX. The water from the 
 pump enters the left-hand branch pipe leading into the foot of 
 the accumulator cylinder, and forces up the accumulator ram with 
 its cross head or top T piece, and the attached weight or dead 
 load, until the ram has reached nearly to the end of its stroke. Then 
 
254 
 
 LECTUKE XX. 
 
 the top of the T piece or a projecting bracket on the side of the 
 wrought-iron cylinder containing the dead load, engages with 
 and lifts a small weight attached to a chain passing over a pulley 
 fixed to the guide frame or to the wall of accumulator house. 
 This chain is connected directly to the throttle valve of the steam 
 
 engine supply pipe, or to the 
 belt shifting gear (if the 
 pump is driven by belt gear- 
 ing), and being provided with 
 a counter-weight, the motor 
 and pump are automatically 
 stopped by the raising of the 
 weight and the chain in the 
 accumulator house. Should 
 the water which has been 
 forced into the accumulator 
 cylinder be now used by a 
 crane or other machine, the 
 load on the ram causes it to 
 follow up and keep a constant 
 pressure per square inch on 
 the water. The starting 
 weight naturally falls as the 
 receding T piece or bracket 
 descends, thus pulling the 
 starting chain, and opening 
 the steam engine throttle 
 valve, or shifting the belt 
 from the loose to the fixed 
 pulley, and again setting the 
 pump to work. Should the 
 hydraulic machines be work- 
 ing continuously, the& the 
 pump is kept going, for the 
 
 C4/1 111 UJtti. V; V illlUCi \J. VVlWlltlllU n , -| 
 
 iron and suspended from the water from li P^ses directly 
 top of T-piece or crosshead. on to the machines, and only 
 
 the surplus water finds its 
 way into the accumulator cylinder if the pump's supply exceeds 
 the demand of the machines for water. 
 
 The annular cylinder of wrought iron is generally filled with 
 scrap iron, iron slag, or sand, or other inexpensive weighty 
 material. The accumulator cylinder AC has a stuffing- box and 
 gland at its upper end. A coil of hemp woven into a firm rect- 
 angular section and smeared with white lead is placed in the 
 bottom of the stuffing-box. The gland is screwed down on the top 
 
 THE HYDRAULIC ACCUMULATOR. 
 
 INDEX TO PARTS. 
 AC for Accumulator cylinder. 
 
 AP 
 W 
 
 Accumulator plunger or ram. 
 Weight or load contained in an 
 annular cylinder of wronght 
 
THE HYDRAULIC ACCUMULATOR. 255 
 
 of this packing until the normal pressure of the water in the 
 cylinder cannot leak past it. Cup leather packing is seldom used 
 for this simple form of accumulator ; just the ordinary packing 
 that would be used for pump rods is found to answer all re- 
 quirements. This is the simplest form of accumulator which 
 we have described, but it requires the greatest load for a certain 
 hydraulic pressure per square inch. There are several other forms 
 of accumulators, and several most interesting appliances such as 
 capstans, cranes, bridges, punching and riveting machines, &c., 
 are worked by them, which we would have liked to have described 
 here, but the limits of our space and the complexity of tbeii 
 construction necessitate our deferring this pleasure to our Ad 
 vanced Course. 
 
 EXAMPLE IY. Describe and sketch in section a hydraulic 
 accumulator, showing how the ram is kept tight in the cylinder. 
 A hydraulic press, having a ram 16 inches in diameter, is in con- 
 nection with an accumulator which has a ram 8 inches in diameter 
 and is loaded with 50 tons of ballast ; what is the total pressure 
 on the ram of the press ? (S. and A. Exam. 1892.) 
 
 ANSWER. The first part of the question is answered by the 
 previous figure and by the text. 
 
 By Pascal's Law the pressure per square inch in the accumulator 
 is equal to the pressure per square inch in the hydraulic press. 
 Consequently 
 
 Total Pressure on Press Cross Area of Press 
 
 Total Load on Accumulator Cross Area of Accumulator 
 
 qo x 16 x 16 
 '*' P= 8x8 - = 20 tona 
 
256 LECTTJRE XX. QUESTIONS. 
 
 LECTURE XX. QUESTIONS. 
 
 1. Draw a section through a hydrostatic press, showing the cylinder, 
 ram, and force pump, together with the valves. Why is the base of the 
 cylinder of a large press rounded instead of being flat as in a steam 
 cylinder ? If the diameter of the ram is 9 times that of the force pump, 
 and if Q be the pressure on the pump, what is the pressure exerted by the 
 ram, neglecting friction ? Ans. 81 Q. 
 
 2. Explain by aid of a sketch the mode of packing the ram of a hydraulic 
 press and explain how it acts. The force which actuates the force pump 
 is applied at the end of a lever giving a mechanical advantage of 14 to i, 
 and the area of the plunger of the pnmp is i square inch. What pressu/e 
 must be applied to the end of the lever to produce a pressure of i ton per 
 square inch on the water enclosed in the press ? Ans, 160 Ibs. 
 
 3. In the force pump of a press the area of the plunger is of a square 
 inch, the distance from the fulcrum of the lever handle to the plunger is 
 2 inches, and the distance from the fulcrum to the other end of the lever 
 is 2 feet ; what pressure per square inch is exerted on the water under- 
 neath the plunger, when a weight of 20 Ibs. is hung at the end of the lever 
 handle ? Ans. 720 Ibs. per square inch. 
 
 4. In what way do you estimate the theoretical advantage gained by the 
 use of the hydraulic press ? In a small press the ram is 2 inches and the 
 plunger ^ inch in diameter ; the length of the lever handle is 2 feet, and 
 the distance from the fulcrum to the plunger is i inches. Find the 
 pressure exerted on the ram when 10 Ibs. is hung at the end of the lever. 
 Ans. 2560 Ibs. 
 
 5. In an hydraulic press with two pumps the plungers are 2^ and i inch 
 in diameter, and each is worked by a similar lever, which is acted on by 
 the same force. When the larger pump alone is at work the pressure on 
 the ram is 40 tons ; what will it be when the smaller plunger is only work- 
 ing ? Ans. 250 tons. 
 
 6. An hydraulic press, which is used for making lead pipes, has a ram 
 20 inches in diameter, while the ram which presses the lead is 5 inches in 
 diameter. Find the pressure per square inch on the lead when the 
 hydraulic gauge indicates i ton per square inch. Sketch a sectional 
 elevation of the press, and show the packing of the hydraulic ram. 
 An*. 1 6 tons. 
 
 7. How is the pressure taken off the object under compression when 
 required, in a hydraulic press 1 Sketch the arrangement. What is the 
 proportion of the diameters of the plunger and ram when the theoretical 
 advantage gained thereby is TOO to i, neglecting friction ? Ans. i to 10. 
 
 8. Make a rough sketch, and write a short description of the hydraulic 
 lifting jack. It may be arranged on any system that you are acquainted 
 with. Show clearly how the valves act and how the jack is lowered. 
 
 9. Sketch and describe the hydraulic bear or portable punching machine. 
 Explain how the punch is raised and how the tool is handled. 
 
 10. Sketch and describe the construction of a vessel suitable for storing 
 up a supply of water under pressure, and intended for actuating hydraulic 
 machinery. If the plunger of this vessel be 17 inches in diameter, what 
 load will bring the pressure of the water to 700 Ibs. per square inch f 
 JLns. 1 58,950 Ibs. 
 
 11. Sketch and describe the hydraulic accumulator for storing up water 
 
LECTURE XX. QUESTIONS. 
 
 under pressure. If the ram of the accumulator be 6 inches in diameter, 
 what load will be required to produce a water pressure of 500 Ibs. on the 
 square inch ? To what head of water would this pressure correspond ! 
 Ans. 14, 142-8 Ibs. and 1152 feet. 
 
 12. A hydraulic accumulator, with a ram of 16 inches in diameter, is 
 connected with a hydraulic press whose ram is 26 inches in diameter. 
 The load on the accumulator is 80 tons ; what force would the press 
 exert ? Make a vertical section through the accumulator, showing its 
 construction. Ans. 21 1*25 tons. 
 
 13. Make a sectional sketch of a hydrostatic press suitable for giving 
 a pressure of 100 tons, showing the valves and pump and by what con- 
 trivance the leakage of water is prevented. 
 
 The pump for such a press has a cylindrical plunger I inch in diameter 
 with a lever of 10 to I, what should be the least diameter of the ram 
 which would give 100 tons pressure when a force of 56 Ibs. was applied at 
 the end of the pump lever ? What form is most suitable for the base of 
 the ram cylinder, and for what reason is a special form adopted ? 
 
 Ans. 20 inches. 
 
 14. Sketch and describe any tool used by riveters and worked by water 
 pressure. (S. E. B. 1902.) 
 
 15. The pressure of water in a hydraulic company's main is 750 Ibs. per 
 square inch, and the average flow is 25 cubic feet per minute. What 
 horse-power does this represent ? If the charge for the water is twopence 
 per 100 gallons, what is the cost per horse-power hour ? (S. E. B. 1902.) 
 Ans. 8r8; 2'3d. 
 
 1 6. Distinguish between the velocity ra1 io and the mechanical advantage 
 of a machine. In a hydraulic lifting jack the ram is 6" in diameter, the 
 pump plunger is $" diameter, the leverage for working the pump is io to I. 
 What is the velocity ratio of the machine ? Experimentally we find that 
 a force of 23 Ibs. applied at the end of the lever lifts a weight of 8500 Ibs. 
 on the and of the raui. What is the mechanical advantage of the machine ? 
 
 Ans. 470; 425. 
 
 17. A hydraulic crane is supplied with water at a pressure of 700 Ibs. 
 per square inch, and uses 2 cubic feet of water in order to lift 4 tons 
 through a height of 12 feet. How much energy has been supplied to the 
 crane ? and how much has been converted into useful work ? 
 
 Ans. 201,600 ft. -Ibs. ; 107,520 ft.-lbs. 
 
 1 8. Sketch and describe the construction and working of any hydraulic 
 accumulator with which you are acquainted. If an accumulator has a ram 
 20" diameter with a lift of 15', and the gross weight of the load lifted is 
 130 tons, what is the pressure of water per square inch and the maximum 
 energy in ft. -Ibs. stored in the accumulator, neglecting friction 7 (S. E. B. 
 1900.) Ans. (i) 927 Ibs. (2) 4,368,000 ft.-lbs. 
 
 19. A single-acting hydraulic engine has three rams, each of 3 inches 
 diameter ; common crank 3 inches long ; pressure of water above that of 
 exhaust 100 Ibs. per square inch ; 100 revolutions per minute ; no slip of 
 water. What is the horse-power? If this engine does 2-15 horse-power 
 usefully by means of a rope, what is the efficiency? (S. E. B. 1901.) 
 Ans. 3 '2 horse-power, and efficiency "67. 
 
 20. Water at a pressure of 700 Ibs. per square inch is supplied to a 
 hydraulic crane, and n cubic feet are used in lifting 15 tons through a 
 height of 1 8 feet. How much energy has been given to the crane ? How 
 much energy has been wasted? (B. of E. 1903.) 
 Ans. Energy given to crane = i, ioS,8co ft.-lbs. 
 
 Energy which is wasted = 504,000 ft.-lbs. 
 
258 LECTURE XX. QUESTIONS. 
 
 21. The ram of a hydraulic accumulator is 4 inches in diameter ; wfiat 
 is the total weight of the ram and the load upon it in Ibs. if the desired 
 water pressure in the accumulator is i^ tons per square inch, neglecting 
 friction? If, owing to the friction of the ram against the cup leathers, 
 5 per cent, of the load is wasted, what load would be necessary to produce 
 the required pressure ? (B. of E. 1905.) 
 
 Ans. (i.) 18-85 tons ;( ii.) 19-8 tons. 
 
 22. Convert a Horse-Power-Hour into foot-pounds per minute. 
 
 If water under the pressure of 700 Ibs. per sq. inch acts upon a 
 piston or ram I square foot in area through a distance of I foot, what 
 work is done ? Vv hat work is done per gallon of water ? 
 
 If the Hydraulic Company charges 1 8 pence for 1000 gallons of 
 such water, how much is this per horse-power-hour ? (B. of E. 1904.) 
 . i H.P. = 33,000 ft. -Ibs. per minute. M H. P. -hour = 33,000 x 60 
 ft. -Ibs. Work done on ram = 100,800 ft. -Ibs. Work done per gallon 
 of water= 16,128 ft. -Ibs., and cost per H.-P.-hour=2'23 pence. 
 
259 
 
 LECTURE XXI. 
 
 CONTENTS. Morton and Velocity Uniform, Variable, Linear, and Angular 
 Velocity Unit of Velocity Acceleration Unit of Acceleration 
 Acceleration due to Gravity Graphic Representation of Velocities 
 Composition and Resolution of Velocities Newton's Laws of Motion 
 Formulae for Falling Bodies Formulae for Linear Velocity with 
 Uniform Acceleration Atwood's Machine with Experiments Results 
 and Formulae Galileo's and Kater's Pendulum Experiments The 
 Path of a Projected Body Centrifugal Force due to Motion in a Circle 
 Centrifugal Force Machine Experiments I. II. III. Example I. 
 Balancing High-speed Machinery Centrifugal Stress in the Arms of 
 a Fly- wheel -Exam pie 1 1. Enetgy Potential Energy KineticEnergy 
 Accumulated Work Accumulated Work in a Rotating Body The 
 Fly-wheel Radius of Gyration Example III. The Fly Press- 
 Example IV. The Energy Stored in a Rotating Fly-wheel Motion 
 on Bicycle and Railway Curves Momentum Examples VI. to IX. 
 Questions. 
 
 Motion and Velocity. (i) Motion is the opposite of rest, foi 
 it signifies change of position. 
 
 (2) Velocity is the rate at which a body move?!, or rate of 
 motion. It is considered absolute when it is measured from some 
 fixed point, and relative if it refers to another body in motion at 
 the same time. 
 
 (3) Uniform Velocity takes place when the rate of motion does 
 not change i.e., when the body moves over equal distances in 
 equal times. 
 
 (4) Variable Velocity takes place when the rate of motion 
 changes i.e., when a body moves with either a constantly in- 
 creasing or decreasing velocity. For example, a stone pitched 
 into the air rises with a gradually decreasing velocity, but falls 
 with a gradually increasing rate of motion.. 
 
 (5) The Unit of Velocity is the velocity of a body which moves 
 through unit distance in unit time. The British unit of velocity 
 is therefore i foot in i second. In physical problems velocity is 
 generally expressed in feet per second, but for convenience the 
 engineer reckons the piston speed of engines in feet per minute, 
 and the public speak of the speed of a man walking, of a horse 
 trotting, or of a train, in miles per hour. 
 
 (6) Linear Velocity is the rate of motion in a straight line, and 
 is measured, as we have just stated, in feet per second or per 
 minute, or in miles per hour. 
 
 If v = the velocity ; I = the distance ; and t = the time. 
 
 I I 
 
 Then = 7; or I = vt ; or t = - 
 
26O LECTURE XXI 
 
 (7) Angular Velocity is the rate at which a body describes an 
 angle about a given point for example, the number of revolu- 
 tions per minute of a pulley ; but angular velocity may also be 
 measured by the feet per second or per minute which a point at 
 a known distance from the centre of motion moves. 
 
 (8) Acceleration. In the case of variable velocity, the rate of 
 change of the velocity is termed the ace deration, and may be either 
 positive or negative i.e., it may be an increasing or a decreasing 
 rate. 
 
 (9) The Unit of Acceleration is that acceleration which imparts 
 unit change of velocity to a body in unit time ; or in this country 
 it is an acceleration of i foot per second in one second. 
 
 (10) The Acceleration due to Gravity is considerably greater 
 than the above unit, and varies at different places on the earth's 
 surface. At Greenwich it is 32-2 feet per second in one second. 
 In Elementary Applied Mechanics questions we will indicate it by 
 the symbol g, and consider # = 32 feet per second in one second. 
 
 Graphic Representation of Velocities. The linear velocity 
 of a point (such as the e.g. of a body) may be represented in the 
 same way as we have hitherto represented a force. A line drawn 
 from a point with an arrow-head indicates the direction of motion, 
 and the length of the line to scale the magnitude of the velocity. 
 (See p. 3, Lecture I.) 
 
 Composition and Resolution of Velocities. Velocities 
 may be compounded and resolved in exactly the same way as we 
 treated forces by the parallelogram and triangle of forces, &c., 
 in Lecture VIII. 
 
 Newton's Laws of Motion. I. A body in motion, and not 
 acted on by any external force, will continue to move in a straight 
 line and with uniform velocity. 
 
 II. WJien a force acts upon a body in motion, the change produced 
 in the quantity of motion is the same, both in magnitude and 
 direction, as if the force acted on the body at rest. 
 
 The change in the quantity of motion is therefore proportional to 
 the force applied, and takes place in the direction of that force* 
 
 III. If two bodies mutually act upon each other, the quantities of 
 motion developed in each in the same time are equal and opposite* 
 
 Or, Action and reaction are equal and opposite. 
 
 These three laws were first stated clearly by Sir Isaac Newton 
 as the result of inductive reasoning. Having observed certain 
 facts, he set about investigating what would be the consequence 
 if his conjectures as to these facts were applied to particular 
 
 Here * quantity of motion " means " momentum," or mafas x Telocity, 
 . . Quantity of motion or momentum = Wv/g. 
 
LINEAR VELOCITY. 26 1 
 
 cases. Finding that his estimate of the probable result came 
 true, he formulated a general law in accordance with his observa- 
 tions and reasonings. 
 
 The student has already conceived the truth of the first and 
 third laws in the reasonings and applications of force to matter, 
 treated of in the previous Lectures. We will now give in as 
 brief a form as possible the formulae for falling bodies, because 
 they naturally lead on to the formulae for " centrifugal force " on 
 a rotating body, and to the " energy stored " up in a moving body, 
 both of which are of great interest and importance to the young 
 engineer. The experimental and algebraical proofs of these 
 formulae are given in most books on Theoretical Mechanics, so 
 that we might assume that the student had studied these, yet the 
 following will be refreshing. 
 
 FornrulSB for Falling Bodies. If a body falls freely in vacua 
 under the action of gravity from rest through a height h feet ; 
 then (since gravity produces a constant acceleration in the velocity 
 of the body) at the end of each successive second the velocity of 
 the body will be increased by gr, or 32 feet. Let v be the velocity 
 of the body at the end of t seconds, 
 
 Then, . . t? gt; but tf2gh 
 
 2g 
 
 * 
 
 And, . . t-2 -^- /!* 
 
 9 9 V g 
 
 "Formulae for Linear Velocity with Uniform Accelera- 
 tion. Suppose that instead of the uniform accelerating force of 
 gravity we have any other constant force of P Ibs. acting on a 
 body, and if this force moves the body through a distance of ~l feet 
 along a perfectly smooth /horizontal plane, the above formulas 
 naturally become* 
 
 Then, . . v=*at; but i? 2al 
 
 And, 
 
 * We intentionally use the letter I for length or distance, and a fbi 
 acceleration. Most writers nse the word " space " for distance and the 
 symbol s; but space is of three dimensions, and involves the idea ol 
 volume. It cannot therefore be, strictly speaking, used to represent dis- 
 tance or length, which is only of one dimension. The letter /is also often 
 used for acceleration ; but/ naturally represents a force, so we prefer to 
 use, a, for acceleration, in order to be consistent with our notation. 
 
262 LECTURE XXI. 
 
 Atwood's Machine. * General Description. This machine is 
 much used in Physical Laboratories by teachers and students, to prove 
 by experiment the previously mentioned Laws and Formulae for the rela- 
 tion between time, distance, velocity and acceleration. The results have 
 many important applications to practical mechanics which will bedealt with 
 later on in this Manual, and in the author's more advanced text-book on 
 Applied Mechanics, vol. ii. 
 
 In the latest form, (see fig. p. 263) it is provided with both vertical and 
 horizontal adjustments for the scale and plumb-bob, and may be fixed to 
 the wall of a class-room by its brackets. The graduated brass upright is 
 8 feet long, and carries on its upper end a light, accurately adjusted and 
 balanced aluminium wheel, which runs freely in agate bearings to minimise 
 the friction. The inertia of this wheel is equivalent to that of a known 
 mass situated at its circumferencef. Two sliding platforms, one of which 
 is a sliding ring; afine, strong, flexible silk thread ; cylindrical weights with 
 circular overweights of the same diameter Oi and riders r> are also pro- 
 vided. 
 
 The release of the left-hand weight is gently effected by aid of a 
 pneumatic pipe and ball, as shown at the foot of the figure. This permits 
 the other right-hand weight with its rider O to start falling freely 
 without either jerk, swing or vibration. A correction for the friction of 
 the wheel bearings may be made by putting a little fine wii'e on the 
 descending weight when two equal cylindrical weights are used, until no 
 slowing down is observable after setting these weights in motion. The 
 weight of the cord or thread passing over the pulley may be balanced by 
 a length of the same kind and size, attached* to the bottoms of the right 
 and left weights ; or, by hanging therefrom loose, vertical lengths of such 
 thread. Either of these plans enables the experimenter to raise and lower 
 the weights into position for adjustment or starting an experiment. 
 
 Method of Usiny the Machine. If the two weights attached to the ends 
 of the cord are equal, then no motion will ensue, since the downward force 
 of gravity acting upon one weight balances its downward force upon the 
 other through their tensions on the connecting thread. The balanced 
 wheel is unaffected by the action of gravity. But, if the right-hand 
 weight be made heavier than the left-hand weight, by placing a very small 
 wire weight to balance the friction of the wheel bearings, plus a rider o 
 upon it, then, it will be set in motion, solely due to the action of gravity 
 upon the rider, until the rider is caught by the ring. The ring, although 
 intercepting the rider, permits the weight with the little wire weight to 
 pass clear through it without touching. The stop-stand may be clamped 
 at any desired position on the vertical scale. 
 
 Now, referring to the two following figures, let us suppose that only a 
 rider ^ be placed on the right-hand weight whose bottom is level with the 
 
 * George Atwood w.is born in 1745, educated at Cambiidge, where he 
 became a Fellow and Tutor of Trinity College. He published a few 
 treatises ou Mechanics and Engineering, and died in 1807. The author 
 first experimented with the old Atwood's Machine in 1868-9, made by 
 Professor Copeland in 1796, which belonged to the Natural Philosophy 
 Department of Aberdeen University, and again with an identically similar 
 machine at the College of Science and Arts, Glasgow, from 1880-87. He 
 had it fitted with automatic electrical time starting and recording 
 apparatus. 
 
 t This mass may be arrived at by trial as explained under the following 
 case VI. 
 
IMPROVED AT WOOD'S MACHINE. 
 
 263 
 
 LINE DIAGRAM TO REPRESENT 
 THE DISTANCES, TIMES, AND 
 THE VELOCITIES OF THE RIGHT- 
 HAND FALLING WEIGHT. 
 
 WEIGHTS, RIDERS, AND 
 OVERWEIGHTS. 
 
 IMPROVED ATWOOD'S MACHINE. 
 By E. E. Becker of London. 
 
264 LECTURE XXI. 
 
 zero of scale and top level with A on the line diagram. Then if th 
 ring intercepts this rider at B one second after the weights have begun 
 moving, the motion during that time has been uniformly accelerated, 
 and the velocity is represented to scale by line BE.* But, when 
 the right-hand weight passes the ring, its motion will thereafter be 
 uniform until its base is arrested by the stop-stand with the velocity 
 which it had acquired when its top just passed through the ring, viz., CD 
 which is equal to BE. We may thus find the velocity gained in the first 
 second, and, by changing the mass of the weights and riders, the positions 
 of the ring and of the stop-stand, we can try the following and other ex- 
 periments, wherein the rider may be replaced by one or other of the 
 circular overweights which pass clear through the ring, and cause uniform 
 acceleration until they come to the stop-stand. 
 
 Noting the Times in Seconds. These may be done by aid of a simple 
 pendulum beating distinct seconds, or a laboratory clock or a split seconds 
 stop-watch. It is simplest and best to start the right-hand weight with 
 its lower surface at zero upon the beat of the sixtieth second of a minute ; 
 to so set the ring that it catches the rider at an exact number of seconds 
 from the start, and that the base of the right-hand weight is arrested by 
 the stop-stand S, at an exact number of seconds from the time of 
 starting. This saves any confusion and trouble, arising from noting parts 
 of a second, and ensures that the two known distances passed through by 
 the lop of the right-hand weight, viz., from A to B and from A to C, are 
 accomplished in exact whole seconds. The distance CS is simply equal to 
 the height of the right-hand weight. 
 
 I. To prove, for uniform velocity, that the velocity is equal to the dis- 
 tance passed through divided by the time ; that is, v = h/t ; or, h = vt. 
 
 II. To prove, that the distances described from rest are proportional 
 to the squares of the times ; that is, h oc &. 
 
 III. To prove, that the distances from rest during acceleration are half 
 those described in the same timet after the motion has become uniform. 
 
 IV. To prove, that the acceleration (or increase of velocity per second) 
 is equal to the velocity (at any instant) divided by the time from rest ; 
 that is, a = vjt ; or, v = at. 
 
 V. To prove, that the distance moved through in a certain time is 
 equal to half the acceleration multiplied by the square of the time ; that 
 is, A = i a 2 ; or, a = 2A/2 ; and v^ = 2ah = 2al. 
 
 VI. To prove, that when a force produces uniformly accelerated motion 
 in a body, or a system of bodies, then the acceleration is directly propor- 
 tional to the force and inversely proportional to the total mass moved ; 
 
 ra 
 
 or, a = -T, ; but F = M a = m g 
 
 * The scientific meaning of the term acceleration is an increase of 
 velocity per unit of time. And, since velocity is a rate or distance per 
 unity of time, we see that acceleration means an increase of distance 
 passed over by a body per unit of time per unit of time ; or, an increase of 
 distance per second per second. Whereas, when traffic managers of rail- 
 ways or shipping companies speak or write of the speeds of their trains or 
 ships being accelerated, they simply mean that their average velocity has 
 been increased so as to cover their normal distance between certain 
 stations or ports in less time than before. It is only during the times of 
 getting up to their full speed that the train or ship has ( + ) or positive 
 acceleration, and when slowing down that ( - ) or negative acceleration is 
 experienced. 
 
EXPERIMENTS WITH ATWOOD'S MACHINE. 265 
 
 RESULTS OF EXPERIMENTS WITH AN ATWOOD'S MACHINE. 
 
 No. of 
 Experiment. 
 
 Position of 
 Ring B from 
 zero A in ins. 
 
 Position of 
 stop-stands 
 from zero A 
 in inches. 
 
 Time in seconds 
 to ring B from 
 
 rest. 
 
 Time in seconds 
 to stop-stand 8 
 from rest. 
 
 I 
 
 If 
 
 8| 
 
 I 
 
 3 
 
 2 
 
 7 
 
 21 
 
 2 
 
 4 
 
 3 
 
 I5l 
 
 47i 
 
 3 
 
 6 
 
 J 
 
 CASE I. When accelerating force is removed then the velocity become* 
 uniform and is equal to the distance passed through the time ; that w, 
 
 v = ~ ; or, h = v t 
 
 Exp. (i) Distance from ring to stand = 8f - if = 7 inches. 
 Time taken to move =31=2 seconds. 
 
 .. j = l = = 3 "5" = velocity acquired in 1 second. 
 
 Exp. (2) Distance from ricg to stand = 21 7 = 14 inches. 
 Time taken to move =4 2=2 seconds. 
 
 .. Vt =-1 = = 7" = velocity acquired in 2 seconds. 
 
 Exp. (3) Distance fron* ring to stand = 47 - isf = 3ii inches* 
 Time taken to move , =6 3=3 seconds. 
 
 .. v 3 = = s^= 10 "5" = velocity acquired in 3 seconds. 
 t s 6 
 
 CASE II. The distances described from rest are proportional to the square* 
 of the times. 
 
 Exp. (i) Distance in ist second = if inches = = f x i, or : 
 Exp (2) 2nd =7 = 7 = | x 4, : 
 
 Erp.(3) 3rd =i 5 f =*^= x 9, 
 
 32 
 
 CASE III. The distances from rest during acceleration are half those 
 described after the motion becomes uniform, in the same time. 
 
 Exp. (i) Distance passed through from rest in i second = (If x 1) inches. 
 Distance after motion becomes uniform in (31 
 
 = 2 seconds) is 8f if =7 
 
 . * . Distance in I second after motion becomes uniform 
 
 = 3^ inches . = (If x 2) 
 
 Exp. (2) Distance passed through from rest in 2 seconds =(7xl) 
 Distance after motion becomes uniform in (4 2 
 
 = 2 seconds) is 21-7 ... . . = 14 
 
 Distance in 2nd second after motion becomes uniform =(7x2) t, 
 CASE IV. Uniform acceleration is equal to velocity -r time ; or, a = vjt. 
 From the detailed working out of Case I. we see, that the change of 
 Telocity = (v x - v e ) = (t^-t^) = (t>,- a ) = 3-5" per MO. 
 
266 LECTURE XXI. 
 
 Hence, since the change of velocity in 3 seconds is 10-5" ; we get, 
 - = I0 5 = 3-5 = a the acceleration. 
 
 CASE V. Distance moved through = \ acceleration X time 2. 
 Or, h = i a *2. 
 
 Exp. (3) Distance h moved through in time (t) 3 seconds = 15!". 
 And, from Case IV. the acceleration a = 3-5" per sec. per sec. 
 
 Hence, by substituting known values in the equation 
 w A = a <2. 
 
 <e get, A = ^ x 3>5 x 32 = I5 | inches> 
 
 Now, from Case IV. v = a t, . . v* = a2 2 ; or, f2 = v 2/a2, 
 
 But from Case V. h = a 2 ................... . 2 2 A/a. 
 
 Hence, 2A/a = t>2/ a 2. 
 
 Or, 2 a A = 1>2. 
 
 Or generally, t>2 = 2 a Z where Z is the distance passed through by a body 
 having an acceleration a. 
 
 Corollary to Case V. If a body has a certain initial velocity u before it 
 becomes uniformly accelerated, then the final velocity v, will be the sum 
 of the initial and accelerated velocities. 
 
 From the previous definition of uniform acceleration, viz., that it is 
 the increase of velocity per unit of time from the commencement of the 
 acceleration, we see that 
 
 u + a 
 
 . . 
 
 The average or mean velocity will be half the initial and final velocities, 
 
 Or i[u + (u + a )] = u + \ a t. 
 
 And, the distance I moved through or height h through which the body 
 passed will be tlds mean velocity multiplied by the time of its motion from 
 the instant that the noted acceleration a began. 
 
 That is, I = h = (average velocity) x (time of motion) j 
 Or, h = (u + % a t) x t. 
 .'. h = ut + % a t%. 
 
 The above refers to positive acceleration or getting up speed : but the 
 same applies to negative acceleration or retardation of a body's initial 
 velocity if we apply the sign instead of the + sign. 
 
 Here, I = ut - \ a &, 
 Hence generally, / = ut + % a I 2 , 
 
 And, if M = o, or the body's motion is started from rest, we get as in 
 Case IV. 
 
 To prove the above by the Atwood's machine : 
 
 First, place an overweight and then a rider upon the right-hand weight; 
 _jid bring lower surface of the latter opposite the zero mark of the scale. 
 
 Second, set the ring at say 15!" below zero A, so that the rider is caught 
 at B, the end of the 3rd second and calculate the velocity with which the 
 whole moving mass is going at the instant the rider is caught by the ring ; 
 let this velocity = u inches per second. 
 
 Third, note the time t in seconds and the acceleration a per second per 
 second which the overweight O now causes the remainder to possess in 
 passing through any convenient height h inches between the upper edge of 
 the ring, and the upper side of the right-hand weight when its bottom is 
 arrested by the stop-stand S. 
 
 Fourth, check the average of an odd number of trials by substituting 
 the values for h, u, t and a in the formula 
 h = ut + i a t*. 
 
EXPERIMENTS WITH ATWOOD^S alA CHINE. 267 
 
 E VI. When a constant force produces uniform acceleration of a 
 body or system of bodies, the acceleration is proportional to the force and 
 inversely proportional to the total mass being moved by the force. 
 In the previous Experiments with the Atwood's machine : 
 
 If m =- Mass of the rider = ^. 
 
 g 
 
 ,, w = Weight of the rider = | oz. or '5 oz. 
 F = Force produced by gravity on rider = mg = *5 oz. 
 ,, M = Total mass moved = (right + left weight + thread -4- 
 equivalent mass active at circumference of wheel to 
 overcome its inertia + wire weight to balance friction 
 
 W -f to 
 of wheel bearings + mass of rider) = - 
 
 W = Weight due to whole moving mass M (minus mass m of 
 
 weight %) = 3 Ib. 6| oz. = 5475 oz. 
 
 Then W + w = Total weight in motion up f o position of ring. 
 And a = Acceleration produced on Mass M by gravity acting on 
 
 m = 3 '5 inches or (3*5 -f 12 feet) per second per 
 second. 
 
 = Value of the acceleration produced by the force of 
 gravity upon a freely falling body (to be found from 
 the result of the previous experiments). 
 
 Hence, from the above statement we see that the following relations 
 should hold good, viz. : 
 
 F F 
 
 a= B ;andy=- ir , 
 
 Or, Ma = F = mg 
 
 M - A . or g - a ^ a (W + M) 4 g _ a (W + w) 
 'ma' m w -J- g 10 
 
 Inserting the values obtained from the previous experiment*. 
 
 We get, 
 
 , = a = 33-*3 . per se, per sec. 
 
 It has been proved by the most careful experiments carried out 
 at Greenwich Observatory, that the force of gravity there, when reduced 
 to sea level, produces an acceleration of 32*1912 ft. per sec. per sec. But, 
 Aberdeen, where the above experiments were made, is nearer the centre 
 of gravity of the earth, the value may well be nearer 32*2 which is con- 
 sidered the usual average value, although 32 is often taken when only 
 approximate results are desired. 
 
 Regarding Inertia of the Wheel. The value of the small weight x which 
 forms part of W in the last equation and which balances the inertia of 
 the aluminium wheel, may be found from two experiments producing 
 two different accelerations cij and a% by two different riders TOJ and tn., 
 respectively, and substituting their known values in the previous equation, 
 until it is found that uniform velocity is obtained in each of these trials 
 after the riders m^ and m^ have been caught by the ring. 
 
 Let, W = W + Xy Where, W = the combined weight of the left and 
 
 right-hand equal weights + the small wire on 
 
 the latter which is required to balance the friction 
 
 of the axle in its bearings 
 And x = the weight required to balance the inertia 
 
 of the wheel. 
 
268 LECTURE XXI. 
 
 Then, since gravity has a constant value v e get 
 
 W+w (W+x + wJ, 
 
 9 = a ^ or in Exp. (i) g = a x -- ^ 
 
 (W+X + Wo) 
 
 and in Exp. (2)g - a z -- 
 Hence in Exp. (i)-^p 1 = (TT+ x +^) and in Exp. (z)-~ = (W + x + wj 
 
 Or in Exp. (i) a; = -^- 1 -- (W + wj and in Exp. (2) x = -^p = (W + w a ) 
 
 When Experiments (i) and (2) give the same value for x under the above 
 conditions it balances the inertia of the wheel in both cases. 
 
 It will now be interesting to find from the foregoing data what should 
 be the ratio of the mass of a rider to the whole mass moved in order to 
 produce an acceleration of i foot per second per second ; since the 
 dynamical unit of force which is sometimes called the "BritL-h 
 Absolute Unit of Force " called the Poundal, is that force which, acting 
 for i second upon a mass of i lb., imparts to it a velocity of I foot per 
 second. 
 
 Here M = ( W + ").= lib. 
 
 9 
 a = i ft. per sec. per sec. 
 
 }i m -^-, and w has to be found T 
 
 9 
 From Case VI. we see that 
 
 m a i 
 
 But, since the masses are proportional to the weights < 
 We get, M _ W + w _ 32*2 
 
 m w i * 
 
 And, since (W + w) = i lb. 
 
 = f-L-\ lb. = i oz. (nearly). 
 
 Or, a rider of ^ lb., i.e., oz. would be required to be placed upon the 
 right-hand weight of an Atwood's machine where the sum of all the 
 moving weights reckoned as before, was equivalent to i lb. in order to 
 produce an acceleration of i foot per second per second. 
 
 Hence, speaking generally, the value of a poundal (or so-called absolute 
 
 dynamical unit of force) is equal to ^ or- lb. in the ordinary engineers 
 
 gravitation unit of force ; and a force of i pound weight is equal to 32 or 
 g poundals. But, we shall not bother the elementary student in this book 
 with poundals, since, as we said before, the gravitation unit of force or 
 force which will sustain a weight of i lb. of matter is sufficiently absolute 
 for any particular place and for our purpose. 
 
 It will at once be seen from the following proportion, that no rider, 
 however heavy, if it were placed upon any weight, however small (whjen 
 the latter was balanced against the force of gravity) could produce an 
 acceleration of 32*2 feet per second per second. For, the impressed 
 acceleration a would also have to be equal to g or 32*2, when we get 
 
 a W 32'2I 
 
 That is, W could have no weight became u> 
 
GALILEO'S AND KATER'S EXPERIMENTS, 
 
 269 
 
 The application of the foregoing explanations, formula? and reasoning 
 to ordinary applied mechanics questions will be seen, when we come to 
 deal with constant forces acting in any direction upon masses of matter 
 for a known time and producing a certain acceleration ; such as the 
 acceleration produced upon the piston and piston rod, &c., of a steam 
 engine by the pressure of the steam up to the point of cut off. Or, upon 
 the fly wheel due to the mean force upon the crank pin during so many 
 revolutions ; or upon the starting or the stopping of railway trains, centri- 
 fugal machines, dynamos, and many other prime movers, or motors where 
 it is advisable to raise their speed as quickly as possible to a permissible 
 maximum or normal velocity and to stop them quickly. 
 
 Galileo's Experiments. More than 100 years before Atwood was 
 born, the famous Italian philosopher Galileo (about 1583-1630) experi- 
 mented upon falling bodies of different kinds and weights by letting them 
 fall simultaneously but separately from the top of the leaning tower at 
 Pisa and found that they reached the ground at the same instant, or from 
 the same height in practically the 
 same time. He also let spherical 
 balls roll down a straight, smooth, 
 inclined plane (where the friction 
 between the ball and the plane 
 was a minimum) and found that 
 the distances through which the 
 ball passed in successive seconds 
 varied directly as the square of 
 the times from rest ; and that the 
 velocities acquired by the ball 
 varied directly as the times from 
 rest ; or, 
 
 Times in seconds varied as 
 I : 2:3 : t 
 
 Distances passed varied as 
 12 : 22 : 32 : ^ gt 2 
 
 Speeds acquired varied as 
 i :2 :3 :gt. 
 
 Pendulum Experiments. 
 
 Galileo (about 1583), Huyghens 
 (about 1650-70), and Captain 
 Kater, an Englishman (about 
 1813-32), experimented with pen- 
 dulums to determine not only the 
 length of the seconds pendulum 
 but also to ascertain the value 
 of the force of, and acceleration 
 due to gravity at different places. 
 The investigation of -this subject 
 must be deferred until harmonic 
 motion has been explained. It 
 may, however, be stated that for 
 a seconds pendulum 
 
 The length l = (^~) 
 
 = 39*13983 inches at Greenwich 
 (reduced to sea level) where g = 
 32'i9i2feet persecondper second. 
 
 EATER'S REYEESTBLE 
 
270 
 
 LECTURE XXI. 
 
 Rater's Reversible Pendulum. This pendulum consists of a 
 vertical bar carrying a brass weight near each end, together with two 
 similar boxwood weights, between the brass weights which are so attached 
 as to compensate for the air displacement of the brass ones. The pendulum 
 may be hung from one or other of two movable steel knife edges working 
 on agate planes. We now test to see if the lower weight has been cor- 
 rectlj placed, because, the time taken by the pendulum to perform say 100 
 swings about the top movable steel knife edge should be equal to the time 
 taken to perform 100 swings about the other knife edge. If the times are 
 not quite equal then one of the brass weights or its movable knife edge 
 must be readjusted until the time of a swing when suspended from either 
 end must be exactly equal. The length of an isochronous, simple equiva- 
 lent pendulum is precisely equal to the final adjusted distance between the 
 knife edges. Therefore, knowing the length of the pendulum arid its time 
 of oscillation, the acceleration due to gravity can be f..und from the 
 
 gt 2 
 
 previously stated formula I = - With proper care Kater's pendulum can 
 
 be made to give a very accurate result for the value of the? acceleration 
 due to gravity. 
 
 The velocity attained by a body sliding down any 
 straight inclined plane varies directly as the square root 
 of the height.* 
 
 Let AB be a smooth inclined plane of length I making an angle 6 with 
 the horizontal, as in the figure, Eesolve the weight of the smooth-faced 
 
 To PEOVE THAT THE VELOCITY ATTAINED BY A BODY SLIDING DOWN 
 ANY INCLINED PLANE DEPENDS UPON THE HEIGHT OF THE PLANE. 
 
 body W as represented to scale by the dotted vertical line ab, into two 
 components, one cb parallel to the plane and the other ac perpendicular 
 to it. 
 
 Now, the latter component force ac has no effect on the motion of the 
 body, since its surface and that of the plane are supposed to be smooth, 
 and therefore the former component cb is the only force which causes the 
 body to move down the plane. 
 
 The magnitude of this force F = ab sin = W sin 6. 
 
 * It has been proved in the Author's more Advanced Text-Book on 
 Applied Mechanics and Mechanical Engineering, Vol. II., Lecture XXII. > 
 that although the kinetic energy of a spherical ball, or of a cylinder of a 
 Certain mass, after it falls or slides or rolls down any path through a 
 definite height is of the same value ; yet the velocity and momentum are of 
 different values for the cases of rolling and of sliding. 
 
VELOCITY, ETC., IMPARTED TO A FALLING BODY. 
 
 Also, it has been shown from Case VI. in connection with the 
 Atwood's machine, &c.. that the acting force = mass moved x accelera- 
 
 W 
 
 tion attained by body ; that is, F = x a 
 
 Or, Wsin0x# =Wa 
 
 Further, it was shown in connection with Case V. that t>2 = 2 a I. 
 Now, substituting the values for a and I in this equation and figure, we 
 get v2 = 2g sin x BA. 
 
 But, ||= sin 0, or -?5= BA. 
 
 .^ = 20sin0 x j^- e = 2<?BC = 2gh. 
 
 That is, r 2 oc h the vertical height of plane. Or, v oo Vh. 
 (Then multiplying both sides by W and dividing them by zg.), 
 
 We get, (the kinetic energy or stored work) = W^ P tential> i 
 2g \ energy. / 
 
 The Velocity, the Momentum, and the Kinetic Energy 
 attained by a body in passing from a given height along 
 any path are each respectively the same, when the friction 
 between the path and the body is negligible. The following 
 figure illustrates an arrangement to prove, that the velocity acquired by a 
 
 APPARATUS TO PROVE THAT THE VELOCITY, THE MOMENTUM AND 
 THE KINETIC ENERGY IMPARTED TO A FALLING BODY is THE 
 
 SAME WHICHEVER PATH IT FOLLOWS, IF FRICTION BE 
 NEGLIGIBLE OR NEGLECTED. 
 
 body falling freely down any curved path is the same whatever be the 
 shape of the path. It shows one straight and two curved paths. Each 
 path is made of a smooth glass plate, or a pair of parallel polished metal 
 rods, along which a truly spherical billiard ball may be allowed to roll 
 down from a platform. The ball when it reaches the lowest position, 
 i> allowed to travel up an adjustable and very smooth, plate glass, inclined 
 plane. 
 
 It will be noticed that in each instance, the ball ascends to the same 
 height on this plane. Consequently we conclude from the results of the 
 experiment that the velocity, the momentum and the kinetic energy (or 
 energy of motion) of the ball when commencing to ascend this inclined 
 plane must be respectively the same. Further, if the ball experienced no 
 frictional resistance along its pajth, then it would run up the inclined plane 
 until it reached exactly the same level from which it started with no 
 
272 
 
 LECTURE XXI. 
 
 initial velocity on the other side. The resistance of the air and any slight 
 want of smoothness in the paths make this ideal condition unattainable 
 in practice. But, with care, the results are sufficiently accurate to prove, 
 that if A, the height from start to foot of path, W the weight of the 
 ball, and g the acceleration due to gravity remain constant throughout the 
 experiment, and since the ball ascends along the incline always to the 
 same height for a given angle of inclination, the kinetic energy attained 
 at the foot of each path is the same, for nothing else could send it to the 
 same height unless the kinetic energy ^ mvz ; the momentum mv ; and 
 velocity v were the same in each case. Seeing, however, that the body 
 started from rest at a height A above the lowest point of its path, its 
 potential energy before starting = WA, and since no energy is assumed to 
 have been lost we get WA oc Wt> 2 /2. Hence, v2 oc 2 gh, or t;2 oc A since 
 2 g is a constant, i.e. t v oo ^/h. Consequently under these circum- 
 stances, the velocity, momentum and kinetic energy attained by a body 
 falling down or sliding over any inclined or curved path is solely dependent 
 upon the vertical height of the fall and quite independent of the shape or 
 length of the path taken by the body when friction is eliminated or 
 neglected. 
 
 Comparison of Dynamical Formulae for Linear and 
 Angular motion when due to any Force producing a 
 constant Acceleration. 
 
 KINEMATIC FORMULA. 
 Or, formulae dealing with pure motion in the abstract. 
 
 Linear, Angular. 
 
 v=at 
 
 KINETIC FORMULAE 
 
 Or, formulae dealing with the models of real bodies under the action 
 of actual forces. 
 
 F _ F _v a -v 1 _T.M. 
 
 Tr/ - Tir ~ 7~~" . tt j 
 
 ( Angular ) _ Twisting Moment 
 (Acceleration} "Moment of Inertia 
 _ Twisting moment 
 "Mass x (radius) 3 
 T.M.XI 
 
 Momentum = Im- 
 pulse or Force x 
 
 Angular ^ 
 Momentum 
 
 time it acts 
 
 or moment 
 
 . 
 
 of momen- 
 
 
 tum of the 
 
 
 rotating 
 
 1M^-F* 
 
 body. 
 ....ilw 5 
 
 Kinetic Energy = Stored work. 
 
 Or, 
 
 fMoment^ 
 
 i of r 
 
 I Impulse J 
 
 :T.M..x0 
 
 Kinetic Energy = Stored Work. 
 
PATH OF A PROJECTED BODY. 273 
 
 The Path of a Projected Body which then falls under 
 the Action of Gravity is a Parabola. The accompanying illustra- 
 tion shows a suitable form of apparatus for observing the path, which a 
 body, projected horizontally, describes under the combined actions of the 
 projecting force and of gravity. The whole apparatus should be firmly 
 secured to the wall, and the ball should always be allowed to roll lown 
 from the top of the concave quadrant " race," in order that the path traced 
 on the blackboard may be definitely followed. 
 
 When a body has a certain velocity imparted to it, such as a stone 
 thrown at an object, or water issuing under pressure from a pipe, or a 
 bullet fired from a rifle in any direction, other than that of a truly vertical 
 one, it has two different motions imparted to it. The one direction and 
 motion is the result of the initial velocity, while the other is due to the 
 earth's attractive force, termed the force of gravity. In the same way, 
 the ball, when it just leaves the lower end of the quadrant "race," has 
 both a horizontal and a vertical motion. The resultant motion is deter- 
 mined from the relative values of these component motions, by the prin- 
 ciple of the parallelogram of velocities. The ball will not move along a 
 straight line, but will describe a parabolic curve, as shown by the dotted 
 curved line in the blackboard of the figure. If the ball had left the 
 quadrant with a velocity in any direction (other than that of a vertical one) 
 it would still have described a parabolic path. 
 
 Now referring to the two upper left and right figures of a cone, we see 
 that the parabolic curve BVC is the outline of the section of the cone GAL, 
 by a plane passing through VO and parallel to the opposite generating 
 line AL of the cone. 
 
 Generally speaking, a parabola is a curve traced by a point, which 
 moves in such a manner, that the distance from the point at any instant 
 to the focus of the curve is always equal to the normal distance of the 
 same point from the directrix of the curve. 
 
 Directrix and Focus. If a point so move that the ratio of its distance 
 from a given fixed point, to its perpendicular distance from a fixed straight 
 line be a constant, it describes a conic section, of which the fixed straight 
 line is termed the directrix and the fixed point the focus. The constant 
 ratio referred to is termed the eccentricity, and its magnitude determines 
 the nature of the conic. Thus, if in the figure DOD be the directrix and 
 F the focus, and if the point P moves so that the ratio of its distance frocr 
 if, is to its distance PN from DOD, be a constant, then P will trace out a 
 conic section which will be a parabola, an ellipse or a hyperbola, according 
 as the ratio in question is equal to, less or greater than unity That is 
 as PF is equal to, less than or greater than PN or as FV = , <or> VO. 
 
 Proof of Equation to the Parabola. In the right-hand upper figure, let P 
 be a point on the curve BVC, then the distance of P from the focus F is 
 equal to the length of the perpendicular line PN, let fall from the same 
 point P upon the directrix line DOD. 
 
 Any conic section made by a plane, such as VE, which cuts the two 
 extreme generating lines such as AG and AL of the cone GAL is called an 
 ellipse. An example of an ellipse is shown by the curve BLCGB, of 
 which F! and F 2 are the/oc*. 
 
 Any conic section made by a plane such as VH, parallel to the axis AC 
 IB called a hyperbola. 
 
 Referring to the right hand top figure, the equation of a parabola, viz,. 
 y a =4ax may be proved as follow : 
 
 C; 
 
274 
 
 LECTURE XXI. 
 
 INDEX OF PARTS. 
 
 A for apex of cone. 
 AQ ,, axis ,, 
 VC section 
 BVC parabola at VC. 
 DOD ,, directrix of para 
 
 bola. 
 
 V ,, vertex. 
 F focus. 
 
 OUTLINE VIEW OF CONE 
 WITH SECTION PLANE. 
 VC FOR PARABOLA ; 
 VH FOR HYPERBOLA ; 
 AND VE FOR ELLIPSE. 
 
 PROJECTED VIEW OF 
 CONE SHEWING 
 PARABOLIC CURVE 
 BVC, AND ELLIPSE 
 BLCGB. 
 
 APPARATUS FOR DETERMINING THE PATH OF A BO.JY PBOJECTD 
 HORIZONTALLY ASD FALLING FREELY. 
 
CENTRIFUGAL FORCE. 2/5 
 
 By Euclid I. -4 7. 
 
 FP 2 = FM 2 + MP 2 
 
 Or, MP 2 = FP 2 - FM 2 
 That is, MP 2 = P]S T2 - FM 2 
 
 Or, MP 2 - (PT + TN) 2 - (VM - VF) 2 
 
 Tluit i-, MP 2 = 2 TX x 2 PT (But MP = y: 
 
 \ Al-o PT = x 
 .-. T, 2 = 4 aa; I And TN = a. 
 
 Centrifugal Force due to Motion in a Circle. EXPERI- 
 MENT I. When a body such as a stone is attached to a cord 
 and whirled round and round in a circle, the hand experiences a 
 pull in the direction of the string, which is in tension under 
 the action of a force, and the faster the body is moved the 
 greater becomes the stress in the string, just as David of old 
 must have felt it before he let go that pebble from his sung which 
 went so straight for Goliath's brow. The stone is constantly 
 tending to fly off at a tangent, and is only kept moving in the 
 < ircular path by the reaction pulling it towards the centre of 
 motion. The pull from the centre of motion is called the centri- 
 fugal or centre-flying force, and the exactly equal and opposite 
 reaction is termed the centripetal or centre-seeking force. It may 
 be proved by geometry that each of these forces is equal to 
 
 EXPERIENCING THE EFFECT OP CENTRIFUGAL FORCE. 
 
 the weight of the body x the square of the velocity 4- the 
 acceleration due to gravity x the radius of the circle described 
 by the body.* 
 
 * At present the student mast accept the above formula as correct. 
 We shall have occasion to deduce the formula by aid of geometry in the 
 Advanced Course. (See Text-Book on " Applied Mechanics and Mechanical 
 Engineering," Vol. II., lectures xxii. and xxiii.) 
 
276 
 Or, 
 
 LECTURE XXI. 
 
 p - 
 
 gr 
 
 Ibs. 
 
 "Where P = Pull on the cord, or the centrifugal force in Ibs. 
 M W = Weight of the body in Ibs. 
 w v = velocity of the body in feet per second, 
 >* 9 gravity's acceleration = 32' per second in one second, 
 M r = radius from centre of motion to e.g. of body in feet, 
 
 Centrifugal Force Machine. General Description. This machine 
 enables the student to systematically experiment and prove the laws con- 
 nected with centrifugal force. 
 
 A polished wooden beam about 2 ft, in length is rotated by means of 
 
 CENTRIFUGAL FORCE EXPERIMENTAL MACHINE. 
 
 the pulley at any speed. Attached to the centre of this beam is an 
 accurately machined box with one side flexible. The box is filled with 
 mercury or coloured alcohol. An upright glass tube is fitted into the 
 centre of this box, and exactly in the axis of rotation, so that the height 
 of the mercury can be measured whatever be the speed of the machine. 
 The height of the mercury in the tube can be adjusted by means of a 
 screw at the side of the box. Fastened to the centre of the flexible side 
 of the box is a long brass rod, which in turn is attached to one end of a 
 flat brass rod, whilst the other end of the rod is connected to the end of 
 the beam. This arrangement of supporting the long brass rod allows it to 
 
CENTRIFUGAL FORCE MACHINE. 2/7 
 
 move backward and forward with less friction than if it were made to 
 elide on a bearing. This long rod is graduated, and a sliding weight can 
 be clamped at any position from the box. Hence, we see that if the 
 centre of the flexible side of the box be pulled or pushed, the mercury will 
 rise or fall in the vertical glass tube ; although, we cannot observe the 
 yielding of the flexible side. With a fixed vertical scale alongside the 
 glass tube it is easy to measure the rise and fall of the mercury in the 
 latter. The weight shown on the left-hand side can be used for balancing 
 the weight on the movable rod, and in this case it should be clamped at 
 the same distance from the axis of the instrument. 
 
 Prior Adjustments. Before commencing any experiments it is necessary 
 to pull the end of the long movable brass rod with a force of i, 2, or 3 Ibs., 
 &c., by means of a spring balance, to note the respective heights of the 
 column of mercury in the glass tube. Hereby we can afterwards tell the 
 value of our scale measurements. We can also make a number of 
 experiments with the sliding weights removed from the rods, to find the 
 centrifugal force of the remaining parts of the machine when run at 
 different speeds, in order that these readings may be subtracted from 
 subsequent observations to get true results. 
 
 Working of the Centrifugal Machine. One experimenter turns the handle 
 of a separate pulley, which drives the pulley shown in the figure by means 
 of a rope. At the same time he keeps his eye upon the height of the 
 mercury column in order to keep the speed constant, and he counts the 
 number of revolutions per minute which his hand makes. 
 
 It is easy with this instrument to test the law which is usually given, 
 that if a body be compelled to move in a curved path it exerts a force 
 directed outwards from the centre of motion, and the amount in pounds is 
 found by multiplying the mass of the body by the square of the velocity 
 in feet per second, and dividing by the radius of the curved path we get 
 
 Wv^ 
 
 the centrifugal force =P= . Or, if we multiply the mass by the radius 
 
 of the circle and by the square of the angular velocity in radians per second, 
 we get ir 2 =P, the centrifugal force urging the weights from the centre 
 ,of motion in a radial direction. 
 
278 LECTURE XXI. 
 
 EXPERIMENT II. Take a pail and half fill it with water. 
 Attach a rope to the centre of the handle, and swing it round 
 and round your head. The water does not fall out, even if you 
 swing it in a vertical plane, if the velocity be sufficient to cause 
 the centrifugal force to be greater than the force of gravity. 
 
 EXAMPLE I. A small tin pail, containing i Ib. of water, with a 
 rope attached to its handle, is to be whirled in a vertical circle. 
 If the distance from the hand or centre of motion, to the surface 
 of the water be 2 feet, what is the least number of revolutions 
 per minute that you can give it in order not to spill any of the 
 water ? 
 
 ANSWER. Here P must be at least equal to i Ib., for W = i Ib. 
 and r = 2 feet, whilst g = 32. 
 By the formula 
 
 Or, v = 8 x 60 = 480 ft. per minute. 
 
 Now a circle of 2 feet radius = 12*56 feet circumference. 
 
 480 
 
 .*. - 7 = 38*2 revolutions per minute. 
 12*56 
 
 Consequently, if you whirl the pail at 40 revolutions per 
 minute, there will be no fear of any water coming out of it even 
 when it is upside down at the highest part of the circle. 
 
 EXPERIMENT III. Turn a disc of wood with a small barrel on 
 one side of the centre. Fit the wheel and the barrel so truly 
 with a turned axle that when the axle is supported by eye hooks 
 at each end for bearings, a cord wound round the barrel and 
 then pulled sharply, will cause the wheel to revolve freely at a 
 high speed without vibration or oscillation. Now bore a hole 
 through the disc near its circumference, and run in molten lead 
 into this hole. Again spin the wheel rapidly, when it will be 
 found to hobble to such an extent as to shake itself almost out 
 of the bearings. 
 
 The centrifugal force due to the unbalanced piece of lead 
 asserts itself so thoroughly that when it reaches the highest 
 position of its revolution round the axis, it overcomes gravity, and 
 lifts the whole wheel and barrel clean out of the bearings. It 
 thereby creates such a disturbance as to leave a distinct impres- 
 sion on the mind of the student. 
 
 Next bore another hole through the disc of the same size as the 
 former one, and at the same distance from the axle, but diametri- 
 cally opposite to the front hole, and run in the same weight of 
 
BALANCING HIGH-SPEED MACHINERY. 2/9 
 
 lead into it. Again spin the wheel, and it will be found to run 
 smoothly. 
 
 This experiment conveys to the young engineer a most useful 
 lesson, for it not only shows him the effect of centrifugal force 
 due to want of balance, but it also gives him an idea how to rectify 
 the evil. 
 
 Balancing High-speed Machinery. All high-speed machi- 
 nery, whether revolving or reciprocating, should as far as possible 
 be most carefully balanced, in order to prevent centrifugal force 
 coming into play and creating that horrid vibration and noise 
 with which it is always more or less accompanied. There is 
 nothing tends so much to the heating of bearings, and to the quick 
 wearing out of brasses and other bearing surfaces as unbalanced 
 moving parts ; besides which, at very high velocities they become 
 actually dangerous, and have frequently been known to cause 
 destruction to life and property.* 
 
 Centrifugal Stress in the Arms of a Ply-Wheel. If the 
 arms of a fly-wheel or pulley are not properly proportioned to 
 resist the centrifugal force due to the mass of the revolving rim ; 
 or, if tthe casting has been carelessly cooled, so as to set up internal 
 stresses between the arms and the boss or the rim, the wheel may 
 give way. In fact, there is no fly-wheel or pulley made that would 
 not burst, under the very great stress of centrifugal force, if you 
 only ran it fast enough. The student will observe from the 
 formula that the centrifugal force or stress in the arms of a fly- 
 wheel is directly proportional to the square of the velocity, so 
 that by merely doubling the number of revolutions per minute 
 you quadruple the stress in the arms, and if the speed be increased 
 three times, the stress becomes nine times as great. 
 
 EXAMPLE II. Each segment of a fly-wheel, with its correspond- 
 ing arm to which it is attached, weighs 1000 Ibs., and the mass 
 may be taken as collected at a distance of 4 ft. from the axis of 
 the wheel. If each arm has a breaking stress of 100,000 Ibs., 
 what is the maximum number of revolutions per minute that the 
 fly-wheel could be run at without breaking the arms, neglecting 
 the binding strength of the rim of the wheel ? 
 
 ANSWER. By the previous formula for centrifugal force 
 
 1000 X V 2 
 100,000 --= -- .. tf =12,800 
 
 32 x 4 
 
 * See Mr. C. A. Matthey's paper and the discussion on " The Mechanics 
 of the Centrifugal Machine," in the Transactions of the Institution of 
 Engineers and Shipbuilders of Scotland for Session 1898-99. 
 
28O LECTURE XXI. 
 
 113 ft. per second, fully. 
 
 = 113 x 60 = 6780 ft. per minute. 
 
 .*. v = 
 
 Or, v 
 Now, the circumference of a circle of 4' radius = 25 ft. 
 
 6780 . ,. 
 
 ,\ f = 271 revolutions per minute. 
 
 Energy. In applied mechanics energy means the capability 
 of doing work.* 
 
 Potential Energy is that form of energy which a body possesses 
 in virtue of its position or its condition. For example, when a 
 body of 10 Ibs. is lifted 10 ft. high, it has a potential energy of 
 100 ft.-lbs. ; for it takes that amount of work to lift the 10 Ibs. 
 through the 10 ft. ; and if then allowed to fall, it would naturally 
 give out the same quantity of work, either in overcoming friction, 
 or, if it fell freely, it could be usefully employed to that amount 
 and no more. 
 
 Potential energy may also be due to a condition of a body, such 
 as the potential energy in the coiled spring of a watch or clock, 
 which when wound up does work in moving the mechanism. 
 We have also the case of potential energy in a lump of coal which 
 when burned gives out heat, that will raise steam to be used in 
 a steam engine for doing work. Or, in the case of an electric 
 battery, where plates of copper and zinc are respectively placed in 
 solutions of sulphate of copper and zinc, and on being suitably 
 connected by wires to an electric motor, will give out electrical 
 energy, which may be converted into mechanical work by the 
 motor, and thereby effect some useful purpose. 
 
 Kinetic Energy (E^) is energy due to motion. For example, in 
 the first instance of potential energy the weight of 10 Ibs., in 
 falling freely down through 10 ft., had stored up in it, due to its 
 motion, an amount of accumulated work equivalent to 100 ft.-lbs. 
 
 Accumulated Work (E,). If a body of weight W Ibs. be 
 raised to a height h feet above the earth 
 
 The potential energy stored up E^=W& (ft.-lbs.) 
 
 Now, if the body be allowed to fall freely, under the action of 
 gravity, through h feet, it would have a velocity at the end of 
 time t seconds of v feet per second. 
 
 Referring back to the formulas for falling bodies previously 
 given in this lecture we see that 
 
 h - -- .-. Wh = ft.-lbs. 
 *g *g 
 
 * We have specially avoided using this term hitherto, as students are 
 liable to confuse it with force, work, and power. 
 
ACCtTMTTLATED WORK. 281 
 
 Therefore the kinetic energy or accumulated work stored up in a 
 moving body is expressed by the formula 
 
 If a body of weight W Ibs. were impressed forward along a 
 perfectly smooth plane for a distance of I feet, by a force F Ibs., 
 causing an acceleration of, a, feet per second ; then the previous 
 set of formula? for linear velocity would apply when the reaction 
 from the plane cancelled the force of gravity. 
 
 Here, . . F-/ and I - 
 
 But the Wvrk Done through distance I = F x I = E 
 
 / . 
 
 Therefore in this case the accumulated work stored up in the 
 moving body would be expressed by the formula 
 
 Accumulated work in a Hotating Body. If a body of W 
 Ibs. be concentrated at a distance of r feet from the centre of 
 motion, and be rotated so that it has a velocity of v feet per 
 second, then 
 
 The Accumulated Work =- - ft. -Ibs. 
 2 9 
 
 The Energy of a Rotating Ply -wheel is a good example of 
 accumulated work. If the pressure of steam in the cylinder and 
 the point of cut-off be kept constant, and if one or other of the 
 machines which are being driven by the engine be thrown out of 
 circuit or, in other words, if the belt be moved to the loose pulley 
 the load on the engine will be lessened, and the engine will have a 
 tendency to increase in speed. If, however, it be provided with 
 a very heavy fly-wheel, the surplus power of the engine will be 
 stored up in the fly-wheel, so that the increase of speed will not 
 be so great as if it had a light one, or none at all. If a machine 
 should be suddenly brought into circuit again after a short time, 
 then the load on the engine will be as quickly increased ; but the 
 etored-up energy in the fly-wheel will enable it to overcome this 
 sudden demand for power, so that the speed of the engine will not 
 be greatly altered. The fly-wheel, therefore, acts as a regulator 
 of speed, not only for alterations of load, but also for the variable 
 pressures which exist in the cylinder of an engine. This ia 
 
282 LECTURE XXI. 
 
 particularly noticeable in the case of gas engines, where the 
 almost instantaneous explosion of gas in the cylinder at the 
 beginning of a stroke creates an immense force, which would 
 urge the piston forward at lightning speed, if it were not for 
 the very heavy fly-wheel with which the engine is provided. The 
 fly-wheel stores up some of this sudden force and gives it out 
 again during the intervening strokes when there is no explosion, 
 thus tending to a uniformity of speed which would be conspicuous 
 by its absence if the gas engine had only a light fly-wheel, or none 
 at all. In fact, the motion of gas engines would be so erratic 
 without fly-wheels as to prevent their application to many pur- 
 poses for which they are admirably adapted when aided by very 
 heavy ones. 
 
 Radius of Gyration.* It will be evident, almost without ex- 
 planation, that in the case of a fly-wheel or a rotating disc, those 
 parts which are furthest from the centre of motion must accu- 
 mulate more energy than those of the same weight which are 
 nearer to that centre, because they move at a greater velocity. 
 There is, however, for every body a mean radius of rotation, termed 
 " radius of gyration" (k) which is at such a distance from the 
 centre of motion, that if the whole mass of the body were concen- 
 trated there, the same kinetic energy or accumulated work would 
 be developed at the same speed or number of revolutions per 
 minute. The length of this mean radius varies with the shape 
 of the rotating body, and requires a knowledge of higher mathe- 
 matics for its computation ; so we will assume that in the case of 
 a fly-wheel it is at the e.g. of the rim, or that the distance is given 
 in any question requiring solution. 
 
 EXAMPLE III. A fly-wheel weighing 10,000 Ibs. has a mean 
 radius of rotation, k=*r= 5 feet, and turns normally at 100 revolu- 
 tions per minute. Owing to the load being diminished, the speed 
 increases to no revolutions per minute ; what reserve energy is 
 stored up in the fly-wheel, which is fit to overcome any sudden 
 increase of load ? 
 
 ANSWER. Let v l = the velocity in feet per second, at the nor- 
 mal speed n^ revolutions per minute, 
 
 And v 2 = the velocity at the increased speed n a revo- 
 lutions per minute ; 
 
 2 x 22 x q x 100 
 
 Vj = 271-7*^! = ^ = 52*4 ft. per sec. 
 
 7 x6o 
 
 = 2 = 57*6 ft. per sec. 
 
 5 The radius of gyration is called the swing radius by some engineers. 
 
THE FLY-PRESS: 
 
 Stored energy at speed n^ = L 
 
 r> 7 j 
 
 Reserved stored energy = - 2 _ 
 2 g 
 
 W 
 
 2 , 
 
 ' - v 
 
 = 80,375 ft -Ibs. 
 
 The Ply press. This machine is used, in the form shown by 
 the figure, either for embossing or stamping pieces of metal with 
 
 THE FLY-PRESS. 
 INDEX TO PARTS. 
 
 D represents Disc supporting M. 
 M Metal to be stamped. 
 
 Punch or die. 
 F Frame of machine. 
 
 S represents Screw. 
 N Nut for S. 
 L Lever arms. 
 B Balls or weights. 
 
 some design, or for punching thin metal plates. The piece of 
 met :il M, to be embossed or punched, is laid on a disc D, and the 
 die or punch P is caused to come down on M with a large 
 amount of stored-up energy, due to the operator taking hold of 
 
284 LECTURE XXI. 
 
 one or other of the heavy balls B, and giving them a very rapid 
 turn round. The result of this movement is to send the quickly 
 pitched square-double-threaded screw rapidly through its nut N, 
 thereby forcing the guided square carrying the punch straight 
 downwards, and causing the latter to overcome the resistance of 
 the hard metal. Neglecting friction at the screw and the guide, 
 and considering the combined weight of the two balls as = W Ibs., 
 and v = their velocity in feet per second at the instant the punch 
 meets the metal M, then 
 
 The stored energy, or energy of the blow, = - ft.-lbs. 
 
 If . 1= Length the punch or die goes into the metal in feet, 
 And E, = Ilesiotance overcome (mean) in Ibs., 
 
 Then ..... 1^ = ft.-lbs. 
 
 2g 
 
 EXAMPLE IV. Distinguish between energy and power. What 
 is the unit of power in this country ? In a fly-press two balls, 
 each weighing 60 Ibs., are moving with a linear velocity of 15 
 feet per second, what is the measure of the energy existing in the 
 balls (take ^ = 32) ? What is the power required to raise 6600 
 gallons of water up 150 feet in 30 minutes! A gallon of water 
 weighs 10 Ibs. (S. and A. Exam. 1893.) 
 
 ANSWER. (i) Energy is the capability of doing work which a 
 body may possess on account of its position, or condition, or 
 motion. Power is the rate of doing work, or the work done in a 
 given time. The unit of power in this country is the horse-power 
 and is the rate of doing work equivalent to 33,000 ft.-lbs. per 
 minute. 
 
 (2) Here W = combined weight of the two balls =120 Ibs. 
 v = linear velocity of balls = 15 ft. per sec. 
 
 Then energy existing in balls = 
 
 421-87 ft.-lb8. 
 
 2x32 
 (3) Weight of water raised 6600 x 10 
 
 per minute ~^ -- = 220 lbs - 
 
 .". Work done per minute = 2200 x 150' = 330,000 ft.-lbs. 
 ,". Power required 33> 000 _ 10 
 
ENERGY STORED IN A ROTATING BODY. 
 
 285 
 
 To find Experiment ally the Energy Stored in the 
 
 Rotating Mass Of a Fly-wheel. Description of the Experimental 
 Machine. The accompanying figure shows a fly-wheel mounted on an axle. 
 This axle is supported on ball bearings, thereby reducing the friction 
 to a minimum. The diameter of the fly-wheel is 18 inches and weighs 
 about 100 Ibs. If the centre of gravity of the fly-wheel is not exactly in 
 the axis, then it is better to place the wheel as in the figure. One end of 
 a cord is looped over a pin on the axle, and after being wound several 
 times round the axle, the other end is led over an aluminium pivoted 
 pulley, and attached to the 7-lb. weight. 
 
 Object of Experiments. The object of this experiment is to illustrate 
 
 , EXPERIMENTAL APPARATUS FOE DETERMINING THE ENERGY 
 STORED UP IN A ROTATING FLY-WHEEL. 
 
 the "Principle of Work" or the law of the "Conservation of Energy." 
 Since energy cannot be generated or destroyed, the quantity given to a 
 machine can be traced in its transmission through the machine, as clearly 
 pointed out in Lecture V. Hence the amount of energy given to the fly- 
 wheel and the attached weight in this experiment is measured by the pull 
 of the earth on the 7-lb. weight multiplied by the distance through which 
 the latter falls. 
 
 Conservation of Energy. Part of this energy is stored up in the fly-wheel 
 as kinetic energy, part of it is used to turn the wheel against the friction 
 of the bearings, and part is stored as kinetic energy in the falling weight. 
 The last two items of the total energy are converted into heat energy, and 
 
286 
 
 LECTURE XXI. 
 
 gt neral formula. 
 
 
 fEnergy given to^ 
 the apparatus 
 ! during tha 
 
 [ Kinetic } 
 energy E* 
 - = ! stored 
 
 +- 
 
 Energy " 
 converted 
 into heat 
 
 falling of the 
 [ weight. , 
 
 in the 
 ^fly- wheel. ; 
 
 
 by 
 
 friction. 
 
 the several sub-divisions of the total energy may be connected under one 
 
 f Kinetic energy in 
 falling weight at 
 1 the instant when 
 I the cord is released 
 (^ from the axle. 
 
 Determination of Speed of Machine by the Fly-wheel. After setting the 
 mark upon the rim of the fly-wheel opposite the fixed mark, and with the 
 7-lb. weight on the cord to keep it taut, turn the wheel by hand to see as 
 near as possible how many turns the wheel makes in falling a certain 
 number of feet Let this distance through which the weight is to fall in 
 the experiment be h feet, and let n be the number of turns made by the 
 fly-wheel during this fall. Now determine by aid of a stop-watch how long 
 it takes the 7-lb : weight to fall through h feet whilst turning the fly-wheel, 
 and tabulate your results as follows : 
 
 Weight 
 
 Height 
 
 Turns 
 
 Time taken 
 
 Velocity of 
 
 Revolutions 
 
 on cord 
 in Ibs. 
 
 of fall 
 in feet 
 
 of fly- 
 wheel (n) 
 
 foi weight 
 to fall (h) 
 
 weight at the 
 instant when 
 
 per second of 1 
 fly-wheel at 
 
 
 
 
 feet 
 
 cord leaves the 
 
 the instant 
 
 
 
 
 
 axle in feet 
 
 when cord 
 
 
 
 
 
 per second 
 
 leaves the axle. 
 
 Final Calculations. As we have found the total number of revolutions 
 made by the wheel until the cord drops off its pin, and also observed the 
 time taken in seconds. Then, as we know that the speed increases 
 uniformly during tMs interval of time, the mean speed is just half the 
 speed at the end of the time interval ; consequently, if we divide t lie number 
 of revolutions by the number of seconds in which they were performed, and 
 multiply this quotient by 2, we will get the number of revolutions per second made 
 by the wheel when the weight just ceases to act. You can test the accuracy 
 of your result by counting the number of revolutions of the fly-wheel from 
 the time that the cord drops off until the wheel comes to rest due to its 
 own friction and dividing by the time which has elapsed, then multiply 
 this quotient by 2 and this will give you the required aged of the fly- 
 wheel. 
 
MOTION ON BICYCLE AND RAILWAY CUHVES. 287 
 
 Motion on a Curved, Inclined, or "Banked" Track. 
 
 Take as an example a cycling track, and let us suppose that the 
 bicycle is moving with uniform velocity of v feet per second round a 
 smooth circular course of radius OB equal to r feet. Then it is necessary 
 to find at what angle to the horizontal plane the track should be inclined 
 or " banked " in order that the bicycle may keep in its circular path. 
 We see from the diagram that two forces are acting on the bicycle 
 
 Of<- B = Radius r feet 
 
 j Sectional Elevation of Bicycle, \ Course. 
 
 Oj<- -0 B= -Radius r feel - 
 
 Plan of above 
 Bicyde Course. 
 
 SECTIONAL ELEVATION AND PLAN OF A BICYCLE TRACK. 
 
 (i) Its own weight W ; and (ii) the reaction R, which is perpendicular 
 to the smooth track. These two forces have a resultant horizontal force 
 
 P = , which is acting towards the centre of the horizontal circle in 
 
 which the bicycle moves. 
 
 Hence, if we take any vector line to represent the weight W and from 
 the upper end thereof draw a horizontal line, and from the lower end draw 
 another vector line inclined at an angle 6 to the vertical until it meets the 
 horizontal line in a point. The sides of the triangle will represent in 
 
 direction and magnitude the forces W, R, and P, and tan = = = 
 
 f _ Li ^rj v -, from which we can quite easily obtain the required 
 
 angle of the inclination of the track and the height h for the desired 
 width of the same. 
 
 Railway Curves. If the lines of a railway curve be laid at the 
 same level, then the centripetal thrust of the rails on the wheel of trains 
 passing round curves act on the flanges of the wheels, and the centri- 
 fugal thrust of the wheel on the track would tend to push it sideways out 
 
 T 
 
288 -,LECTURE XXI. 
 
 of its place. Bat, in order to have this action and reaction normal to the 
 track, the outer rail is raised and the track thereby inclined to the hori- 
 zontal. The amount of this super-elevation which is suitable for a par- 
 ticular speed is quite easily calculated. 
 
 SECTIONAL ELEVATION OF A RAILWAY CUEVE 
 
 Let G = gauge in inches between the rails. 
 
 v = velocity in feet per second of the train. 
 
 r = radius of curve in feet. 
 
 h = height of super-elevation of one rail above the other in inches. 
 
 6 = the angle of " banking " or inclined plane. 
 Now draw the triangle ABC to represent the three forces, W, K, and P 
 as indicated in the previous case. 
 
 Then AC = AB sin = AB tan 6, since 6 is always very small. 
 
 <y2 
 
 h = G inches. 
 
 gr 
 
 EXAMPLE V. A motor-car moves in a horizontal circle of 300 ft. radius 
 at 30 miles per hour, what is the ratio of its centrifugal force to its 
 weight ? This is the tangent of the angle at which the track ought to be 
 inclined sideways to the horizontal if there is to be absolutely no tendency 
 to side-slip ; find this angle. (B. of E. 1904.) 
 Answer. 
 
 Let v = velocity of the car in feet per second. 
 P = centrifugal force in Ibs. 
 W = weight of the car in Ibs. 
 
 r = radius of horizontal circle of car's motion in feet. 
 g = acceleration due to gravity = 32 ft. per sec. per sec. 
 Then the forces which are acting at the centre of gravity of the car 
 are as follows : 
 
 (i) The weight of the car, W Ibs. acting vertically downwards. 
 
 (ii) The centrifugal force P, equal to - Ibs. acting in a horizontal 
 
 direction. 
 
 (a) To Find the Ratio of the Centrifugal Force to the Weight of the Car. 
 Magnitude of centrifugal force P Wv 2 . w _ ^ 
 Weight of the motor-car = W = ~gr^ ~ ~ gr 
 
 -Z 30 x 5280 x 30 x 5280 
 W ~~ 60 x 60 x 60 x 60 x 32 x 300 
 P 121 1 
 
 = = ; oraboufc 5' 
 
 (b) To find the inclination of the angle 6 at which the track ought to 
 be inclined to the horizontal if there is to be absolutely no tendency to 
 side-slip. 
 
MOMENTOTL 289 
 
 Momentum is the quantity of motion possessed by a body. 
 It is measured by the quantity of moving matter (i.e., its mass) 
 multiplied by its velocity.* 
 
 Or, Momentum = Mass x Velocity 
 
 M m x v 
 
 The momentum of a body w, therefore, that constant forct 
 which, acting for unit time, would stop the body. 
 
 A force of 2 Ibs. acting for unit time, or one second, on a body, 
 will produce a certain amount of momentum; it is therefore 
 obvious that twice the force acting for half the time, i.e., 4 Ibs. 
 acting for half a second, would produce the same momentum. 
 
 EXAMPLE VI. A hammer head of 2\ Ibs. moving with a velocity 
 of 50 feet per second is stopped in - ooi second. Find the average 
 force of the blow. 
 
 ANSWER. Momentum of Hammer Head = mass x velocity. 
 
 = ( 2 i/ 32 )x 50 = 3-906 lbs.-sec. 
 
 Average Force (F) = ma = ( ^y^ 1 ) = ~f ( 'o-ooi ) = 39 6 lbS ' 
 
 The negative signs (before v a and 50) show that the acceleration (a) is in 
 this case a retardation. The momentum 3*906 represents the force in Ibs. 
 which, acting for one second, would stop the hammer head. The average 
 force cf the blow is therefore (3 '906-:- '001) = 3906 Ibs. 
 
 EXAMPLE VIL A ship of 2000 tons, moving at 3 knots, is 
 stopped in one minute ; what is the average retarding force ? 
 Neglect the motion of the water. One knot is 6080 feet per 
 hour. 
 
 ANSWER. Here we may obtain the retarding force in tons and 
 reduce the speed in knots to feet per second. 
 
 Now, 3 knots = 3 x 6080 feet per hour; or, ^ - ft. per sec. 
 
 _ , ... 2000 3 x 6080 
 Hence the momentum = mass x velocity = x ^ 7 units. 
 
 * The mass of a body is its weight in Ibs. divided by the acceleration 
 due to gravity. Hence, if a body is W Ibs. in weight, and if g repre- 
 sents the acceleration due to gravity, or 32 ft. per sec. per sec. then 
 
 TTT 
 
 the mass m = . For example, if a body weighed 3-2 Ibs., then its mass 
 
 is 3- 2 -7-32 or i unit of mass. If this body is moving with a velocity of 
 10 feet per second its momentum is *i x 10 or i. If this momentum be 
 created or destroyed by a force acting for one second only on the body, 
 the force must have a value of i Ib. If it be created or destroyed in ten 
 seconds, then the force is i of a Ib. ; if in ^ second, its value would be 
 10 Ibs. 
 
29O LECTURE XXI. 
 
 r> 
 Since the retarding force acts for one minute, or sixty seconds . 
 
 - The average retarding force = change of momentum -r 60 
 
 2000 3 x 6080 
 
 ** * n 32 x 60 x 60 
 
 Or, = 5 ' 2 7 tons. 
 
 EXAMPLE VIII. A railway train starting from rest along a 
 level line acquires a speed of 30 miles per hour in five minutes. 
 What has been the mean pull between the engine and the train ; 
 the resistances to motion being taken at 10 Ibs. per ton, and 
 weight of train exclusive of engine 150 tons? 
 
 ANSWER. Constant pull due to resistances =10x150=1 500 Ibs. 
 
 ico x 2240 
 Mass of tram = m = j ^--150 x 70 
 
 Velocity of train*. = 30 miles per hour = 3 x S 28 feet P or 
 
 60 x 60 second. 
 
 =* v = 44 feet per second. 
 
 Momentum of train =-m x v**- 150 x 70 x 44 = 462,000. 
 And the time taken to produce this momentum = 5 x 60 seconds. 
 Average pull for change"! __ (mv 9 - mvA _ 462,000 _ n 
 
 nf t.h A TvmmAnt.iim f ~ *~1 ~ TTTTT ~ J 54 lbs - 
 
 of the momentum J y $ I 5 x 60 
 
 .-. Total mean pull between the) ,, 
 
 engine and carriages j = 15 + '54 = 34 Ibs. 
 
 Mass is defined as the quantity of matter in a body. It is 
 measured by the weight of the body in pounds at London divided 
 by the acceleration of gravity at the same place i.e., Mass = 
 t0-r-<7; or, w in Ibs. 4- 32 '2. 
 
 Inertia is defined as that property of matter whereby it tends to 
 remain in a condition of rest or of uniform motion. Hence, we 
 have Newton's first law of motion as stated in the second page 
 of this Lecture, and which is sometimes termed the " Law of 
 Inertia." 
 
 NOTE. The force in Ibs. is the distance-rate at which work is done in 
 foot-pounds, and it is also the lime-rate at which momentum is produced 
 or destroyed. 
 
 Moment of Momentum. Unit moment of momentum or angular 
 momentum is unit momentum at unit perpendicular distance. 
 
WORK DONE BY A VARIABLE FORCE. 
 
 2 9 I 
 
 EXAMPLE IX. A body weighing 3220 Ibs. was lifted vertically by rope, 
 there being a damped spring balance to indicate the pulling force of .Fib. on 
 the rope. When the body had been lifted x ft. from its position of rest, 
 the pulling force was automatically recorded as follows : 
 
 X 
 
 
 
 18 
 
 43 
 
 60 
 
 74 
 
 95 
 
 in 
 
 130 
 
 F 
 
 7700 
 
 7680 
 
 743 
 
 713 
 
 6770 
 
 5960 
 
 5160 
 
 39/0 
 
 Find approximately the work done on the body when it had risen 115 ft. 
 How much of this is stored as potential and how much as kinetic 
 energy ? What is, then, the velocity of the body ? (S. E. B. 1900.) 
 
 Answer. Use squared paper, and plot the values of x along the 
 
 8000 
 
 00 10 20 30 40 50 60 70 80 90 100 110 120 C 
 
 3000 
 
 DIAGRAM OP WORK DONE BY A VARIABLE FORCE. 
 
 abscissae OC, with the corresponding values of F as ordinates. Then draw 
 a curve AB through the several points. Dmde OC into, say, 15 equal 
 parts, and from the middle point of each division read off upon the vertical 
 scale the corresponding values of F. The sum of these ordinates divided 
 by 15 will give the mean force F in Ibs. This force multiplied by the 
 distance through which it acts, viz., 130 ft. gives the total work done by 
 the force or the area of the diagram OAJBC in ft. -Ibs. 
 
LECTURE XXL 
 Total work done by force F=a,rea, of diagram OABO in ft. Ibs. 
 
 Work done by force F in\_ f f . 
 
 passing through 115 ft. J~ " " 
 
 ="6951 xus =799, 360 ft. -Ibs. 
 
 But, work done in lifting the\_ w , -vwisonft Ih* 
 
 weight W through h feet j-WA-322ox 115 -370,300 ft.-lbs. 
 
 Hence, work done by force] ("Work done in^l C K'netic energy, "\ 
 Fibs, in passing through [ = \ lift.ng weight j- -f -j E^, in the body at V 
 115 ft. J [through 1 15 ft. J I end of the lift. J 
 
 Or, 799'36o =370,3 00 
 
 = 799. 3 6 ~ 370.300=429,060 ft.-lbs. 
 ,_ 429, 060 xzg 429,060x64 
 W 3220 ' 
 
 ^=8528. 
 
 v= ^8528 = 92*3 feet per second. 
 
 Poten'ial Energy E^ in body at height of 115 ft. = 370, 300 ft.-lbs. 
 Kinetic Energy E^ in body at height of 115 ft. =4=29,060 ft.-lbs. 
 
LECTURE XXI. QUESTIONS. 293 
 
 LECTURE XXL QUESTION!. 
 
 1. A body moves in a circle with a uniform velocity ; show that It must 
 be acted on by a constant force tending towards the centre, and find the 
 magnitude of the force in terms of the radius of the circle, and of the mass 
 and velocity of the body. 
 
 2. A body weighing 2\ Ibs., fastened to one end of a thread 4 feet long, 
 is swung round in a circle, of which the thread is the radius ; what will be 
 its velocity when the tension of the thread is a force of 20 Ibs. (^=32) ! 
 Ans. 32 feet per second. 
 
 3. When an unbalanced wheel is set in rapid rotation, a considerable 
 amount of shake and vibration is experienced. You are required to explain 
 this result from first principles, and to state the mechanical laws which 
 appear to be at work. How would you calculate the amount of pull that 
 this unbalanced weight exerts ? 
 
 4. What primary law in mechanics asserts itself when some revolving 
 piece of machinery moves at a high velocity, and is unbalanced f A weight 
 of i Ib. is placed on the rim of a wheel 2 feet in diameter, which revolves 
 upon its axis and is otherwise balanced. The linear velocity of the rim 
 being 30 feet per second, what is the pull on the axis as caused by the 
 weight of i Ib. 1 Ans. 28*1 Ibs. 
 
 5. A segment of a fly-wheel, with the arm to which it is attached, weighs 
 3500 Ibs., and the mass of the portion may be taken as collected at a dis- 
 tance of 8 feet from the axis of the wheel, which makes 40 revolutions per 
 minute. What is the force tending to pull away the segment and arm 
 from the boss of the wheel f Ant. 15,365 Ibs. 
 
 6. Define kinetic energy. How does it differ from potential energy ? If 
 a velocity of 300 ft. per second is impressed on a weight of 10 Ibs., what 
 is the measure of the energy now imparted to the weight. 
 
 Ans. 14,062-5 ft.-lbs. 
 
 7. State the rule for finding the amount of work stored up in a given 
 weight when moving with a given velocity. A weight of 6 cwt. moves 
 with a velocity of 20 feet per second ; how many units of work are stored 
 up in it ? Ans. 4200 ft.-lbs. 
 
 8. Write down the formula for the amount of energy stored up in a given 
 weight when moving with a given velocity. Describe, with a sketch, the 
 action of a fly-press. If each ball of the press weighs 50 Ibs., and the work 
 stored up in the balls is 400 ft.-lbs., find the velocity with which they are 
 moving. Take the number 32 to represent g. Ans. 16 feet per second. 
 
 9. Account for the storing up of energy in a rotating fly-wheel. If the 
 weight of the rim be doubled while the rate of rotation remains unchanged, 
 how much is the energy increased ? Ans. Twice. 
 
 10. State the formula for the energy stored up in a fly-wheel, on the sup- 
 position that the whole of the material is collected in a heavy rim of given 
 mean radius. Apply the formula to show (i) the effect of doubling the 
 number of revolutions per minute ; (2) the effect of doubling the weight ; 
 (3) the effect of increasing the mean radius in the proportion of 3 to 2. 
 
 11. A fly-wheel weighs 2j tons, and its mean rim has a velocity of 40 feet 
 per second, If the wheel gives out 10,000 ft.-lbs. of energy, how much is 
 its velocity diminished t An*, x -455 feet per second. 
 
2 9 4 
 
 LECTURE XXI. QUESTIONS. 
 
 12. Explain the use of the fly-wheel in any machine with which yon are 
 acquainted. To what class of machines is such a wheel usually applied T 
 What is the kinetic energy in a wheel revolving at 150 revolutions per 
 minute, if the wheel loses 5000 ft.-lbs. of energy when its speed is reduced 
 to 147 revolutions per minute ? Ans. 126,263 ft.-lbs. 
 
 13. A fly-wheel of a shearing machine has 150,000 foot-pounds of kinetic 
 energy stored in it when its speed is 250 revolutions per minute ; what 
 energy does it part with during a reduction of speed to 200 revolutions 
 per minute ? Ans. 54,000 ft.-lbs. 
 
 If 82 per cent, of this energy given out is imparted to the shears during 
 a stroke of 2 inches, what is the average force due to this on the blade of 
 the shears ? (S. E. B. 1902.) Ans. 265,680 Ibs. 
 
 14. A fly-wheel is required to store 12,000 ft.-lbs. of energy as its speed 
 increases from 98 to 102 revolutions per minute ; what is its kinetic energy 
 at 100 revolutions per minute ? (S. E. B. 1900.) Ans. 150,000 ft.-lbs. 
 
 15. A machine is found to have 300,000 foot-pounds stored in it as 
 kinetic energy when its main shaft makes 100 revolutions per minute ; if 
 the speed changes to 98 revolutions per minute, how much kinetic energy 
 has it lost 1 (S. E. B. 1901.) Ans. 11,880 ft.-lbs. 
 
 16. What do you understand by work, potential and kinetic energy? 
 A bullet weighing i oz. leaves the muzzle of a rifle with a velocity of 
 1350 feet per second ; what is the kinetic energy of the bullet in ft.-lbs. ? 
 
 Ans. 1780 ft.-lbs. 
 
 17. If a gun delivers 400 bullets per minute, each weighing 0*5 oz., with 
 2000 feet per second horizontal velocity ; neglecting the momentum of 
 the gases, what is the average force exerted upon the gun ? (S. E. B. 1900.) 
 Ans. 12*94 Ibs. 
 
 1 8. A bullet of O'l lb., with a speed of 2200 feet per second, is fired 
 into the middle of a block of wood of 30 Ibs., which is at rest but free to 
 move ; find the speed of the block and bullet afterwards. What is the 
 loss of kinetic energy in foot-pounds? ' V S. E. B. 1902.) Ans. 7-3 ft. per 
 sec. ; 7537 ft.-lbs. 
 
 19. A man and his bicycle weigh 170 Ibs. ; he has a speed indicator (not 
 a mere counter). When going at 10 miles an hour on a level road he 
 suddenly ceases to pedal, and in 15 seconds finds that his speed is 8 miles 
 an hour. What is the force-resisting motion ? (S. E. B. 1901.) Ans. I lb. 
 
 20. A car weighing z\ tons and carrying 40 passengers, the average 
 weight of each of them being 145 pounds, is travelling on a level rail at 
 the rate of 6 miles an hour. What is its momentum in engineer's units ? 
 If the propelling force be withdrawn, what average force in pounds must 
 be exerted to bring the car to rest in two seconds ? and supposing the 
 force to be constant, what distance would the car travel before it came to 
 rest? Ans. 3135 Ibs. -ft. -sees. ; 1567-5^8.; 8*8 ft. 
 
 21. A car is drawn by a pull of P Ibs., varying in the following way 
 t being seconds from the time of starting : 
 
 P 
 t 
 
 1020 
 
 980 
 
 882 
 
 720 
 
 702 
 
 650 
 
 713 
 
 722 
 
 805 
 
 
 
 2 
 
 5 
 
 8 
 
 10 
 
 13 
 
 16 
 
 19 
 
 22 
 
 The retarding force of friction is constant and equal to 410 lb. Plot 
 P 410, and the time t, and find the time average of this excess force. 
 What does this represent when it is multiplied by 22 seconds ? (S. E. B. 
 1902.) Ans. 366 Ibs. 
 
LECTUEE XXI. QUESTIONS, 
 
 295 
 
 22. A body weighing 1610 Ibs. was lifted vertically by a rope, there being 
 a damped spring balance to indicate the palling foro F Ib. of the rope. 
 When the body had been lifted x feet from its position of rest, the palling 
 force was automatically recorded as follows : 
 
 X 
 
 
 
 ii 
 
 20 
 
 34 
 
 45 
 
 55 
 
 66 
 
 76 
 
 F 
 
 4010 
 
 39i5 
 
 3763 
 
 3532 
 
 3366 
 
 3208 
 
 3100 
 
 3007 
 
 Find approximately the work done on the body when it has risen 
 70 feet. How much of this is stored as potential energy, and how much 
 as kinetic energy 1 Ans. 247,000 ft.-lbs. ; 1 12,700 ft. -Ibs. 5134,300 ft.-lbs. 
 
 (3. E.B. 1901.) 
 
 23. A tramcar, weighing 15 tons, suddenly had the electric current cut 
 off. At that instant its velocity was 16 miles per hour. Reckoning time 
 from that instant, the following velocities, V, and times, t, were noted : 
 
 V. Miles per hour 16 14 12 10 
 t. Seconds - - o 9*3 21 35 
 Calculate the average value of the retarding force and find the 
 average value of the velocity from t = o to t = 35. (B. of E. 1903.) 
 Ans. Retarding Force=264 Ibs.; Average Velocity =1274 miles 
 per hour. 
 
 24. A projectile has kinetic energy = 1,670,000 foot-pounds at a 
 velocity of 3000 feet per second. Later on its velocity is only 2000 feet 
 per second, how much kinetic energy has it lost ? What is the cause of 
 this loss of energy 1 Ans. 927,778 ft. -pounds. (B. of E. 1903.) 
 
 25. A man weighing 160 Ib. is in a lift which starts to descend with an 
 acceleration of 2 feet per second per second. What force is exerted by the 
 man upon the floor of the lift ? What would the force be if the lift were 
 descending at a uniform speed ? Ans. 150 Ibs. ; 160 Ibs. (B. of E. 1903.) 
 
 26. In a gun, of which the internal diameter is 6 inches, a projectile 
 weighing 100 Ib. has imparted to it in a distance of 12 feet a velocity of 
 2500 feet per second. Find the average pressure of the gases on the base 
 of the projectile up to the time it leaves the gun. (Neglect friction and 
 the energy of rotation of the projectile.) 
 
 Ans. Average Pressure = 7200 Ibs. per sq. inch. (B. of E. 1903.) 
 
 27. A weight of 120 Ib. falls to the ground from a height of 18 feet and 
 just rebounds. If the time of contact between weight and ground be 
 the energy of rotation of the projectile.) 
 
 Ans. Average Pressure = 28, 800 Ibs. per sq. inch. (B. of E. 1903.) 
 
 28. A train weighing 250 tons is moving at 40 miles per hour, what is its 
 momentum in engineers' units ? If this momentum is destroyed in ten 
 seconds, what is the average force acting on the train during these ten 
 seconds. Define what is meant by force by people who have to make exact 
 calculations. Ans. Average Force = 456 tons ; Momentum = 14 670 ton- 
 feet seconds ? (B . o f E. 1904.) 
 
 29. A fly wheel weighs 24,000 Ib., its mean radius (or rather radius of 
 gyration) is 10 feet, it revolves at 75 revolutions per minute what is its 
 kinetic energy ? 
 
 If suddenly disconnected from its engine, in how many revolutions 
 will it come to rest, if we know that in each revolution the energy 
 wasted in overcoming friction is 3000 foot-pounds ? 
 Ans. E* = 2,315,051 ft.-lbs. ; and 772 revs. (B. of E. 1904.) 
 
296 
 
 LECTURE XXI. QUESTIONS. 
 
 30. An ordinary steam engine has a stroke of 18 ins., and the connecting 
 rod is 36 ius. long. The crank shaft makes 400 turns per minute. Find 
 the velocity of the piston, in feet per minute, when it has moved through 
 one-quarter of the stroke, reckoned from the back end. 
 
 Ans. ^=1760 feet per minute. (C. & G., 1905, 0., Sec. A.) 
 
 31. A cricket ball, weighing 0-28 lb. f reaches the batsman when it is 
 travelling horizontally at 96 feet per second ; what is its momentum in 
 engineers' units ? The batsman drives the ball straight back to the 
 bowler with the same speed ; what has been its change of momentum ? 
 If the time of the blow is one-thirtieth of a second, find the average 
 magnitude of the force exerted by the bat upon the ball. (B. of E. 1905.). 
 
 Ans. Change of momentum = i'68 units ; Average Force = 50*4 Ibs. 
 
 32. A casting is bolted to an angle plate on the face plate of a lathe. 
 The casting, angle plate, and bolts are equivalent to 75 Ibs. at a radius of 
 4^ inches. In what position must a weight of 20 Ibs. be fixed to the face 
 plate to effect a balance 1 Ans. i6| inches from centre of plate. 
 
 (B. of E. 1905.) 
 
 33. The angular position D of a rocking shaft at any time t is measured 
 from a fixed position. Successive positions at intervals of 1/50 second 
 have been determined as follows : 
 
 Time t, ) 
 seconds j 
 
 o"oo 
 
 0*02 
 
 0-04 
 
 o'o6 
 
 o'o8 
 
 O'lO 
 
 0*13 
 
 0-14 
 
 o'i6 
 
 0-18 
 
 Position D, ) 
 radians J 
 
 o'io6 
 
 o"2o3 
 
 o'337 
 
 0-487 
 
 0*651 
 
 0-819 
 
 0-978 
 
 I'm 
 
 I '201 
 
 1-222 
 
 Find the change of angular position during the first interval from 
 t = o-o to t = o - o2 ; calculate the mean angular velocity during this 
 interval in radians per second, and set this up on a time base as an 
 ordinate at the middle of the interval. Eepeat this for the other 
 intervals, tabulating the results, and drawing the curve showing approxi- 
 mately angular velocity and time. Kead off the angular velocity when 
 t = 0-075 second. (B. of E. 1905.) 
 
297 
 
 LECTUKE XXII. 
 
 CONTENTS. Some Properties of Materials employed by Mechanics Essen- 
 tial Properties Extension Impenetrability Contingent Properties 
 Divisibility Porosity Density Cohesion Compressibility and 
 Dilatability Rigidity Tenacity Malleability Ductility Elas- 
 ticity Fusibility Load, Stress, and Strain Total Stress and Inten- 
 sity of Stress Tensile Stress and Stress Example I. Compressive 
 Stress and Strain Example II. Limiting Stress or Ultimate Strength 
 Safe Loads and Elasticity Limit of Elasticity Hooke's Law 
 Factors of Safety Modulus of Elasticity Ratio of Stress to Strain- 
 Examples III.-V. Resilience or Work Done in Extending or Com- 
 pressing a Bar within the Elastic Limit Examples VI.-IX. Single 
 Riveted Lap Joints Example X. Questions. 
 
 Some Properties of Materials employed by Mechanics. 
 The properties of matter are almost innumerable, but they may 
 be divided into two classes: (i) Essential properties; (2) Con- 
 tingent properties. The essential properties are those without 
 which matter cannot possibly exist. The contingent properties 
 are those which we find matter possessing, but without which we 
 could conceive it to exist. 
 
 Essential Properties 1. Extension means that property by 
 which every body must occupy a certain bulk or volume. When 
 we say that one body has the same volume as another, we do not 
 mean that it has the same quantity of matter, but only that it 
 occupies the same space.* 
 
 2. Impenetrability means that every body occupies space to 
 the exclusion of every other body, or that two bodies cannot 
 exist in the same space at the same time. 
 
 Contingent Properties. 1. Divisibility means that matter 
 may be divided into a great but not an infinite number of parts. 
 The ultimate particles of matter are termed atoms, derived from a 
 Greek word signifying indivisible. 
 
 2. Porosity signifies that every body contains throughout its 
 mass minute spaces or insfcerstices to a greater or less extent. 
 This has been proved to be the case with every known substance. 
 These spaces are supposed to be filled with a highly elastic fluid 
 called ether. 
 
 For example, when the steel or cast-iron cylinder of a hydraulic 
 
 * For Simple Rules of Mensuration see the Author's Elementary Manual 
 on "Steam and the Steam Engine," Lectures I., II., III. 
 
9 8 LECTURE xxn. 
 
 press is subjected to enormous pressure, water will ooze through 
 the metal from the interior to the outside. 
 
 3. Density is that property by which one body differs from 
 another in respect of the quantity of matter which it contains. * 
 
 Let M p M, = Masses of two bodies 
 Let Vj, V, = Volumes of two bodies. 
 Let Dj, D s = Densities of two bodies. 
 
 If V, - V,, then gi = S ; if D, = D,, then - 
 
 ,, M. 
 If both vary, then ^ 
 
 4. Cohesion is that property by which particles of matter 
 mutually attract each other at insensible or indefinitely small dis- 
 tances. It is therefore different from gravitation, since the latter 
 acts at all distances. It is evident that without this property 
 we could not have a solid, for if a solid body be lifted by one 
 part, the remainder sticks to it, and the whole is kept together by 
 cohesion. 
 
 5. Compressibility and Dilatability are properties common 
 to all bodies, by which they are capable of being compressed like 
 a sponge or extended like a piece of india-rubber in a greater or 
 less degree. 
 
 6. Rigidity signifies the stiffness to resist change of shape 
 when acted on by external forces. Unpliable materials which 
 possess this property in a large degree are termed hard, whilst 
 u hose which readily yield to pressure, without disconnection, are 
 called soft. Substances which cannot resist a change of shape 
 without breaking are termed brittle, whilst those that do resist 
 and at the same time change their form are said to be tough. 
 
 7. Tenacity is the resistance (due to cohesion) which a body 
 offers to being pulled asunder, and is measured by the tensile 
 strength in Ibs. per square inch of the cross section of the body. 
 We will consider this property in the case of metals, &c., when 
 lealinsr with stress and strain. 
 
 8. Malleability is that property by which certain solids may 
 be pressed, rolled, or beaten out from one shape to another with- 
 out fracture. It is therefore a property depending upon the 
 softness, toughness, and tenacity of the material. Gold possesses 
 this property in a higher degree than any other metal, and con- 
 
 * The Density of a substance is either the number of units of mass in a 
 unit of volume, in which case it is equal to the heaviness (i.e., weight of 
 unit volume of substance in unit weight) ; or it is the ratio of the mass of 
 a eriven volume of the substance to the mass of an equal volume of water, 
 m which case it is equal to the specijic gravity. 
 
PROPERTIES OF MATERIALS. 299 
 
 Bequently sheets of gold are procurable of less than one-thousandth 
 of an inch in thickness. Copper is one of the most useful of the 
 malleable metals, and it may be beaten out into most elaborate 
 shapes from the solid ingot. The Swedish iron of which horse- 
 shoe nails are made is also very malleable, and is therefore highly 
 prized by the blacksmith. Lead, although possessing softness, is 
 not sufficiently tenacious to be considered a very malleable metal, 
 but still it finds one of its most useful applications in the form of 
 rolled lead sheathing for roofs of houses and interiors of water 
 tanks, fec. 
 
 9. Ductility* is" that property by which some metals may 
 be drawn down through a die-plate into wire or tubes. This pro- 
 perty depends chiefly on toughness and tenacity. For example, we 
 find that the very fine pianoforte wire used with Lord Kelvin's 
 deep-sea sounding machine is both hard and rigid, but possesses 
 great toughness and tenacity. The copper wire used for electrical 
 conductors becomes harder and harder as it gets drawn down to 
 smaller and smaller sizes, and it has therefore to be annealed in 
 order to comply with the many bendings and unbendings which it 
 has afterwards to undergo in winding and unwinding it upon 
 bobbins whilst twisting it into a stranded conductor or in 
 covering it with a dielectric of cotton, silk, gutta-percha, or 
 india-rubber, <kc. Solid-drawn copper pipes are frequently used 
 for conveying steam and liquids where a sound light job is required 
 to resist great pressures. This flowing property of metals is now 
 taken great advantage of by the engineer in a variety of ways. 
 For example, lead and tin, when subjected to great hydraulic 
 pressure, and properly guided through a die, can be squirted into 
 long continuous rods or pipes, or squeezed on to insulated electric 
 light conductors, so as to form a water-tight protecting sheathing 
 thereto, just as if these metals were composed of so much plastic 
 dough. In fact, all you have to do in order to cause many harder 
 and stronger metals, such as copper, iron, and steel, to flow cold 
 into almost any shape of mould, is to apply sufficient pressure and 
 to give sufficient time for them to retain their natural homogeneous 
 and isotropic structure, or to adopt means for rp^toriner thp 
 structure should they have departed therefrom during any part 
 of the process. A metal is said to be homogeneous when it is of 
 the same density and composition throughout its mass. It is 
 isotropic when it has the same elastic properties in all directions. 
 
 * Refer to the description of the Lever Testing Machine, illustrated in 
 Lecture IV., and to Lord Kelvin's Hydrostatic Wire Testing Machine, 
 illustrated in Lecture XVII., as examples of machines whereby the 
 comparative ductility of certain materials may be ascertained by their 
 percentage elongation. 
 
300 
 
 LECTURE XXII. 
 
 1O. Elasticity is that property, possessed by different solids in 
 a greater or less degree, of regaining their original size and shape 
 after the removal of the force which caused a change of form. 
 We shall see later on that there are limits of plasticity oeyond 
 which the bodies will not regain their exact normal size or shape. 
 
 12. Fusibility is that property whereby metals and many other 
 substances, such as resins, tallows, &c., become liquid on being 
 raised to a certain temperature. The following table shows in 
 round numbers the melting-points of a few of the commoner 
 metals : 
 
 MELTING POINTS OF METALS IN DEGREES FAHRENHEIT. 
 
 Mercury . 
 Tin . . 
 
 . . . - 38 
 
 Copper 
 German silver . 
 
 200O 
 2OOO 
 
 Bismuth . 
 Lead . 
 
 ... 500 
 
 , 600 
 
 Gold . 
 Cast iron . 
 
 2OOO 
 22OO 
 
 Zinc . . 
 
 
 Steel . 
 
 25OO 
 
 Antimony 
 
 . . 800 
 . . 1800 
 
 Nickel, also AluminiuE 
 \Vrouerht iron 
 
 a 2800 
 
 23OO 
 
 
 
 Platinum 
 
 3SOO 
 
 Load, Stress, and Strain. When force is applied to a body 
 so as to produce either elongation or compression, bending, torsion, 
 shearing, or a tendency to any of these, the force applied is 
 termed the load, the corresponding resistance or reaction in the 
 material is termed the stress due to the load. Any alteration pro- 
 duced in the length or shape of the body is termed the strain. 
 
 DEFINITIONS. Load is the force or forces applied to the body. 
 /Stress is the reaction in the body duo to the load. 
 Strain is the alteration in shape as the result of the load. 
 
 The load is called a dead load when it produces a steady or a 
 gradually increasing or diminishing stress. For example, the 
 weight of a roof on the walls of a building is a steady or dead 
 load. The gradually increasing pull produced on the specimen 
 in the lever-testing machine, illustrated by the fourth figure in 
 Lecture IV., is also a dead load. 
 
 The load is termed a live load when it varies from instant to 
 instant. For example, a regiment of soldiers, or a series of 
 vehicles, or a train passing over a bridge creates a live load on the 
 bridge. 
 
 Total Stress and Intensity of Stress. The total stress is 
 the total reaction due to the total load. The intensity of stress, 
 or simply the word stress, expresses the reaction per unit area of 
 the cross section. Thus, if P be the total force applied in lbs. r 
 and A be the total cross section in square inches, then the 
 
 p 
 Mean Intensity of Stress on the section = ~-r Ibs. per square inch. 
 
STRESS AND STEAIN. 30 1 
 
 Tensile Stress and Strain. If the line of action of a load 
 oe along the axis of a bar, tie-rod, or beam, so as to tend to elon- 
 gate the same, the reaction per square inch of cross section is 
 termed the tensile stress, and the elongation per unit of length is 
 called the tensile strain. 
 
 EXAMPLE I. A wire ^ square inch in cross section, and 
 10 feet long, is fixed at its upper end. A load of 1000 Ibs. is 
 hung from the lower end, and then the wire is found to stretch 
 i inch, (i) What is the stress ? (2) What is the strain ? 
 
 ANSWER. (i) Here P= 1000 Ibs., and A = T ^ sq. in. 
 
 Let p = stress or pull per square inch in Ibs. 
 
 p / 
 
 .-. The stress, or p =~^= Ioo / TV = 10 > 000 lbs - P 61 " S Q- ^h- 
 
 (2) Original length = L = 10' = 120", and the increase of 
 length = I = i". 
 
 Let e * strain or extension per unit of length, i.e., per inch in 
 this case, 
 
 increase of length I i" 
 
 .-. The Strain, or e = i i Jh = f = " " * 0083 
 
 original length L 120 
 
 Compressive Stress and Strain. If the line of action of a 
 load be along the axis of a bar, shore, strut, or pillar, so as to 
 tend to compress or shorten the same, the reaction per square 
 inch of cross section is termed the compressive stress, and the 
 diminution per unit of length is called the compressive strain. 
 
 EXAMPLE II. A vertical support in the form of a hollow 
 pillar, having 2 square inches cross section of metal, is 10 feet long. 
 With a load of 10,000 Ibs. resting on the top, it is found to be 
 compressed ^ of an inch in length, (i) What is the stress? 
 (2) What is the strain ? 
 
 ANSWER. (i) Here P = 10,000 Ibs., and A =* 2 sq. inches. 
 Let p = stress or compression per sq. in. of cross section in Ibs. 
 
 P 10,000 
 
 .. The stress, or p = -r- = = 5000 per square inch. 
 
 A. 2 
 
 (2) Original length = L = 10' = 120", and the diminution of 
 length = J = A" 
 
 Let e =- strain or compression per unit of length, t.e., per inch 
 in this case, 
 
 diminution in length -i" 
 or . - origiDaI lengtl f = ^ - -00083 
 
 Limiting Stress or Ultimate Strength. For every kind 
 of material and every way in which a load is applied, there must 
 be a value, which, if exceeded, causes rupture or fracture of tn 
 
302 
 
 LECTURE XXII. 
 
 body. The greatest stress which the material is capable of 
 withstanding is called the limiting stress or ultimate strength per 
 square inch of cross section of the substance, for the particular 
 way in which the load is applied. 
 
 Factors of Safety. The ratio of the ultimate strength or 
 limiting stress to the safe working load is called the factor of safety. 
 This factor of necessity varies greatly with different materials, 
 and even with the same material, according to circumstances. 
 For materials which are subjected to oxidation or to internal 
 changes of any kind, the factor of safety must of necessity be 
 larger than in those which are always kept dry or are well painted 
 and carefully handled. There is no condition in engineering 
 structures which requires a more careful calculation, or estimate 
 of the necessary factors of safety, than that of railway bridges, 
 which are exposed to all sorts of weathers and to extremely 
 variable live loads. The skill of the engineer is therefore brought 
 out, when he designs structures so as to include all possible 
 circumstances to which they may be subjected, and so proportions 
 the material at his disposal, that there shall be a minimum of 
 internal stress and strain, with a maximum resistance to dead or 
 live loads for a minimum cost of material and workmanship.* 
 TABLE OF ULTIMATE STRENGTH AND WORKING STRESS OP MATE- 
 RIALS WHEN IN TENSION, COMPRESSION, AND SHEARING. 
 
 
 i 
 
 
 Ultimate Strength. 
 
 Working Stress. 
 
 
 Tons per sq. inch. 
 
 Tons per sq. inch. 
 
 Materials. 
 
 Ten- 
 
 Com- 
 
 Shear- 
 
 Ten- 
 
 Com- 
 
 Shear- 
 
 
 sion. 
 
 pression. 
 
 ing. 
 
 sion. 
 
 pression. 
 
 ing. 
 
 Cast iron . 
 
 7-5 
 
 45 
 
 14 
 
 i-5 
 
 9 
 
 3 
 
 Wrought-iron bars 
 
 25 
 
 20 
 
 20 
 
 5 
 
 3-5 
 
 4 
 
 Steel bars . 
 
 45 
 
 70 
 
 3 
 
 9 
 
 9 
 
 5 
 
 Copper bolts 
 
 15 
 
 25 
 
 
 3 
 
 5 
 
 
 Brass sheet . . 
 
 14 
 
 
 
 
 3 
 
 
 
 
 Safe Loads and Elasticity. As a rule, however, the object 
 of the engineer is not to put such a stress on his materials of 
 construction as will cause rupture or destruction, but rather to 
 
 * For other tables relating to the Strength of Materials in Engineering 
 Constructions, Factors of Safety, &c., refer to Rankine's "Rules and 
 Tables," Molesworth's "Pocket Book of Engineering Formulae," D. K. 
 Clarke's "Rules and Tables," "The Practical Engineer's Pocket Book;" 
 and for Electrical Engineering Materials to Munro and Jarnieson's " Pocket 
 Book of Electrical Rules and Tables." 
 
LIMIT OF ELASTICITY HOOKE'S LAW. 303 
 
 make machines and raise structures that will withstand all rea- 
 sonable forces likely to be brought to bear upon them. Conse- 
 quently, he is quite as much interested in what may be termed 
 safe loads as in ultimate or destructive ones. He therefore 
 requires to know what loads can be safely applied to materials 
 under different circumstances, so as to comply with that most 
 useful property termed elasticity, which we again define as the 
 capability of regaining their original size, shape, and even strength, 
 after the removal oftht forces which caused a change of form in 
 them. 
 
 Limit of Elasticity Hooke's Law. So long as the 
 stress or reaction per square inch of cross section does not exceed 
 a certain limit, called the limit of elasticity, then the material 
 will return to its original shape, size, and strength, after the 
 removal of the load. This limit has been ascertained for most 
 materials of construction by elaborate experiments, which are to 
 be found tabulated in the Proceedings of the Institutions of the 
 Civil and Mechanical Engineers, and in such books as Rankine's 
 " Rules and Tables," Molesworth's " Pocket Book of Engineering 
 Formula?," and D. K. Clark's " Rules and Tables." For example, 
 with a bar of good wrought iron the elastic limit is only reached 
 after a stress of 24,000 Ibs. per square inch has been brought to 
 bear upon it, and in a similar degree every other material has a 
 corresponding limit, beyond which it is not safe to stress it, for 
 tear that it should be overstrained, and thus lose, to a certain 
 extent, its property of recuperation or restitution, or take a 
 permanent set. 
 
 Within this limit, Hooked Law holds good for metal bars under 
 the action of forces tending to elongate or compress them. This 
 law states that : 
 
 (1) The amount of extension or compression for the same bar 
 is in direct proportion to the stress. 
 
 (2) The extension or compression is directly proportional to 
 the length. 
 
 (3) The extension or compression is inversely proportional to 
 the cross sectional area ; consequently, if the area be doubled the 
 extension or compression will be halved, or the resistance to the 
 load will be doubled. 
 
 Let P = Pull, push, or load in Ibs. on the bar. 
 
 A = Area of cross section of the bar. 
 
 L = Length of the bar before the load was applied. 
 
 n I = Length by which the bar is extended or compressed. 
 
 w p = Stress or load per square inch of cross section = P/A. 
 
 P 
 
 Then, so> long as Y- does not exceed the elastic limit, I varies directly 
 
 U 
 
JO4 LECTURE XXIL 
 
 I P 
 
 as P for the same bar ; or ^ varies directly as -^> for different bari 
 
 of the same material and subjected to the same conditions. 
 
 In other words, so long as the stress does not exceed the elastic 
 limit, the strain will be proportional to the stress. 
 
 Modulus of Elasticity, or Ratio of Stress to Strain. As 
 we have just indicated, by HOOKE'S LAW, if a metal under test be 
 gradually subjected to a stress, and if the load does not exceed 
 the limits of elasticity of the material, the strain will be in pro- 
 portion to the load. 
 
 Consequently, the ratio of the stress to the strain is a constant 
 
 Quantity for eaoh particular substance within the limits of Hooke s 
 ^aw, and is termed the Modulus of Elasticity of the substance. 
 
 But . . Stress oc Strain 
 
 Stress = Ex Strain.* 
 
 Where E represents a constant number or modulus depending on 
 the natural elasticity of each material 
 
 JP 
 
 . F Stress ^ PL^ 
 Strain " l_ '' Al 
 L 
 
 .., ; . . . PL = AZE 
 
 Or, imagine it is pure imagination that a substance could 
 be elongated to double its length or compressed to zero by sub 
 jecting it to a certain load, we should then have an index value, 
 or constant number, or modulus, by which we could compare it 
 with every other substance which behaved likewise under similar 
 circumstances. This imaginary value is termed the Modulus of 
 Elasticity. 
 
 For example, take a bar of wrought iron of i square inch cross 
 section, which is found to stretch 2 4 a o^ o o o P ar ^ ^ ^ s length 
 under a stress of i lb., and consequently by Hooke's Law twice 
 that amount under a stress of 2 Ibs., and so on ; then this number 
 (24,000,000) is called the Modulus of Elasticity of the iron bar. 
 For, if the elasticity of the bar were perfect, it is evident that a 
 stress of 24,000,000 Ibs. would produce a strain or elongation 
 equal to the length of the bar, or, fjoo = i. In other 
 words, the length of the bar would be doubled under this stress. 
 Consequently, we have the following definition, 
 
 * Since stress is reckoned by so many Ibs. per square inch of cross section 
 of a material, and strain is simply an abstract number, it follows that the 
 Modulus of Elasticity (E) must also be reckoned by so many Ibs. per 
 square inch. 
 
MODULUS OF ELASTICITY. 
 
 30S 
 
 DEFINITION. The Modulus of Elasticity of any substance is 
 that load which would double its length on the supposition that the 
 elongation was proportional to the stress, and that the cross section 
 t>f the bar was of unit area, or one square inch, and supposing the 
 bar to remain perfect during the operation. 
 
 From this we again see that 
 
 Mod ulus of Elasticity = 8tres8 = E = ? / * 
 strain A/ L 
 
 Or, .... PL = AZE 
 
 MODULI OP ELASTICITY TO STRETCHING. 
 (See Rankine's Rules and Tables for complete Data.) 
 
 Material 
 
 Modulus of 
 Elasticity in 
 Ibs. per sq. in. 
 in round 
 numbers. 
 
 Material. 
 
 Modulus of 
 Eia.-<iicity in 
 Ibs. per sq. in. 
 ih round 
 
 numbers. 
 
 
 (Mean values.) 
 
 
 (Mean values.) 
 
 Wood, Elm . . . 
 
 I.OOO.OOO 
 
 Lead (sheet) . . . 
 
 700.000 
 
 Larch . . 
 
 1,100.000 
 
 (wire) . . . 
 
 I.OOO.OOO 
 
 Beech . . 
 
 1,300.000 
 
 Brass (cast) . . . 
 
 9.000,000 
 
 ff Birch . . 
 
 1,400.000 
 
 (wire). . . 
 
 14.000.000 
 
 Mahogany . 
 
 I,4OO,OOO 
 
 Copper (cast) . . 
 
 15.000.000 
 
 Oak ... 
 
 1,500,000 
 
 (wire) . . 
 
 17,000,000 
 
 Pine (yellow) 
 
 1,600.000 
 
 Cast Iron .... 
 
 i8.ooo,coo 
 
 M Ash . . . 
 Teak 
 
 I.600,000 
 2.OOO.OOO 
 
 Wrought Iron . . 
 Steel 
 
 25,000.000 
 35.000,000 
 
 n a 
 
 
 
 
 EXAMPLE III. A steel bar 5' long and 2\ sq. in. in cross section 
 is suspended by one end ; what weight hung on the other end will 
 lengthen it by -016 inch, if the modulus of elasticity of steel is 
 30,000,000 Ibs. per square inch? (S. and A. Exam. 1877.) 
 
 ANSWER. First ask what is wanted ? Viz., stress. 
 
 Now the universal rule is Modulus of Elasticity = 
 
 strain 
 
 Or, stress = modulus x strain. 
 
 For, The strain is the elongation per unit of the length. 
 -016" -016 
 
 Consequently, e * 
 
 5'x 12' 
 
 6o 
 
 = 00026. 
 
3o5 LECTURE xxn. 
 
 .. Tho Stress = Modulus x strain 
 
 = 30,000,000 x -00026 = 8000 Ibs. per sq. in. 
 
 And, The Total Stre88 = Sooo Ibs. x 2-25 sq. in. = 18,000 Ibs. 
 
 Or, we might have applied the formula previously deduced 
 Hz., 
 
 where P is the total pull required in Ibs. 
 
 .. 
 
 L 5x 12 
 
 EXAMPLE IV. What do you understand by stress and strain 
 respectively ? If an iron rod, 50 feet long, is lengthened by J inch 
 under the influence of a stress, what is the strain ? (S. and A. 
 Exam. 1892.) 
 
 ANSWER. Stress is the reaction per unit area of cross section 
 due to the load. Let P = the total tension acting on area A ; 
 
 P 
 Then stress P = ~T 
 
 Strain is the ratio of the increase or diminution of length or 
 volume to the original length or volume. Let L = original length 
 of a bar of the material, I = amount by which the length is in- 
 creased or diminished , then, when the bar is subjected to stress, 
 
 The strain = e = ^ 
 L 
 
 In the example given, L = 50' x 1 2" = 600 inches ; and 1= J inch. 
 
 .. Strain, e=i = -i- = ? ='00083 
 L ooo i 200 
 
 EXAMPLE V. From the above question and answer determine 
 the modulus of elasticity of the iron of which the rod is composed, 
 if the load was 4366 Ibs., and the cross section of the rod 2 square 
 inches. 
 
 Total load 
 ANSWER. (i) Stress =/^ 
 
 x ' Cross area 
 
 P 4366 Ibs 
 Or, , . . p = - = li - =2183 Ibs. per sq. in. 
 
 A. 2 
 
 (2) Modulus of Elasticity = - - ! 
 
 Strain 
 
 Or, ... E = P -= 2T83 . =25,000,000 
 
 6 '00083 
 
 .'. A load of 25,000,000 Ibs. would elongate a rod of the iron 
 to double its length by tensile stress. 
 
RESILIENCE WITH EXAMPLES. 307 
 
 Resilience or Work done in Extending or Compressing a 
 Bar Within the Elastic Limit. Definitions of Resilience. (i) When 
 a bar is strained within the elastic limits either by a compressive or a 
 tensile force, then the work done in extending or compressing it is equal to 
 the amount of compression or extension multiplied by the mean stress 
 which produces the strain. The amount of work thus done when tho 
 stress just reaches the elastic limit is termed resilience. 
 
 (2) Resilience may also be defined as half the product of the stress into 
 the strain, where the stress and the strain are those produced when the 
 elastic limit is reached. 
 
 Let P = push or pull in Ibs. applied gradually. 
 
 A = area of bar in square inches of cross-section. 
 
 L = length of bar in inches. 
 
 1= elongation of the bar in inches due to the force P. 
 
 ,, p= stress per square inch when the elastic limit is reached, =P/A. 
 
 E = resilience of the bar. 
 
 I. Where the Load is Gradually Applied. If the load be gradually in- 
 
 p 
 creased from o up to P Ibs., its mean value will be , and the work 
 
 p 
 done or resilience, R =: | I inch-lbs. 
 
 But P=pA=EAJL 
 
 BAP i /P\ a AL I a AL 
 . . work done or res^ence R= -~ =- = ~ P* - 
 
 2 
 
 p a volume ofbar. rt , 1Via 
 >. i. R= E X - 2 - mch-lbs. 
 
 Work done per unit^ _ p 1 
 of volume j~2E 
 
 This latter equation gives the strain energy stored in the 6ar, since the 
 material is still elastic. 
 
 II. When the Load is Suddenly Applied without Initial Velocity. Let I 
 be the elongation or shortening of the bar when the load P is suddenly 
 applied, but without initial velocity. 
 
 Then, the work done by the external load must be equal to the energy 
 tored up in the bar. 
 
 Let P max be the maximum stress per square inch which is produced in 
 
 the bar. 
 
 p 
 
 The me*n stress will be max as its initial value was o. 
 
 Hence, the work done on the bar = - fnax = -. , 
 and the work done by external load = PZ. 
 Therefore, ^J^^^^pf, 
 
 2 2L 
 
 I P 
 
 or, P,^=c2P; andEj- = 2-^. 
 
 Consequently, the maximum intensity of stress induced in a bar by the 
 sudden application of a constant load, without initial velocity, is double 
 the intensity of the stress produced by the load itself. 
 
 III. When the Load is Suddenly Applied and with an Initial Velocity. 
 Let a weight of W Ibs. be dropped from a height of h inches upon a bar, 
 and that the maximum stress produced in the bar was P max Ibs. per square 
 inch, whilst the elongation or compression was I inches. 
 
3O8 LECTURE XXII. 
 
 Work done by the falling weight = W (A + J) = energy stored up in the bar. 
 
 P J, EAZ 2 
 Therefore, -^= - =W (h + l)=Wh if he value of I 
 
 is small. 
 If v=velocity of load P at moment of impact in ft. per sec., 
 
 then kinetic energy, E* = g^ = ^ = W ( h + *) ft , lbs . 
 2y 2 A. 1 2 
 
 EXAMPLE VI. Calculate the resilience of a steel tie-bar, i inch in 
 diameter and 4 feet long if the elastic limit is reached under a load ot 
 20 tons, and modulus of elasticity = 13,000 tons per square inch. 
 
 Answer. 
 
 Work done or resilience, 11 = 
 
 = - inch-tons. 
 
 EXAMPLE VII. A round bar of steel is 20 feet long and i inch in 
 diameter. Find the tensile load which, if suddenly applied, would cau^e 
 an instantaneous elongation of the bar of 'I inch. 
 
 Taking = 13,000 tous per square inch. 
 . 
 
 T7I \ J 
 
 P= =- where symbol letters represent the values stated 
 2Ll in the" text. 
 
 p _ 1 3,000x785 ix 'I 
 
 2 X 2O X 1 2 
 
 EXAMPLE VIII. Determine the greatest weight that can be dropped 
 from a height of i foot on a bar of steel which is i inch in diameter and 
 10 feet long. The modulus of elasticity, =13,000 tons per square inch, 
 and the elastic limit 18 tons per square inch. Also, find the alteration in 
 the length of the bar. 
 
 Answer. 
 
 p m 
 
 P=A = -i7 
 
 13000x2 
 
RESILIENCE WITH EXAMPLES. 309 
 
 EXAMPLE IX. In a tensile test of a piece of flat wrought iron bar 
 the following results were obtained : 
 
 (i) Original dimensions of cross-section, 2*02 inches by 0-51 inch. 
 (ii) Final dimensions of cross-section at point of fracture. I -49 inches 
 
 by 0*39 inch. 
 
 (iii) Gross load at limit of elasticity, 36,000 Ibs. 
 (iv) Gross load at fracture, 59,000 Ibs. 
 (v) Total final extension on a length of 10 inches, 1*63 inches, extension 
 
 when load was 22,000 Ibs., -0075 inch. 
 
 Find from the above data, (a) the modulus of elasticity, (6) limit of 
 elasticity, and tenacity in Ibs. per square inch ; also (c) the reduction of 
 area per cent., and (d) the approximate work done in fracturing this 
 specimen. (L.U.B.Sc. Eng. 1903.) 
 Answer. 
 
 (a) Young's] pL 
 modulus of V E = -T-J 
 elasticity J ** 
 
 _ 22,OOOX 10 220,000 
 
 2 '02 X5 1 X -0075 ~ -007726 
 
 E = 28,473,440 Ibs. per square inch. 
 
 (b) Stress an 
 
 elastic limit I Load at elastic limit _ 36,000 
 in tons per j ~~ Original area ~~ i -0302 
 square inch J 
 
 =34,944 Ibs. per square inch. 
 
 (c) Percentage] Originalarea-final area 
 contraction of >- = ,. . . : - - x 100 
 area J Original area 
 
 ^(2-02 x -51) -(i '49 x -39) IOQ 
 
 2'O2X '51 
 
 (d) Work done^ 
 
 fnch in frac = Work dope in stretching the test-piece 
 taring ^e | Volume of the test-piece 
 
 specimen J 
 
 /Mean load between starting-point and point of\ 
 
 _\ fracture x distance moved through in inches / 
 
 ""/Area in square inches x length of test-piece in) 
 
 \ inches / 
 
 IO 
 
 _ Final length - original length 
 on" io & inches / ~" Original length 
 
 * I -^^=16-3 per cent, 
 
310 
 
 LECTURE XXII. 
 
 Single-Riveted Lap Joints. (See the author's text-book on Steam, 
 ., re Riveted Joints.) 
 Let p = pitch of rivets in inches. 
 d = diameter of rivet hole in inches. 
 
 t = thickness of plate in inches. 
 \ ft = tearing resistance or tensile stress of the plate in Ibs. per 
 
 square inch. 
 
 / 8 ^ shearing resistance or shear stress of the rivet in Ibs. pel 
 square inch. 
 
 n 
 
 SINGLE-RIVETED" T^AP JOINT. 
 
 Then for a single-riveted lap joint, 
 
 the area of plate under tensile stress = (p - d)t, 
 
 Trd~ 
 and the area of rivet under shear stress = - 
 
 4 
 Hence, for equal strength, the following equation must be true : 
 
 In practice, d = i '2 y , or = - is usually taken for boilers. 
 
 ft(p-d) =/ x i'44 = "36 x 
 
 X/r 
 
 Or, 
 
 The values to be taken for/ t and/ s in any given case depends upon the 
 number of rows of rivets ; upon the material which is used (iron or steel), 
 and on whether the holes have been punched or drilled. But, in practice 
 ft ranges from 35,000 to 67,000 Ibs. per square inch, and / from 43,000 to 
 53>oo Ibs. per square inch. Whilst for iron plates and iron rivets, with 
 
 drilled holes, the ratio may be taken as equal to '94; which would give 
 for the single-riveted lap joint (p-d)o*94= 1*131. 
 
SINGLE-RIVETED LAP JOINT. 311 
 
 EXAMPLE X. In a single-riveted lap joint the thickness of the plate ia 
 | inch, and the diameter of the rivet is I inch. If the tearing resistance of 
 the plate is 60,000 Ibs. per square inch, and the shearing resistance cf 
 the rivet is 50,000 Ibs. per square inch, find the proper pitch of the rivets. 
 
 (C. & G., 1905, O., Sec. B.) 
 
 Answer. Substituting the numerical values given by the question in 
 the above formula, we get : 
 
 (p-<Dr= 1-131 
 
 50,000 
 
 1-131 x 50,000 . . 
 J*- 1 = -- 60^55 -- =? 95 
 
 j> = '95 + i = 1'95 inches. 
 
312 LECTURE XXII. QUESTIONS. 
 
 LECTURE XXII. QUESTIONS. 
 
 1. State and define the essential and contingent properties of matter, 
 and give the names of those engineering materials with which you happen 
 to be practically acquainted, that best exemplify each property. 
 
 2. What is the meaning of the term ductility as applied to wrought iron ? 
 Describe, with sketches, some apparatus for testing a piece of metal as to 
 ductility. If a uniform bar of iron 10 inches long is found to stretch 
 i inches at the time of fracture, what is the measure of the ductility of 
 the material of the bar 1 Ans. 15 per cent. 
 
 3. Give the approximate breaking tensile stress for a bar of cast iron of 
 one square inch sectional area, and the same for a bar of wrought iron 1 
 What is the meaning of the term ductility as applied to wrought iron, and 
 how is the ductility of iron measured ? 
 
 4. What must be the diameter in inches of a round rod of wrought iron 
 in order to sustain a load of 50 tons ? It is given that a bar of iron i square 
 
 inch in section will just support a load of 25 tons. Ans.d=\/ -- = i'6". 
 
 5. What is the modulus of elasticity of a substance ? A round bar of iron, 
 12 feet long and i^ square inches in sectional area, is held at one end and 
 pulled by a force till it stretches inch ; find the force, the modulus of 
 elasticity being 30,000,000, Ans. 39,063 Ibs. 
 
 6. A round bar of steel i" in diameter and 10 feet long, is fi.xed at its 
 upper end, and a load is applied to the bottom end and stretches it -05". 
 Find the load if the modulus of elasticity is 30,000,000. Ans. 9817 '5 Ibs. 
 
 7. Find the dimensions of a transverse section of a square rod of fir to 
 sustain a suspended load of 10 tons, the rod being held vertically. The 
 breaking load of a rod of fir one square inch in section is 6 tons. Ans. 
 1-29 inches. 
 
 8. Find the extension produced in a bar of wrought iron 4 feet long and 
 2 square inches in section by a suspended weight of 4! tons, the modulus 
 of elasticity of the material being 29,000,000 pounds per square inch. 
 
 Ans. -009 inch. 
 
 9. What do you understand by the terms stress, strain, and modulus of 
 elasticity ? A tie-rod, 100' long and 2 square inches cross area, is stretched 
 .75" under a tension load of 32,000 Ibs. What is the intensity of the stress, 
 the strain, and the modulus of elasticity under these circumstances? 
 
 Ans. 16,000 Ibs. per square inch; 0-000625; 
 25,600,000. 
 
 10. Define what is meant by " dead load," " live load," " limiting stress," 
 "limit of elasticity," and " factors of safety." 
 
 11. What do you understand by stress and strain respectively? If an 
 iron rod, 50 ft. long, is lengthened by in. under the influence of a stress, 
 what is the strain ? If the rod is 2 sq. in. in section, and the load 1 1,000 Ibs., 
 what is the modulus of elasticity ? Ans. '000417 ; 13,200,000. 
 
 12. Find the stress produced in * pump-rod 4" diameter, lifting a bucket 
 28" diameter if the pressure on the top of the bucket be 6 Ibs. per square 
 inch in addition to the atmosphere, and the vacuum below the bucket be 
 26" by gauge. Beckon each 2" of vacuum = i Ib. Ans. 931 Ibs. per sq. inch. 
 
 13. If the rod in question is 5' long, find its extension if the modulus of 
 elasticity = 9,000,000. Ans. '006 inch. 
 
LECTURE XXII. QUESTIONS. 313 
 
 14. What do you understand by the terms tensile, compressive and 
 thearing strength respectively of any material? Define "modulus of 
 elasticity." If a wrought-iron bar of I square inch sectional area just 
 breaks under a tensile stress of 60,000 Ibs., what would be the area of the 
 section of a tie-rod which would just support a load of 20 tons? 
 
 Ans. 75 sq. inch. 
 
 15. A wrought-iron tie bar, f inch in diameter, has a modulus of 
 elasticity of 28,000,000 Ibs. per square inch. Its length is 23 inches ; find 
 the load under which the bar will extend '015 of an inch. Find also the 
 stress per square inch. 
 
 Ans. 8067*6 Ibs. and 18,261 Ibs. per sq. inch. 
 
 1 6. How would you find out for yourself the behaviour of steel wire 
 loaded in tension till it breaks ? What occurs in the material ? Use the 
 words "stress" and "strain" in their exact senses. 
 
 17. An iron rod, of i inch diameter and 12 feet in length, stretches 
 3/32-inch under a load of 6 tons suspended at its extremity. Determine 
 the stress, strain, and modulus of elasticity of the bar. Ans. 17, 112*3 Ibs. 
 per sq. in. ; 0.00065 ; E = 26,526,615 Ibs. per sq. in. 
 
 1 8. What do you mean by Stress, Strain, and Modulus of Elasticity T 
 
 A wire 10' long and | sq. inch in sectional area is hung vertically, and 
 a load of 450 Ibs. is attached to its extremity, when the wire stretches 
 015" in length. What are the stress and strain respectively? And also 
 the modulus of elasticity? Ans. 3600 Ibs. per sq. inch; -000125; 
 E = 28,800,000 Ibs. per sq. inch. 
 
 19. An iron wire is loaded with gradually increasing tensile loads till it 
 breaks. We want to know its modulus of elasticity, its elastic limit 
 stress, and its breaking stress. What measurements and calculations do 
 we make? (S. E. B. 1900.) 
 
 20. Sketch apparatus and describe a laboratory experiment by which 
 you could find E, Young's modulus of elasticity, for an iron wire 10 feet 
 long and 0*05 inch diameter. How would you secure the upper end of the 
 wire ? How apply the load ? And how measure the elongation ? How 
 would yon plot your results and how deduce the value of E? 
 
 (S. E. B. 1902.) 
 
 21. Describe an experiment by which you could determine E, Young's 
 modulus of elasticity, by stretching an iron wire. (B. of E. 1903.) 
 
 22. A bolt z\ inches diameter has a tensile load of 30 tons, what is the 
 stress ? What is the strain if Young's modulus of elasticity is 3 x io 7 
 pounds per square inch ? What is the elongation of a part which when 
 unloaded was 102 inches long ? (B. of E. 1904.) 
 
 Ans. Stress = 13,689*8 Ibs. per sq. inch. Strain = '000456. 
 Elongation = '0465 inch. 
 
 23. Two bars of equal length, both;of rectangular section but of different 
 materials, are firmly riveted together at their ends and subjected to a pull 
 so that they are compelled to stretch the same amount. If A 1? A 3 
 represent their sectional areas, and E lf E 2 the values of Young's modulus 
 for the two materials, show that when the pull is P Ibs. the intensities of 
 stress induced in the two bars are 
 
 PE, PE 2 
 
 and 
 
 A! Ej + A 2 B a A x Ej + A 2 E 2 
 
 respectively the limits of elasticity being not exceeded. 
 
 (C. & G., 1004, 0., Sec. B.) 
 
3*4 
 
 NOTES AND QUESTION'S. 
 
 24. A bar of mild steel, of rectangular section, is 2 inches wide and 
 J inch thick, and is 10 inches long. If Young's modulus is 12,500 tons per 
 square inch, find the amount the bar stretches when the load on it is 
 10 tons. How much work is then stored up in the bar ? 
 
 Ans. I = -008 inch ; work stored up in bar = 89*6 in.-lbs. 
 
 (C. & G., 1905, 0., Sec. B.) 
 
 25. An iron column is 12 inches in external diameter, and the metal is 
 i ^ inch thick. The load on the column is 125 tons. What is the com- 
 pressive stress in the metal ? By what amount will the column be 
 shortened, if its length is 15 feet, and if Young's modulus of elasticity is 
 12,500,000 pounds per square inch? 
 
 Ans. f c = 6630 Ibs. per square inch ; I = -095 inch. (B. of E. 1905.) 
 
 26. The following results were obtained during a tensile test of a mild 
 gteel bar f inch in diameter : 
 
 Total load on^| 
 
 
 
 
 
 
 
 
 
 the bar in V 
 
 0-88 
 
 176 
 
 2-64 
 
 S-Sa 
 
 4-40 
 
 5-28 
 
 6-16 
 
 7-04 
 
 tons . . J 
 
 
 
 
 
 
 
 
 
 Elongation on"\ 
 a length of 1 
 8 inches stated j 
 
 0-0012 
 
 O'O024 
 
 0-0035 
 
 0-0047 
 
 0-0061 
 
 0-0075 
 
 0-0088 
 
 0-0102 
 
 in inches . J 
 
 
 
 
 
 
 
 
 
 (a) Plot a curve on squared paper, going evenly through the points, to 
 show the relation between the load and the elongation. 
 
 (b) Find the load necessary to cause an elongation of 0*0040 inch. 
 
 (c) Find the total work done in inch-tons upon this 8" length of the bai 
 during the test. (B. of E. 1905.) 
 
 Ans. (b) Load = 3-0^ tons. 
 
 (c) Total work done -0-036 inch-ton. 
 
NOTES AND QUESTIONS. 
 
LECTURE XXIII. 
 
 CONTENTS. Stresses on Chains Shearing Stress and Strain Example I. 
 Torque or Twisting Moment Torsion of wires Table giving 
 the strength, moduli of Elasticity and Kigidity of various materials 
 Strength of Solid Kound Shafts Example II. Table giving the 
 Horse-Power which steel shafting will transmit at various speeds 
 Strength of Hollow Kound Shafts Kelation between the Twisting 
 Moment and Horse-Power transmitted by shafting, as well as the 
 diameter necessary to transmit a given Horse-Power Examples, III. 
 IV, Questions. 
 
 IN this Lecture we will continue the subject of " strength of mate- 
 rials," and finish the course with reasons for the shapes generally 
 given to sections of cast iron, wrought iron, and steel girders. 
 ^ Stresses on Chains. The only stress to which the sides of the 
 links of chains are subjected under ordinary circumstances, is that 
 of tension. This stress tends to bring the sides of the links closer 
 together, and consequently we find that large chain cables for 
 mooring ships (where very sudden and severe stresses are encoun- 
 tered) have a cast-iron stud or wedge fitted between the inner 
 sides of the links. These studs most effectually keep the sides of 
 the links apart, and prevent any link jamming a neighbouring 
 one. They add materially to the strength of the chain, for they 
 are in compression whilst the sides of the links are in tension. 
 Being composed of cast iron, which offers the immense resistance 
 to compression of fully 45 tons per square inch,* there is not 
 much fear of their giving way before the sides of the links. 
 
 The strength of a stud-link may be taken as equal to double 
 the strength of a rod of wrought iron, of the same diameter and 
 quality of material as that of which the chain is composed, whereas 
 the strength of an open-link chain is only about 70 per cent, of this 
 amount, even with perfect we! ding. f 
 
 In Molesworth's " Pocket-Book of Engineering Formulae," the 
 student will find at page 54 a formula for the safe load on chains 
 viz. 
 
 W=7.i^ 
 
 Where W = Safe load in tons. 
 d = Diameter of iron in inches. 
 
 * See Table of the Ultimate Strengths and Safe Working Loads given 
 In Lecture XXII. 
 t Some well-known authorities give less than 70 per cent 
 
SHEARING STRESS AND STRAIN. 317 
 
 Now, Btich a formula is very easy of application, but the student 
 should never rest content until he finds out how the constants 
 have been arrived at, and what relation the various symbols have 
 towards each other. If he refers back to the short table of 
 " Ultimate Strengths and Working Loads " given in the previous 
 Lecture, he will find opposite wrought-iron bars and under tension, 
 the value 5 tons per square inch as the safe working load. Con- 
 sequently, applying what was said above about perfect stud-link 
 chains, he will see that 
 
 rtwice the load of a rod of the same diameter and 
 quality as that of which the chain is composed. 
 
 .. W = 2 x 5 x cross area of the chain iron. 
 
 W=2 x 5 x-d*=2* 5x^x^^ = 7-8^ 
 4 7 
 
 This is near enough to the constant given by the above empirical 
 formula to enable him to see how it has been obtained. 
 
 Chains which are subjected to many sudden jerks (such as lift- 
 ing chains for cranes and slings) become in time crystalline, or 
 short in the grain, and consequently brittle and unsafe. The 
 best precaution to adopt in order to periodically remove this en- 
 forced internal condition, is to draw them once a year very slowly 
 through a fire, thus allowing them to become heated to a dull red, 
 and then to cool them slowly in a heap of ashes. This method 
 is followed at Woolwich Arsenal and some other Government 
 works. 
 
 Shearing Stress and Strain. The action which is produced 
 by shearing and punching machines on iron, steel, or copper plates, 
 &c., is to force one portion of the metal across an adjacent portion. 
 The shearing stress is the reaction per square inch opposing the 
 load or pressure applied to the shears or punch, and the shearing 
 strain is the deformation per unit length or volume. Rivets 
 holding boiler plates together, fulcra of levers, the pins of the 
 links of the chain of a suspension bridge, the cotter keys of a 
 pump rod, are all subjected to shearing stresses and strains. The 
 ultimate and the working shearing stresses for a few engineering 
 materials were given in a table in Lecture XXII. 
 
 In the case of loaded beams (which we will consider shortly in 
 connection with bending moments) the shearing force at any point 
 or any transverse section thereof is equal to the algebraical sum 
 of all the forces on either side of the point or section. 
 
 EXAMPLE I. A steel punch i" diameter is used in a large 
 shipyard punching machine to make holes in steel plates i" thick. 
 What will be the total shearing stress or least pressure required ? 
 
3r8 
 
 LECTUKE XXI1T. 
 
 ANSWER. Referring to the table in last Lecture, we see that the 
 ultimate shearing strength or shearing stress for steel bars (which 
 we will assume to be the same as for plates) is 30 tons per square 
 inch. Now a hoi 3 i" diameter has a circumference = ird 
 = 3* 14", and since the plate is i" thick, the area of the resisting 
 section must be the circumference of the hole x its depth, or 
 = 3' 14" x i" = 3'i4sq. in., and the total pressure required = 30 tons 
 x 3-14 = 94-2 tons. 
 
 Torque, or Twisting Moment.* In the case of a shaft 
 having a lever, pulley, or wheel fixed to it with a force P Ibs., 
 applied at radius R feet from the centre of the shaft, then 
 The twisting moment T.M. is = P x R Ibs.-feet. 
 
 Or, The torque = P x R x 12 Ibs.-inches. 
 
 * The term torque was devised by the late Professor James Thomson, of 
 Hlasgow University, to signify the twisting or torsional moment. The 
 Ibs. -feet of torque must not be confused with ft. -Ibs. of work ; or with 
 resilience (which is the work done in straining a body, as measured by the 
 elongation or compression in feetxthe mean load causing the strain). It 
 will therefore save confusion, if we take the force applied at the end of 
 the arm in Ibs., and the leverage or arm in inches, and then multiply them, 
 so as to get the torque in Ibs. -inches. 
 
 APPARATUS FOR SHOWING THAT THE ANGLE 
 
 OF TORSION IN ROD is PROPORTIONAL TO 
 
 ITS LENGTH, AND THE VARIATIONS 
 
 IN THE RIGIDITY OF DIFFERENT 
 
 MATERIALS. 
 
TORSION OF WIRES. 
 
 Torsion in Bods. It will be seen from the previous figure, that the 
 front puller carries a cord with known weights suspended from it. The 
 spindle of this pulley moves in ball bearings, and carries at its inner end a 
 three-jaw chuck. This chuck is for holding one end of the rod under test, 
 whilst the ther end is for tightly clamping it to the back bracket of the 
 machine. The torsional couple is applied by means of weights hung from 
 the afore-mentioned cord, and the torsion in degrees is read off on the dial. 
 To eliminate errors the weight should be hung first on one side of the 
 pulley, then on the other side, and the mean of the two readings taken. 
 It will be noticed, that both the pointer and the scale can be readily moved 
 to any distance from the fixed end of the rod. 
 
 It can thus be shown that the angle of torsion in a rod is proportional 
 to its length and that different materials have different rigidities. 
 
 Torsion of Wires. In the figure, AB represents a wire 
 held firmly at the top end of a supporting rod, which is 8 feet 
 long. A pulley is fixed firmly to the wire at B, and this pulley 
 is acted upon by two cords which tend to turn it without moving 
 its centre sideways, i.e., they act on the pulley with a turning 
 moment only. But, the pulley can only turn by giving a twist 
 to the wire. Hence, if a light pointer be fastened to the wire at 
 0, the former moves over a dial, and the angle turned through by 
 the wire and the pointer is called the total angle of twist at 0. 
 Similar pointers fixed to the wire at D and E give the angle of 
 torsion at these points. The dials at D and E are supported upon 
 adjustable sliders, so that they may be moved up or down the 
 vertical rod in order that the torsion of any length of rod may be 
 measured. 
 
 (i) If the length AE be i foot, and the distance between the 
 dials E and D, D and C, be also i foot, then we should find that the 
 angles of twist at E,D and are as i : 2 : 3 respectively. That 
 is, the angle of twist is proportional to the length of wire twisted. 
 
 (ii) By varying the twisting moment and noting the angle of 
 twist which is produced at each variation, you can prove that the 
 angle of torsion or twist is proportional to the twisting moment. 
 
 (iii) If you try different sizes of wires of the same material and 
 length, and apply to each of them the same twisting moment, then 
 you will find that the amount of twist produced in them, will be 
 inversely as the fourth power of the diameter of the wires. 
 
 (iv) If you take wires of the same diameter and length, but of 
 different materials, and apply the same twisting moment to them, 
 then you will find that the amount of twist will be inversely pro- 
 portional to the modulus of rigidity of the material. 
 
 Note, the Modulus of Rigidity of a Material in Ibs. per square 
 inch may be defined as the ratio of shearing stress to shearing 
 strain. This constant or coefficient of shearing elasticity for each 
 particular material is indicated in the following table by the 
 letter 0. 
 
320 
 
 LECTUEE XXIII. 
 
 EXPEEIMBNTAL APPAEATUS FOB MEASURING TH] 
 
 TOESION OF WIRES. 
 
STEENGTH OF SOLID ROUND SHAFTS. 
 
 321 
 
 BTBBNGTH, MODULI OF ELASTICITY, AND RIGIDITY OP VARIOUS 
 MATERIALS. 
 
 Materials. 
 
 Breaking strength 
 to resist tension, 
 n tons per square 
 inch. 
 
 Modulusof elasticity, 
 E, in tons per 
 square inch. 
 
 Modulus of rigidity. 
 C, in tons per 
 
 square inch. 
 
 Aluminium - bronze 
 
 ) 
 
 
 
 (90% copper Und 
 
 40 
 
 6500 
 
 2500 
 
 10% aluminium) . 
 
 J 
 
 
 
 Brass wire . . 
 
 20 to 25 
 
 5000 to 6500 
 
 2OOO tO 23OO 
 
 Oast-iron .... 
 
 5 { o 15 
 
 4500 to 7000 
 
 1700 to 2700 
 
 Charcoal-iron (hard- 
 
 V "?S to 40 
 
 
 
 drawn) .... 
 Charcoal-iron 
 
 I OJ *T 
 
 \ 
 
 12,500 to 13,500 
 
 5000 to 5500 
 
 (annealed) . . 
 
 } 30 
 
 
 
 Copper (cast) . . . 
 
 8 to 12 
 
 5000 to 6000 
 
 1900 to 2300 
 
 (rolled) . . 
 
 13 to 16 
 
 5500 to 7500 
 
 2100 tO 2900 
 
 wire (annealed) 
 
 18 to 20 
 
 
 
 
 
 (hard-drn.) 
 
 26 to 30 
 
 
 
 
 
 Delta Metal (forged) 
 
 22 tO 24 
 
 6350 
 
 234O 
 
 German-silver wire . 
 
 30 
 
 4800 
 
 l8OO 
 
 Gun-metal (90% cop- 
 per and 10% tin) . 
 Muntz metal (rolled 
 or forged) . . . 
 
 V 12 tO 17 
 
 \ 22 
 
 5000 
 
 6350 
 
 IQOO 
 2340 
 
 Phosphor bronze 
 
 16 to 18 
 
 ) 
 
 
 wire 
 (hard-drawn) 
 
 J- 45 to 70 
 
 > 6000 to 7000 
 
 2300 to 2700 
 
 Platinum .... 
 
 
 
 10,500 
 
 4000 
 
 Steel (ordinary) . . 
 (annealed) . . 
 
 7o 
 
 IOO 
 
 \ 13,000 to 14,000 
 
 5200 to 5700 
 
 Strength of Solid Round Shafts. It is evident trom the 
 above, that a shaft subjected to a twisting moment must offer a 
 sufficient resistance thereto, otherwise it would be twisted, or 
 sheared, or ruptured through by the torque. It may be proved 
 that in the case of solid round shafts their resistance to torsion 
 is directly proportional to the cubes cf their diameters when made of 
 the same material and quality.* 
 
 * This is evident from the fact that the shaft must offer a moment of 
 resiitance, or shearing moment, equal to the twisting moment at the instant of 
 rupture. Now, the area to be sheared, is the cross area of the shaft 
 
 =!LD a . where D is the diameter of the shaft. The mean arm or leverage at 
 
 4 
 
 which this resistance acts is equal to half the radius of the shaft, for at the 
 centre the arm is=o, and at the circumference it is=r, the radius of the shaft. 
 
 The mean arm is therefore= 
 
 And, if the shearing resistance per 
 
 tauwre inch of cross section of the material be=/, the product of these 
 three quantities will be the total thearing moment, and must equal the 
 twitting moment viz. =P x R, where P is the force applied at the end of the 
 
322 LECTURE XXIII. 
 
 Let Dj, !> D, - Diameters of three shafts, i", 2", and 3" dia- 
 
 meter respectively. 
 Tj, T r T 3 Torques which they will respectively resist 
 
 when stressed to the same extent. 
 Then, . . T^T.rT.iiD/iD/.-D/ 
 Or, . . . T^T,:!,:: i : 2' : 3 
 
 ::i : 8 : 27. 
 
 In -other words, the strengths of the three solid shafts will be as 
 1:8:27. 
 
 A good wrought- iron shaft of i" diameter has been found to 
 withstand a torque of 800 lbs,-ft., or 9600 Ibs.-inches, which means 
 that they will resist 800 Ibs. force at i foot, or 12" leverage, or 
 400 Ibs. at 2 feet, or 24", and so on. 
 
 Or, . P x R' = 800 Ibs.-feet of torque 
 i.e., . P x R" = 9600 Ibs. -inch torque. 
 EXAMPLE II. On the above basis, what force acting at the 
 circumference of a pulley 20" diameter will break a wrought- 
 iron shaft 2" diameter? 
 
 ANSWER. By the above rule we have the proportion : 
 
 T I: T, -D^iD/ 
 
 But Tj - P! x R/' = 800 Ibs x i a" 
 AndT,= P 8 x R 8 " - P, x 10" 
 
 ... P^: P 2 R, :: J>*:J>f 
 i.e., P a R a x D^ = P^ x D, 1 
 
 P.R. x D, 8 800 x 12" x 8 
 
 <0r > p . - iy x R, -nr^" ' 76801bs 
 
 lever or circumference of the pulley, and K the length of the arm or radius 
 of the wheel or pulley. 
 
 Consequently, P x E =-J^( ^-D 2 x - \ =f g "- ' W 
 
 But S is a constant quantity for any particular material. Also, TT and 16 
 are constants. .. P x R vary as D 3 . 
 
 At the instant of rupture the strength of the shaft just balances or is 
 equal to the twisting moment P x R. 
 
 .*. The strength of shaft varies as D 8 . 
 
 This is the same as the general statement in the text above. Without 
 Borne such algebraical explanation, students are sorely puzzled how the 
 cube of the diameter crops up ; or still more so when they see the following 
 which appears in some text-books. 
 
 :Such a statement is, however, quite evident after the above analysis. (We 
 must leave the consideration of hollow shafts, tubes, &c.. to our Advanced 
 Course.) 
 
H.P. TRANSMITTED BT STEEL SHAPTa 
 
 323 
 
 
 
 5-1 C? 
 
 02 
 
 I 
 
 JOd 'SAag 
 
 O W Tj-vO OO O CM 'i-vO OO O CM **-$ 00 OvoOvoQVOOvoOvoO VO < 
 MMMMIH<ieteiaCICQCOCOCOrO*Ttn vovO VO tx 1x00 00 Ov OM 
 
 ONVOMVO WOO TfO VOM fx CO ON *}- O vO O vo ON r*- ON COOO C* tx, C1 <O O 
 
 8? 5,3,8 #cT2 3S3 |&lf i&SfHf5' 
 
 MMtHHMHMMOOIOiOIOIMCOCO^^-^LOVO vovO vO vO Cx 
 
 * tx O\ CI 1-v> O\ H -<j-\ 
 IH M M CH W M CJ COCO 
 O tvOO CS O i-i CO ^*- 
 
 00 M CO "ICO 
 
 <J- OvO C1 00 -<1- O VO 00 <* O 
 
 OvO M K.WOO ^ON'OO^O 01 
 
 CO OOO O CO lOOO O COVO OO H 
 
 oi cocococoTi-^-Ti-Ttvn 
 
 CO C1 O Ov CxVO vo CO O Ov fxvO VO CO CO vovO 00 O\ M C1 TJ- LO CxOO O 
 *MOO * MOO VOW 0\vO N ONO COO CXTj-MOO VON O tx-t-MOO VOCO 
 CO TJ- TJ- vovO vO txOO OOO\OOHCOCOvo C^.00 O N Tj-iOtxOvO N J* 
 
 aize. 
 
 "g I "8 
 
 \r> O lOO ")O OO *O O OO lOO lOO CO OOO O CO U10O O CO lOOO O 
 
 ci tot^O w lotxo M iotxO oi lOt^OO woo o M r^cogvb ooo ^o 
 
 M H M CI 0* 01 COCOCOCO-'i-Ti-rl-Tj-iO lOvO \O txCO 00 Os O O w HI M 
 
 ' 
 
 Ot 01 COCOCO^-^trfvO' 
 
 & 
 
 Scl 
 
 3 
 
 1 
 
 I 
 a 
 
 \ 
 
 vp op ON p ci ^ jovp oo p\ p M co t ^ yo co M ON txvp .""t- w p op vp ^ ji 
 
 VOOO M vooo M Vf Vx O COvb O COvb ON C1 O 00 vo CO HI O\ Cx, vo W O 00 vb 
 M HI C1 01 COCOCO^^-^vovovo vovO tx txOO ONO O HI Ot COrfrfvo 
 
 ONM ti 
 
 ' . M f* y^^p <p P f P i p 7*" r r* j 1 *p p > ^ ?*& 
 
 For 
 
 co^yoyovp fxoo ON ONO M H p -<f ^ M oo joeioo yoei ONVO ON\p 
 
 *O<O "^00 ON O HI C1 CO vovb txOO ON O M Vfvb ON 01 V Vx O C1 VOOO O CO 
 SMMMMMI-IIHHIOICOCOCO^-^^TJ-VOVO 
 
 fO O Jx jf O tx ;< M t . rj- M op JO 00 JO ONvp CO ONvp O Cx .* M t* 
 CO * <* vovfl VO txOO WONOOMC1C1CO vovO 00 O HI CO VO 1x00 O Ct CO 
 
 S 8 
 
 18 
 
324 LECTURE XXIII. 
 
 Strength of Hollow Bound Shafts. The shafts of large land 
 and marine engines are sometimes made hollow. The ratio between the 
 torsional strength of a solid round shaft and a hollow one i as follows : 
 
 Let D = outer diameter of either shaft. 
 
 d = inner diameter of the hollow shaft. 
 
 D 4 -d 4 
 Then, their strengths are as D 8 : - jj- , 
 
 and their weights are as D : D a - &. 
 
 Relation between the Twisting Moment and Horsa 
 Power Transmitted by Shafting, as well as the Diameter 
 Necessary to Transmit a given Horse Power. 
 
 Let P=the force of the twisting couple, either constant or the 
 mean value if variable, in pounds. 
 
 R=the length of lever arm of the twisting couple in feet. 
 N = revolutions of shaft per minute. 
 ,,T.M. = mean twisting moment = PxR Ib.-feet. 
 /, = shearing stress per square inch on cross-section of shaft. 
 
 Then, T.M. = resisting moment = d s f s = -i<)6d%. 
 
 But, work done per minute = P x 27rRN = T.M. x 2irN ft. -Ibs. 
 Also, work done per minute = H.P. x 33,000 ft. -Ibs. 
 .-. T.M.X27rN = H.P. x 33,000 
 
 Or, T.M. = H.P.x 33.000 ^^ = 12 x 33,000 xH.P. lbB< . inoheB< 
 
 2?rN 
 
 Hence, T. M. = 63, 030*^1 Ibs.-inches = 63 3 x H ' T P - = 28 H ' F> ton-inches. 
 N 2240 x N N 
 
 But, T.M.=^|&. 
 
 Hence, ^'=63,03^ 
 
 If we assume the safe values of the stress f t to be as follows : cast-Iron 
 3600 Ibs. per square inch ; wrought-iron = 9000 Ibs. per square inch ; 
 and steel * 13,500 Ibs. per square inch, then the diameter in inches for a 
 round shaft in terms of the horse-power to be transmitted is, for Oast-iron 
 
 Of course, the twisting moment is here assumed to remain constant at its 
 mean value. In practice the twisting moment varies in many cases, and 
 to allow for this it is usual to take the maximum twisting moment from 
 i -3 to 1*5 times the mean twisting moment, thus the values of diameter d 
 as found above are slightly increased. 
 
 It should also be borne in mind that shafts in practice are subjected to 
 bending as well as twisting, owing to the loads due to the weights of 
 
STRENGTH OF SHAFTS. 325 
 
 pulleys and the pulls of the belts. Hence, if the usual rule for the dia- \ 
 
 3 /TT p 
 
 meter of a wrought-iron shaft is <*=3'3/y ^7- - when torsion only is con- 
 sidered, then it wiU be d=ex3'3/y S^ 1 when tending is taken into 
 account. Some values of the coefficient c are given in the following table : 
 
 Kind of Shaft. 
 
 Value of c. 
 
 Propeller shafts of steamships, and shafts with similar 
 load ... 
 
 ria 
 
 Line shafting in mills, etc 
 
 i '3 
 
 Crank-shafts and shafting subjected to shocks, such as 
 
 i -42 
 
 
 
 EXAMPLE III. (a) Find the diameter of a solid steel propeller shaft to 
 transmit 12,000 H.P. at 80 revolutions per minute. (6) If the shaft is to 
 be hollow, find its external diameter, from strength considerations, when 
 its internal diameter is two-thirds of its external diameter. 
 Answer. Let D 1= r diameter of the tolid steel propeller ihaft. 
 
 D= outer diameter of the hollow steel propeller shaft. 
 a = inner diameter of the hollow steel propeller shaft. 
 N= number of revolutions per minute of shaft. 
 (a) Keferring to the previous article on the relation between the dia- 
 meter of shaft necessary to transmit a given horse-power, we deduced the 
 following formula for a solid round steel shaft, when the stress // was 
 assumed as 13,500 Ibs. per square inch. 
 
 Substituting the numerical values given by the question, we get 
 . =2-9 //?=2-9 / i5o= 15-4 inches. 
 
 But, to allow for variations in stress, bending moments, and shocks to 
 which the propeller shaft may be submitted, we find from the table that 
 the value of the coefficient c is 1*13. 
 
 Hence D l = cd= 1-13 x 15*4= 17*4 inches for the solid round shaft. 
 
 II P 
 (6) From the above, we see that <P=(2-g)* - Also, from the relation 
 
 of the strength of solid to hollow round shafts, we get, d? : 
 
 ~ 
 
 ~ = (2-9? 5^- for hollow steel shafts subjected to torsion onfy~> 
 D 4 -d 4 H.P. 
 
326 LECTURE XXIII. 
 
 Jfow, substituting the values given by the question, we get 
 
 Or, iD a *= 
 
 .. P a 8 = 3x2158-5 =6475-5. 
 
 . . D a = #6475-5 =18-64 inches. 
 If bending is taken into account, then 
 
 D 2 = 1-13 X 18-64 =21 inches for the hollow round shaft. 
 
 EXAMPLE IV. The screw shaft of a marine engine is 10 ins. diameter, 
 and the revolutions 100 per minute. It is replaced by twin screw shafts 
 rotating 500 times a minute. If the total horse-power developed in the 
 two cases be the same, and the working stress is also the same in the twin 
 screw shafts as in the single screw shaft, find the proper diameter of 
 shafts in the second case and compare their weights. 
 
 (C .&(*., 1905, O.3a,fr, 
 
 Answer. 
 
 Let D! = diameter of single screw shaft. 
 n D 2 = diameter of one of the twin screw shafts. 
 ,, N! = number of revs, per minute of single shaft. 
 N 2 = twin screw shafti. 
 
 H.P. = total horse-power to be developed in each case. 
 ft. = working stress in Ibs. per square inch. 
 
 Now, Dj = \/ $'i x ^ ; ~ * 
 
 and, D, = // x " x 33-o J*f. 
 
 5-1 x 12 x 33,000 x H.P. 
 
 .-. IV = ZTTNj/ 
 
 D a 8 5-1 x 12 x 33,000 x H.P. 
 
 Or 
 
 in" 
 
 i.e., D x = 1-442 D 2 . . . D a = ~ = 7" 
 
 But, the weights of shafts are proportional to their cross sectional areas 
 if their lengths are equal. 
 
 | Dj a (1-442 D 2 ) a _ 2-08 1-04 
 
 Hence, ^-, = -^r~ ~T ~ ' 
 
 Therefore, the weights of the shafts in the two cases are approximately the 
 same, but the diameter of th^ single screw shaft is nearly i times the 
 diameter of one of the twin screw shafts. 
 
LECTURE XXm. QUESTIONS. 327 
 
 LECTURE XXIII. QUESTIONS. 
 
 1. An open link chain is constructed of round wrought-iron rod, inch 
 in diameter ; calculate what is the probable breaking load of the chain. 
 Wrought-iron chains are liable to deterioration by constant use ; what 
 change do they undergo, and what precaution is taken to prevent their 
 breaking 1 Ans. io tons. 
 
 2. A steel punch inch in diameter is employed to punch a hole in a 
 plate f inch in thickness. What will be the least pressure necessary in 
 order to drive the punch through the plate when the shearing strength 
 of the material is 35 tons per square inch ? An*. 51*56 tons. 
 
 3. Define what is meant by ** shearing etress and* strain," "torqifc. n- 
 j~?1tiDg moment." Show by an example that a shaft subjected to torqu* 
 iaars a shearing stress tending to sever it afe right angles to its axis. 
 
 4. What is meant by the twisting moment " of a shaft ? If a wrought 
 iron shaft i inch in diameter breaks in torsion by a force of 800 Ibs. at the 
 end of a lever i foot long, what force at the end of a lever 2 feet long 
 will break a shaft of the same material, but 2 inches in diameter 1 Find 
 also the diameter of a wrought-iron shaft to resist a force of 2 tons at a 
 distance of 18 inches from its centre. Ans. 3200 Ibs. 2 inches full. 
 
 5. If a shaft, 2 inches in diameter, is found equal to the transmission of 
 4 horse-power, what amount of power can be transmitted by a shaft 4 
 inches in diameter, all other questions remaining the same ? 
 
 Ans. 32 horse-power. 
 
 6. If a revolving shaft, which is 2 inches diameter, is found sufficiently 
 strong to transmit 4 horse-power, how much power may be transmitted 
 by a shaft which is 3 inches in diameter, supposing all the other conditions 
 to be the same, and that the iron of both shafts is subjected to the same 
 stress? Ans. 13*5 H.P. 
 
 7. If 8co Ibs. at the end of a 12-inch lever be a safe stress to apply to a 
 wrought-iron bar one square inch in section, find the effort which a shaft 
 2 inches in diameter can transmit at the circumference of a pulley one 
 foot in diameter, and making 300 revolutions per minute. Find also the 
 horse-power transmitted. Ans. 8893 Ibs. ; 254 H.P. 
 
 8. If a wrought-iron shaft of i inch diameter is broken by the torsion of 
 a load of 800 Ibs. acting at the end of a 12-inch lever, find the weight 
 which, when applied to the end of the same lever, would break a*shaft of 
 the same material, but 3 inches in diameter. State, in general terms, the 
 reasoning by which you arrive at the result. Ans. 21,600 Ibs. 
 
 9. Suppose that a shaft of i inch diameter may be safely subjected to 
 a torque of 2000 Ib. -inches; what torque will a 2j inch shaft safely 
 resist ? Calculate the horse-power which may be safely transmitted by the 
 latter shaft if its speed is 150 revolutions per minute. (B. of E., 1902.) 
 
 Ans. 22,780 Ib. inches; 54 horse-power. 
 
 10. A wire of Siemens' steel O'i inch diameter is to be twisted till it 
 breaks. Sketch the arrangement and show how the angle of twist and 
 the twisting moment are measured, how the results may be plotted on 
 squared paper, and the sort of results that may be expected. In what 
 way may a wire of twice this diameter be expected to behave f 
 
 (B. of E. 1901.) 
 
328 LECTURE XXIII. QUESTIONS. 
 
 n. If a shaft 4 inches in diameter will safely withstand a torque of 
 120,000 lb. -inches, what torque would a g-inch shaft take? What H.P. 
 would the former transmit at 200 revolutions per minute, and what would 
 the latter transmit at 50 revolutions per minute ? (B. of E., 1903). 
 
 Ans. Torque = 1,367,000 lb. -inches ; H.P a = 3847 and H.P. 2 = uio. 
 
 12. Compare the strengths and weights of a solid wrought-iron shaft 
 and a hollow steel shaft of the same external diameter assuming the 
 internal diameter of the hollo w shaft half the external, the working stress 
 of steel ij times that of iron, and the densities of wrought-iron and steel 
 to be the same. (C. & G-., 1903, O. , Sec. B.). 
 
 ^ ns Strength of solid W.I. shaft_32_ i 
 
 Strength of hollow steel shaft ~ 45 ~ 1^4* 
 
 Weight of solid W.I, shaft _ 4 = ij 
 Weight of hollow steel shaft" 3 T~" 
 
 13. The propeller shaft of a vessel, whose engines develop 1000 horse- 
 power at 60 revolutions per minute, is 8" dia. Assuming the shaft sub- 
 jected to pure torsion, and that the maximum twisting moment on the 
 haft is I J times the mean, estimate the maximum shear stress induced in 
 the shaft (C.&G., 1904,0., Sec. B.) 
 
 Ans. Max shear stress = 13, 050 Ibs. per sq. inch. 
 
 14. The figure shows the skeleton mechanism of a direct-acting steam- 
 engine. A is the cross -head, B is the connecting-rod, and C is the 
 crank. 
 
 The connecting rod is 4 cranks long, and in the position shown in the 
 figure, the crank has turned through an angle of 45 from the dead centre 
 in a clock-wise direction. 
 
 - A force F, due to the steam pressure on the piston of 12,000 pounds, 
 acts upon the cross-head. Find graphically, or in any other way, the 
 thrust in the connecting rod, and the magnitude of the force It between 
 the cross-head and slide bar. All friction to be neglected. 
 
 (B. of E. 1905.) 
 
 Ans. F = ^12,200 Ibs. ; B = 12,430 Ibs. ; and B = 2220 Ibs. 
 
 15. Describe, with a sketch of the apparatus, how you would experi- 
 mentally determine the law connecting the twisting moment and the angle 
 of twist for a mece of steel wire. (B. of E. 1905.) 
 
LECTURE XXIY. 
 
 CONTENTS. Hooke's Coupling or Universal Joint Double Hooke's Joint 
 Sun and Planet VTheels Cams Heart Wheel or Heart-shaped Cam 
 Cam for Intermittent Motion Quick Return Cam Example 
 Pawl and Ratchet Wheel Reversible Pawl Masked Ratchet Silent 
 Feed Watt's Parallel Motion Parallel Motion Questions. 
 
 IN this and the following Lecture we shall examine a few of the 
 many devices for transmitting circular motion and for converting 
 it into rectilinear motion, or vice versd, together with other 
 miscellaneous mechanisms. 
 
 Hooke's Coupling or Universal Joint. This is a contrivance 
 sometimes used for connecting t\vo intersecting shafts. Each of 
 
 the shafts ends in a iork, F 1? F,, 
 which embraces two arms of the 
 crosspiece, O. The four arms of 
 this cross are of equal length. As 
 Cj rotates, F x and F 2 describe 
 circles in planes perpendicular to 
 their respective axes. Since these 
 planes are inclined to each other 
 the angular velocity of C 2 at any 
 
 HOOKE'S JOINT. instant is different from that of C 1 , 
 
 but the mean angular velocities are 
 
 equal to one another, because at one instant C 2 goes faster than C t , 
 and at another slower. This joint will not M ork when the two 
 shafts are inclined at 90, or any smaller angle, to each other. 
 
 Double Hooke's Joint. The variable velocity ratio obtained 
 with a Hooke's joint may be obviated by the use of two joints 
 instead of one. The forks are connected by an intermediate link, 
 C 2 , which must be carried on corresponding arms of the two 
 crosses, as shown in the next figure. If the intermediate shaft 
 be equally inclined to the other 
 two shafts, the irregularities caused 
 in the motion by its transmission 
 through the first coupling are 
 exactly neutralised by the equal 
 and opposite ones caused by the 
 second joint. The first and third 
 
 DOUBLE HOOKE'S JOINT. 
 
330 
 
 LECTURE XXIV. 
 
 shafts, therefore, revolve with the same velocity at every instai t. 
 The double joint works equally well whether the two extrer e 
 axes are inclined as shown in the figure, or are parallel to ea h 
 other but not in line. 
 
 Both the single and double Hooke's joint are, as a rule, used 
 only for light work, such as for astronomical instruments. 
 
 Sun and Planet Wheels. This device was invented by Watt 
 to convert the oscillatory motion of the beam in his engines into 
 the circular motion of the flywheel. As will be seen from the 
 
 SUN AND PLANET WHEELS. 
 
 first figure, it consists of a wheel D, rigidly fixed to the connect- 
 ing rod D B, and kept in gear with another wheel C, by the 
 link DEC. The wheel C, is keyed to the flywheel shaft. As 
 the beam oscillates up and down, the connecting-rod pulls D up 
 one side of C, and pushes it down the other. It thereby causes 
 C to rotate, and with it the shaft and flywheel.* 
 
 Cams. Cams are usually of the form of discs or cylinders. 
 They rotate about an axis, and give a reciprocating motion to 
 
 * See Vol. I., Lecture XIX., of the author's text-book on "Applied 
 Mechanics " for a description of epicyclic trains and the application of the 
 formula to this case. Watt first applied this motion to his " Double Acting 
 Steam Engine" in 1784. See Lecture XVIII. of the author's elementary 
 manual on "Steam and the Steam Engine." 
 
HEART WHEEL OR HEART-SHAPED CAM. 
 
 331 
 
 Borne point in a rod by means of the form of their periphery or 
 surface, or by grooves in their surface. 
 
 The cam generally revolves uniformly round its axis, whilst the 
 reciprocating motion may be of any nature, depending on the 
 shape of the cam, and may be in a plane inclined at any angle to 
 the axis of rotation. In the following examples, uniformity of 
 rotation is assumed in the case of the cam, and the motion of the 
 reciprocating piece takes place in a plane perpendicular to the 
 axis. 
 
 Heart Wheel or Heart-shaped Cam. Suppose that it is 
 required to give a uniform 
 reciprocating motion to a 
 bar moving vertically be- 
 tween guides, and in a line 
 passing through C, the centre 
 of motion of the cam plate. 
 
 Let the sliding bar be at 
 its lowest position, as shown, 
 and when in its highest 
 position let its extremity be 
 at the point 6. The distance 
 thus moved is called the 
 travel and will be passed over 
 during one-half revolution 
 of the cam. The required 
 curved outline may be ob- 
 tained in the following 
 manner : "With centre C, 
 describe circles passing 
 through the extreme posi- 
 tions of the end of the rod. 
 Divide the travel into, say, 
 six equal parts at the points 
 
 i, 2, 3, (fee. Divide the semi-circumference into the &ame number 
 of equal parts by radial lines C i', C 2', <fec. Then with centre C, 
 draw the concentric arcs i, i'; 2, 2'; <fcc., intersecting these radii 
 in the points i', 2', 3', &c. The dotted line drawn through these 
 points will represent the required curve. 
 
 If the end of the sliding bar rests on this curve it is clear, that 
 for equal angles turned through by the cam, the bar will move 
 outwards through equal distances, and consequently, will have 
 uniform linear motion imparted to it. The return motion will 
 evidently be obtained by the similar and equal curve i", 2", 3", 
 <fec., on the opposite side of the cam. 
 
 , 6" 
 HEART-SHAPED CAM. 
 
332 
 
 LECTURE XXIV. 
 
 A cam so formed would impart the required motion to a point. 
 If the end of the sliding bar be provided with a roller in order to 
 diminish the friction, then the shape of the cam must be altered 
 so that the centre of the roller shall move over the outline of the 
 cam as traced above. To accomplish this, we must draw a curve 
 inside the original one by describing small arcs with centres on 
 the original curve as at i', 2', 3', &c., with a radius equal to that 
 of the roller, and then by drawing a smooth curve touching these 
 arcs, as shown by the heavy line in the figure. 
 
 CAM GIVING AN INTERVAL OF RUST. 
 
 CAM GIVING A QUICK RETURN. 
 
 Cam for Intermittent Motion. Sometimes the motion 
 imparted by a cam is intermittent. For instance, a common 
 form of lever punching machine is fitted with a cam which gives 
 the punch an upward movement, then a period of rest, nnd 
 finally a downward movement during each revolution. As an 
 example of this, let us set out a cam to impart vertical motion 
 to a bar, so that the latter shall be raised uniformly during the 
 first half revolution, remain at rest during the next one-sixth, 
 and descend uniformly during the remainder of the revolution. 
 
 As before, suppose the reciprocation to be in a line passing 
 through C, the centre of motion of the cam plate. Then, with 
 centre C, draw circles passing through the extreme positions of 
 the end of the bar. Divide the circumference into three parts 
 corresponding to the periods of one-half, one-sixth, and one-third 
 revolution, by drawing radial lines making angles of 180, 60, 
 and 120. Since the motion is to be uniform, divide the travel 
 
QUICK RETURN CAM. 
 
 333 
 
 into a convenient number of equal parts, say twelve ; and the 
 
 circumference into the same number of equal parts by radial lines. 
 
 Draw the concentric arcs 2, 2"; 4, 4"; &e., and 3, 3'; 6, 6'; 
 
 as shown. The curves through 
 
 the points so determined will 
 
 give the required motions. The 
 
 interval of rest will evidently be 
 
 given by the circular portion 
 
 from 12" to 12'. The complete 
 
 outline is represented by the 
 
 heavy line in the diagram. 
 
 Quick Return Cam. The 
 student will readily understand 
 from the right-hand figure, that 
 if two-thirds of a revolution be 
 occupied in raising the motion 
 bar and the remainder in lower- 
 ing the same, the return stroke 
 will be performed in half the 
 time of forward stroke. The 
 curves of this cam are found in 
 the same way as in the previous 
 examples. 
 
 EXAMPLE. A vertical bar, 
 moving in guides, is driven by 
 a circular cam plate having a 
 centre of motion in the centre 
 line of the bar. The distance 
 from the centre of motion to the 
 centre of the plate is 2 inches, 
 and the bar exerts a pr. ssure of 
 10 Ibs. when rising, but falls by cc p 
 its own weight. Find the work 
 done in 100 revolutions of the 
 olate. 
 
 VB 
 
 SR 
 
 CP 
 CM 
 
 CIRCULAR CAM PLATE. 
 
 INDEX TO PARTS. 
 
 G represents Guides. 
 
 Vertical bar. 
 Sliding roller. 
 Circular cam plate. 
 Centre of plate. 
 Centre of motion. 
 
 ANSWER. Since the distance between the roller S R, and the 
 centre of the plate C P, remains constant as the plate revolves, 
 it is evident that the bar will move as if it were actuated by a 
 crank of length equal to the distance between C M and C P, and 
 a connecting-rod of length equal to the radius of the plate. Hence, 
 the stroke of the bar will be 4 inches, or J foot i.e., twice the 
 length of the equivalent crank. Neglecting friction, the work 
 done in raising the bar by one revolution of the plate, will be : 
 
334 LECTURE XXIV. 
 
 Pressure x distance moved 10 x J ft.-lbs. 
 .*. Work done in 100 revolutions = 100 x ioxj = 3383 ft.-lbs. 
 
 Pawl and Ratchet Wheel. A toothed wheel which is acted 
 upon by a vibrating piece, termed a click or pawl, is called a ratchet 
 wheel. Ratchet wheels are made in many different forms, and an* 
 
 PAWL AND KATCHET. 
 
 used for a variety of purposes. For instance, clocks and watches 
 are usually provided with ratchet wheels to allow the spring or 
 weight to be wound up, without disturbing the rest of the works, 
 and they are used to drive the feeding arrangements of many 
 machines. When, as in the latter case, the click or pawl drives 
 the ratchet wheel, it is carried on a vibrating arm. In the first 
 figure, A B is the vibrating bar which drives the ratchet wheel, 
 by means of the click B C, and teeth C C, when moving in the 
 direction shown by the arrow. When A B moves back to A B', 
 the click slides over the top of the next tooth and drops behind it. 
 It is then ready to drive the wheel through the space of another 
 tooth when A B again moves forward. While the pawl is moving 
 back from B to B', the wheel is prevented from moving with it 
 by another pawl or detent, b C. In this case, the vibrating bar is 
 on the same axis as the ratchet wheel ; but this is not always 
 430. The reactions between the teeth and the pawl keep ihern in 
 
REVERSIBLE PAWL. 
 
 335 
 
 contact with each other. The resultant pressure of the teeth on 
 the pawl must therefore be such, that its moment tends to turn the 
 pawl towards A, the centre of the ratchet wheel. This condition 
 evidently is satisfied if C D, the direction of the resultant pressure 
 at C, passes between A and the axis B, about which the pawl 
 turns. Similarly, the moment of the resultant pressure on the 
 detent must trend to turn it towards A, but its direction, d C (not 
 C d), must lie outside A 6, because this detent ends in a hook* 
 
 REVELSIBLE CLICK. 
 
 Both pawls might have been like B C, which acts by pushing, or 
 both hooks, which act by pulling, like b C. The pawls are pressed 
 against the ratchet by their own weight, or by springs, according 
 to circumstances. When a ratchet wheel is used only to prevent 
 the recoil of the axis on which it is fixed, the vibrating arm is, of 
 course, not required, and only the detent is used. 
 
 Reversible Pawl. The above figure shows a form of click used 
 in the feed motion of shaping and other machines. The ratchet 
 wheel is here an ordinary toothed wheel, and the click B C is so 
 shaped as to be able to drive it either way. When the click is in 
 
336 LECTURE XXIV. 
 
 the position shown in full lines, it drives the ratchet wheel in 
 the direction of the arrow. When the wheel is required to rotate 
 the other way, the click is lifted over to the dotted position ; and, 
 if it be desired to stop the feed motion without stopping the 
 machine, the click is put in an upright position. A portion of 
 the pin at B, which turns with the click, is triangular in section, 
 A spring presses on this part and so keeps the click in any one 
 of its three positions. The ratchet wheel is keyed to A, the axis 
 of the screw which moves the slide carrying the cutter, and the 
 friction between this screw and its nut is sufficient, without any 
 detent, to prevent the ratchet from moving back. The vibrating 
 arm A B, which carries the click is driven by a small eccentric 
 or crank. The pawl may, of course, be made to move the ratchet 
 more than one tooth at a time by adjusting the angle through 
 which A B vibrates. 
 
 Masked Ratchet. In numbering machines it is often necessary 
 to print the same number twice, as in cheques and their counter- 
 foils. The ratchet which shifts the type wheels must therefore 
 be moved at every alternate back-stroke of the printing machine. 
 This may be accomplished by putting a second ratchet, running 
 free on the shaft, alongside the driving one and making the pawl 
 broad enough to move both. The second ratchet has the same 
 number of teeth as the other, but its teeth are made alternately 
 deep and shallow. It is also a little larger than the driving 
 ratchet, so that the pawl passes over the top of the teeth of the 
 latter, without moving it, when in a shallow tooth. Next stroke 
 the pawl drops into a deep tooth. This allows it to catch the 
 teeth of the main ratchet and so shift the type wheel. This 
 arrangement is called a masked ratchet. 
 
 Silent Peed. A ratchet wheel is always more or less noisy in 
 action, and the wear caused by the sudden drop of the pawl ig 
 considerable. To avoid this, a friction catch is sometimes sub* 
 gtituted for the pawl and a grooved wheel for the toothed one. 
 The pawl and ratchet then becomes a silent feed* The action of 
 this arrangement will be easily understood by a reference to the 
 figures. E C is an eccentric cam tapered at its edge to fit the 
 groove in the grooved wheel G W. When E moves in the 
 direction of the arrow the friction causes it to turn about its axis, 
 and, since the axis is not concentric with the circular part of its rim, 
 it gets wedged in the groove. Hence, for the rest of the stroke, the 
 lever carries G W round with it. At the beginning of the return 
 stroke, E C turns in the opposite direction, and so gets released 
 from the groove. A detent E D, precisely similar to E C, but 
 carried on * fixed arm, prevents the wheel from moving back- 
 
VERTICAL SAWING MACHINE. 
 
 337 
 
 .VERTICAL SAWING MACHINE, BY JOHN M'DOWAL & SONS OP 
 JOHK STONE, SHOWING SILENT FEED. 
 
338 LECTURE XXIV. 
 
 wards. The lever L L, is worked by an eccentric, and the length 
 of its stroke may be adjusted by altering the position of the end of 
 the eccentric rod E R, in the slot. The full-page illustration shows 
 a sawing machiDe, with this feed motion at the right-hand side. 
 
 CW 
 
 EC 
 
 INDEX TO PARTS. 
 
 G W for Grooved wheel. 
 EC,, Eccentric cam. 
 
 L Lever. 
 
 ED ,, Eccentric detent. 
 E R Eccentric rod. 
 
 WORSSAM'S SILENT FEED. 
 
 Watt's Parallel Motion. Referring to the illustration of 
 s 
 
 WATT'S APPROXIMATE STRAIGHT- LINE 
 MOTION. 
 
 CONSTRUCTION FOR LENGTHS 
 OP LINKS. 
 
 Watt's double-acting engine, previously mentioned in this Lecture, 
 the student will notice that the beam and piston-rod are connected 
 
PARALLEL MOTION. 339 
 
 by a set of links. This system of links has been called Watt's 
 Parallel Motion. The first figure will serve to show the principle 
 on which an approximate rectilinear motion is obtained. Part 
 of the beam of the engine is shown in three different positions, 
 C Tj, C T a , and C T,. The point, T, in it is connected by the 
 link, T t, to the end of a lever or radius rod, c t, pivoted at c. 
 In their mid positions, C T r c t v these two levers are usually 
 parallel to each other, and perpendicular to the line Pj P, P r 
 The point, T, describes an arc of a circle round C, and t round c. 
 As these arcs curve in opposite directions, we should expect 
 some intermediate point on the link T , to curve in neither 
 direction, but to describe an approximate straight line. This 
 
 "P t O T 1 
 point P may be found by making -p- = . The actual path 
 
 of P is like the figure 8, and the parts which cross are very nearly 
 exact straight lines for a short distance on either side of the 
 crossing. 
 
 Prof. Rankine gives the following construction for the lengths 
 of the links in his Machinery and Millwork . Let A be the centre 
 of the beam, G D the centre line of the piston-rod's motion, and B 
 the mid position of its end. Draw A D perpendicular to G D. 
 Make D E equal to one-fourth of the stroke, and join A E. Draw 
 E F perpendicular to A E, and meeting A D in F. A F is the 
 length of the beam. If G be the point where the radius rod cuts 
 G D, draw G K at right angles to G D, and make D H equal to 
 G B. Join A to H, and F to B, and produce A H and F B to 
 meet G K in K and L, Then, F L is the connecting link, K L is 
 the radius rod, and B is the point on the link F L, to which the 
 piston-rod must be attached. 
 
 Parallel Motion. In the accompanying figure A B T t is a 
 parallelogram, and c is a point in A t produced. In the meantime 
 we will leave the links C T and B D out of account and consider 
 the parallelogram only. Join B c and we have two similar 
 triangles, B A c and P t c. ' 
 
 s4 = T^. Or, P * = B A - = a constant. 
 13A Ac Ac 
 
 That is, in every position of the parallelogram, the point P 
 remains in one fixed position in the link T t. Moreover the ratio 
 Be 
 p-^ is constant, and, therefore, whatever patk P traces out, B will 
 
 trace out a similar one This is the principle of the pantograph, 
 which is used for enlarging or reducing drawings. N<m, we have 
 
340 
 
 LECTURE XXIV. 
 
 just seen how we may make P move in an approximate straight 
 line by the link C T. B will, therefore, also move in an approxi- 
 mate straight line. We might have guided B instead of P with 
 a radius rod, but this would have necessitated longer and heavier 
 links and would have occupied more space. 
 
 In applying this motion to his engine, Watt made A t c the 
 beam, and attached the piston-rod to B and the air pump-rod to 
 
 ---V 
 
 PARALLEL MOTION. 
 
 PARALLEL MOTION FOR RICHARD'S 
 INDICATOR. 
 
 P. The lengths A c, t c were, therefore, proportional to the strokes 
 of the piston and pump bucket respectively. Sometimes a third 
 link was added so as to get a second parallelogram and a second 
 point moving parallel to P, and this was used to drive the feed- 
 pump. 
 
 The right-hand figure shows the parallel motion of Richard's 
 steam engine indicator.* The student will at once see that it is 
 a modification of Watt's parallel motion. In this case the piston- 
 rod P E, is guided by a collar so as to move vertically, and is 
 attached by the link E F to the bar C D, between D and the 
 centre C. The motion of p, to which the pencil is attached is, 
 therefore, a magnified copy of the piston's motion. 
 
 * See Lecture XVI. of the author's "Elementary Manual of Steam and 
 the Steam Engine " for a description of this indicator. 
 
LECTURE XXIV. QUESTIONS 34 1 
 
 LECTUBE XXIV. QUESTIONS. 
 
 1. Describe Hooke's joint for connecting two axes whose directions meet 
 in a point. 
 
 2. Sketch and describe the double Hooke's joint, and explain why it is 
 used in certain cases in preference to the single joint. 
 
 3. Explain the manner in which Watt used the so-called Sun and Planet 
 Wheels as a substitute for a crank and connecting-rod, and account for the 
 result which he obtained. 
 
 4. Sketch a cam for giving a bar a uniform reciprocating motion, and 
 explain how you find the form of its periphery. 
 
 5. Set out a form of cam which, when acting on a bar by uniform 
 rotation, will cause the backward and forward motion of the bar to have 
 an interval of rest between each. 
 
 6. Describe, by the aid of the necessary sketches, how the circular 
 motion of the driving pulley is converted into the recipiocating motion of 
 the punch in an ordinary machine for punching holes in metal plates. 
 Calculate the approximate maximum pressure in pounds at the end of a 
 punch in cutting a hole I inch in diameter through a steel plate | inch 
 thick, the resistance of the plate to shearing being taken as 50,000 Ibs. per 
 square inch of section. Ana. 98,175 Ibs. 
 
 7. Describe the nature of a cam, and give any examples you know of for 
 which it is used. 
 
 8. What is a cam ? Show, with a sketch, how-to obtain by means of a 
 cam a motion in a direction parallel to the axis of rotation. 
 
 9. What is a cam ? For what purposes in mechanism are cams generally 
 used ? Sketch and describe the construction and actual form of a cam in 
 use in any machine with which you are acquainted. Sketch a cam which 
 would give a slow forward and quick return motion to a reciprocating 
 piece, with an interval of rest between the two motions. 
 
 10. How is a cam in the form of a heart set out, so that when the cam 
 rotates uniformly it may cause a sliding piece to move to and fro with a 
 uniform velocity ? 
 
 11. Determine a form of cam which, by rotating uniformly, will com- 
 municate a reciprocating movement to a sliding bar, but with an interval 
 of rest at the beginning and end of each stroke. 
 
 12. Sketch a pawl and ratchet wheel as used for preventing the recoil of 
 the gear. 
 
 13. Describe a ratchet wheel, and the manner in which it is held and 
 driven. Show its application in a ratchet brace where it is combined with 
 a lever and screw. Sketch the contrivance, and point out the manner in 
 which the drill is advanced. 
 
 14. Sketch a ratchet feed motion, such as is suitable for a planing 
 machine, and explain the manner in which the amount of feed is regulated. 
 
 15. What is a ratchet wheel, and how is it driven T In what way can it 
 be arranged that a ratchet wheel shall advance half the space of a tooth at 
 each stroke of the driver ? 
 
 1 6. By what contrivance may a ratchet wheel with 60 teeth be made to 
 -act as if it had 120 teeth ? 
 
3 P LECTURE XXIV. QUESTIONS. 
 
 17. Sketch and describe some form of pawl which will drive a ratche 
 wheel during both the forward and backward strokes. 
 
 1 8. It is sometimes useful to advance a ratchet wheel at every alternate 
 forward stroke of the driver, instead of at every stroke, as is commonly 
 the case ; describe and sketch a mechanical contrivance which will give 
 such a movement. 
 
 19. Describe, with the necessary sketches, gome form of silent-feed 
 arrangement commonly used instead of a ratchet wheel, for advancing the 
 timber in sawing machines. Explain the principle of the friction grip upon 
 which such a contrivance depends. 
 
 20. A pinion with 20 teeth works in a straight rack, the distance from 
 centre to centre of the teeth in the pinion being | inch. The pinion is 
 driven by a ratchet wheel with 40 teeth fixed on the pinion shaft ; find 
 the advance of the rack in inches for each tooth moved through by the 
 ratchet wheel. Ans. '25 inch. 
 
 21. Sketch and describe a vertical sawmill, showing how the silent feed 
 is applied. 
 
 22. Explain the principle of Watt's approximate straight-line motion, 
 commonly called a "parallel motion " By what combination of linkwork 
 I* an exact straight-lime motion obtained ? 
 
 23. Sketch and describe the action of a ratchet wheel. For what 
 purposes are such wheels used in machine tools ? Show how a ratchet 
 wheel of 30 teeth can be used to give a feed through ^th of a revolution 
 at each stroke of the arm of the pawl. (S. E. B. 1900.) 
 
 24. Design a cam to lift vertically A sliding piece at a nniform speed 
 through 2 ins., the return motion being also at a uniform rate, but at half 
 the speed ; having given that the line of stroke, produced, of the slider 
 passes through the axis of the cam shaft, that the nearest approach of the 
 centre of the roller to the cam centre is 2 ins., and that the diameter of 
 the roller is in. (C. & G., 1904, 0., Sec. A.) 
 
 25. Describe, with the aid of a sketch, Watt's parallel motion, and show 
 liow to find the position of the point in the coupler which most nearly 
 describes a straight line. Sketch the path which the tracing point 
 describes for all possible positions. (C. & Gr., 1905, 0., Sec. A.) 
 
( 343 ) 
 
 LECTURE XXV. 
 
 CONTENTS. Reversing Motions Planing; Machine Reversing by Friction 
 Cones and Bevel Wheels Whitworth's Reversing Gear Quick Return 
 Reversing Motion Whitworth's Quick Return Motion Whitworth'fc 
 Slotting Machine Common Quick Return Horizontal Shaping 
 Machine Quick Return with Elliptic Wheels Vertical Slotting 
 Machine Questions. 
 
 Reversing Motions. Planing Machine. In Lecture XI. 
 we illustrated and described a belt reversing motion for uniform 
 forward and return speeds and also for a quick return as applied 
 
 
 HORIZONTAL PLANING MACHINE WITH QUICK RETURN BELT 
 GEARING BY MESSRS. J. ARCHDALE & Co., BIRMINGHAM. 
 
 to planing machines. The above figure will serve to show the 
 manner in which the quick return and slower forward motions 
 
344 
 
 LECTURE XXV. 
 
 are applied to a modern planing machine. Here, the smaller 
 fixed and loose pulleys are situated to the extreme left, whilst the 
 larger ones are placed alongside of them. The pulleys are 
 connected by underneath toothed gearing with the rack of the 
 moving table upon which the material to be planed is bolted. 
 The under side of this table has two straight, truly planed and 
 scraped V-shaped surfaces, which accurately fit corresponding 
 V grooves on the upper surface of the strong heavy bed plate. 
 Two vertical standards, with planed and scraped surfaces on their 
 front faces, serve to guide the cross-piece which carries two tool- 
 holders. An up and down motion is given to this cross-piece by 
 means of the uppermost handle, spindle, two pairs of bevel 
 pinions and vertical screws actuating nuts connected to the back 
 of it. A horizontal motion is given to the tool-holders by the 
 lower right-hand handle and horizontal screws. The tool-holders 
 themselves are adjustable up and down by handles and screws as 
 shown. They may also be so set as to cut at any angle, or a 
 horizontal tool-holder can be fixed upon one or other of the 
 upright standards to plane vertical surfaces. All the above- 
 mentioned motions may be actuated from either side of the 
 machine. 
 
 We shall now illustrate and describe several other forms which 
 are frequently used in connection with machinery of different 
 kinds. 
 
 Reversing by Friction Cones and Bevel Wheels. In the 
 first of the two following figures, the 
 two cones B and C are fixed to the 
 shaft DD, which can be moved up 
 or down so that the cones B or C 
 
 . .,,.,. ,. ..... maybe alternately brought into con- 
 
 jiM I A tact with the other cone A, which is 
 
 I iiJL fixed to its shaft. Suppose A to be 
 
 10 llllj I g7 the driver ; then, when it is in con- 
 | IF 3 ^ tact with B, the shaft D D will be 
 .1 III turned in one direction, and when in 
 
 contact with C it will be rotated in. 
 the opposite way. The spindle or 
 shaft D D may, however, be driven 
 by a belt or toothed gearing, and con- 
 
 FEICTION CONE REVERSING sequently when A is in contact with 
 GEAR. B, it will be turned in one direction 
 
 and when in contact with C it will 
 
 be rotated in the opposite way. This device is frequently used 
 in connection with governors for engines, in order to lengthen or 
 
WHITWORTH S REVERSING GEAR. 
 
 345 
 
 shorten the rod to the throttle valve or cut-off gear and thus 
 enable the governor balls to regain their normal position, without 
 altering the quantity of steam being admitted. 
 
 If we require to transmit more force than the friction between 
 the cones will effect without 
 slipping, then we must substi- 
 tute toothed bevel wheels for 
 plain cones. In such a case, if 
 the speed and stress were con- 
 siderable the sudden engagement 
 of the wheels on one side or the 
 other would be apt to damage 
 the teeth; hence, it is usual 
 with steam cranes, winches, 
 windlasses, screwing gear, &c., 
 to have B and G free to rotate 
 upon their shaft D D, and 
 al A-ays in gear with A, as shown 
 by the second figure. The re- 
 versal in this case is effected by 
 means of the clutch E, which 
 can be slid along a feather on 
 the shaft by the lever, so as to 
 engage B or C and thus make 
 it turn with the shaft. It will 
 be easily seen that the direction in which A will rotate depends 
 upon whether B or C is locked to the shaft D D. 
 
 Whitworth's Reversing Gear. Another modification of this 
 gear is that made by Sir Joseph Whitworth & Ob. for planing 
 machines. In this case, the reversal is effected by shifting the 
 driving belt by the fork B F, from the forward pulley F P to the 
 backward one B P. The latter is cast in one piece along with or 
 keyed to the boss of the bevel wheel B "Wj, which runs loose on 
 the shaft S ; whilst, the former and also B W 2 are rigidly con- 
 nected to this shaft. A loose pulley L P is placed between the 
 forward and backward ones in order to facilitate the shifting of 
 the belt from the one to the other and to carry the belt when the 
 machine is not at work. The table T upon which is placed the 
 casting or other material to be operated upon, has a strong nut 
 fixed upon its under side, and is moved along the bed or slide T S 
 by the driving screw DS. This screw is keyed to the bevel 
 wheel B W 3 and is consequently driven in a forward or backward 
 direction according as the belt is on FP or BP. In order to 
 save the time that would otherwise be wasted if the cutting tool 
 
 BEVEL WHEEL AND CLUTCH 
 REVERSING GEAR. 
 
346 
 
 LECTUKE XXV. 
 
 was so fixed as to cut in one direction only, Sir Joseph Whitwoi th 
 designed a cylindrical revolving tool-holder, which automatically 
 turns the tool half round at the end of each stroke of the table 
 The reversal of the motion of the table is automatically effected 
 by its pushing the reversing stops R S t (and R S., not shown in the 
 figure) as it nears the end of a stroke. These stops are fixed iii 
 
 r.s 
 
 WHITWOETH EEVERSING GEAE. 
 
 such positions upon a rod connected to the reversing levers R L, 
 as to shift the belt fork at the proper time and thus give the 
 required length of stroke. 
 
 Quick Return Reversing Motion. Another form of re- 
 versing motion, used for planing machines which only cut one 
 way, is made up of a train of spur wheels and a rack R. As in 
 the previous case, there are three belt pulleys, each of the outer 
 ones being connected to a separate pinion. One of these pinions, 
 D P 2 , drives the rack on the planing table through a spur wheel 
 F 2 and pinion P fixed on an intermediate shaft ; the other, D P p 
 which is connected with F P, transmits its motion through another 
 pair of wheels F x and D r This will cause the rack and table to 
 move in the opposite direction, and as these wheels are made of 
 unequal sizes, the motion is also slower than when driving 
 through D P 2 . This slower motion is used for the cutting stroke 
 
WHITWORTH'S QUICK RETURN MOTION. 
 
 347 
 
 and the quicker one for the return stroke. The reversal is 
 effected by shifting the belt from FP to BP, or rice versa, as in 
 the Whitworth gear. 
 
 B.R 
 
 QUICK RETURN REVERSING GEAR FOR PLANING MACHINES. 
 
 Whitworth's Quick Return Motion. In a shaping or 
 slotting machine the table carrying the work is fixed and the tool 
 moves over it, cutting in one direction only. In such a case, the 
 tool usually obtains a reciprocating motion from a compound 
 crauk, so arranged as to give a quicker return stroke in order to 
 
 Return Strok 
 
 QUICK RETURN MOTION. 
 
 save time. Looking at the above diagrammatic figure, A and B 
 are two fixed points and C is a point connected to the tool-holder, 
 so guided as to move along L M at right angles to A B. A crank 
 
348 LECTURE XXV. 
 
 D B E, centred at B, has its outer end joined to C, by the 
 connecting-rod EC. A second crank A D rotates round A and 
 drives the first by having a pin at D which moves in a slot B D. 
 Now, when D is at the position G, C will be at the extreme left 
 of its stroke ; and when D is at F, C will be at the other end of 
 its stroke. Hence, if A D rotates uniformly in the direction of 
 the arrows ; will make its cutting stroke from left to r.ight 
 while D is moving round G K F ; but during its return stroke D 
 only requires to move round F H G. The return stroke will 
 therefore occupy less time than the cutting stroke in the same 
 ratio as the arc F H G is less than the arc G K F. 
 
 Sir Joseph Whitworth applied this principle to shaping 
 machines in the manner shown by the next figure, which has the 
 
 WHITWORTH'S QUICK RETURN MOTION FOIL SHAPING MACHINES. 
 
 same lettering for corresponding parts as the previous one. 
 Here, the crank A D is obtained by putting a pin D on a toothed 
 wheel which rotates freely on a, fixed shaft S, whose centre is at A ; 
 and the crank D B E is supported by a pin in a hole bored at B 
 in the end of this shaft. In the back of this crank-piece there is 
 a slot B D in which D can slide, and in the front another slot B E 
 in which E can be clamped in any position so as to adjust the 
 length of B E and thus give the required travel to the tool. The 
 large wheel is driven uniformly by a pinion P connected with the 
 belt pulley.* 
 
 * Students who desire further information on machine tools such as 
 drilling and milling machines or on measuring appliances should refer to 
 Professor Shelley's "Workshop Appliances," Professor Goodeve's "Elements 
 of Mec' anism," and Lineham's " Text-booV of Mechanical Engineering:," &c. 
 
WHITWORTH'S SLOTTING MACHINE. 
 
 349 
 
 Whitworth's Slotting Machine.* The following side 
 elevation of Sir Joseph Whit worth <fe Co.'s slotting machine 
 will serve to show how their quick return motion is practically 
 applied. The driving stepped cone D C receives its motion 
 from a corresponding overhead cone fixed to the workshop 
 
 WHITWORTH'S SLOTTING MACHINE. 
 
 shafting. Its motion is in turn communicated to the com- 
 pound crank E B D (just described and illustrated by the two 
 preceding figures) through the toothed drivers and followers 
 Dp F! ; D 2 , F 2 ; D 3 , F 3 . The quick up and slow down motion of 
 the vertical slide bar V S, with its tool-holder T H, is obtained 
 from the compound crank through the connecting-rod C E. The 
 table T, upon which the metal to be slotted is bolted, may be 
 shifted by hand levers or automatically moved to and fro, cross- 
 
 * The above figure is reduced from a lithographed drawing which 
 appears in Mr. I.ineham's book on " Mechanical Engineering," to which 
 students may refer for further views and details. 
 
350 
 
 LECTURE XXV. 
 
 wise and turned at pleasure, by the feed gearing F G, actuated by 
 a ratchet at B. This ratchet has a reversible click of the kind 
 illustrated in the previous lecture and it is driven by a red from 
 the feed cam F 0. The whole of the 
 moving parts are supported by a strong 
 heavy bed plate, cast in one piece with 
 the upright framing so as to prevent 
 vibration in the material being oper- 
 ated upon or chattering of the cutting 
 tool. 
 
 Common Quick Return. Another 
 form of quick return very often used 
 for shaping and slotting machines is 
 based on the mechanism of the oscillat- 
 ing steam-engine. A crank A C rotates 
 uniformly and imparts motion to a 
 slotted arm B F. This arm will have 
 its extreme positions at B H and B K. 
 It will therefore make its forward and 
 back swings while C is moving round 
 H D K and K E H respectively, and 
 hence it has the quick return motion 
 COMMON QUICK RETURN desired for the tool. The tool-holder 
 FOE SLOTTING MACHINE, is connected by a rod to such a point 
 
 in B F as will give the required travel. 
 
 Horizontal Shaping Machine. The following illustration 
 shows a shaping machine with this gear for driving the cutting 
 
 SHAPING MACHINE WITH QUICK RETURN BY MESSES. 
 SELIG SONNENTHAL & Co., LONDON. 
 
QUICK RETURN WITH ELLIPTIC WHEELS. 
 
 351 
 
 tool. The work is fixed in a vice mounted on a table at the front 
 of the machine, and this table- can be moved up and down by a 
 rack or screw, and traversed along the bed in either direction, 
 backwards and forwards by ai screw. The tool-holder moves 
 inwards and outwards over this vice and is provided with a 
 screw for moving the tool vertically- and a worm for adjusting 
 its inclination. The gear for driving the tool is actuated by a 
 sliding pinion on a shaft which lies along the whole length of the 
 back of the machin and is driven by a stepped cone pulley. The 
 feed of the tool-holder towards the left or right, is effected by a 
 nut and screw underneath. This screw is worked by a ratchet 
 seen to the right, in a similar way to that described for the 
 previous machine. 
 
 Quick Return with Elliptic Wheels. Vertical Slotting 
 Machine. We illustrate another slotting machine which has a 
 different method 
 
 of obtaining a ^^ "^\ D 
 
 quick return. In 
 this machine, the 
 moving bar is ac- 
 tuated by a simple 
 crank, but this 
 crank is driven 
 at different speeds 
 for the cutting and 
 return strokes by 
 the following de- 
 vice. One portion 
 of the circumfer- 
 ence of the wheel 
 on the crank-piece 
 is part of an ellipse 
 
 , B D A. ELLIPTIC WHEELS FOB QUICK RETUEN. 
 
 circular. The wheel for driving this has an elliptical part EOF, 
 and a complete circle E H F G. Suppose B and F to be in contact. 
 Then, as the lower wheel moves round, the two ellipses keep in 
 contact until A and E come together. The two shafts thus 
 make about half a revolution in the same time. The circular 
 wheels now come into gear, and the lower wheel must make one 
 and a half turns while the upper rotates through its second half. 
 This will bring B and F aarain into contact and the whole process 
 repeats itself. The quicker motion with the elliptical wheels is 
 used for the upward stroke and the slower motion from the 
 
352 
 
 LECTURE XXV. 
 
 circular wheels for the downward or cutting stroke. By looking 
 at the side elevation, and at the view of the complete machine, 
 it will be seen that the circular parts of the wheels D JB and H Gr 
 
 VERTICAL SLOTTING MACHINE BY MESSES. T. SHANKS & Co., 
 JOHNSTONE. 
 
 are placed to the left of the elliptical parts, B C and C F ; and 
 how, at each junction B and F of the two portions, one tooth 
 stretches right across .both of them so as to give a steady connected 
 motion. Also, how the smooth part A K B of the upper elliptic 
 wheel is cut away to clear the point of the other ellipse during 
 
QUICK RETURN WITH ELLIPTIC WHEELSv 353 
 
 thoso revolutions in which the circles are in gear. From what 
 has already been said about the other machines the student will 
 easily understand the working of this one. He should, however, 
 notice that the moving bar has a counterbalance to relieve the 
 gearing from the weight of the sliding bar, also to reduce the 
 driving force required for the upstroke and to prevent any sudden 
 drop of the tool on the work which may be placed upon the table. 
 
354 LECTURE XXV. QUESTIONS. 
 
 1. Illustrate and explain the form of reversing gear you would employ 
 for a steam winch in which the engine shaft always runs one way. 
 
 2. Show a method of applying a self-acting motion for reversing the 
 motion of the table of a planing machine when a screw is employed for 
 driving it. 
 
 3. Describe, with a sketch and index of parts, some form of quick 
 return gearing suitable for a planing machine, the movement being 
 obtained by a combination of pulleys and spur wheels. 
 
 4. Sketch and describe an arrangement for driving the table of a planing 
 machine by means of a screw, so that the table may travel 50 per cent, 
 faster in the return than in the forward or cutting stroke. Why is a 
 square threaded screw employed in such a machine? 
 
 5. Sketch and describe an arrangement of mechanism for reversing the 
 table in a screw driven planing machine. In what way can a quick return 
 of the table be obtained in such a machine ? 
 
 6. Sketch and describe a good form of slow forward and quick return 
 for a shaping machine. 
 
 7. Describe, with sketches, a planing, a slotting, or a shaping machine, 
 showing clearly how the cutting and feeding motions are effected. 
 
 8. Sketch and describe the mechanism for feed motions: (i) In a 
 machine where there is a reciprocating movement, as in a planing 
 machine ; (2) where there is a continuous movement, as in a machine for 
 boring cylinders. 
 
 9. Sketch and describe a vertical slotting machine with quick return 
 elliptical gear. Give a separate diagram and explanation of the elliptical 
 gear. 
 
 10. Sketch and describe the arrangement of mechanism by which the 
 tool of a planing machine is traversed across the slide of the machine at 
 each stroke of the table. (B. of E., 1900.) 
 
 11. Describe, with sketches, the mechanism for giving an automatic 
 feed to the cutting tool of a lathe or shaping machine, and how it is put 
 in or oat of action, and the amount of feed varied. (B. of E., 1902.) 
 
 12. Describe, with sketches, two quick return motions for driving the 
 table of a planing-machine one in which the quick return is obtained by 
 belting, and the other in which it is obtained by ordinary gearing. 
 
 (C. &G., 1903, O., Sec. A.) 
 
 13. The mechanism of the ordinary direct-acting engine is used as a 
 quick return motion for a small shaping machine by simply placing the 
 crank shaft centre below the line of stroke, instead of in the line of stroke 
 produced. The crank radius is if ins., the connecting rod is 7 ins. long, 
 and the centre of the crank shaft is 3^ ins. below the line of stroke. Find, 
 graphically, the stroke and the time ratio of the return and cutting strokes 
 the crank shaft being supposed to rotate uniformly. 
 
 Ans. Length of stroke = 4^ inches ; Time ratio = i to i -2. 
 
 (0. & G., 1903, 0., Sec. A.) 
 
 14. Describe, with the aid of sketches, the mechanism known as the 
 crank and slotted lever used to give a quick return motion to the tool of a 
 shaping machine ; and if the crank radius is half the distance between 
 the fixed centres, find the ratio of the times occupied in performing the 
 cutting and return strokes. The crank shaft may be supposed to rotate 
 uniformly, and the obliquity of the connecting rod to the line of stroke of 
 the tool-box neglected. Ans. (0. & G., 1904, 0., Sec. A.) 
 
( 355 ) 
 LECTURE XXVI. 
 
 CONTENTS. Measuring Tools and Gauges Limit Ganges Micrometer 
 Screw Gauge Sir Joseph Whitworth's Early Realisations of Mechanical 
 Accuracy Improved Equivalents Micrometer Gauge A New Set of 
 English Gauges Whitworth Millionth Measuring Machine Improved 
 Standard Workshop Measuring Machine Construction and Uses of 
 the Tangentometer Questions. 
 
 Measuring Tools and Gauges. A few years ago? a pair of 
 calipers and a foot-rule graduated to ^-inch, formed an outfit for the 
 British working mechanic. Anything more accurate than this was ex- 
 pressed by the somewhat elastic term full or bare ! As far as each 
 individual piece of fitting was concerned, the work turned out was pro- 
 bably of a good sound nature, and the same workman was able, through 
 practice, to make a number of parts to within a fair degree of accuracy, 
 when we consider the tools which were at his disposal. In this method 
 of working, however, discrepancies due to individuality occur, and modern 
 practice demands the suppression of these as far as possible. Again, 
 good manual labour is not so cheap nowadays as it used to be. The result 
 is, that machines, which lessen the cost of production and turn out work 
 of a well-finished, accurate, and interchangeable nature, have greatly 
 superseded the old rough-and-ready methods when dimensions were 
 indicated by "sooking fits," " hair-breadths," and suchlike terms. 
 
 The introduction of the micrometer no doubt altered things to some 
 extent ; but this instrument, being of a delicate nature, requires careful 
 handling, and that means lost time and consequently lost money. The 
 micrometer is rather a tool-room instrument for the purpose of checking 
 and comparing the various gauges and tools used in the shops. 
 
 Limit Gauges. It is now found that the quickest and most accurate 
 method of turning out interchangeable work is to make use of some 
 system of limit gauges. In reality, a limit gauge consists of two gauges, 
 one of which is larger and the other smaller by a minute fraction than 
 the size intended. Thus, if we employ an internal limit gauge of the form 
 
 not go In 
 
 INTERNAL LIMIT GAUGE. 
 
 shown in tne figure, for measuring a Ig-ineh hole, and, if we fin1 that the 
 right-hand end goes into the hole, but that the left-hand end does not, 
 then we know for certain that the diameter of the hole cannot possibly 
 differ from ig-inch by more than -0012 inch too large, or '0004 inch to* 
 small. 
 
356 
 
 LECTURE XXVI. 
 
 The difference limit varies, of course, with the class of work which it is 
 intended to produce. It is quite evident, therefore, that for a machine 
 which runs in a dust-laden atmosphere, a much greater difference limit 
 will be required in the bearings, than for another similar machine which is 
 designed to work in a comparatively clean atmosphere. 
 
 But, no matter how tight or how slack the working fits may be, it will 
 be found, that work constructed on some system of limit gauges will have 
 a much greater all-round efficiency than that made in the old haphazard 
 trial and error fashion with an ordinary rule and calipers. 
 
 EXTERNAL LIMIT GAUGE. 
 
 The above or second figure is an illustration of an External Limit 
 Gauge for a 2|" or 2-25 inch shaft. One end is marked to "go on" and is 
 only o'OOoS" or 8/10,000 of an inch less than the 2*25 in. shaft ; whilst 
 the other or smaller caliper end is marked to "not go on," for it is 0*0014 in. 
 or i4/ro,ooO of an inch less than the 2^25 in. shaft. Consequently, the 
 maximum difference between the "go on" fit and the "not go on" fit is 
 (-2 -25 in. - 0.0008 in. ) - (2*25 in. - 0-0014 in -) or simply (0*0008 in - O'OOi4in. ) 
 r=o'OOo6in., that is 6/10,000 in., or say 1/1666 in. which is less than 
 i/ioooth of an inch. If the wider end of the gauge just goes on, the shaft 
 will be a " tiyhljit," but if the narrower end just goes on, the shaft will be 
 a " running f<,t " in a bush bored exactly to 2-25 in. diameter. 
 
 STANDARD INTERNAL AND EXTERNAL 
 CYLINDRICAL GAUGES. 
 
MICROMETER SCREW GAUGE. 
 
 357 
 
 The third figure shows a Standard Internal and External Cylindrical 
 Gauge. This type of gauge is chiefly used as a standard o'f reference. 
 It is hardened, ground, lapped by hand, and is accurate to within -oooi 
 inch. 
 
 The Micrometer Screw Gauge. This instrument is very useful 
 for measuring diameters or thicknesses to within 'ooi inch. On turning 
 the milled cap shown at the right-haud side of the illustration, the 
 
 operator turns the screw of 40 
 threads to the inch, which also 
 
 i | Q fr[ may be seen from the figure. It 
 
 \ j IJjgi ~ is clearly evident, that if one 
 
 complete turn be given to the 
 milled cap, the screw will advance 
 or recede fa inch. Therefore, if 
 the edge of the sleeve (which 
 forms part of the cap) be divided 
 into 25 equal parts along its 
 circumference, and the cap be 
 rotated by a part of a revolution 
 corresponding to one of these 
 divisions, the screw will have 
 advanced by -fe of fa inch, i.e., 
 
 175*07 or >o01 mcn 
 
 It will be seen from the two figures that micrometer screw gauges 
 
 MICROMETER SCREW GAUGE. 
 
 FOR MEASURING ALL SIZES LESS THAN 
 
 0-3 INCH BY THOUSANDTHS 
 
 OF AN INCH. 
 
 STARRETT MICROMETER GAUGE. 
 
 are made of various ranges and styles. If a mere portable gauge is 
 required to suit different sizes, then one of the best forms is that made by 
 the Starrett Co., U.S.A. In which the position of the movable end is 
 determined by inserting a hardened steel tapered -pin into hardened 
 steel bushed holes. For fixed measuring machines of great accuracy the 
 Whitworth Millionth Measuring Machine is still considered the standard 
 in this country. 
 
358 LECTURE XXVI. 
 
 Sir Joseph Whit worth's Early Realisations of Me- 
 chanical Accuracy. "Whitworth's Standard Measuring Machine ' r 
 will be illustrated and described near the end of this Lecture. But, it may 
 be mentioned here, that the whole subject of accurate, scientific, mechanical 
 measurement and its standardisation had remained in great confusion and 
 uncertainty until Whitworth first carefully considered and then made, 
 about the year 1840, mechanically perfect flat-surface plates. Second, 
 he made standard screw threads, screw taps and dies, as well as parent 
 leading screws for lathes, &c. Lastly, a standard or parent Measuring 
 Machine, which was to be an instrument of such extreme precision that it 
 could detect the difference of one one-millionth of an inch in the end 
 measurement of short standard bars. 
 
 These three early steps in his career were realised in succession. In 
 fact, he could never have made the standard measuring machine if he had 
 not previously made and drilled his men into producing the two former 
 sets of tools. 
 
 All the present-day accurate machine-shop surface-plates, screw threads, 
 standard bars, gauges and fixed measuring machines, may be said to be 
 the outcome of Whitworth's skill, perseverance and forethought, in 
 systematising the production of standard tools. He also advertised the 
 results and sold accurate copies of his correct parent tools not only 
 amongst British engineers but throughout the civilised world. 
 
 Sir Joseph Whitworth, before commencing the afore-mentioned difficult 
 tasks, was satisfied that the most practical means of workshop measure- 
 ment was to be founded on the truth of surface and the sense of touch. He 
 maintained to his dying day, that the most delicate sense of the mechanic 
 and mechanician was that of touch. In confirmation thereof, he showed 
 that when a piece of metal had parallel end faces and was so held between 
 the two fair-in line and parallel measuring planes of his machine, that the 
 piece being tested just gravitated slowly downwards, due to its own weight 
 between these two faces. Then, any good mechanic could at once detect 
 the difference in bringing these two plane parts or distance faces nearer 
 together or further apart by the minute difference of one one-millionth of 
 an inch ! Of course, such extreme accuracy is not required in ordinary 
 tool-making and engine -building works, but it is required in some works. 
 
 Improved Equivalents Micrometer Gauge. This new instru- 
 ment has certain advantages over the ordinary screw Micrometers. By 
 means of the screw on the hub and the two divided discs, readings can be 
 taken up to '300 inch with the various equivalents. Or, measurements may 
 be made in the same way as with the older style of gauge for all measure- 
 ments less than one inch by TTTOTT f an inch between the jaws. 
 
 On one side of the disc appear decimals of an inch, decimals of mtlli-- 
 metre standard wire gauge and Stubs' round hole wire gauge. On the 
 other side, fractions of an inch by -fa inch, and screwing sizes for B.A. 
 threads. 
 
 Equivalents to other gauges, such as the "British Standard Wire 
 Gauge," "Stubs' round hole wire gauge," and fractions of an inch on the 
 metric standard, can be read off without the necessity of reference tables ; 
 whilst at the same time the jaws are set to give an exact size in decimals 
 of an inch. It will thus be seen, that by turning the disc until any 
 decimal or other number appears, several equivalents can be read 
 simultaneously, which will be especially convenient for telegraph and 
 electrical engineers. 
 
 The hub is fitted with a small milled thumb piece, projecting beyond 
 the ordinary hub, by means of which the speed of turning the screw 
 
IMPROVED EQUIVALENTS MICROMETER GA-'GE. 
 
 359- 
 
 Tnay be increased. The anvil end of the Micrometer is flush with the 
 screw spindle, thus allowing of the close calipering of projections or 
 ledges. 
 
 NOTE. I am indebted for the figure and description of this improved 
 gauge to the patentees and makers, Messrs. Grimshaw and Baxter, of 
 29 Goswell Road, London. 
 
3'60 LECTUEE XXVI. 
 
 A New Set of English Gauges (Windless Patents). Descrip- 
 tion of the Following Six Figures. Fig. I shows the lower half of a pocket- 
 case^" x 5" x i ") containing a complete set of these Caliper Gauges. 
 It will be observed that both the external and the internal measuring holders 
 are divided in the centre, to allow of the insertion of one or more of the 
 steel blocks (illustrated on the right hand of the case) between either of 
 these two holders. 
 
 Fig. II indicates the arranging of the External Gauge to measure inch. 
 This can be effected in a very short time by means of the " quick-grip 
 lock-nut device." 
 
 Fig. Ill gives a photographic proof of the perfectly flat surfaces and 
 the very superior finish of the faces of the seven cast-steel blocks, by their 
 clinging together after having been wrung together. 
 
 Professor Tyndall was the first scientist to prove that* perfectly flat 
 surfaces adhere together, due to the mere molecular attraction between 
 the great number of bearing points when brought into close contact. He 
 entirely disposed of the previously held theory that the adherence (of, 
 say, two good surface plates) was due to the exclusion of the air between 
 them, and, therefore, to atmospheric pressure. 
 
 Fig. IV shows the internal measuring holder arranged to measure 3! 
 inches. It is usually finished with flat ends, but it can be supplied with 
 spherical ends having a radius slightly smaller than the smallest cylinder 
 of 2\ inches which the gauge will enter. 
 
 Fig. V exemplifies how the f ", ", ft* y and & blocks, when wrung 
 together, are used to check the i* block by the external gauge. This process 
 may be reversed by placing the i" piece between the jaws and the five 
 smaller pieces in the holder. This method of self -checking is one of the 
 best points of these gauges. 
 
 Fig. VI is a view of the new 1910 pattern of a " Combined Limit and 
 Double-ended Caliper Gauge." When the jaws marked (+, +) and 
 { ) are opposite each other, as in the photo, the gauge is set for 
 limit measurements, the (+, -f) gap being slightly larger and the 
 { > ) gap slightly smaller than the exact measurement aimed at. 
 But, when slackened, turned round and then readjusted (so that the 
 (+ > & ) markings come opposite each other), both ends become exact 
 caliper or snap gauges. They now indicate the precise size due to 
 the number and breadths of the round steel standard-sized blocks, 
 which have been inserted in the centre, between the two lock-nuts. 
 
 It may be mentioned, that all the parts of these several gauges are made 
 of the best British cast-steel, properly hardened and finished. 
 
 Range of Measurements by the External and In- 
 ternal Holders. In the up-to-date standard sets there are but 
 seven steel blocks, viz., i", ", f", ", ^ %' and T V*. This gives the ex- 
 ternal measuring holder a range from T y to z\\ by T V of an inch, or forty 
 different sizes ; and the internal measuring holder a range from 2%" to 4", 
 also by r Vths of an inch, or twenty-four different sizes. In addition to this, 
 forty end sizes from r V" to 2 J" are obtainable by wringing the blocks together 
 in combinations. Thus, we get a range of over 100 separate standard gauge 
 sizes with one small pocket instrument. 
 
 Mr. H. M. Budgett, of the Crown Works, Chelmsford, informs the 
 author that he is now (1910) adding ^" and ^" measuring pieces to 
 the above-mentioned seven blocks. This will greatly increase the range 
 of the sizes obtainable with this instrument. In fact, the outside gauge 
 will then give 160 sizes and the internal one 96 sizes, whilst the pieces 
 
FOR DESCRIPTION AND DETAILS SEE PROFESSOR JAMIESON'S MANUAL 
 OF APPLIED MECHANICS. 
 
 FIG. I. 
 
 SHOWING COMPLETE SET IN CASE, HALF ACTUAL SIZE. 
 These Gauges are made by the Crown Works. Chelmsford, Engla 
 
FIG. II. 
 SHOWING EXTERNAL GAUGE BEING ARRANGED TO MEASURE --in. 
 
 FIG. Ill, 
 SHOWING HOW THE VARIOUS PIECES WILL ADHERE WHEN WRUNG TOGETHER 
 
FIG IV. 
 
 SHOWING THE INTERNAL GAUGE ARRANGED TO MEASURE 34 in. 
 
 
 FIG. V. 
 
 SHOWING HOW THE GAUGES ARE SELF-CHECKING, 
 
FOR DESCRIPTION AND DETAILS SEE PROFESSOR JAMIESON'S MANUAL 
 OF APPLIED MECHANICS. 
 
 FIG. VI. 
 SHOWING COMBINED LIMIT AND DOUBLE ENDED CALIPER GAUGE. 
 
WHITWORTH MILLIONTH MEASURING MACHINE. 361 
 
 themselves will give 163 sizes ; so that 419 different exact sizes could 
 thereby be measured ! 
 
 Uses of the Gauges with a Surface-Plate. In every up- 
 to-date tool-making and engineering workshop there should be a set of 
 standard surface-plates. Then, the previously described gauges may be 
 used in many ways with such a surface -plate. For example, an exact 
 height gauge may be formed by wringing one arm of the external measuring 
 holder with any desired number of blocks, and then using them on the 
 surface-plate. 
 
 The National Physical Laboratory Certificate. We 
 
 finish the description of these British-made gauges in reproducing from a 
 photograph the Certificate granted in August 1909 by The National Physical 
 Laboratory, with two objects in view : 
 
 First, to show the extreme accuracy of the gauges, where none of the 
 seven blocks had an error of one one hundred-thousandth of an inch ! 
 
 Second, to bring to the notice of Engineering Students, that if ever they 
 should be fortunate enough to devise anything new in regard to ther- 
 mometers, pyrometers, barometers, micrometers, cyclometers, galvano- 
 meters, electrometers, or steam, vacuum and mechanical gauges, &c., 
 they may have the accuracy of their invention or improvement tested by 
 an absolutely impartial and reliable judge at a comparatively small cost. 
 
 Certificate of Examination of Seven End Gauges. 
 By the National Physical Laboratory, Teddington. 
 
 For : H. M. Budgett, Crown Works, Chelmsford. 
 Form : Cylindrical, i" in diameter with a " hole through the centre. 
 The end faces are perpendicular to the axis of the cylinder. 
 These gauges have been compared with the Laboratory Standards, and 
 the mean lengths at 62 F. have been found to be : 
 
 Gauge. Length at 62 F. Gauge. Length at 62 F. 
 
 t 0.062 50 inch. 0.25000 inch. 
 
 0.12500 ,, 0.37500 
 
 T <V 0.18750 ,, $ 0.50000 
 
 Gauge i was i.ooooo inch at 62 F. 
 
 Each gauge has been tested at several points, and the end faces in each 
 case have been found to be parallel within o".ooooi. 
 
 J. A. HABKEB, 
 August 25, 1909. For the Director. 
 
 The Whitwprth Millionth Measuring Machine. From 
 
 what has been said in this Lecture about Sir Joseph Whitworth's early 
 realisations of mechanical accuracy, the student will be anxious to learn 
 how he accomplished the wonderful task of constructing a machine capable 
 of detecting a difference of only one-millionth of an inch in the length 
 of two bars. Another object which he then had in view was to produce 
 an exact fractional part of the " British Parliamentary Standard of Length," 
 viz. the legal yard, or 36 inches. Consequently, he first of all made 
 a machine to receive between its two measuring noses a bar of only i inch 
 long, and thereafter he constructed the machine which we now illustrate 
 and describe to measure bars of 3* and 4" in length, so that twelve of the 
 
362 LECTURE XXVI. 
 
 former or nine of the latter placed end to end would make up the British 
 standard yard. 
 
 In 1851 he was awarded the Council Medal of the first Great Exhibition 
 (which had been proposed and successfully carried through under the 
 patronage of the late Queen Victoria and Prince Consort) for a precisely 
 similar machine to take in a bar of 40" in length. 
 
 Referring to the figure, we see that it consists of a strong, rigid cast- 
 iron framing F, which stands upon three legs, two at the right-hand end 
 and one leg at the left-hand end. This frame F not only forms the bed 
 of the machine, but it also includes the bases of the two fixed head-stocks, 
 of which C 1 and C 2 are the upper or removable caps. Precisely in the centre- 
 line of the two head- stocks the casting is moulded into two hollow right- 
 angled or V grooves. These V grooves are then carefully planed and 
 scraped perfectly fair and square and a rectangular steel bar O is so very 
 carefully fitted into each of them that they bear and can be moved evenly 
 throughout the length of each end head-stock base, along the centre line. 
 The ends of these two steel bars of o or square section are then faced at 
 right-angles to their length, and a truly central hole is bored and screwed, 
 to receive a long steel screw at their outer ends and shorter sere wed- in 
 noses Nj, N 2 as shown in the figure. The inner faces of these two noses 
 are so carefully turned and scraped that they are " dead parallel " to 
 each other. It is between these noses N x and N 2 that the short standard 
 bar or its copy is to be placed and its length measured in the following 
 way. 
 
 It will be observed, that on the extreme left of the framing a graduated 
 wheel QW lt has been fitted to the outer turned end of the screw spindle 
 
 GW 
 
 WHITWORTH MILLIONTH MEASURING MACHINE, 
 
 whose nose is Nj. Since the pitch of the screw has twenty threads to the 
 inch, and the flat periphery surface of the graduated wheel GW, is divided 
 into 250 equal parts, with a fine pointer at P, it will be readily understood, 
 that if this wheel GW a is so turned that the pointer P shows a movement 
 through one division, then the inner face of the nose Nj will onlij move 
 one five-thousandth of an inch forward or backward, according to the 
 direction in which the wheel is turned. This screw and wheel are there- 
 fore only suitable for comparatively rough to and fro adjustments of the 
 nose N lf since *Vx*i v = 7< fo*th of an inch. 
 
WHIT WORTH MILLIONTH MEASURING MACHIN 7 E 363 
 
 Now look at the other head-stock where the screw which actuates the 
 nose N 2 has also a pitch of twenty threads to the inch, and the wheel WW 
 has 200 teeth. This wheel gears with the worm-screw WS, whose spindle 
 is fixed to the centre of the large graduated wheel GW 2 , which has 250 
 equal divisions marked upon its flat-faced periphery with a pointer not 
 shown [in [the figure. Consequently (since & x ?fa X ^ = j^^, 
 if the wheel GW 2 be so carefully turned that its pointer shows a movement 
 of one division, the inner face of the nose N 2 will only move to or fro by the 
 one -millionth of an inch. 
 
 Such a small movement might bind or let the short bars i*, 2", 3* or 4* 
 in length drop down as the case may be, and thus to a certain extent hamper 
 the " finer feelings " of the operator. So Whitworth introduced what he 
 called a "feeler " (which consists of a thin parallel slip of hardened polished 
 steel) between one face of the standard bar (or the end of the bar to be 
 measured and compared with the standard one) and the flat face of the 
 nose N! or N 2 . When the adjustment of Nj and N 2 is considered perfect, 
 this " feeler " should just glide down gently by its own weight, between 
 one of these nose faces and the nearest end of the bar under test. 
 
 Such an accurate machine as we have just described must not only be 
 brought to and kept at a fixed standard temperature, say 62 Fahr. for 
 British measurements and 70 Fahr. for United States of America tests, 
 but care must be taken to protect it from dust, moisture, and currents of 
 ir. 
 
 Improved Standard Workshop Measuring Machine. 
 For engineering workshops, a much less expensive and simpler fixed machine 
 is now made. It is, however, sufficiently accurate to detect a difference 
 of ^^ of an inch in the length between one bar or gauge [and another. 
 It consists of a strong frame and bed- or lathe-shears, with a fixed head- 
 stock at one end and another one which is movable. The latter is 
 moved to or fro by means of a screw placed longitudinally between the 
 whole length of the shears. This screw is rotated by an end wheel, and 
 the screw gears with a nut fixed into the tongue of the movable head- 
 stock. The spindle of this " poppit " or movable head is often round and 
 parallel throughout. Consequently, all to-and-fro adjustments of the 
 right-hand nose-face N 2 are made in the same way as shown by the 
 previous figure in connection with the fixed head-stock, 
 p Now, supposing that you have to compare the length of a bar with that 
 of a standard bar (whose length had been previously adjusted between the 
 noses N! and N 2 and the pressure thereon noted), you place the new bar be- 
 tween Nj and N 2 . If this bar is, say, longer than the standard bar, the free 
 spindle of the movable head-stock will be pressed slightly outwards. Then, 
 since its outer end now presses harder against the elastic centre of a circular 
 metal case containing water (with a vertical graduated, clear glass tube 
 extending therefrom), the water will rise higher in the tube than in the 
 former case, that is when the standard bar was under measurement. 
 
 The graduated wheel of the fixed head-stock is now turned until the 
 height or " head " of water in the glass tube is brought to the same position 
 as in the first case. Then, the difference between the two readings upon 
 the divided wheel at once indicates the difference in ten-thousandth of an 
 inch between the two bars. We know, that the pressure upon the ends of 
 each bar was the same because the hydrostatic " head " was the game. Th 
 replaces the personal error of the operator with his feeler. 
 
3^4 
 
 LECTURE XXVI. 
 
 Construction of The Tan?3nt >met9r. This instrument con- 
 ists of a horizontal metre rule, divided into inches and parts of an inch 
 along its upper scale, and into centimetres and millimetres along its lower 
 scale. At the centre of this rule there is attached one end of a second 
 rule, half a metre in length and perforated at various points. A small slot 
 is cut in the second rule to enable the degrees marked on the semicircle 
 to be read. Attached to this second rule by means of a pin, are (i) a 
 third rule, (2) a plumb-line. The third rule is graduated into cm. and mm., 
 and has a fine slot cut down its centre to enable the divisions on the first 
 or horizontal rule to be read. In order to avoid parallax a cursor or fine 
 dividing line connects the vertical with the horizontal rules. The whole 
 instrument is made of boxwood and fixed on a firm central base. 
 
 811k thread 
 
 Plumb-Line. 
 
 Plumb 
 Weight 
 
 BERRIDGE'S TANGENTOMETER. 
 (As made by Harris & Co,, Ltd., Birmingham.) 
 
 Uses of The Tangentometer. This instrument einblea teachers 
 to explain graphically the meanfng of the several trigonometrical ratios to 
 technical students. It also enables a student to measure for himself 
 various angles and to prove his answers to questions by noting the ratio 
 of the actual lengths of the sides of a right-angled triangle. He can 
 then compare his results with the Table of Trigonometrical values for 
 "Functions of Angles " at the end of this book. 
 
 FOR EXAMPLE, taking the above scale figure of the full-sized instrument, 
 we see, that the centre line of the second rule (which is 50 cms. long) lies at 
 an angle of 48 to the centre line of the first or horizontal rule. And, the 
 third or vertical rule hangs at right angles to the first rule. The length 
 of the plumb-line of the third rule is 37 cms. to where it cuts the centre 
 line of the first rule at right, angles at 33-5 cms. from its centre or zero 
 mark. 
 
USES OF THE TAMJENTOMETER. 
 
 365 
 
 Hero, rhe first or horizontal rule is called the base; the second or 50 cm. 
 rule i- th.j hypothenuse ; and the third or vertical rule is called the per- 
 pend cid r. Then, we have the following ratios for the acuie angle of 48* 
 in the lull-sized instrument.* 
 
 Sine of the 
 
 Tangent 
 Cotangent 
 Secant 
 Cosecant 
 
 Perpendicular 
 HyputbtiuUae 
 
 _ 5^ _ 
 Hjputuenuse 
 
 PerriPTidicnlar 
 
 p , 
 
 Perpendicular 37 cms. 
 
 Hvpofhennse ^o cms. 
 
 Table for 8 * 
 
 j 
 Perpendicular 37 cms. 
 
 i Radian = Unit angle in circular measure = = 57'29. 
 
 .-. 48 -i- 57 -29 = 8378 of a radian. See Table for the radian of 48*. 
 
 * When resolving a force into its two component forces, the student 
 must pay particular attention to the ( + ) or { ) sign of the value of the 
 angle. See Functions of Angles in Castle's "Practical Mathematics for 
 Beginners," chapter xvi., or other book on Trigonometry. 
 
 NOTE. I am indebted to Ludw. Loewe and Co. Ltd., 30 Farringdon 
 Road, London, for the use of their Electros of " Limit Gauges." 
 
 I have to thank Mr. H. M. Budgett, of the Crown Works, Chelmsford r 
 for six excellent views of his company's new English Gauges, and the 
 Louis Gassier Co., Limited, for the liberty to reproduce the Micrometer 
 Gauges and the W'hitworth measuring machine which appeared in Cassier's 
 Magazine, September 1901. The attention of Students who are interested 
 in gauges and accurate work is directed to " The Specification of the 
 Engineering Committee on Standards " ; and to comments upon the 
 same in The Electrician for Aug. 1906. 
 
366 LECTURE XXVI. QUESTIONS. 
 
 LECTURE XXVI. QUESTIONS. 
 
 1. Describe any micrometer screw gauge with which you are acquainted 
 suitable for measuring to the rgV* of an inch. Sketch and describe care- 
 fully the method of graduation and the position of the gauge when set to 
 measure -374 inch. 
 
 2. Sketch and describe the construction and use of external and internal 
 workshop gauges, by means of which the size of a spindle (say 2 inches 
 diameter), and that of a hole into which it fits, may be ensured within 
 specified limits of accuracy. State any advantages due to this system of 
 working. (B. of E., 1902.) 
 
 3. Sketch a cylindrical one inch external gauge, and describe generally 
 the measuring machine which you would require to employ, and the 
 manner of using the same, in order to construct another gauge of the like 
 kind, but measuring i -005 in diameter. How is the gauge worked down 
 to the right size and finished ? 
 
 4. Sketch and describe the construction and action of a *' Whitworth 
 Millionth Measuring Machine." For which purposes have you seen it 
 used? 
 
 5. Sketch and describe an " Equivalents Micrometer Gauge " and state 
 the advantages which it possesses over an ordinary gauge. 
 
 6. Sketch and describe fully an " External Limit Gauge " for a 2} inch 
 shaft, such as the one illustrated by the second figure in this Lecture. 
 Explain concisely and clearly how you would use such a gauge. Show 
 your calculations in full for the percentage error in 2^ inches diameter, if 
 a turner just makes a shaft as a tight fit for the "go on " end of the gauge. 
 Also for the percentage error if he should so reduce the diameter of a 
 shaft, to let the "not go on " end of the gauge fit the shaft. 
 
 7. Describe by aid of sketches Sir Joseph Whitworth's early attempts 
 towards and realisations of mechanical accuracy with standard surface- 
 plates, screws, gauges, and measuring machines. 
 
 8. Sketch and describe the new set of English Gauges as made by the 
 Crown Works, Chelmsf ord, according to Windley's Patents. 
 
 9. Show how the gauges in the previous question are self-checking. 
 Also, show by calculation, the exact range or number of different measure- 
 ments which can be made with seven blocks and the external holders. 
 If two extra blocks, -fa* and -fa", were added, prove how many more measure- 
 ments could be made. 
 
 10. Sketch neatly and describe concisely Sir Joseph Whitworth's Millionth 
 Measuring Machine. State the pitch of screw, number of teeth on worm- 
 wheel, and number of divisions on the graduated wheel to measure with 
 such accuracy. 
 
 11. Sketch, index and describe, any fixed good workshop measuring 
 machine and state how you would use it. 
 
 12. Find the sine, cosine, tangent, cotangent, secant and cosecant of an 
 angle in a right-angled triangle whose base is 40 cms, and perpendicular 
 20 cms. Also express 27 in radian measure. 
 
( 367 ) 
 
 APPENDIX A (pages 368 to 370). 
 
 (I) General Instructions by the Board of Education for their Examina- 
 tions on Applied Mechanics, Stage 1. 
 
 (ii) General Instructions by the City and Guilds of London Institute for 
 their Examination on Mechanical Engineering, Ordinary Grade. 
 
 (iii) Rules and Syllabus of Examinations by the Institution of Civil Engi- 
 neers for Admission of Students. 
 
 APPENDIX B (pages 371 to 401). 
 
 Board of Education's Exam. Papers in Applied Mechanics, Stage I. 
 The City and Guilds of London Institute's Ordinary Exam. Papers in 
 Mechanical Engineering. And the Institution of Civil Engineers' Exam. 
 Papers * Elementary Mechanics, arranged in the order of the Lectures. 
 
 APPENDIX C (pages 403 to 410). 
 
 The latest Exam. Papars pertaining to Mechanics and set by the govern- 
 ing bodies enumerated under Appendix A. 
 
 APPENDIX D (pages 411 to 413). 
 
 (i) Units of Measurement and their Definitions ; Practical Electrical 
 Units and their Symbol Letters. 
 
 (ii) Examination Tables, Useful Constants, Logarithm?, Antilogarithra 
 and Functions of Angles. 
 
 2 A 
 
( 368 ) 
 
 Appendix A. 
 
 May Examination on Subject VII. 
 APPLIED MECHANICS.* 
 
 BY THE BOARD OF EDUCATION, SECONDARY BRANCH, 
 SOUTH KENSINGTON, LONDON 
 
 Stage 1. 
 
 GENERAL INSTRUCTIONS. 
 
 If the regulations are not attended to, your paper 
 will be cancelled. 
 
 Immediately before the Examination commences, the following 
 REGULATIONS ARE TO BE READ TO THE CANDIDATES. 
 
 Before commencing your work, you are required to fill up the numbered 
 slip which is attached to the blank examination paper. 
 
 You may not have with you any books, notes, or paper other than that 
 supplied to you for use at this examination. 
 
 You are not allowed to write, draw, or calculate on your paper of 
 questions. 
 
 You must not, under any circumstances whatever, speak to or com- 
 municate with another candidate. Those superintending the examina- 
 tion are not at liberty to give any explanation bearing upon the paper. 
 
 You must remain seated until your papers have been collected, and then 
 quietly leave the examination room. No candidate will be allowed to 
 leave before the expiration of one hour from the commencement of the 
 examination, and none can be re-admitted after having once left the room. 
 
 All papers, not previously given up, will be collected at 10 o'clock. 
 
 If any of you break any of these regulations, or use any unfair means, 
 you will be expelled, and your paper cancelled. 
 
 Before commencing your work, you must carefully 
 read the following instructions. 
 
 Put the number of the question before your answer. 
 
 You are to confine your answers strictly to the questions proposed. 
 
 Such details of your calculations should be given as will show the 
 methods employed in obtaining arithmetical results. 
 
 The value attached to each question is shown in brackets after the 
 question. 
 
 A table of logarithms and functions of angles and useful constants and 
 fc rriiulffi is supplied to each candidate. 
 
 The examination in this subject lasts for three hours, 
 
 * See Appendix C for the latest Exam. Papers." 
 
APPENDIX A. 369 
 
 CITY AND GUILDS OF LONDON INSTITUTE. 
 DEPARTMENT OF TECHNOLOGY. 
 
 Technological Examinations. 
 MECHANICAL ENGINEERING.* 
 
 OBDINARY GRADE PART L 
 (FIRST YEAR'S COURSE.) 
 
 INSTRUCTIONS. 
 
 No Certificates will be given on the results of this Examination (First 
 Year's Course), but the Candidates' successes will be notified to the Centre 
 where they were examined. 
 
 To obtain a Certificate, it is essential that Candidates should pass both 
 in Part I. and Part II. ; the Examination in Part II. witt be hdd on 
 Thursday, May 3 Jot 7 p.m. Candidates may take both Parts I. and IL in 
 the same year. 
 
 The class of Certificate and the order of Prize will be determined by the 
 results of the Examination in Part II. only. 
 
 The maximum number of marks obtainable is affixed to each question. 
 
 The number of the question must be placed before the answer in the 
 worked paper. 
 
 Three how* allowed for thit paper. 
 
 The Candidate is at liberty to use divided scales, compasses, set squares, 
 calculators, slide rules, and tables of logarithms. 
 
 A piece of squared paper to be given to each Candidate, if required. 
 
 The Candidate is not expected to answer more than vine question*, 
 wSaich must be selected from (too Sections only. 
 
 * See Appendix C for the latest Exam. Papers. 
 
 t This date is only approximate, and subject to a slight alteration eaoh 
 year. 
 
APPENDIX A. 
 
 The Institution of Civil Engineers* Rules for 
 Admission of Students. 
 
 SYLLABUS OF THE EXAMINATIONS.* 
 
 1. ENGLISH (one Paper, time allowed, 3 hours). A general Paper com- 
 prising questions in Geography, History and Literature. 
 
 2. MATHEMATICS (two Papers, time allowed, 3 hour* for each). Papers 
 comprise questions in Arithmetic; Algebra; Geometry (Euclid I. -IV. ) ' 
 and Trigonometry. 
 
 3. Two subjects, to be selected by the Candidate from the following 
 ten : a language is not compulsory, but in any case not more than one 
 language may be taken (time allowed, 3 hours for each Paper) : 
 
 LATIN, GREEK, FRENCH, GERMAN, ITALIAN, SPANISH. 
 
 Elementary Mechanics of solids and fluids, t ELEMENTARY 
 PHYSICS, including heat, light, Electricity and Magnetism. 
 ELEMENTARY CHEMISTRY. GEOMETRICAL AND FREEHAND DRAWING. 
 
 * See Appendix C for the latest Exam. Papers. 
 
 f My Manuals on Applied Mechanics, Magnetism and Electricity, are 
 suitable for Young Engineers desiring to prepare by correspondence or 
 otherwise for these subjects, of which the most recent examples in Mechanics 
 are given in Appendix C, whilst those on Electricity and Magnetism are 
 printed in the Appendix of the Eighth Edition of the latter work. Candi- 
 dates should write direct at once to the Author of this book for his C.E. 
 Prospectus, which gives full details of Tuition for these Examinations. 
 
 RULES OF THE EXAMINATIONS. 
 
 The Examinations are held in London in February and October annually, 
 on four days beginning on the second Tuesday in each of those months. 
 The February Studentship Examination may, in the discretion of the 
 Council, be held also in Manchester, Glasgow and Newcastle-on-Tyne. 
 
 The Council will consider an Application from a person who is duly 
 recommended for Admission as a Student of the Institution, to present 
 himself for the Studentship Examination. 
 
 , Applications to attend the Associate Membership Examination will be 
 received from Students of the Institution who are not less than 21 years 
 nor more than 26 years of age on the last day for entry. 
 
 Arrangements may be made for the examination of Candidates In India 
 
 1 or in the Colonies, after submitting duly completed proposals for Election. 
 All applications for Rules, Forms and Admission, &c., must be made 
 through the Secretary, the Institution of Civil Engineers, Great George 
 Street, Westminster, London, S.W. 
 
ORDINARY QUESTIONS. 
 
 371 
 
 Appendix B. 
 
 See Appendix D for Practical Electrical Units 
 LECTURE II. ORDINARY QUESTIONS. 
 
 1. In a shale mine in order to drain one of the pits a treble ram pump, 
 driven by an electric motor, is employed. The rams are 9^ inches in 
 diameter by 1 2-inch stroke, they each make 3475 strokes per minute, and 
 the height to which the water is lifted is 393 feet. 
 
 Fnd: 
 
 (a) How many gallons of water this pump can lift per minute. 
 (6) How many foot-pounds of useful work are done per minute, 
 (c) The useful horse-power when the pumps are running steadily. 
 
 (B. of E., 1906.) 
 
 2. A windmill is employed to drive a pump which has to lift water from 
 a well and deliver it into an overhead tank. It was found that when the 
 windmill works steadily under the action of a uniform wind for a period 
 of i hour, 5000 gallons of water are raised from the well and delivered 
 into the tank the average height of lif t is 60 feet. What under these 
 conditions is the useful horse-power of the windmill ? (B. of E., 1907.) 
 
 3. An electrical hoist is employed in raising coal from the hold of a 
 ship and delivering it into railway cars, the amount of lift being 125 feet. 
 If the coal is raised at the rate of 2400 Ibs. per minute, what is the useful 
 horse-power ? 
 
 Convert this into watts. 
 
 If the current is supplied at a voltage of 250, and if the efficiency of the 
 whole arrangement is 50 per cent., how many amperes of current must be 
 supplied to the motor working the hoist ? (B. of E., 1907.) 
 
 4. Two closely coiled spiral springs were made out of round steel wire, 
 J-inch diameter. The one spring, A, had a mean diameter of coil of 4 inc^ es 
 
 and the other, B, had a mean diameter of coil of 5 inches ; both springe 
 had 12 complete coils. These two springs were tested by loads extending 
 them axially, and the results of the tests are shown in the table below : , , 
 
 Axial load \ 
 in pounds / 
 
 2 
 
 4 
 
 6 
 
 8 
 
 10 
 
 12 
 
 H 
 
 16 
 
 18 
 
 20 
 
 Extension "| 
 
 
 
 
 
 
 
 
 
 
 
 of the 
 spring A. I 
 Inches. J 
 
 0-26 
 
 0-52 
 
 079 
 
 1*06 
 
 1-32 
 
 i'59 
 
 1-86 
 
 2'12 
 
 2'39 
 
 2-66 
 
 Extension ) 
 
 
 
 
 
 
 
 
 
 
 
 of the I 
 spring B. f 
 Inches. ) 
 
 0-51 
 
 I '02 
 
 i'53 
 
 2-04 
 
 2 '5 5 
 
 3-06 
 
 3'57 
 
 4-09 
 
 4'6o 
 
 5-12 
 
 Plot the results on squared paper. 
 
 Given that the law connecting the extension of these springs 
 with their mean diameter of coil is of the form 
 
 Extension of B /Mean diameter of coil of B\n 
 Extension of A ~ \Mean diameter of coil of A) 
 what is the probable value of n ? (B. of E., 1908.) 
 
APPENDIX B. 
 
 5. Show how to determine the work done by a variable force[~moving"in 
 its own direction. A cage weighing 1200 Ibs, is raised 300 ft, by a windlass 
 having a wire rope weighing i Ibs. per foot run. Show, by a diagram to 
 scale, the work done at any stage, and mark on it the numerical values for 
 a lift of 100, 200, and 300 feet respectively. (C. & G., 1909, O., Sec. A.) 
 
 6. An engine cylinder, fitted with a Joy valve gear, has a stroke of 
 26 inches, and a connecting-rod 74 inches long. In the accompanying 
 
 LINE DIAGEAMS OF PROBLEM ON JOY'S VALVE GEAR. 
 
 sketch the link AB is pivoted to the connecting-rod at A, and to the free 
 end B of a swinging link BC centred at C. The motion for operating the 
 valve is taken from a point D on the link AB. Draw the path of thefpoint 
 D for a complete revolution of the crank. Take a scale of lin. = i ft. 
 
 (C. & G., 1909, 0., Sec. A.) 
 
 LECTURE IV. ORDINARY QUESTIONS. 
 
 i.l The right-angled bell crank lever, centred at A, shown[in the[sketch 
 is attached to a spring by one of its arms, and to another^lever, centred^at 
 B, by the other arm. 
 
 If the spring requires a direct pull of 20 Ibs. hi order to stretch it 2[inches, 
 find what force P, applied as shown, will stretch the spring this amount.C ! 
 
 (B. of E., 1906.) 
 
 24" ~~2-'-* 
 
 22 
 
 oj 
 
 1 
 
 J 
 
 '--4'-* A 
 
 B 
 
 ? 
 
 -> 
 ( 
 
 nT* 
 
 BELL CRANK LEVER AND SPIRAL SPRING. 
 
ORDINARY QUESTIONS. 
 
 373 
 
 2. An ordinary bell-pull, shown in the sketch, is in equilibrium. Deter- 
 mine in any way you please the magnitude of the force Q and the magni- 
 tude and direction of the resultant thrust upon the supporting pin A. 
 
 (B. of E., 1907.) 
 
 P=10lbs. 
 
 ORDINAEY BELL-PULL LEVEB. 
 
 3. The figure shows the mechanism known as a knuckle joint.^A^force 
 of 50 Ibs. is applied at the point A, its line of action being perpendicular 
 to the line BC. Determine graphically, or in any other way, the vertica 
 thrust delivered by the block D. Neglect friction. 
 
 How will this thrust vary as the block D descends ? (B. of E., 1908.) 
 
 TOGGLE OR KNUCKLE JOINT. 
 
3/4 APPENDIX B. 
 
 LECTURE V. ORDINARY QUESTIONS. 
 
 i. Describe how you would determine experimentally the coefficient 
 of sliding friction between two pieces of metal of any convenient size when 
 the speed of rubbing is low. (B. of E., 1906.) 
 
 LECTURE VI. ORDINARY QUESTIONS. 
 
 f I. Describe any one form of lifting tackle with which you are acquainted 
 and explain with reference to it the terms " velocity ratio," " force ratio," 
 and " efficiency." Explain how you would determine their numerical 
 values for all loads up to the full capacity of the tackle. (n 
 
 (C. & O., 1908, 0., Seo. A.) 
 
 LECTURE VII. ORDINARY QUESTIONS. 
 
 I. The sheave of a differential pulley block consists of two parts which 
 have diameters of 8 and 9 inches respectively. What is the velocity ratio 
 of the mechanism when a load is being raised ? 
 
 If the mechanical efficiency of the pulley is 32 per cent., what pull must 
 be exerted in order to raise a load of 2 tons ? (B. of E., S. 1, 1909.) 
 
 LECTURE VIII. ORDINARY QUESTIONS. 
 
 1. The weight of a span of telegraph wire is 127 Ibs. At one end the 
 wire makes an angle of 5 and at the other an angle of 7 with the horizontal, 
 what are the pulling forces at these ends ? (B. of E., S. i , 1909. ) 
 
 2. A simple Warren girder is as sketched. Loads of 3 and 4 tons are 
 carried at the two joints of the top member. Find, analytically or other- 
 wise, the forces in the different members, (C. & G., 1906, 0., Sec. B.) 
 
 1 3 TONS 14 TONS 
 
 SIMPLE WARREN GIRDER. 
 
 3. Explain one method of determining the stresses in the members of 
 a pin-jointed frame, j 
 
ORDINARY QUESTIONS. 
 
 \4 Tons 
 
 375 
 
 5 Tons 
 
 ^ ___. 
 
 LOADED ROOF TRUSS. 
 
 Determine the stresses in all the members of the roof truss loaded, as 
 shown in the figure. (C. &. G., 1908, O., Sec. B.) 
 
 4. A simple crane is of the form shown in the diagram, and carries 5 tons 
 at C. Determine the stress in AB, AC, and BC, and find the magnitude 
 of the balance weight W so that there shall be no bending moment on the 
 post BD. (C. & G., 1908, O., Sec. D.) j 
 
 5 Tons 
 
 LINE DIAGRAM OF A SIMPLE CRANE. 
 
 5. "A crane of 'the form shown by the accompanying sketch carries a 
 load of 3 tons. The reaction at the upper bearing ia horizontal. Deter* 
 
 Horizontal, * 
 
 Reaction, 
 
 LINE DIAGRAM OP A WALL CRANE. 
 
376 APPENDIX B. 
 
 mine the resultant pressure on the footstep bearing, and the stresses in the 
 members of the crane, assuming that all the connections are pin-joints. 
 
 (C. & G., 1909, 0., Sec. D.) 
 
 LECTUBE X. ORDINARY QUESTIONS. 
 
 1. A machine weighing 8 tons is dragged slowly along a horizontal floor. 
 If the coefficient of friction between the base of the machine and the 
 floor is 0*35, find in pounds the magnitude of the pull, and the normal 
 pressure on the floor when (a) the line of pull is horizontal, (6) the line of 
 pull makes an upward angle of 30 with the horizontal. (B. of E., 1907.) 
 
 2. The length of a journal is 9 inches, and diameter 6 inches, and it carries 
 a load of 3 tons. What horse-power is absorbed when making 100 revo- 
 lutions per minute, taking the coefficient of friction as "015, and how many 
 thermal units are radiated away per minute when the temperature of the 
 bearing remains constant ? (C. & G., 1907, 0., Sec. A.) 
 
 3. An electric locomotive draws a train of 700 tons up an incline of 
 i in 100 at a steady speed of 10 miles per hour. If the frictional resistances 
 are equal to 15 Ibs. per ton, what is the total pull exerted on the train and 
 what is the horse-power ? Find the current consumption in amperes, if 
 the voltage is 625, and if 60 per cent, of the electric energy supplied to the 
 locomotive is spent in hauling the load.*] (B. of E., S. i, 1909.) 
 
 LECTURE XI. ORDINARY QUESTIONS. 
 
 1. Hemp ropes are employed to transmit power from the engine shaft to 
 the driving pulleys on the different floors in a spinning factory. The 
 maximum tension in a rope is twice the minimum tension, the breaking 
 strength of one rope is 5700 Ibs., and it is desired to have a factor of safety 
 of 30. 
 
 Find the maximum horse-power which can be safely transmitted by one 
 of these hemp ropes at a speed of 70 feet per second. (B. of E., 1906.) 
 
 2. It is required to transfer 3 horse-power to a pulley 16 inches diameter 
 by belting. The revolutions per minute are 100, the tension in the tight 
 side of the belt is i times that in the slack side, the thickness of the belt 
 is -/TJ- inches, and the maximum working stress allowable is 320 Ibs. per 
 square inch. Find the least width of belt. (C. & G., 1906, 0., Sec. A.) 
 
 3. In a rope brake on a fly-wheel 8 feet diameter, the ropes being i inch 
 diameter the load is 500 Ibs., and the pull on the spring balance varies 
 from 10 to 20 Ibs. during a test. Find the brake horse-power, the revolu- 
 tions being 105 per minute. (C. & G., 1907, 0., Sec. A.) 
 
 4. Two shafts, which are not parallel, and do not intersect, are to be 
 connected by a belt passing over suitably placed pulleys. Explain what 
 are the necessary conditions to be observed in order that the belt shall 
 remain on the pulleys. A horizontal shaft, running along one side of a 
 machine shop, drives another horizontal shaft at right angles to the first 
 shaft and 20 ft. blow. Sketch a suitable arrangement for the belt-drive 
 if both shafts are to revolv at the same speed. 1 
 
 (C. & G., 1908, 0., Sec. A.) 
 
ORDINARY QUESTIONS. 
 LECTURE XII. ORDINARY QUESTIONS. 
 
 377 
 
 I. An engine having a stroke of 12 inches, and a connecting-rod 24 
 inches long, centre to centre, makes 300 revolutions per minute. Find 
 graphically the velocity of the piston at six intermediate positions of the 
 stroke, and draw a curve showing the velocity of the piston at any instant. 
 
 (C. & G,, 1908, O., Sec. A.) 
 
 LECTURE XIII. ORDINARY QUESTIONS. 
 
 I. A crane, tested in the usual way, and in which the velocity ratio is 
 40, gave the following results : 
 
 Weight lifted (W) 
 
 ICO 
 
 300 
 
 500 
 
 700 
 
 Force applied (P) 
 
 8-5 
 
 17-0 
 
 25-6 
 
 34 '2 
 
 Plot a curve showing the relation between P and W on a W base, and, OIL 
 the same base, plot a curve of efficiency. (C. & G., 1906, 0., Sec. A.) 
 
 LECTURE XV. ORDINARY QUESTIONS. 
 
 i. Taking the mean diameter of the thread of a i-inch bolt to be 0*92- 
 inches, the number of the threads to the inch being 8, and the coefficient 
 of friction O'I7 ; find the turning couple required to overcome an axial 
 force of 2i tons, and the efficiency under this load. 
 
 (C. & G., 1906, O., Sec. A.) 
 
 LECTURE XVI. ORDINARY QUESTIONS. 
 
 i. To do the cutting work in a small screw cutting lathe it is found: 
 that 0-47 H.-P. is required, and that the frictional losses in the gearing, 
 bearings, &c., absorb another 0*21 H.-P. How many foot-pounds of work 
 per minute is the driving-belt giving to the lathe ? j 
 
 The countershaft is driven by an electric motor, and the countershaft 
 and belts absorb 0-17 H.-P. How many watts must the motor give off in 
 order to keep the lathe running ? 
 
 If the voltage is 220, how many amperes will the motor require, assuming; 
 that its own efficiency is 89 per cent. ? 
 
 i H.-P. = 746 watts, and amperes multiplied by volts = watts. 
 
 (B. of E., 1906.) 
 
3/8 APPENDIX B. 
 
 2. The table of a drilling machine is raised by a hand- wheel, to the spindle 
 of which is attached a single-threaded worm which meshes with a worm- 
 wheel having 40 teeth. Compound with the worm-wheel is a spur-pinion, 
 having 19 teeth of i-inch pitch, which meshes with a rack on the frame of 
 the machine. Sketch the arrangement and find how many turns of the 
 handle are required to raise the table through 2 feet. 
 
 (C. & G., 1906, 0., Sec. A.) 
 
 3. In the feed gear of a drilling machine, in which a rack is used to give 
 the traverse of the spindle, the spindle is rotated by a bevel- wheel of 18 
 teeth keyed on the driving shaft, gearing with one of 32 teeth on the sleeve 
 surrounding the spindle. The greatest and least diameters of the pulleys 
 on the speed cone for the driving shaft are 7 inches and 4 inches, and this 
 cone drives a similar speed cone on the horizontal speed shaft. On this 
 shaft is a single-threaded worm which gears with a worm-wheel of 45 teeth 
 on the vertical feed shaft. A single-threaded worm on this shaft gears 
 with a wheel of 30 teeth, turning on a horizontal stud, and to which is 
 attached a pinion of 15 teeth gearing with, a rack of -inch pitch, which 
 gives the required feed. Find the least and greatest number of revolutions 
 of the drill spindle per inch of feed. (C. & G., 1906, O., Sec.A.) 
 
 4. The traverse shaft of a lathe is driven from the headstock mandrel 
 by belting, the greatest diameter of the speed cone at the extremity of the 
 mandrel being 5 inches. This drives a similar cone on the transverse shaft, 
 and the smallest diameter is 2 inches. A worm on the traverse shaft meshes 
 with a single-threaded worm-wheel, having 40 teeth, turning on a stud 
 carried by the saddle. At the front end of this spindle is. a spur-wheel of 
 15 teeth, meshing with a wheel of 45 teeth, which turns on a stud carried 
 by the apron ; and compound with this last wheel is a pinion of 12 teeth, 
 which meshes with the rack of -inch pitch, attached to the lathe bed. 
 Sketch the mechanism and find the traverse of the saddle per revolution 
 of the headstock mandrel. (C. & G., 1907, 0., Sec. A.) 
 
 5. A lathe is driven by a belt running on the 1 2-inch diameter pulley of 
 its speed cone, which then revolves at 200 revolutions per minute, and the 
 back gear of the lathe reduces this speed in the ratio of 9 to i. Under 
 these working conditions it is found that when a certain cut is being taken 
 
 off a bar, 6 inches diameter, the horse-power transmitted by the belt is 
 0*60. What is the pressure on the cutting tool in a direction tangential 
 to the turned surface, if we assume that 75 per cent, of the power trans- 
 mitted through the belt is lost in frictional and other wasteful resistances ? 
 
 (B. of E., 1908.) 
 
 6. Show how screws, differing in pitch from the leading screw, can be 
 cut in a lathe. If the leading screw of a lathe has three threads per inch, and 
 is right-handed, what arrangements of change would you use to cut (i) a 
 right-handed screw of four threads to the inch, (ii) a left-handed screw of 
 eleven threads to the inch ? You may assume that you have a set of change 
 wheels with teeth varying from 20 to 100 by differences of five teeth 
 
 (C. & G., 1908, 0., Sec. A.) 
 
 7. Describe, with the help of neatly-drawn sketches which should be 
 roughly to scale, any form of loose head-stock or poppet-head for a small 
 lathe, with which you have had practical experience. Show how the 
 spindle or poppet is advanced or withdrawn, and how it is clamped. 
 
 (B. of E., S. i., 1909.) 
 
 ! 8. A small machine tool is driven direct by an electric motor. How 
 would you determine the horse-power absorbed in the process of cutting 
 *he material ? (B. of E., S. i, 1909.) 
 
ORDINARY QUESTIONS. 
 
 379 
 
 9. The countershaft of a drilling machine makes 240 revolutions per 
 minute, and it carries a stepped pulley, the diameters of which are 12, 9 
 and 6 inches respectively. This drives an intermediate shaft by a belt and 
 pulley with similar steps. The intermediate shaft drives the drill spindle 
 by a bevel wheel of 30 teeth gearing with one of 40 teeth on the drill spindle. 
 Calculate the possible speeds of the drill spindle, and also determine the 
 diameter of the largest drill you can use if the circumferential cutting 
 speed is limited to 240 inches per minute. 
 
 (C. & G., 1909, 0., Sec. A.) 
 
 LECTURE XVIII. ORDINARY QUESTIONS. 
 
 I. The following results were obtained during an experiment to deter- 
 mine the quantity of water which would be discharged through a small 
 circular orifice in the side of a tank. The diameter of the orifice, which 
 had sharp edges, was I inch. 
 
 Number of 
 experiment 
 
 Duration of 
 experiment 
 
 Actual 
 discharge 
 
 Head of water above 
 centre of orifice 
 
 
 Minutes. 
 
 Lbs. 
 
 Inches. 
 
 I 
 
 15 
 
 57 6 
 
 1-5 
 
 2 
 
 15 
 
 660 
 
 2'0 
 
 3 
 
 15 
 
 733 
 
 2'5 
 
 4 
 
 15 
 
 827 
 
 3'27 
 
 5 
 
 15 
 
 915 
 
 4-01 
 
 6 
 
 15 
 
 1,011 
 
 5*o 
 
 7 
 
 10 
 
 737 
 
 6-0 
 
 8 
 
 10 
 
 788 
 
 7'0 
 
 jj- Plot on squared paper a curve to show the relation between the dis- 
 charge in Ibs. per minute, and the head of water above the centre of the 
 orifice. 
 
 From your curve determine the discharge in gallons per hour when the 
 head of water was 5^ inches. (B. of E., 1907.) 
 
 2. A straight balk of timber is 20 feet long and 12 inches square in cross- 
 section : its weight per cubic foot is 43*5 Ibs. If a weight of 112 Ibs. is 
 placed on the centre of the balk when it is floating in water, find the depth 
 to which the balk will be immersed. (B. of E., S. i, 1909.) 
 
 LECTUEE XIX ORDINARY QUESTIONS. 
 
 I. The rim of a turbine is going at 50 feet per second ; 100 Ibs. of fluid 
 enter the rim each second, with a velocity in the direction of the run's 
 motion of 60 feet per second, leaving it with no velocity in the direction 
 O f the wheel's motion. What is the momentum lost per second by the 
 fluid ? This is force. _^What work is done per second upon the wheel ? 
 
 (B. of E., 1907.) ' 
 
APPENDIX B. 
 
 2. A hydraulic press has a ram 6 inches in diameter : water is supplied 
 to the press from a single-acting pump, which has a plunger i inch in 
 diameter with a stroke of i inches. Neglecting frictional and other 
 losses in the pump and press, find the average rate (in foot-pounds per 
 minute) at which the pump works, if it makes 100 working strokes per 
 minute, while the press is exerting a force of 70 tons. (B. of E., 1907.) 
 
 3. A centrifugal pump, driven by an electric motor directly coupled to 
 it, is found during a test to deliver 320 gallons of water per minute into 
 ,n overhead tank, the mean height of lift being 65 feet. What useful 
 horse-power is the motor doing ? 
 
 If during the test the electric motor takes 35 amperes of current at 
 440 volts, what is the combined efficiency of the whole plant ? 
 
 (B. of E., 1908.) 
 
 4. A turbine, which gives off 50 horse-power to a belt running on a pulley 
 on its shaft, is supplied with water which, as it enters the turbine, is under a 
 head of 125 feet. If 75 per cent, of the total energy of the entering water 
 is thus utilised, what work is done per pound of water, and how many 
 gallons of water pass through the turbine per working day of 10 hours ? 
 
 (B. of E., S. i, 1909.) 
 
 LECTURE XX. ORDINARY QUESTIONS. 
 
 i. Describe, with the help of neatly drawn sketches, which should be 
 roughly to scale, a hydraulic jack, showing clearly all valves. 
 
 (B. of E., 1908.) 
 :, the wa1 
 
 2. In a hydraulic crane, with a ram 8 inches diameter, the water pressure 
 is 800 Ibs. per square inch, and the velocity of lift is increased eight-fold 
 by the use of a four-sheaved pulley block. What load can this crane lift 
 if its mechanical efficiency is 40 per cent. ? 
 
 How many gallons of power water will be used in lifting the load 50 feet ? 
 
 (B. of E., 1908.) 
 
 LECTURE XXI. ORDINARY QUESTIONS. 
 
 1. A cycle track is approximately elliptical in shape, the maximum 
 radius of curvature being 1 50 yards and the minimum 50 yards. 
 
 Find at each of these two places, the ratio which the centrifugal force 
 bears to the weight, if the speed of the racing cyclist is 25 miles per hour. 
 What would be the two inclinations of the track to the horizontal if the 
 track is laid so as to be perpendicular to the resultant force in each case ? 
 
 (B. of E., 1906.) 
 
 2. A railway truck weighing 10 tons starts from rest down an incline 
 mile long of i in 250. If the frictional and other resistances are equiva- 
 lent to 8 Ibs. per ton weight of the truck, with what velocity will the truck 
 be moving when it gets to the end of the incline ? 
 
 How far would it then run along a level stretch of the line before coming 
 to rest ? (B. of E., 1906.; 
 
ORDINARY QUESTIONS. 
 
 381 
 
 3. Two adjacent positions, G lt G 2 of the centre of mass G of a balance 
 weight were obtained by geometrical construction from a skeleton diagram 
 of the mechanism. These positions, measured in feet from two perpen- 
 dicular axes, were found to be as follows : 
 
 
 X 
 
 y 
 
 01 
 
 0-167 
 
 0*078 
 
 o, 
 
 0-352 
 
 0-146 
 
 The displacement O l G 2 took place in 1/50 second. Find the x and y 
 components of the mean velocity of G for this interval. 
 
 Plot the points G lt G. 2 , on squared paper. (B. of E., 1906.) 
 
 4. A traction engine travels at 6 miles per hour ; the road wheels are 
 6 feet in diameter and are driven through 5 to i gearing. Find the angular 
 velocity in radians per second of the fly- wheel on the engine shaft. 
 
 (B. of E., 1907.) 
 
 5. The rim of a cast-iron pulley has a mean radius of 12 inches; the 
 rim is 6 inches broad, and inch thick, and the pulley revolves at the rate 
 of 1 50 revolutions per minute ; what is the centrifugal force on the pulley 
 rim per inch length of rim ? 
 
 One cubic inch of cast-iron weighs 0-26 Ib. (B. of E., 1907.) 
 
 7. With an automatic vacuum brake a train, weighing 170 tons and 
 going at 60 miles an hour on a down gradiant of I hi 100, was pulled up in 
 a distance of 596 yards. Find the total resistance per ton in pounds, and 
 the time taken to stop the train. (C. & G., 1907, 0., Sec. A.) 
 
 8. A horizontal jet of water issues at a velocity of 20 feet per second 
 fiom the 2 -inch diameter nozzle of a hose pipe, and strikes a vertical wall. 
 What is the mass of water, in engineers' units, which strikes the wall per 
 second ? What is the momentum of this quantity of water ? What is 
 the force on the wall ? It is assumed that no water splashes back. 
 
 (B. of E., 1908.) 
 
 9. A man, whose weight is 14 stone, stands on the floor of a lift. What 
 force does he exert on it (i) when the lift is stationary, (ii) when it is des- 
 cending with an acceleration of 10 feet per second per second, and (iii) when 
 it is ascending with the same acceleration ? (B. of E. 1908.) 
 
 10. A train is running round a circular curve of 2000 feet radius at a 
 speed of 50 miles per hour. A weight is suspended by a thin cord from 
 the roof of one of the carriages ; at what inclination to the vertical will 
 the cord hang ? (B. of E., 1908.) 
 
 11. The fly-wheel of a punching machine weighs i tons, and has a 
 radius of gyration of 3 feet. It is turning at the rate of 130 revolutions 
 per minute when the punching of a hole is started, but at the completion 
 of the operation of punching it is found to be turning at the rate of only 
 125 revolutions per minute. How many foot-pounds of work have been 
 expended in punching the hole and overcoming the frictional resistances 
 of the machine ? 
 
 Note. If k is radius of gyration, this means that we may imagine the 
 mass of the whole wheel to be at the distance k from the axis, and this 
 enables us to calculate its kinetic energy. (B. of E., 1908.) 
 
382 
 
 APPENDIX B. 
 
 12. Explain how to determine the velocity of a moving body by con- 
 sidering the space described in a given time. A motor car, starting from 
 rest, travels a distance of s feet in t seconds, in accordance with the following 
 table : 
 
 t 
 
 
 
 I 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 8 
 
 
 
 5 
 
 16 
 
 33 
 
 56 
 
 85 
 
 1 20 
 
 161 
 
 208 
 
 Draw a curve showing the distance travelled at any time within this period, 
 and from this curve determine the velocity of the car at the ends of the 
 third, fourth, and fifth seconds respectively. (C. & G., 1908, O., Sec. A.) 
 
 13. A pulley, 3 feet in diameter, has a peripheral speed of 2000 feet per 
 minute. It is unbalanced to an amount which may be represented by a 
 mass of 0*5 Ibs. at a radius of I foot. Calculate the unbalanced force on 
 the pulley-shaft, and determine the positions at the pulley-rim of two 
 masses of 0-4 Ibs. to give a perfect balance. (C. & G., 1908, O., Sec. A.) ^ 
 
 14. An experiment with a small Pelton water-wheel gave results shown Jn 
 the annexed table : 
 
 Mean 
 Revs, 
 per Min. 
 
 Cubic Feet 
 of Water 
 passing 
 through 
 Wheel 
 per sec. 
 
 Speed of 
 Jet. Feet 
 per sec. 
 
 V. 
 
 Peripheral 
 Speed 
 of Vane. 
 Feet per 
 Sec. 
 
 V. 
 
 Ratio. 
 
 V 
 
 T 
 
 Efficiency 
 of Wheel. 
 Per Cent. 
 
 1090 
 
 0870 
 
 
 
 
 44-0 
 
 975 
 
 0824 
 
 
 
 
 63*8 
 
 885 
 
 0834 
 
 
 
 
 72-4 
 
 645 
 
 0840 
 
 
 
 
 72-8 
 
 540 
 
 0840 
 
 
 
 
 66-8 
 
 460 
 
 0847 
 
 
 
 
 6i-5 
 
 385 
 
 0834 
 
 
 
 
 56-1 
 
 - 265 
 
 0860 
 
 
 
 
 40-1 
 
 The cross-sectional area of the nozzle in square feet was 0*001043. Th 
 mean diameter of the bucket was 107 inches. 
 
 Fill in the third, fourth, and fifth columns of the table. 
 
 Plot a curve to show the variation of efficiency with variation of ratl 
 
 y taking efficiency in vertical ordinates and -y in horizontal abscissae. 
 
 T (B. of E., S. i, 1909.) 
 
ORDINARY QUESTIONS. 
 
 383 
 
 15. Define the term " horse-power." The tension on the draw-bar of a 
 locomotive is 800 Ibs. when the speed is 45 miles per hour. The weight of 
 the train, excluding the locomotive, is 200 tons. If the efficiency of 
 the locomotive is 65 %, find the horse-power of the engine. Also find 
 the accelerating force exerted at the draw- bar to change the speed from 
 45 to 50 miles per hour in one minute. (C. & G., 1909, O., Sec. A.) 
 
 16. Define the terms " velocity " and " acceleration," and show how to 
 determine the acceleration of a body moving in a straight line when its 
 velocity is known at each instant. The accompanying diagram is to be 
 
 6 
 
 CUEVE REPRESENTING THE VELOCITY OF BODY. 
 
 drawn so that each unit square is of I inch side. The curve ABC represents 
 the velocity of a body during a given interval of time, it is circular from 
 A to B, with centre at D, and from B to C it is a straight line parallel to 
 the horizontal axis. Draw to scale a curve showing the acceleration of 
 the body at any instant, and mark on it the numerical values of the ac- 
 celeration at the end of each second. (C. & G., 1909, O., Sec. A.) 
 
 ft 17. Show how the resultant velocity of a body may be obtained when the 
 component velocities impressed upon it are known. A goods engine 
 moving at the rate of 20 miles per hour, has equal coupled wheels 54 inches 
 in diameter, and the crank-pins move in circles 24 inches in diameter. 
 Determine the velocities of the crank-pins relative to the rails for three 
 equally spaced positions reckoned from the lowest positions of the crank- 
 pins. (C. & G., 1909, O., Sec. A.) 
 
 ^ 1 8. Assuming the expression for the law of centrifugal force, show how to 
 obtain a formula for the stress /in the rim of a fly-wheel in the form 
 
 where w is the weight of a unit volume of the rim, v is the velocity, and g 
 is the gravitation constant. Determine the limiting speed in revolutions 
 per minute of a cast-iron fly-wheel rim having a mean diameter of 10 feet, 
 when the allowable stress is 2400 Ibs. per square inch. The weight of one 
 cubic inch of cast-iron may be taken as 0*28 Ibs. 
 
 (C. & G., 1909, O., Sec. A.) 
 
 2 B 
 
APPENDIX B. 
 
 19. A r hammer head weighing 3-22 Ibs. moving at 30 feet per second is 
 stopped in o'ooi second : what is the average force of this blow in pounds ? 
 
 (B. of E., S. i, 1909.) 
 
 20. A locomotive is travelling at 60 miles per hour. The driving-wheels 
 are 6 feet 6 inches in diameter. What is the angular velocity of the driving- 
 wheels in radians per second ? ' 
 
 If the stroke of the piston is 26 inches, what is the mean speed of the 
 piston relatively to the cylinder in feet per minute, assuming there is no 
 slip of the wheels ? (B. of E., S. i, 1909.) 
 
 21. A fly-wheel, which weighs 18 tons, when mounted on its axis and 
 rotated, is found to be out of balance, and, in order to bring it into balance 
 it is found necessary to fix a counter-weight of 420 Ibs. to the wheel, at 
 a distance of 90 inches from the axis of the shaft. What was the distance 
 of the centre of gravity of the unbalanced wheel from the axis of the shaft ? 
 
 Show by a sketch where you would fix the counter-weight. 
 
 (B. of E., S. i, 1909.) 
 
 22. A fly-wheel weighs 8 tons, its radius of gyration is 5 feet 5 inches, 
 and it is rotating at a speed of 90 revolutions per minute. How many 
 foot-pounds of energy are stored up in it ? (P- ' 
 
 If this wheel were supported in two bearings, each 12 inches in diameter, 
 and if the coefficient of friction were 0*01, how much energy is wasted hi 
 overcoming friction in one revolution, and how many revolutions would 
 this fly-wheel make before coming to rest after the turning force was cut 
 off ? ' (B. of E., S. i, 1909.) | 
 
 LECTURE XXII. ORDINARY QUESTIONS. 
 
 1. A tie-bar in a roof is made of steel angle bar ; the section of the steel 
 angle bar is 4 inches by 4 inches by inch, and the tie- bar when finished 
 in the workshop is 20 feet in length. When in position in the roof the tie- 
 bar may during a gale have to resist a total pull of 22^ tons ; what is the 
 tensile stress per square inch in the metal of the tie-bar under these con- 
 ditions, and how much would the tie-bar lengthen under this load ? ,-, 
 
 Young's modulus of elasticity is 12,500 tons per square inch. ^ T 
 
 (B. of E., 1906.) 
 
 2. A knuckle joint is required to withstand a tensile force of 10 tonsi 
 The safe working stress, both in tension and shear, may be taken as 9000 Ibs. 
 per square inch. Find the diameters of the rod and pin, and sketch 
 the joint, roughly, to scale. (C. & G., 1906, 0., Sec. B.) 
 
 3. A rectangular test-bar, in tension, gave the following results : 
 
 Total 1 
 load > 
 in Ibs. J 
 
 8,000 
 
 16,000 
 
 24,000 
 
 32,000 
 
 34ooo 
 
 40,000 
 
 48,000 
 
 56,000 
 
 60,000 
 
 SS.ooo 
 
 Exten-^ 
 
 
 
 
 
 
 
 
 
 
 
 sionin f 
 inches. J 
 
 '002 
 
 0044 
 
 0070 
 
 '0103 
 
 'Oi6 
 
 Tgc 
 
 P470 
 
 r 3 fi 
 
 2'S 
 
 2-9 
 
 Sketch a curve showing the relation between force and extension on any 
 suitable scale squared paper may be used and infer the stress at the 
 
ORDINAEY QUESTIONS. 385 
 
 elastic limit and the maximum stress, the original dimensions of the bar 
 being 1763 inches by '611 inches. If the distance between the gauge 
 points is 10 inches, find the coefficient of elasticity (E) of the bar.^, - ,-- 
 
 (C. & G., 1906, p., Sec. B.)^? 
 
 4. Explain what you mean by the efficiency of a riveted joint, and point 
 out on what it depends. In a marine boiler the diameter is 12 feet, the 
 working pressure is 200 Ibs. per square inch, and the longitudinal joints 
 are butt joints with double straps treble riveted. If the ultimate stress is 
 62,000 Ibs. per sq. inch, the factor of safety 5, and the efficiency of the 
 joint -8, find the thickness of the plate required, and make a rough sketch 
 of the joint. (C. & G., 1906, O., Sec. B.) 
 
 5. In order to connect together the two halves of a long tie-rod, an 
 eye is forged at the end of one half, and a fork (into which the eye enters 
 at the end of the other half, and a pin is passed through the two sides of 
 the fork and through the eye. If the total pull in the tie-rod is 16 tons, 
 and if the shearing stress in the metal of the bolt is not to exceed 8000 Ibs. 
 per square inch, what diameter would you make the pin ? 
 
 (B. of E., 1907.) 
 
 6. A strut is built up out of two pieces of T-steel, each 6 inches by 3 
 inches by f inch, riveted back to back. If this strut supports a load of 
 22-3 tons, what is the compressive stress per square inch ? 
 
 If a total load of 105 tons would destroy this strut, what is the factor of 
 safety ? (B. of E., 1907.) 
 
 7. A bar, of rectangular section, 1 75 inches wide and 0*6 1 inches thick, 
 is found under a load of 20,000 Ibs. to have stretched 0*0056 inch. Find 
 the stress induced, and, if the length be 10 inches, find Young's modulus. 
 
 (C. & G., 1907, O., Sec. B.) 
 
 8. Find the thickness of the plates of a cylindrical boiler 50 inches in 
 diameter to sustain a pressure of 50 Ibs. per square inch, the working stress 
 being 4000 Ibs. per sq. inch and the efficiency of the joint being o - 6o. 
 
 (C. & G., 1907, O., Sec. B.) 
 
 9. If in the last question the joint is a lap joint double riveted, and the 
 diameter of the rivets is f inch, find the pitch, the shear stress of rivets, 
 being 4000 Ibs. per square inch. (C. & G., 1907, 0., Sec. B.) 
 
 10. A piece of steel is to be tested in tension ; show how you would 
 proceed to make a test, and indicate, by means of a diagram, how the 
 force and extension vary with each other. (C. & G., 1907, O., Sec. B.) 
 
 11. A copper trolley wire, which is 0*45 inch in diameter and 60 feet in 
 length, is found to elongate 0-075 m ch under a certain pulL If the mod- 
 ulus of elasticity (Young's modulus) of this quality of copper is known 
 to be 15,000,000 Ibs. per square inch, what is the total pull in the trolley 
 wire ? (B. of E., 1908.) 
 
 12. Explain the meanings of the terms " stress," " strain," and " modulus 
 of elasticit y, " by reference to the case of a rod under tensional stress. A 
 piece of bo iler plate, 2 inches by f inch in cross-section, has a load of 12,000 
 Ibs. applied to it in a testing machine. The modulus of elasticity of the 
 material e xpressed in inches and pounds is 31,000,000. Calculate the 
 values of the stress and strain and determine the increase of length,in 
 a len gth of 12 inches, due to the applied load. (C. & G., 1908, 0., Sec. B.) 
 
 I 3 . Make a sketch of a knuckle joint connecting an eccentric rod to a 
 val ve spindle, and assuming that the total load on the latter is 4000 Ibs., 
 dete nnine the dimensions of the various parts, and design the joint. Show 
 yo ur calculations clearly, and state what working stesses you have assumed 
 
 (C. & G., 1908, 0., Sec. B.) 
 
386 APPENDIX B, 
 
 14. A tension member, 8 inches by J inch in cross section, has a riveted 
 butt-joint with cover plates on each side. The total load on the member 
 is 72,000 Ibs. Design and draw a joint for this member, and show all 
 your calculations. (C. & G., 1908, O. t Sec. B.) 
 
 1 5. A hollow cylinder, 10 inches mean diameter, 10 feet long, and ij inches 
 thick, is to be oast with its axis vertical. Taking the specific gravity of cast 
 iron as 7-5, find the pressure on the bottom of the mould when it is full of 
 metal. One side of a mould for a cast-iron casting is a rectangle, 3 feet deep 
 by 2 feet wide. Find the whole pressure on the side of the mould. 
 
 (C. & G., 1908, 0., Sec. D.) 
 
 1 6. In an experiment with a hollow cast-iron column, 6 feet long,'* 5 
 inches in external diameter, and 4 inches in internal diameter, it was 
 found that under a compressive load of 30 tons the column shortened by 
 0*063 inch; what is the value of Young's Modulus (E) in pounds per 
 square inch for this cast-iron ? 
 
 J^When the load was increased to 192 tons, the column broke; what was 
 the compressive stress in tons per square inch at the instant of fracture. 
 
 (B. of E., S. i, 1909.) 
 
 LECTUEE XXIII. ORDINARY QUESTIONS. 
 
 1. Describe how you would determine experimentally the modulus of 
 rigidity of either a block of india-rubber or a steel rod. (B. of E., 1906.) 
 
 2. In a direct-acting steam-engine mechanism the stroke of the piston 
 is 2 feet and the crank shaft makes 1 50 revolutions per minute. 
 
 What is the speed of the crank shaft in radians per second ? What is 
 the speed of the crank pin in feet per second ? What is the mean speed 
 of the piston in feet per minute ? (B. of E., 1906.) 
 
 3. A solid cylindrical shaft is 5 inches in diameter. Find the external 
 diameter of a hollow shaft of same material, the internal diameter of which 
 is two-thirds the external and which shall have the same strength. Com- 
 pare the weights in the two cases. If the safe working stress be 4 tons per 
 square inch, and the revolutions per minute zoo, find the greatest horse- 
 power which can be safely transmitted. (C. & G., 1906, O., Sec. B.) 
 
 4. A motor having a turning moment T is coupled directly to a shaft 
 making N revolutions per minute. Show how to calculate the work trans- 
 mitted by the shaft, and obtain a formula for the horse-power transmitted 
 in^ terms of N and T and a constant. Determine the horse-power trans- 
 mitted by a shaft making 800 revolutions per minute if the turning moment 
 
 16,000, measured in pounds and inches. 
 
 (C. & G., 1908, 0., Sec. B.) 
 
 5. A chain is to be used for lifting a load of 5 tons. Assuming a safe 
 working stress on the chain of 4 tons per square inch, find the diameter of 
 the iron of the chain. (C. & G., 1908, 0., Seo. D.) 
 
 6. A shaft 3 inches in diameter transmits a twisting moment of 66,ooo-Ib. 
 inches and the flange couplings are bolted together by four bolts spacedi 
 on a circle of 5 inches diameter. Determine the nature and amount of stress 
 on each bolt and determine its diameter if the allowable stress is 
 
 12,000 Ibs. per square inch. (C. & G., 1909, O., Sec. B.) 
 
 7. Explain what kind of stress is produced in a shaft by a twisting 
 
OKDINARY QUESTIONS. 387 
 
 moment, and make a diagram showing how the stress varies across the 
 section of a shaft.'* A piece of tubing, 2 inches in external diameter, and 
 inch thick, is'used as a shaft. Assuming that the stress upon it is uniformly 
 distributed, determine the twisting moment it will transmit if the allowable 
 stress is_ 1 2,000 Ibs. per square inch. (C. & G., 1909, 0., Sec. B.) 
 
 LECTURE XXIV. ORDINARY QUESTIONS. 
 
 i. Sketch~a"single Hooke's joint, and explain in general terms/ how the 
 angular velocity ratio varies during a revolution. What is the object of a 
 double Hooke's joint ? (C. & G. t 1906, 0., Sec. A.) 
 
 LECTUBE XXV. OBDINABY QUESTIONS. 
 
 1. Sketch, ^andjdescribe the action of, the pin and slot mechanism aa 
 applied to shaping machines. 
 
 In such a mechanism, the distance between the two centres of rotation 
 is 3"inches, and the time ratio has to be two. If the line of stroke produced 
 pass'through the centre of the variably rotating crank, and is perpendicular 
 to' the line of centres, find the length of the crank radius, and also of the 
 slotted-link, for a stroke of 10 inches. (C. & G., 1906, O., Sec. A.) 
 
 2. InTthe four-bar mechanism shown in the sketch, the bar A is"a fixed 
 bar ; the* bars B and D rotate about the fixed centres OAB and OA D, and 
 they are coupled together at their outer ends by the bar C ; the bar B 
 
 FOUR BAR MECHANISM. 
 
 revolves~with uniform velocity round its fixed axis" OAB at 50 revolutions 
 per minute. Find in any way you please the position of the bar D when 
 
388 APPENDIX B. 
 
 the bar B is turned in a clockwise direction through angles of 30, 60, and 
 90 from the position shown in the sketch. 
 
 Prepare a table similar to the one shown, and obtain and enter up the 
 results required to complete the table : 
 
 Angle turned through 
 by the bar B. 
 
 Angle turned through 
 by the bar D. 
 
 Mean angular velocity of 
 the bar D in radians per 
 second during each interval . 
 
 30. . . 
 
 
 
 60. 
 
 
 
 
 90. 
 
 
 
 (B. of E., S. i, 1907.) 
 
 3. Sketch the arrangement in a planing machine in which bevel gears 
 are used, explaining how the motion is reversed, and how a quick-return 
 motion is obtained. Sketch also a " shipper " mechanism in which cams 
 or lugs are used. (C. & G., 1907, 0., Sec. A.) 
 
 P 4. The piston of a vertical engine has a travel of 18 inches, and the con- 
 necting-rod is 36 inches in length between centres of bearings. The line of 
 action of the piston cuts the horizontal position of the crank 4 inches from 
 the centre of the crank-shaft. Draw the curve of position of the mid-point 
 of the connecting rod for a complete revolution of the crank. Use a scale 
 of th. (C. & G., 1908, O., Sec. A.) 
 
 W$- Describe, with the help of neatly-drawn sketches which should be roughly 
 to scale, a belt gear for giving a slow cutting speed and a quick return 
 motion suitable for use in a planing machine, the table of which is traversed 
 to and fro by a screw or rack. (B. of E., S. i, 1909.) 
 
STUD. INST. C.E. QUESTIONS. 389 
 
 LECTUEE II. STUD. I. C.E. EXAM. QUESTIONS. 
 
 1. Explain the use of diagrams for determining the results of experiment. 
 Plot the following experimental values of P and W and obtain the relation 
 between them : 
 
 P pounds ... 8 16 24 40 
 
 W o 200 400 800 
 
 (Stud. I. C. E., Feb. 1906.) 
 
 2. Show how the work done by a variable force can be represented 
 graphically. Assuming that the resistance of a spiral spring", is propor- 
 tional to its extension, and that a load of 24 Ibs. extends the spring 
 0-25 inch, determine the work done in extending the spring i inch. 
 
 (Stud. L C. E., Oct. 1906.) 
 
 3. Define the terms " force " and " work" 
 
 A spring is compressed and the relation between the compressive force 
 and the compression is as given below. Find graphically the work done 
 in the process. 
 
 Force in Ibs. . . I o I 15 I 35 | 65 I no I 170 I 300 
 
 Compression in inches . | O | I j 2 | 3 | 4 | 5 | 6 
 
 (Stud. I. C. E., Oct. 1907.) 
 
 LECTURE III. STUD. I. C.E. EXAM. QUESTIONS. 
 
 1. A metre rule (weight 50 grams) rests on the edge of a table with 20 
 centimetres projecting over the edge. On the other end rests a 2O-gram 
 weight. How far from the edge of the table may a 5oo-gram weight be hung 
 before the rod tilts ? (Stud. I. C. E., Feb. 1905.) 
 
 2. Prove that the moment about any axis of three forces in equilibrium 
 is zero, and extend the theorem to any number of coplanar forces in 
 equilibrium. 
 
 Determine the tension of the rope wound on a capstan 2 feet in diameter 
 when 10 men, each weighing 12 stone, are pushing horizontally on the 
 capstan- bars 4 feet from the deck at a radial distance of 8 feet, the vertical 
 through a man's centre of gravity overhanging his toes a distance of 
 I -foot. (Stud. I. C. E., Oct. 1905.) 
 
 3. Prove that the centre of gravity of a triangular plate of uniform 
 thickness is on the line joining an apex to the centre of the opposite side, 
 and at a distance from this apex of two-thirds of the length of this median 
 line. Also show how to determine by experiment the centre of gravity of 
 an irregular plate of uniform thickness. (Stud. I. C. E., Feb. 1906.) 
 
 4. Show how to determine the resultant of two parallel forces. A 
 horizontal bar, 6 feet long, is supported at each end by rings depending 
 from spring balances. Determine the position of the centre of gravity of 
 the bar if the spring balances indicate loads of 40 Ibs. and 50 Ibs. respectively. 
 
 (Stud. I. C. E., Oct. 1906.) 
 
 5. Explain what is meant by the moment of a force. Show that the 
 sum of the moments of two forces in a plane with respect to a point in 
 that plane is equal to the moment of their resultant. Also show that the 
 moment of a couple, with respect to any axis at right angles to the plane 
 of the couple is invariable. (Stud. I. C. E., Oct. 1906.) 
 
 6. Show how to determine experimentally the centre of gravity of an 
 irregular body, and in particular explain how you would proceed to deter- 
 mine the centre of gravity of a metal plate shaped to the section of a tram- 
 way rail (Stud I. C. E., Feb. 1907.) 
 
39 APPENDIX B. 
 
 7. What do you understand by the " centre of gravity " of a body ? 
 
 A balk of timber weighs 800 Ibs. One end rests on the ground,and the 
 other on a " V " support placed on a weigh-bridge. The weight recorded is 
 320 Ibs. The weigh-bridge is then moved so that the " V " is I foot nearer 
 the end that is resting on the ground and the weigh-bridge registers 360 
 Ibs. Find how far the centre of gravity of the balk is from that end. 
 
 (Stud. I. C. E., Oct. 1907.) 
 
 8. What is a " couple " ? How is a couple specified, and how can it be 
 represented ? 
 
 A pair of compasses is opened so that the legs are at 90. Couples are 
 applied to the legs whose moments are respectively 3 Ib.-foot units and 
 4 Ib.-foot units, and they twist in opposite ways. Find what couple must 
 be applied at the hinge to equilibrate the two, and the axis of that couple. 
 
 (Stud. I. C. E., Oct. 1907.) 
 
 9. Define the term " centre of gravity. " A cylindrical vessel is 5 feet deep 
 and weighs 100 pounds ; when it is empty its centre of gravity is 2 feet 
 above its base. It is gradually filled with water. Plot to scale a curve 
 showing the relation between the depth of water in the vessel and the height 
 of the new centre of gravity, if the vessel when just full can contain 500 
 pounds of water. (Stud. I. C. E., Feb. 1908.) 
 
 LECTURE IV. STUD. I.C.E. EXAM. QUESTIONS. 
 
 1. Answer, giving reasons, the following questions on the balance : 
 
 (i) What conditions must be satisfied in order that a balance may 
 
 be true ? 
 
 (ii) If the scale-pan knife-edges are above the middle knife-edge 
 show that the sensitiveness of the balance increases with the 
 load, 
 (iii) Why must the scale-pans be suspended freely from the beam ? 
 
 (Stud. I. C. E., Feb. 1905.) 
 
 2. Explain the action of one form of lever weighing-machine. A loo-ton 
 testing machine, using a single lever for weighing the pull on a test-piece, 
 is arranged so that the line of action of the pull is 4 inches distant from 
 the fulcrum, and this pull is balanced by a weight of 5000 Ibs. on the 
 long arm of the lever. Calculate the distance the weight moves from its zero 
 position to balance the full load of 100 tons. (Stud. I. C. E., Feb. 1907. ) 
 
 LECTURE V. STUD. I.C.E. EXAM. QUESTIONS. 
 
 i. Explain how work is computed when a force moves its point of appli- 
 cation in any direction. 
 
 Find the work done per minute by a force pulling a body weighing 
 400 Ibs. over a rough plane at the rate of 5 miles an hour, if the coefficient 
 of friction is 0-25. (Stud. I. C. E., Feb. 1906.) 
 
 LECTURE VI. STUD. I.C.E. EXAM. QUESTIONS. 
 
 I. Write a short essay on the use and principle of a machine, bringing 
 in the meaning of the terms : velocity or displacement ratio, effort, load, 
 advantage, efficiency. 
 
 Describe some experiment you have made with a machine, and illustrate 
 the meaning of the above terms by numerical examples. 
 
 (Stud. I. C. E., Feb. 1905.) 
 
STUD. INST. C.E. QUESTIONS. 
 
 39* 
 
 2. Give sketches of three systems of pulleys, and state their mechanical 
 advantage. 
 
 Explain which system is preferable for a long pull, as in hoisting a 
 weight ; and which is to be preferred for a strong pull, as in setting up a 
 backstay. (Stud. I. a E. f Oct. 1905.) 
 
 LECTURE VTI. STUD. I.C.E. EXAM. QUESTIONS. 
 
 I. Explain the principle of action of the " Weston " or differential 
 pulley, and show how to determine the displacement or velocity ratio. In 
 such a machine the velocity ratio was found to be 20, and in order to lift 
 a weight of 420 Ibs. a pull of 40 Ibs. was exerted. Determine the efficiency- 
 of the machine for this load. (Stud. L C. E., Feb. 1906.) 
 
 LECTUBE VITE. STUD. I.C.E. EXAM. QUESTIONS. 
 
 1. A rod AB, the weight of which may be neglected, is hinged at A, andl 
 a weight of 14 Ibs. is suspended from its middle point. A string is fastened 
 to the end B, and when the system is at rest the rod is inclined at 30 to- 
 the horizontal, and the string makes an angle of 90 with the rod. Find 
 the pull along the string. (Stud I. C. E., Oct. 1904.) 
 
 2. Two rods AC, BC are freely jointed together at C, and a load of 28 Ibs. 
 is suspended from C. The two ends A and B are connected by a horizontal 
 string. If the system be placed vertically with A and B on a smootk 
 
 A B 
 
 Two RODS AC AND BC WITH STRING AB. 
 
 floor, find by a graphic method the thrusts along AC, BC, and the pull of 
 the string, when AB = 5 feet, AC = 4 feet, BC = 3 feet. * I 
 
 (Stud. I. C. E., Feb. 1905.) ' 
 
 3. Show in a diagram the forces which maintain equilibrium in a book 
 held horizontally between a finger and thumb, and draw a graphical deter- 
 mination of their magnitude. 
 
 A boat is propelled by two sculls, each 9 feet long and 6 feet from the 
 rowlock to the blade, and the sculler pulls each hand with a force of 20 Ibs. 
 Prove that the thrust on each rowlock is 30 Ibs., but the propulsive torch- 
 on the boat is 20 Ibs., and that the boat moves about double as fast as tee- 
 hands pull. L (Stud. I. C. E., Oct. 1905.) 
 
392 
 
 APPENDIX B. 
 
 ^4.: A flexible oord is carried by two pegs A and D in the same horizontal 
 line and 24 inches apart. A weight of 8 Ibs. hangs at B and an unknown 
 weight, at C,J thereby, causing the cord to assume the form shown in the 
 tflgure. ( 
 
 A *s 6' *~ tO' * 8" +D 
 
 FLEXIBLE CORD WITH ATTACHED WEIGHTS. 
 
 Find by a graphical construction the magnitude of the unknown weight 
 srad the tensions in the parts AB, BG and CD of the cord. 
 
 (Stud. I. C. E., Feb. 1906.) 
 
 5. Show how to determine the resultant of a number of forces meeting 
 at a point. Six forces acting at a point are parallel to the sides of a regular 
 hexagon taken in order, and their magnitudes are 4, 6, 7, 9, 8 and 3 pounds 
 respectively. Find the resultant, assuming that all the forces are directed 
 towards the point. (Stud. I. C. E., Oct. 1906.) 
 
 6. Obtain the graphical condition for the equilibrium of a number of 
 tf orces acting at a point. A jointed frame is loaded as shown in the figure. 
 .Determine the stresses in the members of the frame. 
 
 (Stud. I. C. E., Feb. 1907.) 
 
 ,4000 
 
 3000 Ibs 
 
 A JOINTED LOADED FRAME. 
 
STUD. INST. C.E. QUESTIONS. 393 
 
 7. Give the conditions that forces acting on a rigid body may be hi 
 equilibrium. ! I ' 
 
 A uniform plate, weighing 5 Ibs., is made in the form of a right-angled 
 area whose sides are 3, 4, and 5 feet. It is hung up by means of a string 
 and a peg so that the 5 -foot side is horizontal The peg is at the angle 
 joining the 4- and 5-foot sides; the string is at the other acute angle and 
 makes an angle of 60 with the 5 -foot side. Find graphically the tension 
 in the string. (Stud. I. C. E., Oct. 1907.) 
 
 8. What condition must be fulfilled in order that a system of forces 
 acting on a body, which are all in one plane, but not acting at one point, 
 may be hi equilibrium Show how the construction can be used to find 
 the supporting forces required for a structure when loaded. 
 
 (Stud. I. C. E., Oct. 1907.) 
 
 9. If three forces act on a rigid body, what conditions must be fulfilled 
 in order that equilibrium may be maintained ? 
 
 A uniform beam of timber weighs 200 pounds. One end rests on the 
 ground, the other has a cord attached to it. This cord is pulled till the 
 beam makes 30 with the horizontal, and then the cord makes 60 with the 
 horizontal Find (graphically or otherwise) the tension hi the cord and 
 the force on the ground. Ans. : Tension in cord = 100 Ibs ; reaction = 
 173 2 Ibs. (Stud. L C. E., Feb. 1908.) 
 
 LECTURE IX. STUD. I. C.E. EXAM. QUESTIONS. 
 
 1. Define horse-power, and prove that a locomotive of H horse-power 
 can draw a train of W tons against a resistance of r Ibs. per ton at a speed 
 
 of f iff miles an hour. 
 
 Calculate the horse-power of a locomotive drawing a train of 200 tons 
 up an incline of i in 200 at 50 miles per hour, taking the road and air re- 
 sistance at this speed as 28 Ibs. per ton. (Stud. I. C. E., Oct. 1905.) 
 
 2. Define the terms " work " and " power." A car weighing 3 tons is 
 running at 20 miles per hour up a uniform slope of i in 50. The frictional 
 resistances are 50 pounds per ton. Find the H.-P., and the work done in 
 20 minutes. Ans., H.-P. = 13*8; work done in 20 mins. = 9, 100,800 ft. Ibs. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
 LECTURE X. STUD. I.C.E. EXAM. QUESTIONS. 
 
 i. What is meant by coefficient of friction ? Show how the coefficient 
 of friction between wood and wood can be determined. 
 
 (Stud. I. C. E., Oct. 1904.) 
 
 If a machine, such as a screw-jack, does not overhaul, show that the 
 work done against the friction must be more than 50 per cent, of the total 
 work the man does. (Stud. I. C. E., Oct. 1904.) 
 
394 APPENDIX B. 
 
 2. Define the coefficient of limiting friction, and prove that it is the 
 tangent of the slope of the incline on which the body is on the point of 
 sliding. 
 
 Determine geometrically the greatest slope on which a four-wheeled 
 carriage can be held by the brakes, applied to the hind pair of wheels. 
 
 (Stud. I. C. E., Oct. 1905.) 
 
 LECTUHE XI. STUD. I.C.E. EXAM. QUESTIONS. 
 
 1. A belt-pulley has a diameter of 5 feet, and it is delivering 12 H.-P. to 
 a line of shafting. If the pulley make 120 revolutions per minute, find the 
 force along the tight part of the belt. (Stud. I. C. E., Oct. 1904.) 
 
 2. How can the brake horse-power of an engine be determined. If the 
 diameter of the brake-pulley is 3 feet, speed 250 revolutions per minute, 
 load 24 Ibs., and the spring-balance which takes up the rope at the slack 
 end registers 4 Ibs., find the horse-power. (Stud. I. C. E., Feb. 1905.) 
 
 3. Explain the meanings of the terms " work " and " energy," and define 
 the practical unit of work used by British engineers. 
 
 Find the brake horse-power of an engine making 300 revolutions per 
 minute if the tensions on the tight and slack sides of the brake strap are 
 72 Ibs. and 6 Ibs. respectively, and the brake wheel is 5 feet in diameter. 
 
 (Stud. I. C. E., Feb. 1907.) 
 
 4. Define the term " horse-power." 
 
 A pulley whose diameter is 4 feet is making 250 revolutions per minute. 
 A belt is put on the pulley and a weight of 50 Ibs. is hung from one end. 
 A spring balance attached to the other end reads 12 Ibs. Find the horse- 
 power delivered to the pulley. If the angle of embrace of the belt is 120 
 find the extra force on the bearing due to the brake. 
 
 (Stud. I. 0. E., Oct. 1907.) 
 
 LECTURE XIII. STUD. I.C.E. EXAM. QUESTIONS. 
 
 P I. Describe any experiment made by you to determine the efficiency of 
 a machine. 
 
 In a test of a hand-crane, with gear having a velocity ratio of 1 50 : I , it 
 was found that an effort of 25 Ib. at the handles raised a load of I ton. 
 Determine the efficiency of the machine for this load. 
 
 (Stud. I. C. E., Feb. 1907.) 
 
 2. Explain the terms " velocity ratio," " force ratio," and " efficiency," 
 as applied to a machine. A lifting crab has the following relation between 
 the force on the handle in pounds and the weight lifted in tons. Draw 
 curves connecting the force ratio and the efficiency with the load, and find 
 the efficiency of the crab at loads of 5 tons and 1 5 tons. The velocity ratio 
 is 500. 
 
 Force in pounds . . . 30 60 90 120 
 Load in tons . . . 3-3 ^ 7-4 12*5 17*5^. 
 
 Ans. ^Efficiencies are 0*558 and O'6987 respectively. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
STUD. INST. C.E. QUESTIONS. 395 
 
 LECTURE XVII. STUD. I.C.E. EXAM. QUESTIONS. 
 
 1. Show that the total pressure upon a flat plate immersed in water is 
 proportional to the depth of its centre of figure below the surface. Calculate 
 the pressure per square inch on a horizontal plate at a depth of 10 feet, 
 assuming that a cubic foot of water weighs 62*4 Ibs. 
 
 (Stud. I. C. E., Oct. 1906.) 
 
 2. Show how to find the total pressure on any submerged plane sur- 
 face. 
 
 A sluice gate is 4 feet wide and 6 feet high ; the bottom is 20 feet below 
 the. surf ace of the water ; find the total pressure on the gate. 
 
 [One cubic foot of water weighs 62$ Ibs.] (Stud. I. C. E., Oct. 1907.) 
 
 LECTUBE XVIII. STUD. I.C.E. EXAM. QUESTIONS. 
 
 1. At the bottom of a barometer there is a bubble of air, of which the 
 density is 0-0013 grams per centimetre, and diameter o'3 millimetre. If 
 this rises up the tube its volume increases. Explain why. At what height 
 will its volume be doubled? (Stud. I. C. E., Feb. 1905.) 
 
 2. Show how to determine the total pressure and the centre of pressure 
 of a vertical rectangular plate immersed in water with one edge at the 
 surface. Find the total pressure and the depth of the centre of pressure 
 for a plate 12 feet square forming one side of a tank full of water. 
 
 (Stud. I. C. E., Feb. 1906.) 
 
 3.| Define" the term "specific gravity," and show how to determine the 
 specific gravity of (i) a solid, (ii) a liquid. (Stud. I. C. E., Oct. 1906.) 
 
 4.' Explain how you would proceed to demonstrate experimentally that 
 a body floating in water displaces a quantity of liquid equal to itself in 
 weight. 
 
 The displacement of a tug-boat in sea-water was found to be 1560 cubic 
 feet. Calculate the weight of the boat, assuming that a cubic foot of sea- 
 water weighs 64 Ibs. (Stud. I. C. E., Feb. 1907.) 
 
 5. Define the term "centre of pressure" for a surface immersed in a 
 fluid. 
 
 A rectangular sectioned water-channel has a board put vertically across 
 it, which is held up against the water-pressure by two horizontal bars, 
 one at the bottom of the channel, the other 18 inches up the sides. Find 
 the depth of water in the channel which will just overset the board, and 
 the pressure then existing on the board if the breadth of the channel is 
 5 feet. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
 LECTURE XIX STUD. I.C.E. EXAM. QUESTIONSI 
 
 1. Describe by aid of a sketch the action of an ordinary suction pump 
 for raising water from a well. Calculate the work required to raise 80 
 gallons of water through a height of 20 feet, if the pump has an efficiency 
 of 64 per cent. (Stud. I. C. E., Feb. 1907.) 
 
 2. Sketch and describe a force pump suitable for raising water from a 
 well. 
 
396 APPENDIX B. 
 
 It is required to raise water through a total height of 80 feet by a pump 
 of 6 inches stroke with a barrel 3 inches in diameter. If 20 strokes per 
 minute are made, find the gallons pumped per hour, and the horse-power 
 required. (Stud. I C. E., Oct. 1907.) 
 
 3. Describe, with sketches, some kind of double-acting force-pump. 
 
 A pump is delivering water into a boiler in which the pressure is 120 
 Ibs. per square inch above atmospheric pressure. Find the work done in 
 foot-pounds per pound of water delivered to the boiler. Find also the 
 horse-power of the pump if it delivers 2,000 gallons per hour and its efficiency 
 is 60 per cent. Ans. : Work done per Ib. of water delivered = 276 ft.lbs. ; 
 and H.-P. of pump = 47. (Stud. I. C. E., Feb. 1908.) 
 
 LECTURE XX. STUD. I.C.E. EXAM. QUESTIONS. 
 
 i. The pressure of water in a high-pressure main is 700 Ibs. weight per 
 square inch. A load of 2 tons is to be lifted by means of a ram driven 
 from the main. Find the sectional area of the ram, and the cubic feet of 
 water used per H.-P.- hour. (Stud. I. C. E., Oct. 1904.) 
 
 | LECTURE XXI. STUD. I.C.E. EXAM. QUESTIONS. 
 
 i. What is the difference between acceleration and velocity ? 
 
 Plot to scale the following velocities in terms of the time, and write 
 down the times from the start, when the acceleration is zero ; also state 
 the period during which the acceleration is negative. Find by the method 
 of equidistant co-ordinates the average velocity. 
 
 Time in Velocity in Feet Time in Velocity in Feet 
 
 Seconds. per Second. Seconds. per Second. 
 
 5-0 
 
 1 I0'0 
 
 2 13-0 
 
 3 H-S 
 
 4 IS'S 
 5- iS'8 
 
 6 15-8 
 
 7 IS'S 
 
 8 167 
 
 9 18-3 
 
 10 20-3 
 
 11 22-5 
 (Stud. I. C. E, Oct. 1904.) 
 
 2. A train of 120 tons is found to increase its speed from 20 miles per 
 hour to 40 miles per hour in ten minutes. If the frictional resistance is 
 2000 Ibs. weight, find the force which must be pulling the tram assuming 
 it to be a constant force. Find average and final horse-power. 
 
 " (Stud. I. 0. E., Oct. 1904.) 
 
 3. Water is projected horizontally from a nozzle. If the point at which 
 it strikes the floor is 6 feet below the nozzle and 5 feet from the vertical 
 line drawn through the nozzle, find the velocity with which the water is 
 projected, (g = 32*2 feet per sec. per sec.) 
 
 (Stud. I. C. E., Oct. 1904.) 
 
 4. A 2-ton fly-wheel drops in speed from 100 revolutions per minute to 
 90 revolutions per minute. If the mean radius is 5 feet, find the work given 
 up by the fly-wheel. (Stud. I. C. E., Oct. 1904.) 
 
STUD. INST. C.E. QUESTIONS. 397 
 
 5. A lo-ton truck, moving at the rate of 4 feet per second strikes an 
 8-ton truck which is standing at rest. If the two move off together after 
 the impact, find the velocity they start with. If the resistance due to- 
 friction, &c., is 200 Ibs. weight, find how far they will run before coming: 
 to rest. (Stud. I. C. E., Oct. 1904.) 
 
 6. A ball weighing 10 Ibs. is making 50 revolutions per minute in a 
 horizontal circle of 5 feet radius. Find the force in Ibs. weight acting: 
 upon it towards th*e centre of the circle. 
 
 (Stud. I. C. E., Oct. 1904.) 
 
 7. The distances passed over from rest in I, 2, 3, 4, 5 ... etc., units of 
 time are respectively 0-05, 0-3, 0-58, 0*95, 1-4, 2'O, 27, 3-5, 4-4, 5 35, &S+ 
 77, 9-0, 10-4, i r6, 13-2, 14-9, 16-6, 18-4, 20-3, 22-3, 24-3, 267, 28-8, 31-2,. 
 337 units of length. 
 
 Plot the space-time curve. The unit of time is T fo second, and the unit 
 of length i centimetre. Find the acceleration and show how far it is 
 constant. (Stud. I. C. E., Feb. 1905.) 
 
 8. Give examples of how a diagram can be used 
 
 (i) to correct the observation or measurement of an experiment 
 
 (See, for example, Question i.) 
 (ii) to find the relation between two quantities, e.g., the effort and 
 
 load in a machine, 
 (iii) to find the average value of a quantity, e.g., to find the average 
 
 velocity of a body. (Stud. I. C. E., Feb. 1905.) 
 
 9. A shot is projected from a gun. Explain why 
 
 (i) The momentum of the shot is equal (under certain conditions) 
 
 to the momentum of the gun. 
 (ii) The energy of the shot when it leaves the gun is greater by far 
 
 than the energy communicated to the gun. 
 
 Example : shot 100 Ibs. ; gun i ton ; velocity of shot 1,200 feet per 
 second. Find velocity of recoil of the gun, and the energy of the gun. 
 
 (Stud. I. C. E., Feb. 1905.) 
 
 10. In the car of a balloon a piece of iron is hung from a spring balance. 
 The balance registers 5 Ibs. when the car is at rest. What will it register 
 when the car is rising with an acceleration of 2 feet per second per second ? 
 
 (Stud. I. C. E., Feb. 1905.) 
 
 11. Give some account of the advance made in dynamics by either 
 Galileo or Newton. 
 
 Show how to find the acceleration towards the centre of a circle of a 
 body moving with uniform velocity in the circle. 
 
 (Stud. I. C. E., Feb. 1905.) 
 
 12. Continuous brakes are now capable of reducing the speed of a train 
 of 3f miles an hour every second, and take 2 seconds to be applied ; show 
 in a tabular form the length of an emergency stop at a speed of 3f , 7^, 
 1 5, 30, 45, 60 miles an hour. 
 
 Compare the resistance with gravity ; express the resisting force in Ibs. 
 per ton ; calculate the coefficient of adhesion of the brake-shoe and rail 
 with the wheel, and sketch the mechanical arrangement. 
 
 (Stud. I. C. E., Oct. 1905.) 
 
 13. If W tons is transported from rest to rest a distance feet in t seconds, 
 being accelerated for a distance s 1 and time ^ by a force of P 1 tons up to 
 
398 APPENDIX B. 
 
 velocity v feet per second, and then brought to rest by P 2 tons acting'f or 
 Jt 2 seconds through * 2 feet, prove the formulas 
 
 3?-- 
 oo T? 
 
 s, 
 
 (iii) V = 2- 
 
 Supposing i in m is the steepest incline a train can crawl up, and I in n 
 Is the steepest incline on which the brakes can hold the train, prove that 
 the quickest run up an incline of i in p from one station to stop at the 
 next, a distance of a feet, can be made in 
 
 V- 
 
 ( 
 
 I I\ 
 
 m + n) 
 
 V second^ 
 
 fc- 
 
 5) (5- 
 
 i\ H 
 
 f ? J 
 
 Calculate for w = 50, n = 5, p = 100, a = 5280. 
 
 (Stud. I. C. E., Oct. 1905.) 
 
 14. Determine the motion of a circular hoop of radius a feet, whirling in 
 -a vertical plane on a round stick held horizontally, if released when the 
 centre is moving with velocity V feet per second at an angle a with the 
 
 horizon, and prove that it will make - - revolutions per second in the 
 
 air. 
 
 V 
 Prove that the tension in the hoop will be the weight of a length 
 
 feet of the rim. (Stud. I. C. E., Oct. 1905.) 
 
 1 5. Explain how velocities may be compounded. Determine the apparent 
 velocity and direction of rain-drops failing vertically with a velocity of 
 20 feet per second with reference to a bicyclist moving at the rate of 12 
 miles an hour. (Stud. I. C. E., Feb. 1906.) 
 
 1 6. Show that if a body starting from rest and moving in a straight line 
 is accelerated / feet per second per second it will describe a distance s in 
 t seconds expressed by the formula 
 
 s = \ft* 
 
 A train starting from rest receives a uniform acceleration of 0*25 foot per 
 .-a econd per second for one minute. Calculate the distance travelled. 
 
 (Stud. I. C. E., Feb. 1906.) 
 
 17. Define potential and kinetic energy. 
 
 Find the gam of potential energy of a train weighing 320 tons after 
 mounting an incline 4 miles long of i in 200, and find its kinetic energy 
 -when moving at 30 miles an hour. 
 
 (Stud. I. C. E., Feb. 1906.) 
 
 1 8. Explain how forces are measured, and distinguish between the mass 
 of i Ib. and the weight of i Ib. 
 
 Determine what force will be necessary to change the velocity of a masa 
 -of 400 Ibs. from 15 to 25 feet per second in 8 seconds. 
 
 (Stud. I. C. E., Feb. 1906.) 
 
STUD. INST. C.E. QUESTIONS. 
 
 399 
 
 19. Prove that the acceleration a of a body moving with velocity v in a 
 circular path of radius r is expressed by the formula 
 
 Calculate the force required to constrain a locomotive weighing 50 tons 
 to move in a circle of 400 feet radius when its velocity is 30 miles an hour. 
 
 (Stud. I. C. E., Feb. 1906.) 
 
 20. Explain the terms " moment of inertia," and " radius of gyration," 
 and determine their values for the case of a circular disk of mass m and 
 radius r, when rotating about an axis passing through the centre and 
 perpendicular to the plane of the disk. Give numerical values when the 
 mass is 20 Ibs. and the radius is 2 feet. (Stud. I. C. E. , Feb. 1906.) 
 
 21. Define the terms " velocity " and " acceleration " in the case of a 
 body moving in a straight line. A motor-car starting from rest and 
 uniformly accelerated acquires in 2 minutes a velocity of 30 miles an hour. 
 Find the acceleration. (Stud. I. C. E., Oct. 1906.) 
 
 22. A heavy ball, attached to a string 30 inches long, is whirled round in 
 a horizontal circle with constant velocity. Make a diagram showing the 
 forces acting on the ball, and calculate the velocity when the string is BO- 
 inclined that the ball moves in a circle of 24 inches radius. 
 
 (Stud. I. C. E., Oct. 1906.) | 
 
 23. Explain what is meant by kinetic energy and deduce an expression 
 for the kinetic energy of a circular disk, rotating about an axis passing 
 through its centre of figure and perpendicular to the plane of the disk. 
 Calculate the kinetic energy of a disk having a radius of 2 feet and weighing; 
 400 Ibs. when revolving at the rate of 240 revolutions per minute. 
 
 (Stud. I. C. E., Oct. 1906.) 
 
 24. Show how velocities may be compounded. A stone is dropped from 
 a balloon 80 feet above the ground and moving horizontally at the rate of 
 12 miles an hour. Determine the velocity and direction of the stone when 
 it strikes the ground. (Stud. I. C. E., Feb. 1907.) 
 
 25. Define the terms " velocity " and " acceleration." A tram-car start- 
 ing from rest covers s feet in t seconds in accordance with the following; 
 Table : 
 
 t 
 
 s 
 
 ' 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 4 
 
 II 
 
 21 
 
 34 
 
 5Q 
 
 69 
 
 9i 
 
 no 
 
 144 
 
 175 
 
 Plot the space-time curve and from it determine the velocity of the body 
 at the end of each second, and show your results to scale upon a time 
 base. Explain how to determine the acceleration from this latter curve,, 
 and determine its value at the end of the fifth second. 
 
 (Stud. I. C. E., Feb. 1907.) 
 
 26. Define " angular velocity " and " angular acceleration " for a body 
 revolving about a fixed axis, and deduce a formula for the angle^turned 
 through by a shaft starting from rest and accelerated uniformly. 
 
 2C 
 
4OO APPENDIX B. 
 
 The spindle of a dynamo is uniformly accelerated, and in 10 seconds 
 from starting it is found to be revolving at the rate of 600 revolutions per 
 minute. Find the acceleration and the number of revolutions it has made. 
 
 (Stud. I. C. E., Feb. 1907.) 
 
 27. Prove that the acceleration (a) of a body moving in a circle of radius 
 
 y2. 
 r with velocity v is expressed by the formula a = A pulley is found to 
 
 be out of balance to an amount which may be represented by a mass of 
 4 oz. at a radius of i foot. Determine the unbalanced force when the shaft 
 is "making 1200 revolutions per minute. 
 ' (Stud. I. C. E., Feb. 1907.) 
 
 28. Explain what you understand by " acceleration." 
 
 A railway carriage is accelerating at 3 feet per sec. per sec. Find the 
 acceleration possessed by a stone dropping from its roof. If the carriage 
 is 8 feet high, find the time taken to fall and the distance the stone travels. 
 
 (Stud. I. C. E., Oct. 1907.) 
 
 29. Define " angular acceleration," and show how it is related to linear 
 acceleration. 
 
 A hoop whose diameter is 3 feet, is rolling along the ground and comes 
 to rest in 10 seconds, after rolling 240 feet. If it is retarded uniformly, 
 nd the value of the angular retardation. (Stud. I. C. E. Oct. 1907.) 
 
 30. A weight is suspended by a string and rotates in a horizontal circle. 
 Find the forces acting on the weight. 
 
 Such a weight rotates at 20 revolutions per minute when the radius of 
 its circular path is 3 feet. Find the length of the suspending string. 
 
 (Stud. I. C. E. Oct. 1907.) 
 
 31. Define " kinetic energy " and " potential energy." 
 
 A steamer weighing 2000 tons is proceeding at 20*1 miles per hour. 
 When steam is cut off its speed drops to 19*9 miles per hour after it has 
 moved through 200 feet. Find the mean force retarding it. If it speeds 
 up again to its first speed in half a minute, find approximately the work 
 done in foot-tons, and the horse-power required. 
 
 'r (Stud. I. C. E., Oct. 1907.) 
 
 ' "1 3 2. ^Explain how velocities can be represented and combined. 
 Y Two men, A and B, are 5 miles apart, A being due west of B. They 
 start walking at the same moment : A walks to the south-east at 4 miles 
 per hour, and B walks at 3 miles per hour in such a direction as to meet A 
 on his road. Find graphically the two possible times taken for the two 
 men to meet. Ans. : 53 minutes and i hour 46 minutes. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
 33. Define the term acceleration, and show that if a curve be plotted 
 -connecting the velocity of a body and the time, the area under the curve 
 is the distance traversed, and the slope of the curve measures the accelera- 
 tion. 
 
 A body starts from rest with a uniform acceleration. After the lapse of 
 a certain time it is found that in successive intervals of 5 seconds and 7 
 seconds it traverses 62^ feet and 129*- feet respectively : find the accelera- 
 tion at the above time Ans. : 3 feet per sec. per sec. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
 34. Explain what you understand by kinetic and potential energy. 
 A weight of 420 Ibs. is lifted by a force which varies as follows : 
 
 o I 2 3 4 5 6 7 
 Force in pounds 700 610 490 390 380 450 650 800 
 
STUD. INST. C.E. QUESTIONS. 40 1 
 
 Plot a curve connecting the force and height, and hence find ^"poten- 
 tial and kinetic energy of the body, and the work done by the force when 
 the body is 6 feet from the ground. Ans. : EP = 2730 ft-lb. : EK = 
 2359 ft.-lb., and work done by the force = 5089 ft.-lb. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
 35. Explain how energy is stored in a fly-wheel, and obtain an expression 
 for this energy. If such a wheel stores 1,000 foot-lb. when rotating at 
 1 revolution per second, find the work that must be done to change its 
 speed from 10 revolutions per second to 20 revolutions per second. 
 Ans. : EK = *Iw2. Work done in changing the speed = 300,000 foot-lb. 
 
 (Stud. I. C. E., Feb. 1908.) 
 
 36. Explain what you understand by the term " centripetal force." 
 
 A weight of 20 pounds is hung by a string 10 feet long. It is pulled 
 to one side so as to be 6 feet horizontally away from the vertical If the 
 weight is then let go, find from the energy equation the velocity of the 
 weight at the moment it passes through the lowest point, and deduce the 
 total tension in the string at that moment. Ans.: V = ii'32 feet per 
 sec. ; F = 8 Ib. (Stud. I. C. E., Feb. 1908.) 
 
 LECTURE XXII. STUD. I. C.E. EXAM. QUESTIONS. 
 
 1. Prove that the increase of pressure per foot vertically downwards in 
 a liquid of specific gravity s is 0^433 x s Ibs. per square inch. 
 
 Mining in ground of uniform density at a depth of h feet, determine the 
 percentage of coal that can be won, leaving sufficient as pillars for the 
 support of the roof, supposing the coal to crush' under its own weight in a 
 column k feet high. (Stud. I. C. E., Oct. 1905.) 
 
 2. Investigate the mechanical advantage of the smooth screw, and 
 explain generally how the wind drives a windmill, and a screw propeller 
 propels a steamer. 
 
 Prove that a platelayer who can apply a force of 28 Ibs. will be apt to 
 break the screw-bolts if provided with a lever more than 3 feet long ; the 
 screw having 8 threads to the inch, and the breaking tension of the bolt 
 being 30 tons per square inch. (Stud. I. C. E., Oct. 1905.) 
 
 3. Explain the meanings of the terms " stress," " strain " and " modulus 
 of elasticity " with reference to a bar in tension. A tie-bar, 3*5 square 
 inches in section and 16 feet long, stretches 0*05 inch under a load of 
 28,000 Ibs. Find the values of the stress, strain, and modulus of elasticity. 
 
 (Stud. I. C. E., Oct. 1906.) 
 
 LECTURE XX1IL STUD. I. C.E. EXAM. QUESTIONS. 
 
 i. A machine is operated by a shaft making N revolutions per minute 
 and transmitting a twisting moment T. Deduce an expression for the 
 horse-power delivered to the machine and calculate its numerical value if 
 the shaft makes 1 10 revolutions per minute, and the twisting moment is 
 2000, the units being pounds and feet. (Stud. I. C. E., Feb. 1907.) 
 
4O2 NOTES AND QUESTIONS. 
 
B. OF E. QUESTIONS, MAY igiO. 403 
 
 May, 1910, Examination on Subject VII. 
 APPLIED MECHANICS. 
 
 STAGE I. 
 GENERAL INSTRUCTIONS. See APPENDIX A. 
 
 You must not attempt more than EIGHT questions ; EITHER No. i OR No. 22 
 : must be one of these eight, but not both. The remaining seven questions may 
 be selected from Nos. 2 to 21. The questions in Series A are framed to be 
 more particularly suitable for the Building Trades, and those in Series B 
 /or Mechanical Engineers. 
 
 SERIES A. 
 
 1. Describe, with the help of good sketches, onlyonz of the following, (a), 
 (6),(c),or(d): 
 
 (a) A mortar-mixing machine. 
 
 (&) The method of securing the cutting chisels into the cutter blocks 
 of a wood-planing machine. 
 
 (c) Any form of friction clutch suitable for use with a speed cone or 
 
 reversing pulleys. 
 
 (d) Any form of vernier calliper suitable for measuring the dimensions 
 
 of a test-bar to the nearest thousandth of an inch. 
 
 (B. of E., 1910.) 
 
 2. Describe, with sketches, an apparatus to verify the rule for finding the 
 compressive and tensile forces in the jib and tie of a crane. Do the ex- 
 perimental results exactly agree with the rule, and if not, what is the 
 probable reason ? (B. of E., 1910.) 
 
 3. Answer only one of the following, (a), (6), or (c) : 
 
 (a) Two of the tests specified in order to determine the quality of 
 Portland cement are the determination of the tensile strength of 
 (i) a mortar of neat cement, (ii) a mortar with sand. Describe 
 carefully how the specimens would be made and tested. 
 
 (&) You are supplied with a length of steel-wire one-eighth of an inch 
 in diameter. You are asked to find (i) Young's Modulus for the 
 material, (ii) the limit of elasticity, (iii) the breaking stress. 
 Explain how you would carry out the test. (See my Adv. Vol. IT.) 
 
 (c) You wish to know the strength and stiffness of a large timber 
 beam. It is to be built in at the ends ; to be about 20 feet long 
 between supports ; to be, say, 10 inches broad and 12 inches 
 deep, and it is to be loaded uniformly all over. You therefor 
 test a small beam of the same kind of timber ; describe exactly 
 how you would make the test, and how would you use your 
 results ? (See my Adv. Vol. II.) (B. of E., 1910.) 
 
44 
 
 APPENDIX C. 
 
 4. A wooden beam is built into a wall at one end. Eight feet from the 
 wall there is a hook hi the beam, and from this hook is suspended a weight 
 of i ton. What is the bending moment in the beam (i) at the wall, (ii) at 
 3 feet from the wall ? Describe the nature of the compressive and tensile 
 stresses throughout any section. (B. of E., 1910.) 
 
 5. Let the length of a strut divided by the diameter of its section be 
 called x. W is the maximum load carried. Tests were made on a set of 
 cast-iron struts all of the same section but of different lengths, with the 
 following results : 
 
 X 
 
 10 
 
 15 
 
 20 
 
 25 
 
 30 
 
 W 
 
 64,000 
 
 53,500 
 
 44,800 
 
 33,700 
 
 24,IOO 
 
 Plot a curve showing how the strength depends upon x. What is the 
 maximum load when the length is 18 times the diameter ? (See my Adv. 
 Vol. II.) (B. of E., 1910.) 
 
 6. There is a triangular roof-truss ABC ; AC is horizontal and 10 feet 
 long. The angle EC A is 25 and BAG is 55 ; there is a vertical load of 
 
 5 tons at B. What are the compressive forces in BA and BC. What 
 are the vertical supporting forces at A and C. Find these answers 
 any way you please. (B. of E., 1910.) 
 
 7. A man's hand on the handle of a crane moves 120 feet when the 
 weight is lifted i foot ; 35 per cent, of the total energy given by the man 
 is wasted in friction. A load of 1.5 tons is being lifted. What force is 
 being exerted by the hand ? (B. of E., 1910.) 
 
 8. Roughly, what is the weight of a cubic foot of brickwork ? There is 
 a brick building 80 feet long and 50 feet wide. The foundations carry 
 the following weight : First, the volume of brickwork is 24,000 cubic feet, 
 the roof and floors weigh altogether 200 Ib. per horizontal square foot of 
 the area ; the machinery weighs altogether 1 50 tons. What is the total 
 weight to be carried ? What is the breadth of the foundation wall at the 
 footings, if the load there is not to exceed i tons per square foot ? 
 
 (B. of E., 1910.) 
 
 9. It is necessary to keep the " surface level " of water in a shaft at a 
 depth of 30 feet. When left to itself the level rises 4 feet in i minute. 
 The shaft is circular, 6 feet in diameter. What is the weight of water 
 entering per minute ? This water is lifted by a pump whose efficiency 
 is 0.45. What horse-power must be supplied to the pump ? . 
 
 (B. of E., 1910.) 
 
 10. Answer only one of the following questions, (a) or (6) : 
 
 (a) Describe briefly, sketches are hardly needed, how Portland cement 
 is manufactured. Give a reason for each part of the process. 
 What is your notion of what occurs (i) when cement sets, (ii) when 
 it slowly hardens as it gets older ? 
 
 (6) Describe briefly, sketches are hardly needed, how any kind of 
 steel used for girders is manufactured. Give a reason for each 
 part of the process. (B. of E., 1910.) 
 
B. OF E. QUESTIONS, MAY 19 IO. 
 
 405 
 
 11. You are given a 4-ton screw-jack. How would you experimentally 
 determine its efficiency under various loads ? What sort of results would 
 you expect to obtain ? (B. of E., 1910.) 
 
 SERIES B. 
 
 12. The depth of water outside the gate of a dry dock is 25 feet. What 
 is the total water pressure on the gate if the width of the gate is 40 feet ? 
 The weight of i cubic foot of salt water is 64 Ib. (B. of E., 1910.) 
 
 13. In a hydraulic cylinder, i square foot in cross-section, the piston 
 moves through a distance of i foot. The pressure of the water is 1400 Ib. 
 per square inch. What work is done on the piston ? What is the work 
 done per gallon of water used ? (B. of .,1910.) 
 
 14. The pull in the draw-bar between locomotive and train is 13 Ib. per 
 ton when on the level ; the train weighs 200 tons, what is the total force ? 
 If the tram is being pulled up an incline of i in 80 (a vertical rise of i foot 
 in a rail distance of 80 feet), what is the additional pull required ? What 
 is the total pull ? The speed is 2,500 feet per minute. What is the horse- 
 power exercised in drawing the train ? (B. of E., 1910.) 
 
 15. As a particle of water flows without friction, its height h feet above- 
 datum level, its pressure p (in Ib. per sq. ft.), and its speed v (in feet per 
 second), may all alter, but the sum 
 
 20 W I 
 
 remains constant. Here g = 32.2 and w = 62.3. The particle flows from 
 a place A where v = O, p = O, and h = 80, to a place B where p = O,. 
 and h = 40 ; what is the value of v at the place B. (B. of E., 1910.) 
 
 16. The motion of a body of 3,220 Ib. is opposed by a constant frictional 
 resistance of 2,000 Ib. It starts from rest under the action of a varying 
 force F Ib. whose value is here given at the instants at which the body 
 has passed x feet from rest : 
 
 F 
 
 S.HO 
 
 2,870 
 
 2,630 
 
 2,700 , 
 
 X 
 
 
 
 5 
 
 IO 
 
 15 
 
 As more work is being done upon the body than what is being wasted 
 in friction, what is the speed of the body when it has moved 1 5 feet from 
 rest ? (B. of E., 1910.) 
 
 17. An electric motor is employed for lifting purposes. In lifting 80 tons 
 of grain 100 feet high it is found that 20 Board of Trade units of Energy 
 have to be paid for. A Board of Trade unit is i kilowatt for i hour, and 
 a horse-power is 0.746 kilowatt. The cost is twopence per unit. What 
 is the cost of i horse-power hour usefully done ? What is the ratio of useful 
 work to the electric energy supplied ? (See Appendix D of this book.) t^, 
 
 (B. of E., 1901.) 
 
406 APPENDIX C. 
 
 1 8. The radial speed of the water in the wheel of a centrifugal pump is 
 6 feet per second. The vanes are directed backwards at an angle of 
 35 degrees to the rim. What is the real velocity of the water relatively to 
 the vanes ? What is the component of this which is tangential to the rim ? 
 
 (B. of E., 1910.) 
 
 ip.'A projectile leaves the muzzle of a gun at 2,000 feet per second, its 
 path being inclined at 20 upwards. What are the horizontal and vertical 
 components of its velocity ? In 3 seconds how far has it travelled hori- 
 zontally ? What is its vertical height above the gun ? Neglect resistance 
 of the atmosphere. (B. of E., 1910.) 
 
 2O.*A block of cast-iron, 3 inches by 4 inches by 3 inches, is fastened to 
 the arm of a wheel at the distance of 3 feet from the axis. The wheel 
 makes 2,000 revolutions per minute. What is the force tending to 
 fracture the fastening ? One cubic inch of cast-iron weighs 0.26 Ib. 
 
 (B. of E., 1910.) 
 
 .,f an drives air vertically downward through an opening, 8 feet in 
 diameter, with a velocity of 30 feet per second ? The air weighs 0.08 Ib. 
 per cubic foot. 
 
 What weight of air is driven downward per second ? What is its mo- 
 mentum ? (B. of E., 1910.) 
 
 22. Describe, with sketches, only one of the following, (a), (b), (c), 
 or (d) : 
 
 (a) The shaft bearing of any modern fast running machine such as a 
 
 water turbine or dynamo machine. 
 
 (6) An hydraulic appliance in use by hydraulic companies or their 
 customers, such as an accumulator, or a motor, or a force pump. 
 
 (c) Any form of sensitive drilling machine of the " pillar " type. What 
 
 are the advantages of this type of machine for small accurate work ? 
 
 (d) Any form of quick return motion suitable for use on a shaping 
 
 machine. Show how the stroke of the tool may be varied. 
 
 (B. of E., 1910.) 
 
STUD. INST. C.E, QUESTIONS, OCT. 1909, 
 
 407 
 
 Appendix C. 
 
 The Institution of Civil Engineers' Examinations 
 for Admission of Students, October 1909. 
 
 (vii.) ELEMENTARY MECHANICS. 
 
 Not more than EIGHT questions to be attempted by any Candidate. 
 
 1. Define the terms relative velocity and absolute velocity. 
 
 A bicycle has 28-inch wheels, and is being ridden at 20 miles an hour. 
 Find the velocity of a point on the rim 14 inches from the ground, (i) 
 relative to the rider, (2) relative to the ground. (Stud. I. C. E., Oct. 1909.) 
 
 2. Explain the term acceleration, and show how it is to be measured 
 when non-uniform. The velocity of a body at given times is as given in 
 the schedule below. Draw the curve connecting the two quantities, and 
 find the space the body has moved. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
 Time in seconds. . 
 
 o 
 o 
 
 i 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 Velocity in feet per'l 
 second . . . .J 
 
 0'95 
 
 3'8o 
 
 5-00 
 
 4*60 
 
 3-15 
 
 1-65 
 
 075 
 
 o 
 
 3. Give the equations for the motion of a body which has a uniform 
 acceleration. 
 
 An airship is travelling in a horizontal line at 30 miles per hour towards 
 an object on the ground on which it is desired to drop a shell. The ship 
 is 1600 feet above the ground. Find where it must let the shell go in order 
 to hit the object. (Stud. I. C. E., Oct. 1909.) 
 
 4. Show how to find the acceleration produced by a force acting on a 
 body. 
 
 A tramcar whose weight is 14 tons is being pulled horizontally along a 
 track by a force of 2000 Ibs. ; the track friction is 20 Ibs. per ton. Find 
 the acceleration produced in miles per hour per second. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
 5. Enunciate the conditions that must be fulfilled in order that three 
 non-parallel forces may be in equilibrium. 
 
 A uniform plank AB is pivoted smoothly at the end A, and has a rope 
 attached to the end B. The plank is 10 feet from A to B, and weight 
 100 Ibs. The rope is 8 feet long and is fixed to a point C, such that C is 
 
408 APPENDIX C. 
 
 at the same height as A and the angle ABC is a right angle. Determine 
 graphically the tension in the rope and the pressure on the pivot. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
 6. Show that a body which is moving with uniform angular velocity w 
 in a circle of radius r has an acceleration u?2>. towards the centre of the 
 circle. 
 
 A flywheel has an internal trough turned on it to contain cooling water. 
 Find the least possible number of revolutions per minute that will permit 
 the retention of the water if the diameter of the trough is 8 feet. 
 
 (Stud. I. C. E.. Oct. 1909.) 
 
 7. What is the moment of a force ? A timber balk is 42 feet long : a 
 small cross-bar is placed underneath, and 2 feet from one end ; the force 
 required to lift the other from the ground is 600 Ibs. : the cross-bar is 
 moved to 6 feet from the end, and the force is then 500 Ibs. ; find the 
 weight of the balk, and the distance of its centre of gravity from the end. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
 8. Give the condition that a body may stand in equilibrium on a plane. 
 A triangle ABC is cut out of a thick board : the side AB is 10 inches, and 
 the angle ABC is 30. Find the other sides if the triangle is just stable 
 when standing on the side AB on a horizontal plane. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
 9. Define the terms work, power, horse-power. 
 
 A car weighs i^ ton and its engine is working at a constant horse-power, 
 It is running on a road for which the total frictional resistance is 50 Ibs 
 per ton. A five-mile run takes 20 minutes, and in the 5 miles the total rise 
 is 400 feet. The car's speed is the same at each end of the run. Find its 
 HP. (Stud. I. C. E., Oct. 1909.) 
 
 10. What is the meaning of the efficiency of a machine ? 
 
 A screwjack has a screw whose pitch is inch, and the force is applied 
 by a lever 15 inches long. It is found that 5 tons is lifted by a force of 
 56 Ibs. applied at right angles to the lever. Find the efficiency of the 
 jack. (Stud. I. C. E., Oct. 1909.) 
 
 1 1 . What is the specific gravity of a body ? 
 
 A thin metal tube is 32 inches long, and one square inch in sectional 
 area. The bottom inch is filled with a metal. When the tube is put in a 
 vessel of water, it floats with 2 inches out of the water. Find the specific 
 gravity of the metal if the tube weighs 0*6 Ib. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
 12. A plane surface is immersed in a fluid : find an expression for the 
 pressure exerted on the same. 
 
 A cube of 3 feet side is placed in water 4 feet deep, with one face of the 
 cube horizontal. Find the pressure on each side. 
 
 (Stud. I. C. E., Oct. 1909.) 
 
STUD. INST. C.E. QUESTIONS, FEB. 1910. 409 
 
 Appendix C. 
 
 The Institution of Civil Engineers' Examinations 
 for Admission of Students, February, 1910. 
 
 ELEMENTARY MECHANICS. 
 
 Not mare than EIGHT questions to be attempted by any Candidate. 
 
 (The weight of a cubic foot of water may be taken as 62% Ibs. and g as 32.) 
 
 1. Explain how two velocities can be combined. 
 
 A cyclist is riding along a straight road which runs at 30 to a straight 
 piece of railway line. He sees an engine on that line when he is looking 
 in a direction making a constant angle of 45 with his direction of motion. 
 He is travelling at 12 miles an hour : find the velocity of the engine. 
 
 2. Establish the equation for the space traversed by a body moving 
 with constant acceleration. 
 
 A balloon has a total weight of I ton and is at rest 900 feet above the 
 ground. It suddenly lets fall i cwt. of ballast : neglecting friction, find 
 how high it will have risen when the ballast reaches the ground. 
 
 3. A curve is drawn connecting time and velocity of a moving body : 
 show how the space traversed in a given interval can be found from it. 
 
 Such a curve is drawn with a horizontal scale of I inch = I second and 
 a vertical scale of i inch = i foot per second. It is a semi-circle of 
 3 inches radius with its centre at the 3 seconds point on the horizontal 
 line. Find the total space traversed, and the acceleration at the end of 
 the first second. 
 
 4. Give the theorem known as the triangle of forces. 
 
 A uniform heavy bar is 5 feet long and weighs 20 Ibs. Two strings 
 3 and 4 feet long are attached to the ends, the other extremities of the 
 strings being fixed to a peg. Draw the position when the bar hangs- 
 freely ; determine the triangle of forces and the tensions in the strings. 
 
 5. If a body he moving in a circular path of radius r with a velocity v,. 
 
 t'2 
 
 show that there must be a constraining force of the amount acting: 
 
 along the radius. 
 
 A cyclist and his machine together weigh 12 stone; he is turning a 
 corner with a velocity of 15 miles per hour and the radius is 80 feet. 
 Find the force acting. 
 
 6. What is the condition that must be fulfilled in order that a body- 
 acted on by parallel forces shall be at rest ? 
 
410 APPENDIX 0. 
 
 Two heavy uniform rods are fixed at an angle and pivoted at the 
 junction point. The shorter rod is 6 feet long and weighs 10 Ibs. : the 
 longer rod is 8 feet long and weighs 15 Ibs. When suspended by the 
 pivot the shorter rod is horizontal. Find the angle at which the rods are 
 fixed together. 
 
 7. Explain the meaning of the terms worlt and power, and give the 
 usual units in which they are measured. 
 
 A locomotive is steadily pulling a train weighing 500 tons (total) at 
 25 milts an hour up a slope of I in 200. The frictional resistance is 
 10 Ibs. a ton. Find the HP. exerted in traction. 
 
 8. Explain the meaning of the term momentum. 
 
 A vessel of 2000 tons is starting to tow one of 1000 tons. The instant 
 before the rope becomes taut, the vessels are moving on the same line 
 with respective velocities of 7 and 3 knots. Find the common velocity at 
 the moment the rope becomes taut. 
 
 9. What do you understand by the velocity ratio and the efficiency of a 
 machine ? 
 
 A hand crane has a handle radius of i foot. It is found that forty-two 
 turns are required to raise the load one foot, and that a tangential force 
 of 37 pounds will just raise 5 tons. Find the efficiency of the crane. 
 
 10. Define the centre of gravity of a body, and show how you would 
 experimentally find the centre of gravity of a lamina, 
 
 A triangle has sides which are respectively 5, 4 and 3 inches. To each 
 a square is attached externally, and in the plane of the triangle, the 
 squares being all cut from the same sheet. Draw the figure to scale and 
 mark on it the centre of gravity of the area formed by the three squares. 
 
 11. Define the term specific gravity and show how you would measure 
 it for a body. 
 
 A yard of wire weighs 70 grams in air and 61 grams in water. Find 
 the specific gravity and the section of the wire [yard = 91 \ centimetres, 
 i cubic centimetre of water weighs I gram.] 
 
 12. How would you find the pressure exerted by a head of water on a 
 dam? 
 
 A dam is 50 feet high and has a vertical internal face. Find the total 
 pressure per foot run and the position at which it may be considered to 
 act. 
 
APPENDIX D. 
 
 THE CENTIMETRE, GRAMME, SECOND, OR C.G.S. SYSTEM OF 
 UNITS OF MEASUREMENT AND THEIR DEFINITIONS.* 
 
 I. Fundamental Units. The C.G.S. and the practical electrical 
 units are derived from the following mechanical units. 
 
 The Centimetre as a unit of length ; the Gramme as a unit of mass ; and 
 the second as a unit of time. 
 
 The Centimetre (cm) is equal to 0*3937 inch in length, and nominally 
 represents one thousand-millionth part, or I000 1 OOOOQ of a quadrant of the 
 earth. 
 
 The Gramme (gm) is equal to 15 '432 grains, and represents the mass of a 
 cubic centimetre of water at 4 C. Also, I Ib. of 16 oz. is equal to 45 3 '6 
 grammes. Mass (M) is the quantity of matter in a body. 
 
 The second (s) is the time of one swing of a pendulum making 
 86,i64'09 swings in a sidereal day, or the 1/86,400 part of a mean solar day. 
 
 II. Derived Mechanical Units. 
 
 Area (A or cm2). The unit of area is the square centimetre. 
 
 Volume (V or cms). The unit of volume is the cubic centimetre. 
 
 Velocity (v or cm/s) is rate of change of position. It involves the idea 
 of direction as well as that of magnitude. Velocity is uniform when equal 
 distances are traversed in equal intervals of time. The unit of velocity is 
 the velocity of a body which moves through unit distance in unit time, or 
 the velocity of one centimetre per second. 
 
 Momentum (Mv, or gm x cm/s) is the quantity of motion in a body, and 
 is measured by mass x velocity. 
 
 Accleration (a or cm/s2) is the rate of change of velocity, whether 
 that change takes place in the direction of motion or not. The unit of 
 acceleration is the acceleration of a body which undergoes unit change of 
 velocity in unit time, or an acceleration of one centimetre-per-second per 
 second. The acceleration due to gravity is considerably greater than this, 
 for the change of velocity imparted by gravity to falling bodies in one 
 second is about 981 centimetres per second (or about 32*2 feet per second). 
 The value differs slightly in different latitudes. At Greenwich the value 
 of the acceleration due to gravity is g = 981*17 ; at the Equator / = 978*1, 
 and at the North Pole g = 983*1. 
 
 * The Author is indebted to his Publishers, Charles Griffin and Co., for 
 liberty to abstract the following pages on this subject from the latest 
 edition of Munro and Jamieson's " Pocket-book of Electrical Rules and 
 Tables for Electricians and Engineers," to which the student is referred 
 for further values and definitions, and values of Practical Electrical Unit* 
 of Measurement and Testing Rules, &c. A.J. 
 
412 APPENDIX D. 
 
 Force (F or /) is that which tends to alter a body's natural state of rest 
 or of uniform motion in a straight line. 
 
 Force is measured by the rate of change of momentum which it produces, 
 or mass x acceleration. 
 
 The Unit of Force, or Dyne, is that force which, acting for one second on 
 a mass of one gramme, gives to it a velocity of one centimetre per second. 
 The force with which the earth attracts any mass is usually called the 
 *' weight " of that mass, and its value obviously differs at different points 
 of the earth's surface. The force with which a body gravitates i.e. its 
 weight (in dynes), is found by multiplying its mass (in grammes) by the 
 value of g at the particular place where the force is exerted. 
 
 Work is the product of a force and the distance through which its acts. 
 The unit of work is the work done in overcoming unit force through unit 
 distance, i.e. in pushing a body through a distance of one centimetre 
 against a force of one dyne. It is called the Erg. Since the " weight " 
 of i gramme is I x 981, or 981 dynes, the work of raising one gramme 
 through the height of one centimetre against the force of gravity is 981 
 rgs or g ergs. One kilogramme-metre = 100,000 (g) ergs. One foot-pound 
 as 13,825 (g) ergs = 1-356 x 10? ergs. 
 
 Energy is that property which, possessed by a body, gives it the capa- 
 bility of doing work. Kinetic energy is the work a body can do in virtue 
 of its motion. Potential energy is the work a body can do in virtue of its 
 position. The unit of energy is the Erg. 
 
 Power or Activity is the rate of working. The unit is called the Watt(Wp) 
 = io7 ergs per second, or the work done at the rate of one Joule (J) per 
 second. 
 
 One Horse-power (H.P.) = 33,000 ft.-lbs. per minute = 550 ft.-lbs. per 
 second. But as seen above under Work, i ft. Ib. = i'356 x 10? ergs, and 
 under Power, i Watt 10? ergs per second. 
 
 Hence, a Horse-power = 550 x 1*356 x 10? ergs per sec. = 746 Watts. 
 If E = volts, = amperes, and R = ohms. ; then, by Ohm's Law C = E/R, 
 Also, EC = C2R = E2 R = Watts. 
 
 Therefore, H.P. = = = . 
 
 746 746 74 6R- 
 
UNITS OF MEASUREMENT AND THEIR DEFINITIONS. 413 
 
 PRACTICAL ELECTRICAL UNITS. 
 
 1. As a Unit of Resistance (R), th International Ohm (ohm or w), 
 based upon the ohm which is 10$ units of resistance in the C.G.S. system 
 of electro -magnetic units, is represented by the resistance offered to an 
 unvarying electric current by a column of mercury at the temperature^ of 
 melting ice, 14-4521 grammes in mass, of a constant cross-sectional area 
 and of the length of 106*3 centimetres. 
 
 2. As a Unit of Current (C), the International Ampere (A), which 
 is one-tenth of the unit of current of the C.G.S. system of electro-magnetic 
 units, and which is represented sufficiently well for practical use by the 
 unvarying current which, when passed through a solution of nitrate of 
 silver in water, and in accordance with the International specifications, 
 deposits silver at the rate of o'OOinS grammes per second. 
 
 3. As a Unit of Electro-motive Force (E) the International Volt 
 (V), which is the E.M.F. that, steadily applied to a conductor whose 
 resistance is one International Ohm, will produce a current of one Inter- 
 national Ampere, and which is represented sufficiently well for practical 
 use by $$$ of the E.M.F. between the poles or electrodes of the voltaic 
 cell known as Clark's cell, at a temperature of 15 Centigrade, and prepared 
 in the manner described in the International specification. 
 
 4. As the Unit of Quantity (Q) the International Coulomb (A xs), 
 which is the quantity of electricity transferred by a current of one Inter- 
 national Ampere in one second. 
 
 5. As the Unit of Capacity (K) the International Farad (Fd), which 
 is the capacity of a conductor charged to a potential of one International 
 Volt by one International Coulomb of electricity. 
 
 6. As a Unit of Work the Joule (J) (or Watt-second (W x $),) which 
 is ID? units of work in the C.G.S. system, and which is represented 
 sufficiently well for practical use by the energy expended in I second in 
 heating an International Ohm. 
 
 7. As the Unit of Power (P w ) the International Watt (W P ), which is 
 equal to 10? units of power in the C.G.S. system, and which is represented 
 sufficiently well for practical use by the work done at the rate of one Joule 
 per second. The Kilowatt (K.W.) = 1000 Watts = i horse-power. 
 
 8. As the Unit of Induction (L) the Henry (Hj ), which is the 
 induction in the circuit when the E.M.F. induced in this circuit is one 
 International Volt, while the inducing current varies at the rate of one 
 ampere per second. 
 
 g. The Board of Trade Commercial Unit of Work or (B.T.U.) 
 is the Kilowatt-hour (K.W.-hr.) = 1000 Watt-hours = i H.P. working for 
 one hour. Or say 10 amperes flowing in a circuit for i hour at a pressure 
 of 100 volts. 
 
 Note. For further simple explanations with Examples, see 7th Edition 
 of Prof. Jamieson's " Manual of Magnetism and Electricity," pp. 87 to 94, 
 and 222 to 224. Also latest Edition of Munro and Jamieson's Electrical 
 Pocket Book, both published by Charles Griffin & Co. Ltd., London. 
 
414 APPENDIX D. 
 
 EXAMINATION TABLES. 
 
 USEFUL CONSTANTS. 
 
 1 Inch a 25*4 millimetres. 
 
 t Gallon = -1606 cubic foot = 10 Ibs. of water at 62* F. 
 
 I Naut = 6080 feet. 
 
 I Knot = 6080 feet per hour. 
 
 Weight of 1 Ib. in London = 445,000 dyne*. 
 
 One pound avoirdupois = 7000 grains = 453-6 gramme*. 
 
 1 Cubic foot of water weighs 62*3 Ibs. at 65* F. 
 
 1 Cubic foot of air at 0* 0. and 1 atmosphere, weighs *0807 Ib. 
 
 1 Cubic foot of Hydrogen at 0* 0. and 1 atmosphere, weighs -00557 Ib* 
 
 1 Foot-pound 1-8562 x 10' ergs. 
 
 1 Horse-power-hour = 83000 x 60 foot-pounds. 
 
 1 Electrical unit 1000 watt-hours. 
 
 *V. Kquiv^t to .nit Bgnlf. H, I. {, gj ; } ** 
 
 1 Horse-power = 38000 foot-pomnds per minute = 746 watts. 
 Volts x amperes = watts. 
 
 1 Atmosphere = 14*7 Ib. per square inch = 2116 Ibs. per square foot a 
 760 m.m. of mercury = 10* dynes per sq. cm. nearly. 
 
 A Column of water 2*3 feet high corresponds to a pressure of 1 Ib. per 
 square inch. 
 
 Absolute temp., t * V 0. + 273*-7. 
 
 Regnaulfs H = 606'6 + '305 6 0. m 1082 + -805 8* F. 
 
 fu 1.0*46^479 
 
 log M l> = 61007 -f- 
 
 where log ^B = 3-1812, log W C a fi-0871, 
 
 p is in pounds per square inch, t is absolute temperature Centigrade^, 
 
 u is the volume in cubic feet per pound of steam. 
 = 3-1416. 
 
 One radian = 57 '3 degrees. 
 
 To convert common into Napierian logarithms, multiplj by 2-3021 
 The base of the Napierian logarithms is = 2-7183. 
 The value of g at London. 32-182 feet per sec. per sec. 
 
TABLE OF LOGARITHMS. 
 
 415 
 
 10 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 123 
 
 456 
 
 789 
 
 0000 
 
 0043 
 
 0086 
 
 0123 
 
 0170 
 
 0212 
 
 0253 
 
 0294 
 
 0334 
 
 0374 
 
 4 8 12 
 
 17 21 25 
 
 29 33 37 
 
 11 
 
 12 
 13 
 
 14 
 15 
 16 
 
 17 
 13 
 19 
 
 0414 
 0792 
 1139 
 
 0453 
 0328 
 1173 
 
 0492 
 0864 
 1206 
 
 0531 
 OS99 
 1239 
 
 0569 
 0934 
 1271 
 
 0607 
 0969 
 1303 
 
 0645 
 1004 
 1335 
 
 0682 
 1038 
 1367 
 
 0719 
 1072 
 1399 
 
 0755 
 1106 
 1430 
 
 4 8 11 
 3 7 10 
 3 6 10 
 
 15 19 23 26 30 34 
 14 17 21'24 28 31 
 13 16 19J23 26 29 
 
 1461 
 1761 
 2041 
 
 1492 
 1790 
 2068 
 
 1523 
 1818 
 2095 
 
 1553 
 1847 
 2122 
 
 1534 
 1875 
 2148 
 
 1614 
 1903 
 2175 
 
 1644 
 1931 
 2201 
 
 1673 
 1959 
 2227 
 
 1703 
 1987 
 2253 
 
 1732 
 2014 
 2279 
 
 369 
 368 
 358 
 
 12 15 18 
 11 14 17 
 11 13 16 
 
 21 24 27 
 20 22 25 
 18 21 24 
 
 2304 
 2553 
 
 2788 
 
 2330 
 2577 
 2810 
 
 2355 
 2601 
 2833 
 
 2380 
 2625 
 2356 
 
 2405 
 2648 
 2878 
 
 2430 
 2672 
 2900 
 
 2455 
 2695 
 2923 
 
 2480 
 2718 
 2945 
 
 2504 
 2742 
 2967 
 
 2529 
 2765 
 2989 
 
 257 
 257 
 247 
 
 10 12 15 
 9 12 14 
 9 11 13 
 
 17 20 22 
 16 19 21 
 16 18 20 
 
 20 
 
 21 
 22 
 23 
 
 3010 
 
 3032 
 
 3054 
 
 3075 
 
 3096 
 
 3118 
 
 3139 
 
 3160 
 
 3131 
 
 3201 
 
 246 
 
 8 11 13J15 17 19 
 
 3222 
 3424 
 3617 
 
 3243 
 3444 
 3636 
 
 3263 
 3464 
 3655 
 
 3284 
 34S3 
 3674 
 
 3304 
 3502 
 3692 
 
 3324 
 3522 
 3711 
 
 3345 
 3541 
 3729 
 
 3365 
 3560 
 3747 
 
 3385 
 3579 
 3766 
 
 3404 
 3393 
 3784 
 
 246 
 246 
 
 246 
 
 8 10 1214 16 18 
 8 10 1214 15 17 
 7 9 11113 15 17 
 
 24 
 25 
 26 
 
 27 
 28 
 29 
 
 30 
 
 31 
 32 
 33 
 
 3S02 
 3979 
 4150 
 
 3820 
 3997 
 4166 
 
 3833 
 4014 
 4183 
 
 3356 
 4031 
 4200 
 
 3S74 
 4048 
 4216 
 
 3392 
 4065 
 4232 
 
 3909 
 40S2 
 4249 
 
 3927 
 4099 
 4265 
 
 3945 
 4116 
 4281 
 
 3962 
 4133 
 4298 
 
 2.45 
 235 
 235 
 
 7 9 11 12 14 16 
 7 9 10'l2 14 15 
 7 8 1011 13 15 
 
 4314 
 4472 
 4624 
 
 4330 
 4487 
 4639 
 
 4346 
 4502 
 4654 
 
 4362 
 4518 
 4669 
 
 4378 
 4533 
 4683 
 
 4393 
 4548 
 4698 
 
 4409 
 4564 
 4713. 
 
 4425 
 4579 
 4728 
 
 4440 
 4594 
 4742 
 
 4456 
 4609 
 4757 
 
 23568 9J11 13 14 
 23568 9>11 12 14 
 1 3 4 1 6 7 9 10 12 13 
 
 4771 
 
 4786 
 
 4800 
 
 4814 
 
 4S29 
 
 4843 
 
 4S57 
 
 4871 
 
 4886 
 
 4900 
 
 134 
 
 679 
 
 678 
 578 
 568 
 
 10 11 13 
 
 10 11 12 
 9 11 12 
 9 10 12 
 
 4914 
 5051 
 5185 
 
 4923 
 5065 
 5193 
 
 4942 
 5079 
 5211 
 
 4955 
 5092 
 5224 
 
 4969 
 5105 
 5237 
 
 4933 
 5119 
 5250 
 
 4937 
 5132 
 5263 
 
 5011 
 5145 
 5276 
 
 5024 
 5159 
 5289 
 
 5038 
 5172 
 5302 
 
 134 
 134 
 134 
 
 34 
 35 
 36 
 
 5315 
 5441 
 5563 
 
 5328 
 5453 
 5575 
 
 5340 
 5465 
 55S7 
 
 5353 
 5478 
 5599 
 
 5366 
 5490 
 5611 
 
 5373 
 5502 
 5623 
 
 5391 
 5514 
 5635 
 
 5403 
 5527 
 5647 
 
 5416 
 5539 
 5658 
 
 5428 
 5551 
 5670 
 
 134 
 124 
 124 
 
 568 
 567 
 567 
 
 9 10 11 
 9 10 11 
 8 10 11 
 
 37 
 33 
 39 
 
 40 
 
 41 
 42 
 43 
 
 56S2 
 5798 
 5911 
 
 5694 
 5809 
 5922 
 
 57 ;. 
 
 5821 
 5933 
 
 5717 
 5832 
 5944 
 
 5729 
 5843 
 5955 
 
 5740 
 5855 
 5966 
 
 5752 
 5866 
 5977 
 
 5763 
 5877 
 5988 
 
 5775 
 5888 
 5999 
 
 5786 
 5899 
 6010 
 
 123 
 123 
 123 
 
 5 6 7 
 567 
 467 
 
 8 9 10 
 8 9 10 
 
 s a 10 
 
 6021 
 
 C031 
 
 6042 
 
 6053 
 
 6064 
 
 6075 
 
 6035 
 
 6096 
 
 6107 
 
 6212 
 6314 
 6415 
 
 6117 
 
 123 
 
 456 
 
 8 9 10 
 
 6128 
 6232 
 6335 
 
 6138 
 6243 
 6345 
 
 6149 
 6253 
 6355 
 
 6160 
 6263 
 6365 
 
 6170 
 6274 
 6375 
 
 6180 
 6284 
 6385 
 
 6191 
 6294 
 6395 
 
 6201 
 6304 
 6405 
 
 6222 
 6325 
 6425 
 
 123 
 123 
 123 
 
 456 
 456 
 456 
 
 789 
 789 
 789 
 
 44 
 45 
 46 
 
 6435 
 6532 
 6628 
 
 6444 
 
 6542 
 6637 
 
 6454 
 6551 
 6646 
 
 6464 
 6561 
 6656 
 
 6474 
 6571 
 6665 
 
 6484 
 6580 
 6675 
 
 6493 
 6590 
 6634 
 
 6503 
 6599 
 6693 
 
 6513 
 66C9 
 6702 
 
 6522 
 6618 
 6712 
 
 123 
 123 
 123 
 
 456 
 456 
 456 
 
 7 8 9 
 789 
 778 
 
 47 
 43 
 49 
 
 50 
 
 6721 
 6812 
 6902 
 
 6730 
 6821 
 6911 
 
 6739 
 6330 
 6920 
 
 6749 
 6839 
 6928 
 
 6753 
 6848 
 6937 
 
 6767 
 6857 
 6946 
 
 6776 
 6866 
 6955 
 
 6785 
 6875 
 6964 
 
 6794 
 6884 
 6972 
 
 6803 
 6S93 
 6981 
 
 123 
 123 
 
 123 
 
 455 
 445 
 445 
 
 678 
 678 
 678 
 
 6990 
 
 6998 
 
 7007 
 
 7016 
 
 7024 
 
 7033 
 
 7042 
 
 7050 
 
 7059 
 
 7067 
 
 123 
 
 345 
 
 678 
 
 51 
 Pi -7 
 53 
 
 7076 
 7lf:0 
 7243 
 
 7324 
 
 7034 
 7168 
 7251 
 
 7093 
 7177 
 7259 
 
 7101 
 7185 
 7267 
 
 7348 
 
 7110 
 7193 
 7275 
 
 7118 
 7202 
 7284 
 
 7126 
 7210 
 7292 
 
 7135 
 7218 
 7300 
 
 7143 
 7226 
 7308 
 
 7152 
 7235 
 7316 
 
 123 
 122 
 122 
 
 345 
 345 
 345 
 
 678 
 677 
 667 
 
 7332 
 
 7340 
 
 7356 
 
 7364 
 
 7372 
 
 7380 
 
 7388 
 
 7306 
 
 122 
 
 345 
 
 667 
 
416 
 
 TABLE OP LOGARITHMS. Continued. 
 
 
 
 
 i r 
 
 1 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 1 2 3 
 
 456 
 
 789 
 
 55 
 
 7404 
 
 7412 
 
 7419 
 
 7427 
 
 7435 
 
 7443 
 
 7451 
 
 7459 
 
 7466 
 
 7474 
 
 122 
 
 345 
 
 567 
 
 56 
 57 
 
 58 
 
 7482 
 7559 
 7634 
 
 7490 
 7566 
 7642 
 
 7497 
 7574 
 7649 
 
 7505 
 7582 
 7657 
 
 7513 
 7589 
 7664 
 
 7520 
 7597 
 7672 
 
 7528 
 7604 
 7679 
 
 7536 
 7612 
 7686 
 
 7543 
 7619 
 7694 
 
 7551 
 7627 
 
 77a 
 
 122 
 
 1 2 2 
 112 
 
 345 
 
 345 
 344 
 
 567 
 567 
 567 
 
 59 
 60 
 61 
 
 7709 
 
 7782 
 7S53 
 
 7716 
 
 7789 
 7860 
 
 7723 
 7796 
 7868 
 
 7731 
 
 7S03 
 7875 
 
 7738 
 7810 
 
 78S2 
 
 77-^5 
 7818 
 7889 
 
 7752 
 7825 
 7896 
 
 7760 
 79.J3 
 
 7767 
 7C39 
 7910 
 
 7774 
 78-16 
 79. 7 
 
 112 
 112 
 112 
 
 344 
 344 
 344 
 
 567 
 566 
 566 
 
 62 
 63 
 64 
 
 7924 
 7993 
 
 80G2 
 
 7931 
 
 8000 
 8069 
 
 7938 
 8007 
 8075 
 
 7945 
 8014 
 
 80S2 
 
 7952 
 
 8021 
 80i9 
 
 7959 
 028 
 8096 
 
 7966 
 80 5 
 8102 
 
 7973 
 8041 
 8109 
 
 7980 
 8048 
 8116 
 
 7917 
 8055 
 8122 
 
 112 
 112 
 
 112 
 
 334 
 334 
 334 
 
 566 
 556 
 556 
 
 65 
 
 8129 
 
 8136 
 
 8142 
 
 8149 
 
 8156 
 
 8162 
 
 8169 
 
 8176 
 
 8182 
 
 81 9 
 
 112 
 
 334 
 
 556 
 
 66 
 67 
 68 
 
 8195 
 8261 
 8325 
 
 8202 
 8267 
 8331 
 
 8209 
 8274 
 8338 
 
 8215 
 8280 
 8344 
 
 8222 
 8287 
 8351 
 
 8228 
 8293 
 8357 
 
 8235 
 8299 
 8363 
 
 8241 
 83C6 
 8370 
 
 8248 
 8312 
 8376 
 
 SC54 
 8319 
 8382 
 
 112 
 
 112 
 112 
 
 334 
 334 
 334 
 
 556 
 556 
 456 
 
 69 
 70 
 71 
 
 8388 
 8451 
 8513 
 
 8395 
 8457 
 8519 
 
 8401 
 8403 
 8525 
 
 8407 
 8470 
 8531 
 
 8414 
 8476 
 8537 
 
 8420 
 8482 
 8543 
 
 8426 
 8488 
 8549 
 
 8432 
 8494 
 8555 
 
 8439 
 8500 
 8561 
 
 8445 
 85C6 
 8567 
 
 1 1 2 
 
 112 
 112 
 
 234 
 234 
 234 
 
 456 
 456 
 455 
 
 72 
 73 
 
 74 
 
 8573 
 8633 
 8692 
 
 8579 
 8G39 
 8698 
 
 8585 
 8645 
 8704 
 
 S591 
 8651 
 8710 
 
 8597 
 8657 
 8716 
 
 S603 
 8663 
 8722 
 
 8609 
 8669 
 8727 
 
 8615 
 8675 
 8733 
 
 8621 
 8681 
 8739 
 
 8627 
 86S6 
 8745 
 
 112 
 112 
 112 
 
 234 
 234 
 234 
 
 455 
 455 
 
 455 
 
 75 
 
 8751 
 
 8756 
 
 8762 
 
 8768 
 
 8774 
 
 8779 
 
 8785 
 
 8791 
 
 8797 
 
 8802 
 
 112 
 
 233 
 
 455 
 
 76 
 
 77 
 
 78 
 
 8808 
 8S65 
 8921 
 
 8814 
 8871 
 8927 
 
 8820 
 8876 
 8932 
 
 8S25 
 8S82 
 8938 
 
 8831 
 SS87 
 943 
 
 8837 
 8;93 
 8949 
 
 8842 
 8899 
 8954 
 
 8848 
 8904 
 8960 
 
 8854 
 8910 
 8965 
 
 S8f.9 
 8915 
 8971 
 
 112 
 112 
 112 
 
 233 
 233 
 233 
 
 455 
 445 
 445 
 
 79 
 SO 
 81 
 
 8976 
 9031 
 9085 
 
 C9S2 
 C036 
 9090 
 
 S9S7 
 9042 
 9096 
 
 8993 
 9047 
 9101 
 
 89C8 
 90C3 
 9106 
 
 9004 
 90C3 
 9112 
 
 9009 
 90C3 
 9117 
 
 9015 
 9069 
 9122 
 
 9020 
 9074 
 9128 
 
 9025 
 9079 
 9133 
 
 112 
 112 
 112 
 
 233 
 233 
 233 
 
 445 
 445 
 445 
 
 82 
 S3 
 84 
 
 9138 
 9191 
 9243 
 
 9143 
 9196 
 9248 
 
 9149 
 9201 
 9253 
 
 9154 
 92C6 
 9258 
 
 9159 
 9212 
 9263 
 
 9165 
 9217 
 9269 
 
 9170 
 9222 
 9274 
 
 9175 
 9227 
 9279 
 
 91 SO 
 9232 
 9284 
 
 9186 
 9238 
 9289 
 
 112 
 112 
 112 
 
 233 
 233 
 233 
 
 445 
 445 
 445 
 
 85 
 
 9294 
 
 9299 
 
 9304 
 
 9309 
 
 9315 
 
 9320 
 
 9325 
 
 C330 
 
 9335 
 
 9340 
 
 112 
 
 233 
 
 445 
 
 86 
 
 87 
 88 
 
 9345 
 9395 
 9445 
 
 9350 
 9400 
 9450 
 
 9355 
 9405 
 9455 
 
 9360 
 9410 
 9460 
 
 9365 
 9415 
 9465 
 
 9370 
 9420 
 9469 
 
 9375 
 9425 
 9474 
 
 C3GO 
 9430 
 9479 
 
 9385 
 9435 
 
 9484 
 
 9390 
 9440 
 94S9 
 
 112 
 Oil 
 Oil 
 
 233 
 223 
 223 
 
 445 
 344 
 344 
 
 89 
 90 
 91 
 
 9494 
 9542 
 9590 
 
 9499 
 9547 
 9595 
 
 9504 
 9552 
 9600 
 
 9509 
 9557 
 9605 
 
 9513 
 9562 
 9609 
 
 9518 
 9566 
 9614 
 
 9523 
 9571 
 9619 
 
 9528 
 9576 
 9624 
 
 9533 
 9581 
 9628 
 
 9538 
 95S6 
 9633 
 
 Oil 
 
 Oil 
 Oil 
 
 223 
 223 
 223 
 
 344 
 344 
 344 
 
 92 
 93 
 94 
 
 9088 
 96S5 
 9731 
 
 9643 
 9689 
 9736 
 
 9647 
 9694 
 9741 
 
 9652 
 9699 
 9745 
 
 9657 
 9703 
 9750 
 
 9661 
 9708 
 9754 
 
 9666 
 9713 
 9759 
 
 9671 
 9717 
 9763 
 
 9675 
 97-22 
 9768 
 
 9680 
 9727 
 9773 
 
 Oil 
 Oil 
 Oil 
 
 223 
 223 
 223 
 
 344 
 344 
 344 
 
 95 
 
 9777 
 
 9?82 
 
 9786 
 
 9791 
 
 9795 
 
 9800 
 
 9805 
 
 9809 
 
 9814 
 
 9818 
 
 Oil 
 
 223 
 
 344 
 
 96 
 
 97 
 98 
 
 9S23 
 9868 
 9912 
 
 9827 
 9872 
 9917 
 
 9832 
 9877 
 9921 
 
 9836 
 9881 
 9926 
 
 9341 
 
 9886 
 9930 
 
 9845 
 9890 
 9934 
 
 9S50 
 9894 
 9939 
 
 9854 
 9899 
 9943 
 
 9859 
 9903, 
 9948 
 
 9S63 
 9908 
 9952 
 
 Oil 
 Oil 
 Oil 
 
 223 
 223 
 223 
 
 344 
 344 
 344 
 
 99 
 
 9956 
 
 9961 
 
 9965 
 
 9969 
 
 9974 
 
 9978 
 
 9983 
 
 9987 
 
 9991 
 
 9996 
 
 Oil 
 
 223 
 
 334 
 
TABLE OF ANTILOGARITHMS. 
 
 417 
 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 1 
 
 8 
 
 9 
 
 1 2 3 
 
 456 
 
 789 
 
 00| 
 
 1000 
 
 1002 
 
 1005 
 
 1007 
 
 1009 
 
 1012 
 
 1014 
 
 1016 
 
 1019 
 
 1021 
 
 001 
 
 111 
 
 222 
 
 01 
 02 
 03 
 
 1023 
 1047 
 1072 
 
 1026 
 1050 
 1074 
 
 1028 
 1052 
 1076 
 
 1030 
 1054 
 1079 
 
 1033 
 1057 
 1081 
 
 1035 
 1059 
 1084 
 
 1038 
 1062 
 1086 
 
 1040 
 1064 
 1089 
 
 1042 
 1067 
 1091 
 
 1045 
 1069 
 1094 
 
 001 
 001 
 001 
 
 111 
 111 
 111 
 
 222 
 222 
 222 
 
 041 
 
 06 
 
 1096 
 1122 
 1148 
 
 1099 
 1125 
 1151 
 
 1102 
 1127 
 1153 
 
 1104 
 1130 
 1156 
 
 1107 
 1132 
 1159 
 
 1109 
 1135 
 1161 
 
 1112 
 1133 
 1164 
 
 1114 
 1140 
 1167 
 
 1117 
 1143 
 1169 
 
 1119 
 1146 
 1172 
 
 Oil 
 Oil 
 Oil 
 
 1 2 ' 
 1 2 j 
 
 1 2| 
 
 222 
 222 
 222 
 
 07 
 08 
 09 
 
 1175 
 1202 
 1230 
 
 1178 
 1205 
 1233 
 
 1180 
 1-208 
 1236 
 
 1183 
 1211 
 1239 
 
 1186 
 1213 
 1242 
 
 1189 
 1216 
 1245 
 
 1191 
 1219 
 1247 
 
 1194 
 1222 
 1250 
 
 1197 
 1225 
 1253 
 
 1199 
 1227 
 1256 
 
 Oil 
 Oil 
 1 1 
 
 1 2 
 1 2 
 112 
 
 222 
 223 
 223 
 
 10 
 
 1259 
 
 1262 
 
 1265 
 
 1268 
 
 1271 
 
 1274 
 
 1276 
 
 1279 
 
 1282 
 
 1285 
 
 Oil 
 
 112 
 
 223 
 
 11 
 12 
 13 
 
 1288 
 1318 
 1349 
 
 1291 
 1321 
 1352 
 
 1294 
 1324 
 1355 
 
 1297 
 1327 
 1353 
 
 1300 
 1330 
 1361 
 
 1303 
 1334 
 13.5 
 
 1306 
 1337 
 1363 
 
 1309 
 1340 
 1371 
 
 1312 
 1343 
 1374 
 
 1315 
 1346 
 1377 
 
 Oil 
 Oil 
 Oil 
 
 122 
 122 
 122 
 
 223 
 223 
 233 
 
 14 
 15 
 
 1C 
 
 13SO 
 1413 
 1445 
 
 13S4 
 1416 
 1449 
 
 13S7 
 141 
 1452 
 
 1390 
 1422 
 1455 
 
 1393 
 1426 
 1459 
 
 1396 
 14_>9 
 1462 
 
 1400 
 1432 
 1466 
 
 1403 
 1435 
 1469 
 
 1406 
 1439 
 1472 
 
 1409 
 1442 
 1476 
 
 Oil 
 Oil 
 1 1 
 
 1 2 2 
 122 
 1 2 2 
 
 233 
 233 
 233 
 
 17 
 
 is 
 
 19 
 
 1479 
 1514 
 1549 
 
 1483 
 1517 
 1552 
 
 14S6 
 1521 
 1556 
 
 14S9 
 1524 
 1560 
 
 1493 
 1523 
 1563 
 
 1496 
 1531 
 1567 
 
 1500 
 1535 
 1570 
 
 1503 
 1538 
 1574 
 
 1507 
 1542 
 1578 
 
 1510 
 1545 
 1581 
 
 1 1 
 Oil 
 1 1 
 
 122 
 122 
 122 
 
 233 
 233 
 333 
 
 20 
 
 1585 
 
 1589 
 
 1592 
 
 1596 
 
 1600 
 
 1603 
 
 1607 
 
 1611 
 
 1614 
 
 1613 
 
 Oil 
 
 122 
 
 333 
 
 21 
 22 
 23 
 
 1022 
 1660 
 1093 
 
 1626 
 1603 
 1702 
 
 1620 
 1667 
 1706 
 
 1633 
 1671 
 1710 
 
 1637 
 1675 
 1714 
 
 1641 
 1679 
 1713 
 
 1644 
 16^3 
 1722 
 
 1643 
 1687 
 1726 
 
 1652 
 1690 
 1730 
 
 1656 
 1694 
 1734 
 
 Oil 
 1 1 
 Oil 
 
 222 
 222 
 222 
 
 333 
 333 
 334 
 
 24 
 25 
 26 
 
 1738 
 1778 
 1820 
 
 1742 
 1782 
 1824 
 
 1746 
 1786 
 1828 
 
 1750 
 1791 
 1832 
 
 1754 
 1795 
 1837 
 
 1758 
 1799 
 1841 
 
 1762, 
 1803" 
 1845 
 
 1766 
 1307 
 1849 
 
 1770 
 1811 
 1854 
 
 1774 
 1816 
 1858 
 
 Oil 
 Oil 
 Oil 
 
 222 
 222 
 223 
 
 334 
 334 
 334 
 
 27 
 28 
 29 
 
 1862 
 1905 
 1950 
 
 1866 
 1910 
 1954 
 
 1871 
 1914 
 1959 
 
 1875 
 1919 
 1963 
 
 1879 
 1923 
 1968 
 
 1884 
 1928 
 1972 
 
 1888 
 1932 
 1977 
 
 1892 
 1936 
 1982 
 
 1897 
 1941 
 1986 
 
 1901 
 1945 
 1991 
 
 Oil 
 1 1 
 Oil 
 
 223 
 223 
 223 
 
 334 
 344 
 344 
 
 30 
 
 1995 
 
 2000 
 
 2004 
 
 2009 
 
 2014 
 
 2018 
 
 2023 
 
 2028 
 
 2032 
 
 2037 
 
 Oil 
 
 223 
 
 344 
 
 31 
 32 
 33 
 
 2042 
 2089 
 2138 
 
 *2046 
 2094 
 2143 
 
 2051 
 2099 
 2148 
 
 2056 
 2104 
 2153 
 
 2061 
 2109 
 2153 
 
 2065 
 2113 
 2163 
 
 2070 
 2118 
 2168 
 
 2075 
 2123 
 2173 
 
 2080 
 2123 
 2173 
 
 20S4 
 2133 
 2183 
 
 Oil 
 Oil 
 Oil 
 
 223 
 223 
 223 
 
 344 
 344 
 344 
 
 34 
 35 
 36 
 
 2188 
 2239 
 2291 
 
 2193 
 2244 
 2296 
 
 2198 
 2249 
 2301 
 
 2203 
 2254 
 2307 
 
 2203 
 2259 
 2312 
 
 2213 
 2265 
 2317 
 
 2218 
 2270 
 2323 
 
 2223 
 2275 
 2328 
 
 2228 
 2280 
 2333 
 
 2234 
 2286 
 2339 
 
 112 
 112 
 112 
 
 233 
 233 
 233 
 
 4 5 
 4 5 
 4 5 
 
 37 
 38 
 39 
 
 2344 
 2399 
 2455 
 
 2350 
 
 2404 
 2460 
 
 2355 
 2410 
 2466 
 
 2360 
 2415 
 2472 
 
 2366 
 2421 
 2477 
 
 2371 
 2427 
 2483 
 
 2377 
 2432 
 2489 
 
 2382 
 2438 
 2495 
 
 2333 
 2443 
 2500 
 
 2393 
 2449 
 2506 
 
 112 
 112 
 
 112 
 
 233 
 233 
 233 
 
 4 5 
 4 5 
 5 5 
 
 40 
 
 2512 
 
 2518 
 
 2523 
 
 2529 
 
 2535 
 
 2541 
 
 2547 
 
 2553 
 
 2-559 
 
 2564 
 
 112 
 
 234 
 
 455 
 
 41 
 42 
 43 
 
 2570 
 2630 
 2692 
 
 2576 
 2636 
 2698 
 
 2582 
 2642 
 2704 
 
 2588 
 2649 
 2710 
 
 2594 
 2655 
 2716 
 
 2600 
 2661 
 
 2723 
 
 2606 
 2667 
 2729 
 
 2612 
 2673 
 2735 
 
 2618 
 2679 
 2742 
 
 2624 
 26S5 
 2743 
 
 112 
 112 
 112 
 
 2 3 
 2 3 
 3 3 
 
 455 
 456 
 456 
 
 44 
 45 
 46 
 
 2754 
 2818 
 
 28S4 
 
 2761 
 2825 
 2891 
 
 2767 
 2831 
 2897 
 
 2773 
 2833 
 2904 
 
 2780 
 2S44 
 2911 
 
 2786 
 2851 
 2917 
 
 2793 
 2853 
 2924 
 
 2799 
 2864 
 2931 
 
 2805 
 2371 
 2938 
 
 2312 
 2S77 
 2944 
 
 112 
 112 
 112 
 
 3 3 
 3 3 
 3 3 
 
 456 
 556 
 556 
 
 47 
 48 
 49 
 
 2951 
 3020 
 
 :3000 
 
 .2953 
 3027 
 3097 
 
 2965 
 3034 
 3105 
 
 2972 
 3041 
 3112 
 
 2979 
 3048 
 3119 
 
 2985 
 3055 
 3126 
 
 2992 
 3062 
 3133 
 
 2999 
 3069 
 3141 
 
 3006 
 3076 
 3148 
 
 3013 
 30S3 
 3155 
 
 112 
 
 112 
 112 
 
 3 3 
 
 i 3 4 
 3 4 
 
 556 
 566 
 566 
 
TABLE OF ANTILOGARITHMS. continued. 
 
 
 
 
 1 
 
 2 
 
 3 
 
 * 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 1 2 3 
 
 456 
 
 789 
 
 50 
 
 3162 
 
 3170 
 
 3177 
 
 3184 
 
 3192 
 
 3199 
 
 3206 
 
 3214 
 
 3221 
 
 3228 
 
 112 
 
 344 
 
 567 
 
 51 
 53 
 
 3236 
 3311 
 
 3388 
 
 3243 
 3319 
 3396 
 
 3251 
 3327 
 3404 
 
 3258 
 3334 
 3412 
 
 3266 
 3342 
 3420 
 
 3273 
 3350 
 3428 
 
 3281 
 3357 
 3436 
 
 32S9 
 3365 
 3443 
 
 32% 
 3373 
 3451 
 
 3304 
 3381 
 3459 
 
 2 2 
 2 2 
 
 2 2 
 
 3 4 5 
 3 5 
 
 3 5 
 
 567 
 567 
 667 
 
 54 
 55 
 56 
 
 3467 
 3548 
 3631 
 
 3475 
 3556 
 3639 
 
 34S3 
 3565 
 3648 
 
 3491 
 3573 
 3656 
 
 3499 
 3581 
 3664 
 
 3508 
 35S9 
 3673 
 
 3516 
 3597 
 3 81 
 
 3524 
 3606 
 3690 
 
 3532 
 3614 
 3698 
 
 3540 
 
 3622 
 3707 
 
 2 2 
 2 2 
 2 3 
 
 3 5 
 3 5 
 
 3 5 
 
 667 
 677 
 678 
 
 57 
 58 
 59 
 
 3715 
 SS02 
 3S90 
 
 3724 
 3811 
 3899 
 
 3733 
 
 3819 
 3908 
 
 3741 
 3 28 
 39J7 
 
 3750 
 3837 
 3926 
 
 3758 
 3846 
 3936 
 
 3767 
 3855 
 3945 
 
 3776 
 3864 
 3954 
 
 3784 
 3.S73 
 3963 
 
 3793 
 
 3882 
 3972 
 
 123 
 123 
 123 
 
 345 
 446 
 455 
 
 678 
 
 678 
 678 
 
 60 
 
 3981 
 
 3990 
 
 3999 
 
 4009 
 
 4018 
 
 4027 
 
 4036 
 
 4046 
 
 4055 
 
 4064 
 
 123 
 
 456 
 
 678 
 
 61 
 62 
 63 
 
 4074 
 4169 
 4266 
 
 4083 
 4178 
 4276 
 
 4093 
 4188 
 4285 
 
 4102 
 4198 
 4295 
 
 4111 
 4207 
 4305 
 
 4121 
 4217 
 4315 
 
 4130 
 4227 
 4325 
 
 4140 
 4236 
 4335 
 
 4150 
 4246 
 4345 
 
 4159 
 42C6 
 4355 
 
 123 
 123 
 123 
 
 456 
 456 
 456 
 
 789 
 789 
 
 789 
 
 64 
 65 
 66 
 
 4365 
 4467 
 4571 
 
 4375 
 
 4477 
 4581 
 
 4385 
 4487 
 4592 
 
 4395 
 
 4498 
 4603 
 
 4406 
 4508 
 4613 
 
 4416 
 4519 
 4624 
 
 4426 
 4529 
 4634 
 
 4436 
 4539 
 4645 
 
 4446 
 4550 
 4656 
 
 4457 
 4560 
 4567 
 
 123 
 123 
 123 
 
 450 
 456 
 456 
 
 789 
 789 
 7 9 10 
 
 67 
 68 
 69 
 
 4677 
 47S6 
 4898 
 
 46S8 
 4797 
 4909 
 
 4G99 
 4OS 
 49iO 
 
 4710 
 4819 
 4932 
 
 4721 
 4S31 
 4943 
 
 4732 
 4842 
 4955 
 
 4742 
 4853 
 4966 
 
 4753 
 4864 
 4977 
 
 4764 
 4S75 
 4989 
 
 4775 
 
 4887 
 5000 
 
 123 
 123 
 123 
 
 457 
 467 
 567 
 
 8 9 10 
 8 9 10 
 8 9 10 
 
 70 
 
 5012 
 
 5023 
 
 5G35 
 
 5047 
 
 50:^ 
 
 5070 
 
 5082 
 
 5093 
 
 5105 
 
 5117 
 
 1 2 4 
 
 567 
 
 8 9 11 
 
 71 
 "72 
 73 
 
 5129 
 5248 
 5370 
 
 5140 
 5260 
 5383 
 
 5152 
 5272 
 5395 
 
 5164 
 
 5284 
 5408 
 
 517J 
 5297 
 5420 
 
 51S8 
 5309 
 5433 
 
 5200 
 6321 
 5445 
 
 5212 
 63.-J3 
 5458 
 
 5224 
 5346 
 5470 
 
 5236 
 5358 
 5483 
 
 1 2 
 1 2 
 1 3 
 
 567 
 567 
 368 
 
 8 10 11 
 9 10 11 
 9 10 11 
 
 74 
 75 
 76 
 
 5495 
 5623 
 5754 
 
 5508 
 5636 
 5768 
 
 5521 
 5049 
 5781 
 
 5534 
 5662 
 5794 
 
 5546 
 5675 
 5808 
 
 5559 
 56S9 
 5821 
 
 5572 
 5702 
 5S34 
 
 5585 
 5715 
 
 5S48 
 
 5598 
 5728 
 5861 
 
 5610 
 5741 
 5875 
 
 1 3 
 1 3 
 134 
 
 568 
 
 578 
 578 
 
 9 10 12 
 9 10 12 
 9 11 12 
 
 77 
 78 
 79 
 
 5888 
 C026 
 6166 
 
 5902 
 6039 
 6180 
 
 916 
 6053 
 6194 
 
 5929 
 CC67 
 6209 
 
 5943 
 6Gol 
 6223 
 
 5957 
 6095 
 6237 
 
 6970 
 6109 
 6252 
 
 5984 
 (124 
 6266 
 
 5998 
 6138 
 C281 
 
 G012 
 6152 
 6295 
 
 134 
 134 
 134 
 
 578 
 678 
 679 
 
 10 11 12 
 10 11 13 
 10 11 18 
 
 80 
 
 6310 
 
 C324 
 
 6339 
 
 6353 
 
 6368 
 
 6383 
 
 6397 
 
 6412 
 
 6427 
 
 6442 
 
 134 
 
 679 
 
 10 12 13 
 
 81 
 
 82 
 83 
 
 6457 
 6607 
 6761 
 
 6471 
 6622 
 6776 
 
 G-JS6 
 6637 
 6792 
 
 6501 
 6653 
 6i08 
 
 6516 
 6G68 
 SS23 
 
 6531 
 66S3 
 6839 
 
 6546 
 6699 
 6S55 
 
 6561 
 6714 
 
 6S71 
 
 6577 
 6730 
 6887 
 
 6592 
 6745 
 6902 
 
 235 
 2 C 5 
 235 
 
 689 
 089 
 689 
 
 11 12 14 
 11 12 14 
 11 13 14 
 
 84 
 35 
 86 
 
 6918 
 7079 
 7244 
 
 6934 
 7096 
 7261 
 
 6950 
 7112 
 7278 
 
 6C66 
 7129 
 7295 
 
 692 
 
 7145 
 7311 
 
 6998 
 7161 
 7328 
 
 7015 
 7173 
 7345 
 
 7031 
 7194 
 7362 
 
 7047 
 7211 
 7379 
 
 7CC3 
 7228 
 7396 
 
 235 
 235 
 
 235 
 
 6 8 10 
 7 8 10 
 7 8 10 
 
 11 13 15 
 12 13 15 
 ,12 13 15 
 
 87 
 88 
 89 
 
 7413 
 7586 
 7762 
 
 7430 
 7603 
 7780 
 
 7447 
 76:>1 
 7798 
 
 7464 
 7638 
 7816 
 
 7482 
 7656 
 7834 
 
 7499 
 7674 
 7852 
 
 7516 
 7091 
 7S70 
 
 7534 
 7709 
 78t9 
 
 7551 
 7727 
 7907 
 
 75C8 
 7745 
 7925 
 
 235 
 
 2 4 5 
 
 245 
 
 7 9 10 
 7 9 11 
 7 9 11 
 
 12 14 16 
 121418- 
 13 14 16 
 
 90 
 
 7943 
 
 7962 
 
 7980 
 
 7998 
 
 8017 
 
 8035 
 
 8054 
 
 8072 
 
 8891 
 
 8110 
 
 246 
 
 7 9 11 
 
 13 15 17 
 
 91 
 92 
 93 
 
 8128 
 8318 
 8511 
 
 8147 
 8337 
 8531 
 
 8166 
 8356 
 8551 
 
 8185 
 8570 
 
 8204 
 8395 
 8590 
 
 8222 
 8414 
 8610 
 
 8241 
 8433 
 8630 
 
 8260 
 8453 
 8650 
 
 8279 
 8472 
 8670 
 
 8299 
 8492 
 8690 
 
 240 
 240 
 240 
 
 8 9 11 
 8 10 12 
 8 10 12 
 
 [13 15 17 
 14 15 17 
 
 J14 16 18 
 
 94 
 95 
 96 
 
 8710 
 8913 
 9120 
 
 8730 
 8933 
 9141 
 
 8750 
 8954 
 9162 
 
 8770 
 8974 
 9183 
 
 8790 
 8995 
 9204 
 
 8810 
 9016 
 9226 
 
 8S31 
 9036 
 9247 
 
 8851 
 9057 
 9268 
 
 8872 
 9078 
 9290 
 
 8892 
 9099 
 9311 
 
 |2 4 6 
 
 240 
 240 
 
 8 10 12 
 8 10 12 
 8 11 13 
 
 14 If 18 
 15 17 19 
 15 17 19 
 
 97 
 98 
 99 
 
 9333 
 9550 
 9772 
 
 9354 
 9572 
 9795 
 
 9376 
 
 9594 
 9817 
 
 9397 
 9616 
 9S40 
 
 9419 
 9638 
 98C3 
 
 9441 
 9661 
 9886 
 
 9462 
 9683 
 9908 
 
 9484 
 9705 
 9931 
 
 9506 
 9727 
 C954 
 
 9528 
 9750 
 9977 
 
 247 
 247 
 257 
 
 9 11 13 
 9 11 13 
 9 11 14 
 
 15 17 20 
 16 18 20 
 16 18 20 
 
TABLE OF FUNCTIONS OF IHOLM. 
 
 419 
 
 
 
 1 
 
 
 
 
 
 Angle. 
 
 
 
 
 
 
 
 ; 
 
 
 
 Chord. 
 
 fine. 
 
 Tangent. 
 
 Co- tangent. 
 
 Cosine. 
 
 
 
 
 Degrees 
 
 Radians. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 000 C 
 
 
 
 ao 
 
 1 
 
 1-414 
 
 1-5703 
 
 90 
 
 1 
 
 0175 
 
 017 l '0175 
 
 0175 
 
 57-2900 
 
 9998 
 
 1-402 
 
 1-5533 
 
 89 
 
 2 
 
 034D 
 
 035 
 
 0349 
 
 0349 
 
 28-6363 
 
 9994 
 
 1-3S9 
 
 1-5359 
 
 83 
 
 3 
 
 0524 
 
 052 
 
 0523 
 
 0524 
 
 19-0311 
 
 99S6 
 
 1-377 
 
 1-51S4 
 
 87 
 
 4 
 
 0698 
 
 070 
 
 0698 
 
 0699 
 
 14-3007 
 
 9976 
 
 1-364 
 
 1-5010 
 
 86 
 
 5 
 
 OS73 
 
 087 
 
 0872 
 
 0875 
 
 11-4301 
 
 9962 
 
 1-351 
 
 1-4S35 
 
 85 
 
 6 
 
 1047 
 
 105 
 
 1045 
 
 1051 
 
 9-5144 
 
 9945 
 
 1-338 
 
 1-4661 
 
 84 
 
 7 
 
 1222 
 
 122 
 
 1219 
 
 1228 
 
 8-1443 
 
 9925 
 
 1325 
 
 1-4486 
 
 83 
 
 8 
 
 1396 
 
 140 
 
 1392 
 
 1405 
 
 7-1154 
 
 9903 
 
 1-312 
 
 1-4312 
 
 82 
 
 9 
 
 1571 
 
 157 
 
 1564 
 
 1584 
 
 6-3133 
 
 9S77 
 
 1-299 
 
 1-4137 
 
 81 
 
 10 
 
 1745 
 
 174 
 
 1736 
 
 1763 
 
 5-6713 
 
 9848 
 
 1-2S6 
 
 i-rso 
 
 80 
 
 11 
 
 1929 
 
 192 
 
 1903 
 
 1944 
 
 5-1446 
 
 9316 
 
 1-272 
 
 l-37t3 
 
 79 
 
 12 
 
 2094 
 
 209 
 
 2079 
 
 2126 
 
 4-7046 
 
 97S1 
 
 1-259 
 
 1-3614 
 
 73 
 
 13 
 
 2269 
 
 228 
 
 2250 
 
 2309 
 
 4-3315 
 
 9744 
 
 1-245 
 
 1-3439 
 
 i 
 
 H 
 
 2443 
 
 244 
 
 2419 
 
 2493 
 
 4-0108 
 
 9703 
 
 1-231 
 
 1-3265 
 
 Z ' 
 
 w a 
 
 2613 
 
 261 
 
 2588 
 
 8679 
 
 87821 
 
 9659 
 
 1-213 
 
 13090 
 
 5 
 
 16 
 
 2793 
 
 278 
 
 2756 
 
 2S67 
 
 3-4S74 
 
 9613 
 
 1-204 
 
 1-2915 
 
 4 
 
 17 
 
 2967 
 
 296 
 
 2924 
 
 3057 
 
 32709 
 
 9563 
 
 1-190 
 
 1-2741 
 
 3 { 
 
 13 
 
 3142 
 
 313 
 
 3090 
 
 3249 
 
 3-0777 
 
 9511 
 
 T176 
 
 1-2566 
 
 2 ' 
 
 19 
 
 3316 
 
 330 
 
 3256 
 
 3443 
 
 2-9042 
 
 9455 
 
 1-161 
 
 1-2302 
 
 1 
 
 20 
 
 3491 
 
 347 
 
 3420 
 
 3640 
 
 2-7475 
 
 9397 
 
 1-147 
 
 1-2217 
 
 
 
 21 
 
 3665 
 
 364 
 
 3584 
 
 3S39 
 
 2*6051 
 
 9336 
 
 1-133 
 
 "2043 
 
 69 
 
 22 
 
 3840 
 
 382 
 
 3746 
 
 4040 
 
 2-4751 
 
 9272 
 
 1-118 
 
 1S63 
 
 6* 
 
 23 
 
 4014 
 
 399 
 
 3907 
 
 4245 
 
 2-3559 
 
 9205 
 
 1-104 
 
 1694 
 
 67 
 
 24 
 
 4189 
 
 416 
 
 4067 
 
 4452 
 
 2-2460 
 
 9135 
 
 1-089 
 
 1519 
 
 66 
 
 25 
 
 4363 
 
 433 
 
 4226 
 
 4663 
 
 2-1445 
 
 9063 
 
 1-075 
 
 1345 
 
 65 
 
 20 
 
 4538 
 
 450 
 
 4384 
 
 4877 
 
 2-0503 
 
 89S8 
 
 1-060 
 
 1-1170 
 
 64 
 
 "7 
 
 4712 
 
 467 
 
 4540 
 
 5095 
 
 1-9626 
 
 8910 
 
 1-045 
 
 1-0996 
 
 63 
 
 23 
 
 4387 
 
 484 
 
 4695 
 
 5317 
 
 1-SS07 
 
 8829 
 
 1-030 
 
 1-0821 
 
 62 
 
 29 
 
 5061 
 
 501 
 
 4848 
 
 5543 
 
 1-SC40 
 
 8746 
 
 1-015 
 
 1-0647 
 
 61 
 
 30 
 
 5236 
 
 518 
 
 5000 
 
 5774 
 
 1-7321 
 
 8660 
 
 1-000 
 
 1-0472 
 
 60 
 
 31 
 
 5411 
 
 534 
 
 5150 
 
 6009 
 
 1-6643 
 
 8572 
 
 985 
 
 1-0297 
 
 59 
 
 32 
 
 5585 
 
 551 
 
 5299 
 
 6249 
 
 1-6003 
 
 84SO 
 
 970 
 
 1-0123 
 
 58 
 
 33 
 
 5760 
 
 568 
 
 5446 
 
 6494 
 
 1-5399 
 
 8387 
 
 954 
 
 9943 
 
 57 
 
 34 
 
 5934 
 
 585 
 
 5592 
 
 6745 
 
 1-4826 
 
 8290 
 
 939 
 
 9774 
 
 56 
 
 35 
 
 6109 
 
 601 
 
 5736 
 
 7002 
 
 1-4281 
 
 8192 
 
 923 
 
 9599 
 
 55 
 
 36 
 
 G283 
 
 618 
 
 5S78 
 
 7265 
 
 1-3764 
 
 8090 
 
 908 
 
 9425 
 
 54 
 
 37 
 
 6458 
 
 635 
 
 6018 
 
 7536 
 
 1-3270 
 
 79S6 
 
 892 
 
 9250 
 
 53 
 
 33 
 
 6632 
 
 651 
 
 6157 
 
 7813 
 
 1-2793 
 
 7880 
 
 877 
 
 9076 
 
 52 
 
 39 
 
 6S07 
 
 668 
 
 6293 
 
 8098 
 
 1-2349 
 
 7771 
 
 861 
 
 8901 
 
 51 
 
 40 
 
 C9S1 
 
 684 
 
 6428 
 
 8391 
 
 1-1918 
 
 7660 
 
 845 
 
 8727 
 
 50 
 
 41 
 
 71. '6 
 
 700 
 
 6561 
 
 8693 
 
 1-1504 
 
 7547 
 
 829 
 
 8552 
 
 49 
 
 42 
 
 7330 
 
 717 
 
 6691 
 
 9004 
 
 1-1106 
 
 7431 
 
 813 
 
 8373 
 
 43 
 
 43 
 
 7505 
 
 733 
 
 6820 
 
 9325 
 
 1-0724 
 
 7314 
 
 797 
 
 8203 
 
 47 
 
 44 
 
 7679 
 
 749 
 
 6947 
 
 9657 
 
 1-0355 
 
 7193 
 
 781 
 
 8029 
 
 46 
 
 45 
 
 7S54 
 
 765 
 
 7071 
 
 1-0000 
 
 1-0000 
 
 7071 
 
 765 
 
 7854 
 
 45 
 
 
 
 
 
 
 
 
 
 
 Radians. 
 
 Degrees. 
 
 
 
 
 Cosine. 
 
 Co- tangent. 
 
 Tangent. 
 
 Sine. 
 
 Chord. 
 
 
 
 
 
 
 
 
 
 
 Angle. 
 
INDEX 
 
 ACCELERATION, definition, 260, 421 
 ,, due to gravity, 260 
 
 ,, unit of, 260 
 
 Accumulated work, 280 
 
 ,, work in a fly-wheel, 
 
 281 
 ,, work in a rotating 
 
 body, 281 
 
 Accumulator, hydraulic, 253-2^5 
 Action and reaction, 3 
 
 ,, of the common suction- 
 pump, 227-229 
 
 Activity, definition of, 12, 412 
 Actual or working advantage, 67 
 Admission of Students to the Insti- 
 tution of Civil Engineers, 
 rules and syllabus of Exam?, 
 for, 370 
 Air vessel, action of an, 232 
 
 ,, force pump with, 231 
 Ampere, 413 
 
 Angle of repose, io5 ; resistance, 107 
 Angular and linear motion, 272 
 
 ,, velocity, 260 
 Anti-friction wheels, 109 
 Anti-logarithms, Appendix D, 417 
 Applied Mechanics, definition of, I 
 Archimedes, law of, 217-219 
 Atmospheric pressure, 220 
 Atwood's machine, 262-269 
 Axle, wheel and, 55-57 
 
 ,, compound, 72-74 
 
 BA.CKLA.SS, in whael and screw- 
 gearings, 158 
 
 B ick motion gear in a lathe, 1 74-1 8 5 
 Balance, Roman, 35 
 
 ,, bent lever, 43 
 Balancing fast spsed machinery,270 
 Biles, screw for pissing, 165 
 Ball bearings, no 
 Birometer, the mercurial, 221 
 
 Bear, hydraulic, or portable punch- 
 ing machine, 251 
 Bsaringa, ball, 1 10 
 
 ,, roller, 183 
 
 B3d or shears of a lathe, 183, 187. 
 B3ll crank lever, 43 
 Belt-gaaring reversing motions, 122- 
 
 124 
 shape of pulley faces 
 
 for, 127 
 
 ,, ,, velocity ratio of pul- 
 
 leys in, 119-121 
 
 Baits, brake horse-power trans- 
 mitted by, 118 
 
 ,, difference of tension in, 116 
 ,, open and crossed, 121 
 ,, tendency of, to run on high- 
 est parts of pulleys, 127 
 Bench vice, screw, 166-167 
 B^nt lavar balance, 43 
 ,, levers, 42 
 ,, ,, duplex, 43 
 Bavel wheel and clutch reversing 
 
 gsar, 345 
 Bioyole and railway curves, motion 
 
 on, 287 
 Block and tackle, 65-68 
 
 ,, snatch, 65 
 Block, Weston's differential pullev. 
 
 74-77 
 
 BDird of Education Exam., instruc- 
 tions for, 368 
 Baird of Trade unit, 413 
 Bjiies, formulae for falling, 261 
 
 pith of projected badies 
 which fall under the 
 action of gravity, 273 
 ,, proof by Atwood's michine 
 of formulae for falling, 
 262-269 
 BDiler plates, large hydraulic press 
 
 for flanging, 245-247 
 . single riveted lap 
 
 joints, 310 
 
 B)lt, holding down, 156 
 B>n-accord centrifugal pump, 236 
 421 
 
422 INDEX 
 
 Bottle screw- jack, 162-164 
 
 Brake^ horse-power transmitted by 
 
 belts, 118 
 Bramah's hydraulic press, 241-245 
 
 leather collar packing, 243 
 Bucket pump, combined plunger 
 r and, 234 
 
 Buttress screw thread, 155 
 
 c 
 
 CALCULATIONS, note regarding en- 
 gineering, 5 
 Cams, 330-334 
 Capstan, ships, 57-59 
 Carpenter's turkus, 44 
 Centimetre, 411 
 Centre of gravity, 28 
 
 ,, of parallel forces, 26 
 ,, of pressure, 217 
 Centrifugal force, 275 
 
 ,, force machine, 277 
 
 pumps, 236-238 
 
 ^ ,, stress in fly-wheels, 279 
 Centripetal force, 276 
 Chains, stresses in, 316 
 Change wheels in a lathe, 176-179, 
 
 183, 184, 187 
 
 ,, ,, reversing plate or 
 
 quadrant, 183, 187 
 Chinese windlass, 72 
 Circle, definition of pitch, 131 
 Circular discs, velocity ratio of, 130 
 City and Guilds exam., instruction 
 ^ for, 369 
 Civil Engineers, Students' Exam., 
 
 rules and syllabus of, 370 
 Clarke's patent adjustable curve, 
 
 16, 17 
 
 Click, reversible, 335 
 Co-efficient of friction, 105, in 
 Cohesion of matter, 224, 298 
 Combined lever, screw, and pulley 
 
 gear, 160 
 ,, plunger and bucket 
 
 pump, 234 
 
 Comparison of dynamical formulae 
 ^ for linear and angular mo- 
 tion, 2721 
 Components and resultant of forces, 
 
 4 
 
 , 9 of a force at right angles 
 
 to each other, 84 
 
 Composition and resolution of forces, 
 82, 84 ; velocities, 260 
 
 Compound, Weems', screw and 
 hydraulic jack, 249-251 
 
 Compound wheel and axle, 72-74 
 
 Compressibility of matter, 298 
 
 Compressing a bar within the elastic 
 limit, work done in, 307 
 
 Compressive stress and strain, 301 
 
 Cones, stepped speed, 124 
 
 Constants appendix D, 414 
 
 Coulomb, 413 
 
 Couple, definition of, 25 
 
 Coupling joint, Hooke's, 329 
 
 ,, screw, for carriages, 156 
 
 Crab, double purchase, 143, 144 
 single purchase, 140-142 
 
 Crania, stresses in various membciB 
 and jib arrester, 85, 92 
 
 Crank, bell, lever, 43 
 
 Curve, focus and directrix of a, 272 
 
 Curves, motion on bicycle and rail- 
 way, 287 
 
 Cylinder, forming a screw thread 
 on a, 149 
 
 D 
 
 DEAD load, definition of, 500 
 Density of matter, 298 
 Differential pulley blocks, Weston's, 
 
 74-77 
 
 Dilatibility of matter, 298 
 Directrix of a curve, 272 
 Discs, velocity ratio of, 130 
 Distinction, solids, liquids, gases, 
 
 fluids, &c., 224 
 
 Double acting force pump, 234 
 ,, Hooke's joint, 329 
 ,, purchase winch or crab, 143, 
 
 144 
 
 ,, threaded screws, 157 
 Driving belt, difference of tension 
 
 in a, 116 
 
 Ductility, definition of, 299 
 Duplex bent levers, 43 
 Dynamical formulae for linear and 
 angular motion, comparison 
 of, 272 
 
 E 
 
 EFFICIENCY, apparatus for deter- 
 mining, of screw 
 gear, 160 
 
INDEX 
 
 423 
 
 Efficiency of combined lever, 
 screw, and pulley 
 
 fear, 160 
 a machine, defini- 
 tion of, 53 
 
 ,, of screws, 151 
 
 Elasticity, definition of, 300 
 ,, limits of, 303 
 ,, modulus of, 304-306 
 ,, safe loads and, 302 
 ,, table of moduli of, 305 
 Ellipse, 273 
 Elliptic wheels for quick return, 
 
 351-353 
 
 Endless screw and worm-wheel, 168 
 Energy, kinetic, potential, 280,412 
 Eqglish gauges, 360 
 Equilibriant of parallel forces, 25 
 , Jlibriurn, conditions of, In O*M 
 of floating bodies, 
 218 
 Equilibrium, forces in, 3 
 
 ,, graphic demonstration 
 
 of three forces in, 80 
 Erg, 6, 412 
 Exams. B. of E. C and G ; Inst. C.E., 
 
 Appendix C, 403 
 
 Experimentally determining the 
 energy stored up in a rotating 
 fly-wheel, 285 
 Extending a bar within the elastic 
 
 limit, work done in, 307 
 Extension of matter, 297 
 
 FACTORS of safety, 302 
 Falling bodies, formulae for, 261 
 ,, proof of formulae for, by 
 Atwood's machine, 262-269 
 Farad, 413 
 Feed, silent, 340-342 
 Floating bodies, conditions of equi- 
 librium, 218 
 
 Fluids, solids, gases, 224 
 Fly- press, the, 283 
 Fly-wheels, centrifugal stress in, 279 
 ,, energy stored up in, 281 
 
 Fly-wheel, to find experimentally 
 the energy stored up in a 
 rotating, 285 
 Focus of a curve, 272 
 Force^definition of, i, 412 
 .. [elements of a, 2 
 
 Force, moment of a, 21 
 
 ,, pump, single-acting, 229-231 
 ,, pump, double-acting, 234 
 ,, pump with air vessel, 231 
 ,, resolution of a, into two com- 
 ponents at right angles, 84 
 ,, unit of, 2, 412 
 
 Forces, centrifugal, centripetal, 275 
 ,, graphic representation of, 3 
 ,, in equilibrium, 3 
 parallel, 25-28 
 ,, parallelogram of, 82 
 ,, resultant of, 4 
 ,, straight lever acted on by 
 
 inclined, 42 
 ,, three equal, in equilibrium, 
 
 83 
 
 triangle of, 82 
 two, at right angles, 83 
 Forging and hardening luri-lng 
 
 tools, 202 
 
 Friction, angle of, 106 
 ,, anti-, wheels, 109 
 ,, circular discs, velocity 
 
 ratio of, 130 
 
 ,, co-efficient of, 105, in 
 ,, definition of, 101 
 ,, heat developed by, 102 
 ,, inclined plane with, 110- 
 
 112 
 
 ,, inclined plane without, 93 
 .,, laws of, 103-109 
 ,, cone reversing gear, 344 
 Fulcrum, position of, in a lever, 30 
 ,, pressure on, and reaction 
 
 from, 25 
 
 Fuscc, the, 59-61 
 Fusibility, definition of, 300 
 
 G 
 
 GALILEO'S and Kater's pendulum 
 
 experiments, 269 
 Gas gauge, 222 ; gases, 224 
 Gauge limits, micrometer, &c., 357 
 
 ,, Stubs' wire, 358 
 Gauges, low pressure, &c., 221 
 Gear, back motion in lathe, 174- 
 
 192 
 ,, change wheels in lathe, 176- 
 
 195 
 ,, efficiency of combined levr, 
 
 screw and pulley, 160 
 ,, screw and pulley, 160 
 
4 2 4 
 
 INDEX 
 
 Gear, starting and stopping, 124 
 
 ,, worm -wheel lifting, 170 
 Gearing, backlash in wheel and 
 
 screw, 158 
 ,, belt, reversing motions, 
 
 122-124 
 
 ,, belt, shape of pulley faces 
 ,, for, 127 
 
 ,, pitch of teeth in wheel, 132 
 ,, principle of work applied 
 
 to wheel, 135 
 ,, velocity ratio of pulleys 
 
 in belt, 119-121 
 
 ,, velocity ratio in wheel, 123 
 ,, wheel, in jib-cranes, 144- 
 
 146 
 
 Grain, screw for moving, 148 
 Gramme, 41 1 
 
 Graphic representation of forces, 3 
 ,, ,, ,, velocities, 
 
 260 
 
 Gravity, acceleration due to, 260 
 ,, centre of, 28 
 ,, specific, definition of, 214 
 Gyration, radius of, 282 
 
 iH 
 
 HARDENING the tools for a lathe, 
 
 202 
 
 Head or pressure of a liquid, 205 
 Headstock, fast or fixed, of a lathe, 
 
 184, 185 
 ,, movable, for a common 
 
 lathe, 179, 180 
 
 ,, movable, for a screw- 
 cutting lathe, 181 
 Heart-shaped cam, 3^1 
 Heat developed by friction, 102 
 
 ,, relation between, and work, 104 
 Helix of a screw thread, 148 
 Henry, 413 
 Herbert's hexagon turret lathe, 188- 
 
 195 
 Hollow round shafts, strength of, 
 
 324 
 
 Homogeneous materials, 299 
 Hooke's coupling joint, 329 
 ,, double ,, 329 
 
 law, 303 
 Horse-power brake, transmitted by 
 
 belts, 118 
 
 definition of, 12, 412 
 . >, of working agent, 13 
 
 Huyghen's pendulum experiments, 
 269 
 
 Hydraulic accumulator, 253-255 
 bear, 251 
 jacks, 247-251 
 machines, 227-255 
 press, Bramah's, 241-245 
 press, large, 245-247 
 
 Hydraulics, 207-258 
 
 Hyperbola, 273 
 
 IDLE wheel (note on), 138 
 Immersion of solids in fluids, 217 
 Impenetrability of matter, 297 
 Improved Standard Measuring Ma- 
 chine, 363 
 Inclined forces, straight lever acted 
 
 on by, 42 
 
 ,, plane, the screw as an, 149 
 ,, planes, principle of work, 
 
 93-99, 270-272 
 applied to, 97, no, 112 
 Indicator, motion for Richard's, 344 
 Inertia definition, 290 
 Intermittent motion for cam, 332 
 Internal and external limit gauges, 
 
 355-357 
 
 Isotropic material, 299 
 Instructions for Board of Education 
 
 Exams., 368 
 ,, ,, City and Guilds 
 
 E<cam., 369 
 
 ,, ,, Institution of Civil 
 
 Engineers Stu- 
 
 JACK, bottle screw, 162-164 
 ,, traversing screw, 164 
 Jacks, hydraulic, 247-251 
 Jib cranes, stresses in, 85-92 
 ,, ,, wheelgearingin,i44-i46 
 Joint, Hooke's coupling, 329 
 
 ,, double, 329 
 Joints, single -riveted lap, 310 
 Joule's relation between heat and 
 work, 102, 413 
 
 K 
 
 KATER'S and Galileo's pendulum 
 experiments, 269 
 
INDEX 
 
 425 
 
 Kelvin's, Lord, wire-testing ma- 
 chine, 210 
 
 Kilowatt, 413 
 
 Kinetic and potential energy, defi- 
 nitions of, 280 
 
 Kinetic energy imparted to a falling 
 body, 271 
 
 Knuckle-joint, 46-49 
 
 LAP-JOINTS, single riveted, 310 
 Lathe, back motion gear of, 174- 
 
 185 
 
 bed of a, 183, 187 
 ,, change wheels in a, 176- 
 
 179, 183, 187 
 ,, cutting forces and H.P. for, 
 
 199-202 
 ,, forging and hardening the 
 
 tools, 202 
 
 ,, hexagon turret, 188-195 
 ,, ,, ,, feed motion, 
 
 192 
 ,, fixed headstock of a, 184, 
 
 185, 187 
 
 ,, leading screw of a, 175, 1 86 
 ,, mechanism in a screw-cut- 
 ting, 174-188 
 
 ,, motions of saddle and slide- 
 rest of a, 176 
 ,, movable headstock for a 
 
 179, 180, 181 
 ,, reversing plate for change of 
 
 wheels of a, 184, 187 
 ,, saddle of a, 176 
 Law, Archimedes', 217-219 
 ,, Hooke's, 303 
 ,, Pascal's, 208 
 Laws of friction, 103-109 
 
 ,, ,, motion, Newton's, 260 
 Leading screw of a lathe, 175, 186 
 Leather collar packing, Bramah's, 
 
 243 
 
 Lever, bell crank, 43 
 ,, bent, 42 
 ,, combined with screw and 
 
 pulley, 1 60 
 
 ,, definition of a, 22 
 ,, duplex bent, 43 
 ,, machine for testing tensile 
 
 strength of materials, 41 
 ,, position of fulcrum of a, 30 
 
 Lever, practical applications of the, 
 
 35-46 
 ,, pressure on, and reaction 
 
 from, the fulcrum of a, 25 
 ,, principle of moments ap- 
 plied to the, 22 
 principle of work applied to 
 
 the, 53 
 
 ,, safety valve, 38-40 
 ,, straight, acted on by in- 
 clined forces, 42 
 ,, when its weight is taken 
 
 into account, 29 
 Lifting gear, worm-wheel, 170 
 Limiting angle of resistance, 107 
 
 stress, 301 
 
 Limit and micrometer gauges, 355 
 Limits in calculations, 5 
 ,, of elasticity, 303 
 Linear motion, comparison of 
 dynamical formulae for 
 angular and, 272 
 ,, velocity, 259 
 ,, ,, formulae for, with 
 
 uniform accelera- 
 tion, 261 
 
 Liquid, definition of a, 207, 224 
 ,, immersion of solids in a, 217 
 , . pressure due to head of a, 209 
 ,, ,, on any immersed 
 
 surface, 209, 21 1 
 ,, transmission of pressure hv 
 
 a, 208 
 Loads, definition of live and dead, 
 
 300 
 
 ,, safe, and elasticity, 302 
 Lockfast lever and safety valve, 39 
 Logarithms Appendix D, 41 5 
 Low pressure gauges, 22 1 
 Lubrication, 109 
 Lumberer's tongs, 43 
 
 M 
 
 MACHINE, efficiency of a, 53 
 
 measuring, 363 
 ,, modulus of a, 53, 136 
 ,, testing, 40 
 Machinery, importance of balancing 
 
 high speed, 279 
 Malleability, definition of, 298 
 Mass, definition of, 289, 290 
 Materials, machine for testing ten- 
 sile strength of, 47 $2 
 
426 INDEX 
 
 Materials, properties of, 297-311 
 Matter, definition of, I 
 Measuring tools and gauges, 35 5-366 
 Mechanical advantage, 66, 68 
 Mechanics, definition of applied, I 
 Mercurial barometer, the, 221 
 Metals, melting points of, 300 
 Micrometer screw gauges, 357 
 Modulus of elasticity, 304-306, 321 
 ,, of rigidity of a material, 
 
 319 
 
 ,, of a machine, 53, 136 
 Moment of a force, 21 
 
 ,, of momentum, 290 
 Moments, principle of, 21 ; applied 
 to the lever, 22 ; ap- 
 plied to the wheel and 
 axle, 55 
 
 Momentum, definition of 260. 280 
 Uotion and velocity, 250 
 ,, equations of, 261 
 ,, Newton's laws of, 260 
 ,, of saddle and slide rest of a 
 
 lathe, 176 
 ,, on a curved, inclined, or 
 
 " banked " track, 287 
 Motions, reversing, 343 
 
 ,, by belt gearing, 122-124 
 
 N 
 
 NEWTON'S laws of motion, 260 
 Nippers, example of, 45 
 
 o 
 
 Ohm, 413 
 
 PACKING, Bramah's collar, 243 
 Pantograph, 339 
 Parabola, hyperbola, ellipse, 273 
 Parallel forces, centre of, 26 
 
 ,, ,, equilibriant and re- 
 
 sultant of, 25-28 
 ,, motion, 339 
 ,, ,, Watt's, 338 
 
 Parallelogram of forces, 82 
 Pascal's law, 208 
 Passive resistance, 101 
 Path of a projected body which 
 falls under the action of 
 gravity, 273 
 
 Pawl and ratchet wheel, 334 
 
 ,, reversible, 335 
 Pendulum experiments by Galileo, 
 
 Huyghens and Kater, 269 
 Percentage efficiency of a machine, 
 
 63, 67 
 Pincers, 44 
 Pinion, rack and, 132 
 Pitch circle, definition of, 131 
 ,, surface, definition of, 131 
 ,, of rivets, 310 
 ,, of teeth in wheel gearing, 132 
 Piano, inclined, with friction, no- 
 
 112 
 
 ,, without friction, 93, 270-272 
 ,, principle of work applied to 
 
 the inclined, 97 
 ?ka&aof belts and palleya z 
 
 gearing, 125 
 Inclined, 93-99 
 fianing machine, 343 
 Plates, hydraulic press for flanging 
 
 boiler, 245 
 
 Porosity of matter, 297 
 Potential energy, definition of, 280 
 Poundal, the, 2 
 Power, definition of, 12, 412 
 ,, horse, definition of, 12 
 ,, ,, transmitted by belts, 
 
 118 
 
 , , that steel shafting will trans- 
 mit at various speeds, 323 
 ., units of, 12, 412 
 Press, the fly, 283 
 
 ,, Bramah's hydraulic, 241-245 
 
 large, 245-247 
 ,, screw, for compressing bales, 
 
 165 
 
 Presses, packing for hydraulic, 243 
 Pressure, atmospheric, 220 
 ,, centre of, 5, 217 
 . , due to head of a liquid , 209 
 ,, low and vacuum vater 
 
 gauges, 221 
 ,, on fulcrum of a lever, 19, 
 
 2 3-25 
 on ram of a Bramah's 
 
 press, 244 
 
 ,, on sluice gate, 217 
 ,, on surface in liquid, 209 
 
 211 
 ,, measure of, &c., 5 
 
 transmited by liquids, 208 
 total or thrust, 5 
 
INDEX 
 
 427 
 
 Principle of moments, 2 1 ; applied 
 to the lever, 22, 35 ; to 
 the wheel and axle, 55 ; 
 to the wheel and com- 
 pound axle, 73 
 
 of work, 52 ; applied to 
 the leve'r, 53 ; to the 
 wheel and axle, 56 ; to 
 the ordinary block and 
 tackle, 67 ; to the wheel 
 and compound axle, 73 ; 
 to Weston's pulley block, 
 76; to the inclined 
 plane, 97; to wheel 
 gearing, 135 
 Pulley blocks and tackle, ordinary, 
 
 65 
 combined with lever and 
 
 screw, 1 68 
 ,, combined with worm-wheel 
 
 and winch barrel, 168 
 ,, faces for belts, shape of, 127 
 ,, Weston's differential, 74-77 
 Pulleys, 63-65 
 
 ,, arrangement of driving and 
 following, in different 
 planes, 125 
 ,, combinations of fast and 
 
 loose, 122-124 
 ,, tendency of belts to run on 
 
 highest parts of, 127 
 ,, velocity ratio of, in belt 
 
 gearing, 119-121 
 Pump, combined plunger and 
 
 bucket, 234 
 
 ,, common suction, 227-229 
 ,, double acting force, 234 
 ,, force, with air vessel, 231 
 , , plunger force, 229-23 1 
 ,, rods, tension in, 229 
 Pumps, centrifugal, 236 
 
 ,, continuous delivery force, 
 
 without air vessels, 232 
 Punching machine, portable, or hy- 
 draulic bear, 251 
 
 Q 
 
 QUADRANT or reversing plate for 
 
 change wheels, 183-187 
 Quantity (motion or momentum), 
 
 26x3 
 
 Quick return, cam, 334 
 common, 350 
 
 Quick return motion, Whitworth's, 
 
 347 , 
 
 ,, ,, reversing motion, 346 
 elliptic wheels, 351 
 
 R 
 
 RACK and pinion, 132 
 
 Radius of gyration, 282 
 
 Railway carriages, screw coupling 
 
 for, 156 
 ,, curves, motion on bicycle 
 
 and, 287 
 Ratchet, masked, 336 
 
 ,, wheel, pawl and, 334 
 Ratio, velocity, of change wheels in 
 
 a lathe, 176-179 
 ,, ,, definition of, 67 
 
 >t ,, of pulleys in belt 
 
 gearing, 119-121 
 
 ,, of two friction cir- 
 
 cular discs, 130 
 ,, t in wheel gearing, 
 
 133 
 Reaction, action and, 3 
 
 ,, from fulcrum of a lever, 
 
 2 5> 30 
 
 Relation between twisting moment, 
 diameter, and horse power 
 transmitted by shafting, 224 
 Repose, angle of, 106 
 Resilience, definitions, 307, 318 
 Resistance, electrical unit, 413 
 ,, limiting angle of, 107 
 ,, passive, or friction, 101 
 ,, work in overcoming a 
 
 uniform, 6-9 
 ,, work in overcoming a 
 
 variable, 8-1 1 
 
 Resolution and composition of 
 forces, 82-84 ; of velocities, 
 260 
 
 Resolution of a force into two 
 components at right angles, 
 84 
 Resultant and components, 4 
 
 ,, of parallel forces, 25-28 
 ,, two forces at angle and 
 any number at a point, 
 84 
 
 ,, pressure and thrust, 5 
 Reversible pawl, 335 
 
 ,, pendulum, 269 
 
428 INDEX 
 
 Reversing by friction cones and 
 
 bevel wheels, 344 
 ,, gear, bevel wheel and 
 
 clutch, 345 
 
 friction cone, 344 
 
 ,, quick return, 346 
 ,, Whitworth, 345 
 motions, 343 
 for belt gearing, 122-124 
 plate for change wheels, 
 183-187 
 Rigidity, definition of, 298 
 
 ,, modulus of, 319, 321 
 Rivets, pitch, 310 
 
 ,, shearing stress of, 310 
 Rods, tension in pump, 187 
 
 ,, torsion of, 318 
 Roller or ball bearings, no, 183 
 Roman balance, 35 
 Roof truss, stresses in a, 88, 89 
 Rotating body, accumulated work 
 
 in a, 281 
 Rotor, 50 
 
 Rounded screw threads, 155 
 Rules and syllabus of Exams, -for 
 admission of Students to the 
 Institution of Civil Engineers, 
 370 
 
 SADDLE and slide rest of a lathe, 
 
 176, 185 
 
 Safe loads and elasticity, 302 
 Safety, factors of, for materials, 302 
 
 ,, * valve, 38-40 
 Sawing machine, vertical, 337 
 Scalars, 50 
 Screw bench vice, 166, 167 
 
 ,, combined with lever and pul- 
 ley, 1 60 
 
 ,, -coupling for railway car- 
 riages, 156 
 ,, -cutting lathe, description of 
 
 a, 180-188 
 
 ,, -cutting lathe,self-acting, 182 
 ,, -cutting mechanism in a 
 
 lathe, 174-188 
 
 ,, endless and worm-wheel, 168 
 ,, -gauges, micrometer, 357 
 ,, gear, apparatus for demon- 
 strating efficiency of, 160 
 ,, gearing, backlash in, 158 
 ,, -jack, bottle, 162-164 
 
 Screw-jack, compound hydraulic 
 
 and, 249 
 
 ,, -jack, traversing, 164 
 ,, leading, of a lathe, 175, 186; 
 split nut for engaging, 185 
 ,, -press for compressing bales, 
 
 165 
 
 ,, or spiral, for grain, 148 
 ,, pressure or thrust, 5 
 ,, viewed as an inclined plane, 
 
 149 
 
 Screws, right and left-hand, 156 
 ,, single-, double-, and treble- 
 threaded, 157 
 ,, strength, durability, and 
 
 efficiency of, 151 
 Screw thread, 148 
 
 ,, buttress, 155 
 ., ,, rounded, 155 
 
 ,, forming a, on a cylin- 
 der, 149 
 
 ,, square, 154 
 
 Sere \v threads, characteristics of, 151 
 ,, different forms of, 
 
 151-156 
 
 ,, Seller's, 154 
 ,, ,, Whitworth, 151-154 
 
 Second of Time, 411 
 Shafts, strength of hollow, 324 
 
 ,, strength of solid round, 322 
 Shaping machine, 350 
 Shearing strength of rivets, 310 
 Shears or bed of a lathe, 183, 186 
 Ship's capstan, 57-59 
 Silent feed, 336 
 Single-riveted lap joints, 310 
 Siphon, the, 222 
 Slide rest of a lathe, 176, 185 
 Sliding angle, 107 
 Slotting machine, vertical, 341 
 
 ,, ,, Whitworth 's, 349 
 
 Sluice gate, pressure on, 217 
 Snatch block, 65 
 Solid shafts, strength of, 322 
 Solids, immersion, 217 ; defn., 224 
 Specific gravity, 214, 298 
 Speed cones, 124 
 Split nut for engaging leading screw 
 
 o a lathe, 185 
 Squared paper, 14-17 
 Standard measuring machines, 355- 
 
 366 
 
 Starting and stopping gear, 124 
 Starrett micrometer gauge, 357 
 Steel, high speed, 195 
 
INDEX 429 
 
 Steel, specific gravity test, 203 
 Steelyard, 35-38 
 Stepped speed cones, 124 
 Straight levers acted on by inclined 
 
 forces, 42 
 
 Strain, compressive stress and, 301 
 definition of, 300 
 ,, shearing stress and, 317 
 ,, tensile stress and, 301 
 Strength of materials, machine for 
 
 testing, 40-42 
 of materials, ultimate, 301 
 of solid and hollow round 
 
 shafts, 322-324 
 
 Stress, centrifugal, on fly- wheels, 2 79 
 ,, definition of, 300 
 ,, intensity of, 300 
 ,, limiting,or ultimate strength, 
 
 301 
 
 ,, shearing and strain, 317 
 ,, total, 300 
 Stresses in chains, 316 
 
 ,, in jib cranes, 85-89 
 in simple roof truss, 88, 89 
 , , tensile and compressive,3O I 
 Stubs' wire gauge, 358 
 Students' Exams., Institution of 
 Civil Engineers, 389-401, 407 
 Suction pump, common, 227-229 
 Sun and planet wheels, 330 
 Surface, definition of pitch, 131 
 immersed in liquid, 211 
 Swing radius, 282 
 
 TABLE of melting points of metals, 
 
 300 
 
 , , of moduli of elasticity, 305 
 ., of power that steel shafting 
 will transmit at various 
 speeds, 323 
 ,, of ultimate strengths of 
 
 materials, 302 
 Tackle, block and, 65-68 
 Tangentometer, 364 
 Tearing strength of plates, 310 
 Teeth, pitch of, in wheel gearing, 
 
 132 
 
 Tenacity, definition of, 298 
 Tensile strength of materials, 
 
 machine for testing, 40-42 
 ,, stress and strain, 501 
 Tension in driving[belts, 116 
 
 Tension in pump rods, 229 
 
 Test of steel, by specific gravity 
 
 method, 203 
 ,, specimen, work done per cubic 
 
 inch in fracturing a, 309 
 Testing machine, 40-42 
 Theoretical advantage, 66-68 
 Toggle joint, 46-49 
 Tongs, Lumberer's, 43 
 Torsion of rods and wires, 318-320 
 Torque or twisting moment, defini- 
 tion of, 322 
 Transmission of power by belting, 
 
 118; by liquids, 208 
 Traversing screw-jack, 164 
 Treadle lathe, self-acting screw- 
 cutting, 182 
 Triangle of forces, 82 
 Turning tools, forging and harden- 
 
 ing, 202 
 
 ,, tool holders, 194 
 ,, with high speed steel, 195 
 Turkus, carpenter's, 44 
 Turret-lathe, 188-195 
 Twisting moment, 322 
 
 u 
 
 ULTIMATE strength of materials 
 
 301 
 
 Uniform velocity, definition of, 259 
 Unit of acceleration, 260 
 
 force, 2, 412 
 
 ,, horse-power, 12, 412 
 
 ,, power, 12, 412 
 
 ,, velocity, 259, 411 
 
 ,, work, 6, 412 
 Units, C.G.S.,4ii, Practical, 413 
 Universal joint, Hooke's, 329 
 Useful constants Appendix D, 414 
 
 ,, data regarding fresh and 
 salt water, 214 
 
 ,, work in a machine, 52 
 Uses of squared paper, 14-17 
 
 VACUUM water gauges, 221 ^ 
 
 Variable resistance, work done 
 
 against a, 8 
 
 ,, velocity, definition of ,24 59 
 Valve, lever safety, 38-40 
 V-screw-thread, Seller's, 154 
 
 ,, ,, "Whitworth's, 151 
 
430 
 
 Vector's, $a 
 
 Velocities, composition and resolu- 
 tion of, 260 
 ,, graphic representation 
 
 of, 260 
 
 Velocity, and motion, 259 
 ,, angular, 260 
 ,, attained by a body sliding 
 down any smooth in- 
 clined plane, 270 
 ,, definition of, 260,411 
 ,, linear, with uniform ac- 
 celeration, 260 
 ,, linear, 259 
 ,, uniform and variable, 259 
 ,, unit of, 259, 411 
 ,, ratio of change wheels in 
 
 a lathe, 176-179 
 ,, definition of, 67 
 , , pulleys in belt-gear- 
 ing, 119-121 
 two friction circular 
 
 discs, 130 
 
 ,, ,, in wheel-gearing, 
 
 Vessel, action of an air, 232 
 
 ,, ,, force pump with, 
 
 Viscous fluid, 224 
 Volt, 41 3 
 
 w 
 
 WATER gauges, low pressure and 
 
 vacuum, 221 
 ,, useful data regarding fresh 
 
 and salt, 214 
 Watt, 413 
 
 Watt's parallel motion, 338 
 Weems' compound screw and hy- 
 draulic jack, 249-251 
 Weston's differential pulley block, 
 
 74-77 
 Wheel and axle, 55-57 
 
 ,, ,, compound axle, 72-74 
 Wheel gearing, backlash in, 158 
 ,, ,, in jib cranes, 1 44- 1 46 
 
 ,, ,, pitch of teeth in, 132 
 
 ,, ,, principle of work 
 
 applied to. 
 
 INDEX 
 
 Wheel gsiring, velocity ratio in, 133 
 Wheels, anti-friction, 109 
 
 ,, change, in a lathe, 176-179, 
 183, 184, 187 ; quadrant 
 or reversing plate for, 
 183, 187 
 
 ,, fly-, centrifugal stress in, 279 
 ,, pawl and ratchet, 334 
 Wheels, sun and planet, 330 
 Whitworth's millionth measuring 
 
 machine, 358 
 quick return motion, 
 
 347 
 
 ,, reversing gear, 345 
 
 ,, slotting machine, 
 
 349 
 ,, V-screw-threads, 151- 
 
 154 
 Winch barrel, 57 
 
 ,, double purchase, 143-144 
 
 ,, single purchase, 140-142 
 
 Wincli drum combined with pulley, 
 
 worm, and worm wheel, 168 
 Windlass, Chinese, 72 
 Wire gauge, Stubs', 358 
 ,, -testing machine, Lord Kel- 
 vin's, 210 
 
 Wires, torsion of, 319-321 
 Work, accumulated, 280-282 
 ,, definition of, 6, 4 2 
 ,, diagrams of, 9 
 ,, done against variable resist- 
 ances, 8, 17 
 
 ,, done in extending or com- 
 pressing a bar within the 
 elastic limit, 307 
 ,, done on inclines, 110-112 
 ,, done per cubic inch in frac- 
 turing a test-bar, 309 
 ,, principle of, 52 
 ,, relation between, and heat, 
 
 102 
 
 ,, transmitted by belts, 118 
 ,, unit of, 6, 412 
 Workshop measuring machine, 363 
 Worm-wheel, screw and worm, 168 
 
 ,, lifting ffear, 170 
 Worssam's silent feed, 338 
 
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 WILL INCREASE TO 5O CENTS ON THE FOURTH 
 DAY AND TO $1.OO ON THE SEVENTH DAY 
 OVERDUE. 
 
 FED 4 1944 
 
 LD 21-100rn-7,'40 (6936s) 
 
VB I 1 077 
 
 204114