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UNIVERSITY OF CALIFORNIA.

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Class

ELEMENTARY MANUAL

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.

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FELLOW OF THE ROYAL SOCIETY, EDINBURGH; MEMBER OF THE INSTITUTION OF

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NINTH EDITION, THOROUGHLY REVISED

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1910
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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 workf
and headed "Re-written."

REMARKS. — Students are strongly reromme? r e«l 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-
tage—Velocity 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-Wheel—Com-
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-
Power—Examples 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. *.

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 0 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 P1P,P3P4 be any four forces acting in one
direction along one straight line, then their resultant —

R = P! + P, + P3 + P4

(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 Pfff4 be any four forces acting along
one straight line to the right hand or in a positive direction ; and
QiQaQ« ^e any three forces acting along the same straight line,
but in an opposite or left-hand or negative direction, and let

Qi + Q, + Q3 be less than pi + p» + pi + P4

Then the resultant,

R = (P, + P, + P, * P4) - (Q, + Q, + Q8)
and acts in the same direction as PtP2P3P4, 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, + P4) - (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 (PR) is the mean pressure per unit area (p) multi-
plied by the total area (A) under consi Oration. Here, PR=^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 0 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 at 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 io7 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

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.

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

\1
xx

— — ->

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 +pl 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 lo7 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

3rd. .-. 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

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

3'5

Weight on

chain lifted V

0-8

i '4

2'0

2'5

3'S

4 '3

.TO

5 '6

6'2

7-0

in Ibs. J

Corrected ^
values in >
Ibs. J

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 0 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.

100

125

150

180

210

530

525

516

490

425

300

210

160

110

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—

500

•SK

si^ij

i:::::i

F :

- 4-

400

| ::

rJI

rTrrHrr

— 5p

•*-

:l:::::

^ *vj

300

. t : ...:

Tt-f1

^p-::::::::

?00

: jK§:i;:::

"«*

:|:::::

- ---[-

- ^* s

. ,

100

::-:(-

— H — -^

F:|S::|

""] '

= B

•

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 0 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 0 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 0 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.

•I

•5

•6

•8

IX)

Pressure in \
Ibs. per
square inch. J

18-5

I3-5

4'5

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 0
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.

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.,

(oz.)

P x BF

4 (oz.) x 1 6"

BF : AF

BF 16 4

AF 76

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

i6v

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, actingxm 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.

8fl •
j

or,

FB;

or,

E~~

FB
AB

1 6";

or,

16 _

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 E3
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, RI} 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.

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

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. 31

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—

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 Ar Then, for equilibrium we must have

PxFA1==(W+i)xFB . (2)

Subtracting corresponding members of equations (i) and (2)
we get P(FAt - 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,

Or, AAn = n x A At 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

PxAA1 = FB,

_FB
Or, AAt — 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

PxAAM = 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 -Jp — 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 0 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.

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.

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 = -d2.
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.

R BWa

DIAGRAM OF THE LEVERS.

INDEX TO PARTS.

LjL, represent Levers.
F,F2 „ Fulcra.

P „ Pull, or dead
weights.

Where P acts on L,.
Specimen under te.->t.

BpA,, represents Where R acts on

L,,L2.

B., „ Where W acts on S.
M\V „ Movable weight.
BW Balance weight.

HW „ Hand-wheel and
screw for elevat-
ing Flf &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 Av

P x A^\ = E x B^j, and R x A2F2 = W x B3F2
PxA,F, Wx

*i*\ ~ A

P x A1F1 x A2F8
-L

PxA,F, WxB,F2

. . _
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.

Xs
/ \

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
A8F 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.

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.

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,

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 Ibs4 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

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

P7=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, Ht. 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 9J"

,'. 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."

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

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 C1 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.''

of the " Toggles," until the upward pressure is balanced by the
resistance of the " Toggles."

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 Sot 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 B1?

and a force, P, acting vertically upwards at the end Alt 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 At to A^.

The work got out at Bx = W x the vertical distance from Bx 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 A4FA, and BjFBa are similar in every
respect.

LECTURE V.

Therefore,

Hence,
Or,

W '
Af =

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.

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.

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.

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.

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,

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 = Hi6*"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

30 =3300

.*. Mechanical efficiency = lisefnl work . Or ^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

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

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,

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 = 5P; 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 =wei8ht 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.

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.

*,- --

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.

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 shoulderrand 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

* If — iff raised the circumference of the larger circle on one side,

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=£ (2Trrl-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" ; rv= 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 = rr

Hence, P x R = — (R _ r)

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.

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 wrill 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., Wy 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.

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
0 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.

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 0 = §

. E . Sin 0 = 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. :

R2 = P2 + Q2 + 2P 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—

R2 = P2 + P2 + 2P2 Cos a = 2P2 + 2P2 Cos a

or, R2 = 2P2 (i + Cos a)

/. R = 2P Cos 2.

= 4P2 Cos -
(Since Cos a = 2 Cos2--!)

Resultant of any Number of Forces Acting at a Point.— Let
3j, P,2, Pa, &c., be any number of forces acting at a point ; then, by the
parallelogram of forces find a resultant, R,,
for P, and Pa ; and a resultant, Ey for R, and
Ps ; 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.

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 adt 4 tons.

LECTURE VIII.

Force
Diagram

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 S€sveral 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

pressure at B and at 0 of — . It also produces at these points

reactions Kx and K2, 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.

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

— , 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

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

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 1—50 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. & GM 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,

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,

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

h R b , P h

= 7' w=7; andn =; 6

- Sin a ; 5 = Cos a

; and — = Tan a

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

LECTURE IX.

Consequently,

W : P ! R

Or, |

_ *.

= - ; and

R ~

Or,

<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

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. dP_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

?NCLUINED 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 —

— 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 XP" = 7XI=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, lS85> 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

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

2p 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).

and ^ is the co-efficient of friction.

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.

Jbrlcate£
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 jtt

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

Fulcrum.
Small roller.
Scale-pan.
Test-piece.
Cylinder.

H represents Handle.

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
FW2. This enables it to turn them with
great ease at a slow rate in the very small
bearings Bj and B2.
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= ^ AC2 + BC2= V io2+i2= 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 ••• A8-«*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.

0% 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 . . rl= Radius of the axle.

•p _ /Force required to overcome the friction of the axle
t \ when acting at the circumference of the pulley.

„ . . r2 = Radius of the pulley to centre of belt.
Then, Ftxr1= F2xr2.

[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=T«j.

Then taking moments about the centre of the axle, we have —

VVl x r2 + F2 x r2 = W2 x ra
Or, . . T.xrf + Ffxr, = Td x ra

Dividing both sides of the equation by r, we get

T.+F,=Td

.-. F2 = Td-T,

Or, expressed in words, the force F2, acting at the circumference
of the pulley (which is required to overcome the friction of the
axle) is equal to the tension Td 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 Td 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 (Td - Ta) 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
(Td + 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 = (Td — T4) 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 = (Td - 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

7T«P

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*
„ F1? F, = Diameters of the followers.
„ NDi, N"Dj = Number of revolutions per minute of the drivers.
„ Np , Np = 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 Df = circumferential speed, of F,

(Divide both sides by TT)

D,NDj = F,N,,

<* ..... f = ?' ' • • • (2)

J a -^ »3

(But the revolutions of F, and of D2 are the same) .'. Nri = ND
Or, . . . D, N, - F,N, .-. NF> - ?i/?'

X«

_F,NF

Consequently . . ^ -^

• -rr1

/ (Dividing both numerator) ND = _D!_ = Ft x F,
I and denominator by NFi)J ]^pi D2 = DX x D,

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, . NDi x Dj x D, = NFj x Ft 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, Dp is connected by a belt to a follower, Fp
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,
Dt x NDi = Ft x N?i

XT D, x ND 6' x 100
.•. -NPi „ -t_— — * = » 200r.p.m.

*! 3

EXAMPLE IT. — Referring to the previous figure, suppose that
a driving pulley Da (4' diameter), is geared to a follower, Ft
(2' in diameter), and that a second driver D, (4/ diameter), fixed
to the same shaft as Fp 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,
NDi x Dt x D, = Np> x Ft x Ff

** =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 El 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.

HBt,HB2 represent Hangingbrackets
for supporting
shaft, &c.

SCj,SC2 „ Speed cones.
FP „ Fast pulley.
LP „ Loose pulley.

B,,B2 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, Bp from the loose pulley, LP, to the fixed
one, FP ; thus setting the speed cones, SGp SC2, 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, Br 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. Td - 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 Dx = Diameter of the driving disc.
„ Fx = Diameter of the following disc.
„ ND = Number of revolutions per minute of Dr
„ N, = Number of revolutions per minute of Fr

Then, The peripheral velocity of Dl = Peripheral velocity of Ft
i.e ..... irD^-irFjlSr^

Or, .... D1ND1 = F1NF1
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,

The angular velocity of circle A _ 6 r, _ r3
The angular velocity of circle B <£ EC rl

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=2vrnt

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 DI} D,, D, = Diameters of the drivers.
„ Fp Ff, F, = Diameters of the followers.
„ ND , NPt = 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 D3

Or, .

:The speed of the last follower
x the successive diameters
of the followers.

NF3 x FL x F2 x F3

Fs/ T^ v 1R^
i X -C „ X J? -

x D2 X D3

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-

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 rD , rD3, rD = Radii of the respective drivers.
» CD[> Cv CD^ = Circumferences ,,

„ Ti1, nD , n3 = Number of teeth in

3 = Radii of the respective roilowers.

/, CF2, CF3 = Circumferences
S w_, , np* = Number of teeth in

Then,
Or,

x r

J),

, XT*
1 Dl

L x CD x CD x CD = NF x Cr x CF x

123 312

PRINCIPLE OF WORK APPLIED TO WIIEEL-GEAKING. 135

Or, ND x nv x WD 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 —

ND x TIDI x nD^ x n»3 = NF;jX n¥^ x np^nr^
N nFl x np x nF

Substituting the coire-^

spending numerical I ND _ 40 x 80 x 120 4x4x4 64
values for the letters, j ^ ~~ i0 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,, Ds = Diameters of the driving wheels.
„ F,, F8, F,= Diameters of the following wheels.
„ ND = Number of revolutions of the first driver, "Dv or of

the handles, H.
„ NP = Number of revolutions in the same time of the last

follower, F3, 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 ND = W x 2?rr x Np
(Divide both sides of the equation by 2ir)
.% PxRxND =WxrxNP
~ PxR NF P8 NPsxr

= 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,
NF D, x D2 x P,
ND'-^XF.XF,
P x R D! x D2 x P8
"W^Tr=F1xFaxF,

Or, P x R x F! x F2 x F3 = W x r x Dx x D2 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"; D8=2o"; D3=2o"; ^ = 36"; F, = So", and F3= 100".

By the above formula and by substituting the corresponding
numerical values \ve have —

P x R x Ft x F8 x F3 = W x r x DL 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 £0 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"; nD= 12 teeth; n, = 6o

142

LECTUKE

teeth ; WT = the theoretical weight that would be raised if there
were no friction j WA = 600 Ibs. (the actual weight raised) ; h = 20
feet.

( i ) Theoretical advantage = -pr

By the principle of work (neglecting friction.)
P x by its distance* = WT x its distance.*

P X 27rR X ny = WT X 27JT X 71D

P x K, x n¥ = WT x r

(Substituting the above numerical values we get)

ft x

_, WT 800 20

Consequently, . . -^~ = -- = -T—
j- 40 *

WA 600 Ibs. 15

(2) Working advantage =^p~= 40 lbs. =Y

(3) Work put in for every foot WA is raised. From equation
(i) we see that for every foot WA 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 WA is raised

= WAx i' = 6oo Ibs. x i' = 600 ft.-lbs.

. _ . Work got out 600 ft.-lbs.

(<;) The efficiency *= ^TF — r2 . • =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
WTx 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" ; nDL = 1 2 ; nDt =
12; 7iFi = 24; nF> = 48; r=s".

By the formula deduced in the previous lecture from the
principle of work (neglecting friction),

P x R x nf x np = WT x r x nD xnD

324
After cancelling, we get —

8ox3X 2 x 4 = WT= 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. : WA.

Weight actually raised = WA = 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 ~YV2 Ibs. will have a corresponding reaction R2 Ibs.,
and will require a part P2 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 Fr
Imagine the work done to be transferred to the inclined plane,
AB, then any portion of the load having a weight Wt Ibs. will
have a corresponding reaction Rx Ibs., and will require a part
Px 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 Fr Now, these forces act in
identically the same way as the second case of the inclined plane,
which was discussed in Lecture IX., consequently —

* 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

,-g

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.

O'lOO

0'600

2*375

0*125

1 1

0-625

2-500

0*150

0*650

2*625

0*175

0-675

3£

2*750

O'2OO

0-700

2-875

0-225

0-750

3'OOQ

0-250

0-800

3^5

0-275

0-875

3-50

0-300

0-900

375

0-325

I'OOO

4*00

0-350

Is1

1*125

4-25 ,

0-375

1-250

2§

4-50

0-400

1-375

475

0*425

1-500

5-00

0-4^0

1*625

0*475

1*750

5*50

0-500

1*875

575

0-525

4£

2-000

6'oo

0-550

2§

2*I25

0-575

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.

No. of threads per inch

WHITWORTH'S STANDARD NUTS AND- BOLTS. 153

WHITWORTH SCREW NUT.

WHITWORTH'S STANDARD NUTS AXD BOLT-HEADS.

•3

3
CQ

•§ *

"S ~

~Sn

£3

•S £

ts5

— "**

-S S

•£

•g

•^

l»

0-338

0-1093

0-0929

1*8605

0^43

0-942

0-448

l'*

0*1640

0*1341

2*0483

1-0937

•o67

0-525

0-2I87

0*1859

2*2146

I*203I

•1615

0*6014

0-2734

0-2413

i-l

2-4134

I-3I25

•2865

0-7094

0-3281

0*2949

2-5763

I*42l8

•3688

0*8204

0-3828

0*346

27578

1*5312

•4938

0-9191

0-4375

0*3932

3-0183

1*6406

•5904

I 'OH

I9?

0-4921

Q-4557

3'I49I

175

7154

•101

0-5468

0*5085

3-337

i -8593

•8404

•2O I

0-6015

0-571

3'546

1-9687

1*9298

•3012

0-6562

0*6219

375

2*0781

2*0548

'39

0-7109

0*6844

3-894

2-1875

2*1798

•4788

0*7656

07327

4-049

2-2968

2*3048

'5745

0*8203

07952

4*181

2-4062

2-384

I -6701

0-875

0*8399

4-53I

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

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 :L3
L 100x25-45 = 63.6,,

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 0 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=20X2X272Xl8x4~9°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

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 DL (which is keyed to the cone) may
turn the follower Fj ; and the driver Ds (which is keyed to the
same spindle as Ft) 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\\

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 F2 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 pe = Pitch of screw to be cut in inches, or fraction of

inch, between two threads.

„ pa = Pitch of guiding screw »».»». »

„ dr 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 dv 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 dl 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 tc = 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,

*0 PC

Or, . . t0 x ^ x da = t0 x /, x ft

If the train is a simple one, then

i-=/;or,^ x d-iff 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 te = 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.

tg d ' 4 40 4

By using the previous formula, we have pc = J" and pg = J"
... & *s 0r,|=^.../=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-IOX I0-100-/
tg 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 dv 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.

_EH_

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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.

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.

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DESCRIPTION OF A SCREW-CUTTING ..LATHE.

I8S

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, F2, 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".

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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.

\ P6 ,, Pinions.

\-F6 „ Followers or wheels.

IS ,, Intermediate gear stud.

Lj-Ko „ Keys fixed in shafts.

JSj, CS2 „ Friction clutch sleeves.

lv 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.

Cf „ Collar or ruff on shaf tFBS

SS „ Sliding shaft.

^2, L3 „ Levers for working SS.

INDEX TO PARTS.

PP represents Pulley for working force

pump.
LS .. Lathe spindle.

RCj, RC2 ,, Friction ring carriers.

Glf G2 „ Gluts.

TO „ Toggle carrier.

T „ Toggle.

S15 S2 „ Studs.

FRj, FRo „ Friction rings.

S3, S4 „ Shoes for friction r
and toggle carrier.

Lj „ Lever for actuating R

and RC2

BS „ Back gear shaft.

B!, B2 „ Bearings for LS.

B3, B4 „ Bearings for BS.

LECTURE XVI.

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HEXAGON TUKRET LATHE.

LECTURE XVI

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OF THE

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HEXAGON TURRET LATHE.

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SUPPORT
FACE

CUTTER SL/DE G/B SCff£HS&
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 Fv P,,
F2 with P3, the first shaft FBS. As will be seen from the sec-
tional plan the shaft FBS carries two spur pinions P3 and P5,
either of which can be put in or out of gear with follower wheels
F3 and F5 by means of the friction clutches CS^ CS2. The levers
LS, L3 with central spindle SS actuate the friction clutches
CS1? CS2. Hence, for every speed of the motor this shaft FBS
can give two different speeds to the sleeve or followers F3 and F4
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 RC2, which slides along the key K, fixed in the lathe
spindle LS ; or, through double gearing P4, F4, BS, P6, F6 and
friction ring carrier RCj connected to lathe spindle LS by key
K2, the change from the one to the other is made by means
of the lever Lr The friction clutches on the shaft FBS are of
the expanding ring type operated by means of wedge surfaces P
and rollers R1? R2, whilst the clutches RC1? R03 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 Lr

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 nv»ans

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.

FIGt 6. — TURNING TOOLS— VIEW SHOWING RIGHT AND LEFT-HAND

CUTTEES.

feed or traverse shaft FS through the spur wheels SP8 and worm
W, worm wheel WW, and wheels SPp SW1? RPX 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 SS2. 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 P2, 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^
SW6. 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 SW4, SW5 or SW6 to the shaft
SR. The handle or reversing lever RL puts the claw clutch RC
in gear with bevel pinions BPX 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

Ax3

» 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,

208

LECTURE XVII.

a v2ss

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,
Y15 Y2, V3, Y4, 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 Yx, 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 Yp their
springs would have registered pressures corresponding with the
ratio of their areas to the area of valve Yr Or the pressure per
square inch on valve Yx 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 = hawt 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 04 Cr HB is a circular hydrostatic bellows,

3' diameter, with india-rubber
sides. WD2 is the bottom wooden
disc attached by bolts to an iron
tripod T, which is connected at its
centre to the clip C2 ; while WDX 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
WDt 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 Ct 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 WD3, 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 WB2. 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. WDa 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 = Ht = 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 H2 = 4'.
By the formula,

P3 = H2AW.

.'. P2 = 4X 12 x 62.5 = 3000 Ibs.

Consequently, the reduction in pressure is the difference between
these pressures.

Or (Pt - P2) - 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

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., G2 ; second, of the A
HFC, viz., G3 ; third, by joining these two points with a line G»G3,

2l6

LECTUEE XVIII.

and taking a distance along it from G2 towards G3 inversely pro-
portional to the areas of the n DEFH and the A HFC ; this
would give a point G6 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.

1 i

SHOWING POSITIONS OF THE CENTEBS OF GRAVITY.

G, represents centre of gravity of area ABCD
G2 „ „ „ DEFH

G3 „ „ „ HFC

G „ „ „ ENMF

Let Hj, H2, &c., represent depths of G1? G2, &c.
Then H! = JDC = 6' ; H2 = JEF = 3'.

G3 is J of HC below the line HF (see Lecture III., re position of
e.g. of certain areas).

G5 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 = H1A1W= 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.

2I7

H3A3W = 8'x(-, x8')x62-5 = 12000 Ibs.

DEFC = DEFH + HFC - 9000 + 1 2000 = 21000 „
H4A4W = 3'x(6' x 8') x 62-5 = 9000 „
H5A5W = 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 IDS-
Front side = H2A3W = 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 Glf 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.

2I9

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 AjBx 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 m«tacentres, fee., to our Advanoad

220 LECTURE XVIII.

Let Vj = volume represented by A^CD.
» V2 = » „ . „ A2B2CD.

„ 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, V2 = "V\ + v.
Or» V.-V^v.

N ow Va - Vx = volume represented by A^BjBjA,,

= cross sectional area of tank x xt
= 42xa;=i6aj cub. ft.

••• i6x = v or x= ^- ft.

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 wer
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.

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.

= — > 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.

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.

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

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) Px = HXAW =20'

x — x 62*5 = 86-8 Ibs. pull.
144

(2) P8 = 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 DVX, 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, SV2 represent Suction valves.
DV, DV2 „ 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

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.

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 whicv
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 = Weight of w^r ln,lbs' x dlstance m ft- throuSV
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, ijf J Total work done in driving the centrifugal pump.
45 ) 27I2'5 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 27i2'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'96 * 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

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.

242

The guide pillars are securely bolted to the base B by nuts and
iron wash«rs, 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.

„ Pressure on L at A.

, Eelief valve.

„ Fulcrum of L.

, Delivery pipe.

„ L's connection with

, Earn cylinder.

plunger's guide-

, Earn or plunger.

rod.

, Cup leather packing.

,, Pump plunger.

, Top, table, or T piece.

,, Keaction or stress

, Weight lifted, or

on plunger PP.

total pressure on E.

,, Pump barrel.

, Cross girder.

,, Inlet pipe.

, Guide pillars.

,, Suction valve.

, 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.
Wx7rra = Qx7rRa

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 7rRa

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_PxAFxareaof 1 sq. in._336x 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 Tr 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 Tr 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 Tlf 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 Tx 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, 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.

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» caj»e 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 Xi2

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~ "E5

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.

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.

PP
WR
IV
DV
BV

Crank.

Pump plunger.
Water reservoir.
Inlet valve.
Delivery valve.
Belief valve.

HC represents Hydraulic cylinder.
CL Cup leather.

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 0 = 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 tf—2gh

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.

I5l

47i

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 „ „ =3—1=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.

.•. v3 = — = — s^»= 10 "5" = velocity acquired in 3 seconds.
ts 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 „ =i5f „ =*^= £ x 9,

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 (3—1

= 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 = (vx- ve) = (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 = v2/a2,

But from Case V. h = £ a £2 ................... •. £2 — 2A/a.

Hence, 2A/a = t>2/a2.

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 I2,

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 = ^.

,, 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 fo 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 = ax -- ^—

(W+X + Wo)

and in Exp. (2)g - az -- — —
Hence in Exp. (i)-^p1 = (TT+ x +^) and in Exp. (z)-~ = (W + x + wj

Or in Exp. (i) a; = -^-1-- (W + wj and in Exp. (2) x = -^p = (W + wa)

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 : ^gt2

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

gt2

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 0 = 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-

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 0 x BA.

But, ||= sin 0, or -?5= BA.

•.^ = 20sin0 x j^-e = 2<?BC = 2gh.

That is, r2 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 _va-v1 _T.M.

• — T»r/ - ™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.
....ilw5

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 F2 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,.
ya=4ax may be proved as follow :—

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 PBOJECT£D
HORIZONTALLY ASD FALLING FREELY.

CENTRIFUGAL FORCE. 2/5

By Euclid I. -4 7.

FP2 = FM2 + MP2

Or, MP2 = FP2 - FM2
That is, MP2 = P]ST2 - FM2

Or, MP2 - (PT + TN)2 - (VM - VF)2

Tluit i-, MP2 = 2TX x 2PT (But MP = y:

\ Al-o PT = x
.-. T,2 = 4aa; 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 -

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 V2
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.
29

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 vl = the velocity in feet per second, at the nor-
mal speed n^ revolutions per minute,

And v2 = the velocity at the increased speed na 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 _
2g

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.

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.;

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 ag«ed 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<- 0 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 0 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 0 = = =

f _ Li ^rj —v-, from which we can quite easily obtain the required

angle of the inclination 0 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

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 0 = AB tan 6, since 6 is always very small.

<y2

h = G— inches.

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 Wv2 . 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

= = ; oraboufc5'

(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.

„ = (2i/32)x 50 = 3-906 lbs.-sec.

Average Force (F) = ma = — ( ^y^1 ) = ~f ( 'o-ooi ) = 39°6 lbS'

The negative signs (before va 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, „ „ » = 5'27 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*. 0 = 30 miles per hour = 3° x S28° feet Por

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"! __ (mv9 - mvA _ 462,000 _ n

nf t.h A TvmmAnt.iim f ~ *~1 ~ TTTTT ~ J54° lbs-

of the momentum J y $ I 5 x 60

.-. Total mean pull between the) ,,

engine and carriages j = 15°° + '54° = 3°4° 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.

29I

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 :

130

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 =370,300

•£— = 799. 36° ~ 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.

294

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? 'VS. 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

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 :

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. of 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 Mp M, = Masses of two bodies
Let Vj, V, = Volumes of two bodies.
Let Dj, Ds = 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
uhose 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- P61" SQ- ^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 lengtlf = ^ - -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.

Ultimate Strength.

Working Stress.

Tons per sq. inch.

Materials.

Ten-

Com-

Shear-

Ten-

Com-

Shear-

sion.

pression.

ing.

sion.

pression.

ing.

Cast iron .

7-5

i-5

Wrought-iron bars

3-5

Steel bars .

3°

Copper bolts

Brass sheet . .

— —

—

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.

Then, so> long as Y- does not exceed the elastic limit, I varies directly

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 —

. 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 Par^ °^ ^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, fj£°o™o = 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.)

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'

= •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\ aAL I a AL
. • . work done or res^ence R= -~ =- — = ~ P* -»

„ pa volume ofbar. rt, 1Via
>. i. „ R=EX - 2 - mch-lbs.

Work done per unit^ _ p1
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, EAZ2
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.

^(2-02 x -51) -(i '49 x -39) IOQ

2'O2X '51

(d) Work done^

fnch in0 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 /

_ 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.

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 io7
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 A1? A3
represent their sectional areas, and Elf E2 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, PE2

and

A! Ej + A2 Ba Ax Ej + A2 E2

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. fc = 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.

Ans. (b) Load = 3-0^ tons.

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 0 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

6500

2500

10% aluminium) .

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 .

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) . .

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

=!LDa. where D is the diameter of the shaft. The mean arm or leverage at

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, Tr T3 — 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 :

TI:T, -D^iD/

But Tj - P! x R/' = 800 Ibs x i a"
AndT,= P8 x R8" - P, x 10"

... P^: P2R, :: J>*:J>f
i.e., PaRa 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^( ^-D2 x - \ =fg "- ' W

But S is a constant quantity for any particular material. Also, TT and 16
are constants. .•. P x R vary as D3.

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 D8.

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

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S 8

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.

D4-d4
Then, their strengths are as D8 : - jj- ,

and their weights are as D« : Da - &.

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 = dsfs = -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'TP- = 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 ft 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 D1=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 Dl = 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~>
D4-d4 H.P.

326 LECTURE XXIII.

Jfow, substituting the values given by the question, we get—

Or, iDa*=

».«. Pa8= 3x2158-5 =6475-5.

. • . Da= #6475-5 =18-64 inches.
If bending is taken into account, then

D2= 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 D2 = diameter of one of the twin screw shafts.
,, N! = number of revs, per minute of single shaft.
„ N2 = „ „ „ 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/

Da8 5-1 x 12 x 33,000 x H.P.

in"

i.e., Dx = 1-442 D2. . •. Da= ~ = 7"

But, the weights of shafts are proportional to their cross sectional areas
if their lengths are equal.

| Dja (1-442 D2)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~?1«tiDg 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.Pa = 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, F1? F,,
which embraces two arms of the
crosspiece, O. The four arms of
this cross are of equal length. As
Cj rotates, Fx and F2 describe
circles in planes perpendicular to
their respective axes. Since these
planes are inclined to each other
the angular velocity of C2 at any

HOOKE'S JOINT. instant is different from that of C1,

but the mean angular velocities are

equal to one another, because at one instant C2 goes faster than Ct,
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,
C2, 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 ccp
its own weight. Find the work
done in 100 revolutions of the
olate.

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 = 338»3 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 0 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.

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 Ta, 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 Tr c tv these two levers are usually
parallel to each other, and perpendicular to the line Pj P, Pr
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 T1
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 341

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-
| IF3^ 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 W2 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 W3 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 St (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 P2, drives the rack on the planing table through a spur wheel
F2 and pinion P fixed on an intermediate shaft ; the other, D Pp
which is connected with F P, transmits its motion through another
pair of wheels Fx and Dr 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 P2. 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 ; 0 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! ; D2, F2 ; D3, F3. 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-
at«ed 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 fin«1 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 TV*. This gives the ex-
ternal measuring holder a range from Ty to z\\ by TV of an inch, or forty
different sizes ; and the internal measuring holder a range from 2%" to 4",
also by rVths of an inch, or twenty-four different sizes. In addition to this,
forty end sizes from rV" 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 C1 and C2 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, N2 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 Nx and N2 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 QWlt has been fitted to the outer turned end of the screw spindle

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 GWa 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 Nlf since *Vx*iv=7<fo*th of an inch.

WHIT WORTH MILLIONTH MEASURING MACHIN7E 363

Now look at the other head-stock where the screw which actuates the
nose N2 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 GW2, 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 GW2 be so carefully turned that its pointer shows a movement
of one division, the inner face of the nose N2 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 N2. When the adjustment of Nj and N2 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 N2 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 N2 and the pressure thereon noted), you place the new bar be-
tween Nj and N2. 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 = 0 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, Chelmsfordr
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 /

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-'-*

'--4'-* A

->
(

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.

576

1-5

660

2'0

733

2'5

827

3'27

915

4-01

1,011

5*o

737

6-0

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
Of 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, Glt G2 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 :

0-167

0*078

0-352

0-146

The displacement Ol G2 took place in 1/50 second. Find the x and y
components of the mean velocity of G for this interval.

Plot the points Glt 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 :

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.

Peripheral
Speed
of Vane.
Feet per
Sec.

Ratio.

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

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. Ar 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

34»ooo

40,000

48,000

56,000

60,000

SS.ooo

Exten-^

sionin f
inches. J

'002

•0044

•0070

'0103

'Oi6

•Tgc

P470

r3fi

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 mch 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.

0 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 s1 and time ^ by a force of P1 tons up to

398 APPENDIX B.

velocity v feet per second, and then brought to rest by P2 tons acting'f or
Jt2 seconds through *2 feet, prove the formulas —

»3?-™-
oo T?

(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 :—

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.

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.

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.)

4°4

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 :

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 :

S.HO

2,870

2,630

2,700 ,

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.

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

Velocity in feet per'l
second . . . .J

0'95

3'8o

5-00

4*60

3-15

1-65

075

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 I0001OOOOQ 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, 0 = amperes, and R = ohms. ; then, by Ohm's Law C = E/R,
Also, EC = C2R = E2 R = Watts.

Therefore, H.P. = = = .

746 746 746R-

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 (Pw) the International Watt (WP ), 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 B«gn«lf. 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 Ml> = 61007 -f-£

where log ^B = 3-1812, log WC 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

123

456

789

0000

0043

0086

0123

0170

0212

0253

0294

0334

0374

4 8 12

17 21 25

29 33 37

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

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

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

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

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.

1 2

1 2 3

456

789

7404

7412

7419

7427

7435

7443

7451

7459

7466

7474

122

345

567

56
57

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

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

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

8751

8756

8762

8768

8774

8779

8785

8791

8797

8802

112

233

455

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

9294

9299

9304

9309

9315

9320

9325

C330

9335

9340

112

233

445

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

9777

9?82

9786

9791

9795

9800

9805

9809

9814

9818

Oil

223

344

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

9956

9961

9965

9969

9974

9978

9983

9987

9991

9996

Oil

223

334

TABLE OF ANTILOGARITHMS.

417

7 1

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
0 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
0 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

0 1 1
Oil
0 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
0 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
0 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

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
4£OS
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

69£2

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

Angle.

;

Chord.

fine.

Tangent.

Co- tangent.

Cosine.

Degrees

Radians.

0°

000 C

1-414

1-5703

90»

•0175

•017 l '0175

•0175

57-2900

•9998

1-402

1-5533

•034D

•035

•0349

28-6363

•9994

1-3S9

1-5359

•0524

•052

•0523

•0524

19-0311

•99S6

1-377

1-51S4

•0698

•070

•0698

•0699

14-3007

•9976

1-364

1-5010

•OS73

•087

•0872

•0875

11-4301

•9962

1-351

1-4S35

•1047

•105

•1045

•1051

9-5144

•9945

1-338

1-4661

•1222

•122

•1219

•1228

8-1443

•9925

1325

1-4486

•1396

•140

•1392

•1405

7-1154

•9903

1-312

1-4312

•1571

•157

•1564

•1584

6-3133

•9S77

1-299

1-4137

•1745

•174

•1736

•1763

5-6713

•9848

1-2S6

i-rso

•1929

•192

•1903

•1944

5-1446

•9316

1-272

l-37t3

•2094

•209

•2079

•2126

4-7046

•97S1

1-259

1-3614

•2269

•228

•2250

•2309

4-3315

•9744

1-245

1-3439

•2443

•244

•2419

•2493

4-0108

•9703

1-231

1-3265

Z '

w a

•2613

•261

•2588

•8679

87821

•9659

1-213

13090

•2793

•278

•2756

•2S67

3-4S74

•9613

1-204

1-2915

•2967

•296

•2924

•3057

32709

•9563

1-190

1-2741

3 {

•3142

•313

•3090

•3249

3-0777

•9511

T176

1-2566

2 '

•3316

•330

•3256

•3443

2-9042

•9455

1-161

1-2302

•3491

•347

•3420

•3640

2-7475

•9397

1-147

1-2217

•3665

•364

•3584

•3S39

2*6051

•9336

1-133

"2043

•3840

•382

•3746

•4040

2-4751

•9272

1-118

•1S63

•4014

•399

•3907

•4245

2-3559

•9205

1-104

•1694

•4189

•416

•4067

•4452

2-2460

•9135

1-089

•1519

•4363

•433

•4226

•4663

2-1445

•9063

1-075

•1345

•4538

•450

•4384

•4877

2-0503

•89S8

1-060

1-1170

•4712

•467

•4540

•5095

1-9626

•8910

1-045

1-0996

•4387

•484

•4695

•5317

1-SS07

•8829

1-030

1-0821

•5061

•501

•4848

•5543

1-SC40

•8746

1-015

1-0647

•5236

•518

•5000

•5774

1-7321

•8660

1-000

1-0472

•5411

•534

•5150

•6009

1-6643

•8572

•985

1-0297

•5585

•551

•5299

•6249

1-6003

•84SO

•970

1-0123

•5760

•568

•5446

•6494

1-5399

•8387

•954

•9943

•5934

•585

•5592

•6745

1-4826

•8290

•939

•9774

•6109

•601

•5736

•7002

1-4281

•8192

•923

•9599

•G283

•618

•5S78

•7265

1-3764

•8090

•908

•9425

•6458

•635

•6018

•7536

1-3270

•79S6

•892

•9250

•6632

•651

•6157

•7813

1-2793

•7880

•877

•9076

•6S07

•668

•6293

•8098

1-2349

•7771

•861

•8901

•C9S1

•684

•6428

•8391

1-1918

•7660

•845

•8727

•71. '6

•700

•6561

•8693

1-1504

•7547

•829

•8552

•7330

•717

•6691

•9004

1-1106

•7431

•813

•8373

•7505

•733

•6820

•9325

1-0724

•7314

•797

•8203

•7679

•749

•6947

•9657

1-0355

•7193

•781

•8029

45°

•7S54

•765

•7071

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

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,

,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

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

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,

„ 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

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 lev«r,

screw and pulley, 160
,, screw and pulley, 160

424

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

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

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

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

NEWTON'S laws of motion, 260
Nippers, example of, 45

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,

23-25
„ 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

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

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,

25> 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

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

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

Co. LIMITED
TavistoclTSTrEef, Covent Garden , Loudon

THIS BOOK IS DUE ON THE LAST DATE
STAMPED'BELOW

AN INITIAL PINE OF 25 CENTS

WILL BE ASSESSED FOR FAILURE TO RETURN

rms BOOK ON THE DATE DUE. THE PENALTY

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)

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