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KINEMATICS AND KINETICS 
OF MACHINERY 

A TEXT BOOK FOR COLLEGES 
AND TECHNICAL SCHOOLS 



BY 

JOHN A. DENT, M.E. 

ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING, 
UNIVERSITY OF KANSAS 



ARTHUR C. HARPER, M.E. 

HEAD INSTRUCTOR OF MACHINE DESIGN, 
PRATT INSTITUTE 



NEW YORK 

JOHN WILEY & SONS, Inc. 

LONDON: CHAPMAN & HALL, LIMITED 
1921 






\Vb\M 



Copyright, 1921 
BY JOHN A. DENT 

AND 

ARTHUR C. HARPER 



PRESS OF 

BRAUNWORTH ft CO. 

BOOK MANUFACTURERS 

BROOKLYN, N. Y. 



PREFACE 



THIS book is the outcome of several years' experience in teach- 
ing kinematics and kinetics of machinery at the University of 
Illinois. For many years this subject was taught from notes 
prepared by Professor G. A. Goodenough, to which was added 
an article on the gyroscope by Professor F. B. Seely of the Depart- 
ment of Theoretical and Applied Mechanics. These notes were 
several times revised by the authors as experience showed where 
improvements could be made. 

In the fall of 1916 the authors undertook, with the consent of 
Messrs. Goodenough and Seely, to rewrite these notes in text- 
book form. The present volume is the outcome of that under- 
taking. The work was interrupted by the war, which took one 
of the writers into the military service, and imposed on the other 
such a heavy burden of teaching work that further progress ou 
the book was impossible. In the fall of 1919 the work was resumed 
and pushed to completion. 

The introductory chapter on constraint, the chapter on plane 
motion, and the chapter on velocities follow closely Professor 
Goodenough's notes. The chapter on the gyroscope is inserted 
almost without change as written by Professor Seely. The chap- 
ters on accelerations, on inertia forces, on balancing and on gov- 
ernors have been so completely rewritten that little trace remains 
of the original. The chapters on toothed wheels, on cams, on 
wrapping connectors, and on critical speeds have been added by 
the authors. 

It is hoped that this volume will fill a need in the curricula 
of our engineering schools, in that it gives systematic methods of 

ill 



434108 



iv PREFACE 

determining velocities, accelerations, and inertia forces which 
can be applied to practically all mechanisms. These methods 
are in the main graphical; the complicated forms of the equations 
making analytical methods too cumbersome for practical use 
except in some of the simpler types of machines. If the work is 
done to a large scale the results should be accurate enough for all 
practical purposes. 

The book is so arranged that it can be readily adapted to short 
courses as well as to more complete and detailed ones. Thus the 
chapters on gears, cams, and belts may be omitted where these 
subjects are taught in the courses in mechanism or design. The 
chapter on balancing can be profitably studied without the detailed 
analysis given in the chapters on accelerations and inertia forces. 
The chapter on critical speeds and parts of the chapters on gov- 
ernors and gyroscopes involve the use of mathematics which is 
perhaps beyond the range of the average undergraduate. These 
parts may, however, be of great value to the advanced student who 
intends to specialize in scientific design. For the benefit of under- 
graduate students a note on the solution of linear differential 
equations is appended. 

In conclusion the authors wish to extend their thanks to 
Professor Goodenough for valuable suggestions and criticisms in 
the preparation of the work. 

J. A. D. 

A. C. H. 



CONTENTS 



PAGE 

PREFACE iii 

CHAPTER I 

MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS ... 1 
Scope of the Subject Definitions Constrained Motion. Pairs 
Properties of Pairs Higher and Lower Pairs Pairing Elements 
Inversion of Pairs Multiple Contact Links Chains. Uncon 
strained, Locked, Constrained Mechanisms Skeleton Links 
Formation of Constrained Chains Joints Criterion of Constraint 
Application of Criterion Exceptional Cases Analysis of Mechan- 
isms Inversion of Mechanisms Exercises. 



CHAPTER II 

MOTION OF RIGID BODIES * . .29 

Plane Motion of Rigid Bodies Rotation about Fixed Center 
Instantaneous Center Properties of Instantaneous Center Exer- 
cises Instantaneous Center of Relative Motion Law of Three 
Centers Application of Law of Three Centers Higher Pairs 
Special Cases Centrodes Centrodes of Relative Motion Exer- 
cises Motion Produced by Rolling of Centrodes Equivalent 
Mechanisms Gears as Centrodes Pitch Lines of Gears Gear 
Teeth Space Motion. Axodes, 



CHAPTER III 

VELOCITIES OF MECHANISMS 60 

Introductory Analytical Methods, Geared Mechanisms Bev- 
eled Gear Trains Steam Engine Mechanism Method of Instanta- 
neous Centers Angular Velocity Ratios Special Cases Relative 
Angular Velocity Linear Velocity Ratio Special Cases Method 
of Relative Velocities Definitions Special Cases Velocity Images 
Fevolved Velocities Velocity Polygons Examples Velocity 



Vi CONTENTS 

PAGF 

Polygons for Linkages Special Constructions Three-line Con- 
struction Exercises Combined Method of Instantaneous Centers 
and Relative Velocities Four-line Construction Velocity Curves 
Exercises. 



CHAPTER IV 

ACCELERATIONS IN MECHANISMS 86 

Introductory General Principles, Definitions Normal and Tan- 
gential Acceleration Translation Rotation about a Fixed Axis 
Combined Motions. Coriolis' Law Application of Coriolis' Law 
Exercises Relative Accelerations Notation Graphical Construc- 
tions for Normal Acceleration Projection of Acceleration Com- 
ponents Acceleration Image Acceleration Center Acceleration 
Polygon Selection of Scale Ordinary Gear Train Epicycl'c Gear 
Train Four-link Chain Steam Engine Mechanism Klein's Con- 
struction Sliding Pairs Exercises Blake Stone Crusher Exer- 
cises Three-line Construction Four-line Construction Accelera- 
tions in Cams, 

CHAPTER V 

INERTIA FORCES OF MACHINE PARTS 128 

General Statement Acceleration Produced by Single Force Spe- 
cial Cases Kinetically Equivalent System Calculation of Line 
of Action of Resultant Components of Resultant Force Steam 
Engine Mechan sm Forces on Piston Inertia Forces of Piston and 
Rod Turning Effort Atkinson Gas Engine Inertia Forces of 
Piston and Rod Forces on Piston Forces on Flying Link Turn- 
ing Effort Wanzer Needle-bar Mechanism. 



CHAPTER VI 

BALANCING OF ENGINES 145 

Introductory Kinetic Load Due to Unbalanced Mass Centrifugal 
Force Centrifugal Couple Masses in Same Plane of Revolution 
Masses in Different Transverse Planes Balancing of Rotating 
Masses Example General Relations Conditions for Balance 
Analytic Method Inertia Effect of Reciprocating Masses. Har- 
monic Motion Balancing Conditions Engines with Finite Rods 
Acceleration of Piston Secondary Balance Conditions for Pri- 
mary and Secondary Balance Partial Balance Single-crank 
Engine Counterbalancing Two-crank Engine Three-crank En- 
gine, Complete Balance, Partial Balance Engines with More than 



CONTENTS vii 



Three Cranks Four-cylinder Automobile-type Engine Six-cylinder 
Automobile-type Engine Eight-cylinder Automobile-type Engine 
Twelve-cylinder Automobile-type Engine Radial Engine Opposed 
Engine Rotary Engine Offset Engine. 



CHAPTER VII 

GOVERNORS 177 

Purpose Classification, Fly-ball Governor, Shaft Governor Force 
Reduction Mass Reduction Moment of Centrifugal Force 
Elementary Fly-ball Governor Weighted Fly-ball Governor 
Horizontal Spring-controlled Governor Vertical Spring-controlled 
Governor Conditions of Stability Oscillation of Fly-ball Governor 
Conditions of Stability Effect of Change in Speed of Engine 
Damping-out of Oscillations Elementary Centrifugal Shaft Gov- 
ernor Elementary Inertia Governor Inertia Forces in Shaft Gov- 
ernor Moment about Governor Pivot Oscillations in Governor 
Mathematical Analysis of Governor Action Damping-out Oscilla- 
tions. 

CHAPTER VIII 

THE MECHANICS OF THE GYROSCOPE 210 

Introductory Gyroscopic Couple Surging Line of Action of 
Resultant Wheel on Circular Path Angular Momentum Illus- 
trations Analysis of Gyroscope by Method of Angular Momentum 
Precession Loss of Energy in Setting up Precession. 

CHAPTER IX 

CRITICAL SPEEDS AND VIBRATIONS 226 

Introductory Revolving Shaft Loaded at Middle, Critical Speed 
Natural Period of Vibration Uniformly Loaded Shaft Critical 
Speed Natural Period of Vibration Shaft Rotating in Fixed Bear- 
ings, Uniformly Loaded Critical Speed Natural Period of Vibra- 
tion Inclination of Rotating Disc Effect on Critical Speed Effect 
on Period of Vibration Other Systems of Loading References. 

CHAPTER X 

TOOTHED WHEELS 250 

Introductory Rolling Contact Parallel Axes, Rolling Cylinders 
Intersecting Axes, Rolling Cones Axes neither Parallel nor Inter- 
secting, Hyperboloid of One Sheet Friction Gearing Spur Fric- 



yiii CONTENTS 

PAGE 

tion Gearing Grooved Wheels Beveled Friction Gearing Crown 
Friction Gearing Definitions Relation between Circular and 
Diametral Pitch Rectification of Circular Arcs Generation of Cy- 
cloidal Curves Teeth of Wheels Relation between Pitch and 
Arcs of Approach and Recess Size of Describing Circle Inter- 
changeable Wheels Rack and Pinion Annular Gears Approxi- 
mate Method for Laying Out Cycloidal Teeth Grant's Odonto- 
graph for Cycloidal Teeth Involute System Involutes in Sliding 
Contact Path of Point of Contact Teeth of Wheels Relation 
between Height of Tooth and Arcs of Approach and Recess 
Annular or Internal Gears Rack Interference, Limit of Addendum 
Approximate Layout for Involute Tooth Grant's Odontograph 
for Involute Tooth Stub Teeth Fellows' System Nuttall System 
Comparison between Standard and Stub Tooth Gears Propor- 
tions of Cast Teeth Proportions of Cut Teeth Beveled Gears 
Form of Teeth Layout of Beveled Gears Transmission between 
Non-intersecting Shafts Screw Gears, Helical Gears, Worm Gears 
Helical Gears for Parallel Shafts, Herringbone Gears Helical Gears 
for Non-intersecting Shafts Velocity Ratio Worm Gearing, 
Cylindrical Worm, Hindley Worm Circular Pitch, [ Axial Pitch of 
Worm Velocity Ratio Comparison of Hindley and Cylindrical 
Worm Gears Skew Beveled Gears Comparison of Systems. 

CHAPTER XI 

CAMS 303 

General Principles Layout of Cams Definitions Base Curves 
Layout of Base Curves Pressure Angle, Cam Factors Size of 
Cams Rotary Cams, Radial Follower Working Surface of Cams 
Two-step Cams Size of Follower Roller Rotary Cam, Offset Fol- 
lower Rotary Cam, Swinging Follower Positive Motion Cams 
Rotary Cams, Tangential Follower Rotary Cams, Sliding Tangen- 
tial Follower Plane Sliding Cams, Sliding Follower Plane Sliding 
Cams, Pivoted Follower Cylindrical Cam Cylindrical Cam, Slid- 
ing Follower. 

CHAPTER XII 

WRAPPING CONNECTORS 324 

Introductory, Belts, Ropes, Chains Transmission of Motion, 
Equivalent Link Friction between Belt and Pulley Power Trans- 
mitted by Belt Length of Belts, Cone Pulleys Open and Crossed 
Belts Arrangements of Twisted Belts, Reversibility Idlers 
Crowned Pulleys Materials of Belts Rope Drives, Multiple System, 
Continuous System Friction between Rope and Pulley Chain 
Drives Hoisting Tackle Differential Pulley. 



CONTENTS ix 

NOTE A 

PAGE 

IRREGULAR GEARS . . 340 

NOTED 
PROPOSITIONS ON VELOCITY POLYGONS 342 

NOTE C 
Locus OF THE CENTER OF ACCELERATION 344 

NOTE D 

SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS 346 

Definitions Types of Solutions Linear Differential Equations, 
Second Member Zero Character of the Roots Second Member not 
Zero Complimentary Function Particular Integral Summary. 

NOTE E 

INVESTIGATION OF FORCES IN GASOLINE ENGINE 354 

Data Timing Diagram Indicator Card Total Pressure Card 
Wrist-pin Pressure Side Thrust Turning Effort Combined Turn- 
ing Effort Shaking Forces Counterbalancing. 

INDEX . 377 



KINEMATICS AND KINETICS OF 
MACHINERY 



CHAPTER I 

MACHINE MOTIONS, PAIRS, LINKS, CHAINS, 
MECHANISMS 

1. Scope of the Subject. Mechanics is that branch of Physics 
which treats of the motions of material bodies and the forces acting 
on such bodies. That division of mechanics which deals with 
motions is called kinematics, and that division which deals with 
forces is called dynamics. Dynamics is further divided into 
statics, in which the bodies dealt with are considered to be in 
equilibrium, and kinetics, hi which the bodies are acted upon by 
unbalanced forces. 

Mechanics of machinery consists of the study of the application 
of the laws of mechanics to the parts of machines. The subject 
may be divided again into kinematics, statics, and kinetics. Kine- 
matics of machinery consists of the study of the motions of the 
parts of machines without regard to the forces accompanying or 
producing such motions. Statics of machinery consists of the 
study of the forces in machine parts with the assumption that all 
such parts are in equilibrium, that is, disregarding any forces 
which may act to produce accelerations in these bodies. 

In any machine it is impossible that any part shall move 
indefinitely at a uniform speed and in a straight line. It follows 
that there must always be forces producing accelerations in the 
moving parts, or in other words that no moving part can be in 
equilibrium. The province of kinetics of machinery is to take 
into account the accelerating forces. In many instances the 
accelerating forces are quite small, and the problems arising in 
such cases may be treated by static methods. In other cases 



2 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

the accelerating forces are extremely important. This is true of 
all high-speed machinery. In the following pages the subjects 
of kinematics and kinetics of machinery will be treated at length. 
Static problems will be considered only incidentally. 

The study of mechanics of machinery may be approached from 
two different points of view: (1) the motions and forces in 
existing machines may be analyzed; and (2) machines may be 
devised to produce desired motions and forces. It is believed 
that a thorough study and analysis of existing machines will be of 
great value to those who later expect to become designers, and it is 
the purpose of this book to guide students in such study. 

In general, a machine will be regarded as a system of rigid 
bodies, so connected that for a given movement of any one part 
there will be perfectly definite, determinate movement of every 
other part. The assumption of the rigidity of the parts is equiva- 
lent to disregarding any motions due to distortion or vibration of 
the members. Special cases where such distortions are not 
negligible, or where flexible links such as belts and ropes are 
employed, will be given special attention. 

2. Constrained Motion. Pairs. The characteristic of the 
motion of a machine part, as distinguished from that of a free body, 
is that every point of the machine element is constrained to move 
in a fixed predetermined path. In order that this may be the case, 
it is necessary that each machine part must be in contact with one 
or more other parts. The connections between the parts are called 
pairs. Thus in the ordinary steam engine mechanism, Fig. 1, 
there are four pairs: (1) between the crank and bearing, (2) 
between the connecting rod and crank, (3) between the cross- 
head and connecting rod, and (4) between the crosshead and guides. 

3. Properties of Pairs. Since the purpose of pairs is to con- 
strain the relative motion between the pairing bodies, the first 
step in analyzing constrained motions is the study of the properties 
of pairs. The simplest way to constrain a point P to move in a 
given path would be to cut a slot whose center line is the given 
path, and place in the slot a block on which is marked the point 
to be guided. In general, if this block is cut to fit the curvature 
of the slot in one position, it will not fit in some other position 
where the curvature of the slot is different, Fig. 2. If, however, 
the radius of curvature of the slot is constant, the block will fit 



PROPERTIES OF PAIRS 3 

equally well in all positions, Fig. 3. If the radius of curvature of 
the slot is indefinitely increased the circle becomes a straight line, 
Fig. 4. Rectilinear motion may therefore be regarded as a limiting 
case of circular motion. It follows that continuous surface con- 
tact between pairing bodies is possible only when the relative 
motion is circular or rectilinear. For paths which have variable 

_ Crank Pin. 



Counterweight 



Piston Ebd 



. Flu Wheel 

Connect i na Kod 




FIG. 2. 



PIG. 3. 




FIG. 4. 



FIG. 5. 



curvature only line contact is possible. For example, if the sliding 
block, Fig. 2, is replaced by a circular pin, Fig. 5, relative motion 
becomes possible regardless of the curvature of the slot. Pairs 
which permit surface contact are called lower pairs. Those which 
permit only line contact are called higher pairs. 

Another characteristic of lower pairing is that not only the 
point P, but every other point on the block is constrained to move 
in a similar path. Thus in Fig. 6 if an arm be rigidly attached to 



4 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

the block, a point P% on this arm will travel in a circular path 
whose center is 0, the center of curvature of the slot. On the 
other hand, in the case of higher pairing the motions of other 
points on the pin are not constrained. Thus, in Fig. 7, while the 
point PI moves along the slot, the arm attached to the pin may 
revolve around PI, and therefore, a point P^ on this arm may 
move in any one of an infinite number of paths. That is, the 
motions of other points attached to the pin are not constrained. 
A pair which completely constrains the relative motion between 
the bodies connected is called a complete or dosed pair. One 




FIG. 8. 



FIG. 9. 



which does not so constrain the relative motions of the pairing 
bodies is called an incomplete or unclosed pair. All lower pairs 
are, or may be made into, closed pairs. In general higher pairs 
are incomplete. 

If a rod PiP2 has two pins attached so that two points PI 
and P2 must travel along the center line of the slot, Fig. 8, or 
along the center lines of the two distinct slots, Fig. 9, then any 
other point, P 3 , of the rod, Fig. 9, is constrained to move in the 
dotted path AP S B. Thus in general two unclosed pairs furnish 
the same degree of constraint as one closed pair. (A more rigor- 
ous treatment of this subject will be found in Art. 28.) 

Referring again to Fig. 6, it is evident that the circle about 
might be completed, and the block expanded to fill the whole 



HIGHER PAIRS, LOWER PAIRS 

of this circle, without in any way 
altering the character of the motion, 
as shown in Fig. 10. Thus an ordi- 
nary pin-and-eye connection, as for 
example, between the crank and con- 
necting rod of a steam engine, con- 
stitutes a lower pair. 

A third characteristic of lower 
pairs may be described as follows: 
if one of the pairing bodies is held 
stationary and the other caused to 
move, a point on the moving body 
will trace a definite path; if the 
arrangement is now reversed, the 
second part being held stationary 
and the first being moved, a point 
on the moving body will trace a 
second path similar to the first. 
With higher pairing such paths will 
in general be dissimilar. FIG. 10. 

Finally it may be pointed out that lower pairs are easily con- 
structed by means of standard machine tools such as lathes, planers, 
etc., while higher pairs often require special machines such as 
gear cutters. The properties of lower and higher pairs may be 
summarized as follows: 






Exam- 
ples. 


Con- 
tact. 


Constraint. 


Relative Motion. 


Construc- 
tion. 


Lower 
pairs 


Pin and 
eye 

Slide 


Surface* 


Complete 


Same path traced by 
moving point which- 
ever body is held 
stationary 


Lathe, planer, 
etc. 


Higher 
pairs 


Gears 
Cams 


Line 


Incomplete 


Different paths traced 
by moving point ac- 
cording to which 
body is held station- 
ary 


Gear, cutters, 
etc. 



* In some special cases 'lower pairs may have line contact only. An example is a 
round shaft rotating in a square bearing. In such cases extra material may be added to 
give surface contact without in any way altering the character of the motion. As the 
relative motion is the important consideration such pairs are classified as lower. 



6 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

4. Pairing Elements. The geometrical forms placed upon two 
bodies so that they may be connected by a pair are called pairing 
elements. Thus the cylindrical surface of the wrist pin of an 
engine, and the inside surface of the brasses of the connecting rod 
are pairing elements. The surfaces of the crosshead and guides, 
the faces of two gear teeth in mesh, or of two cams in contact, 
furnish other examples. 

5. Inversion of Pairs. In the case of lower pairs the solid and 
hollow elements may be interchanged without changing the char- 
acter of the relative motion. For example, instead of having a 
pin in the crosshead of an engine fitting into an eye in the con- 
necting rod, a pin might be attached to the rod fitting into a hole 
in the crosshead. Instead of having a moving piston in a sta- 
tionary cylinder a moving cylinder might slide over a fixed piston, 




the connecting rod being attached to the moving cylinder as shown 
in Fig. 11. The process of exchanging the hollow and solid 
elements of a pair is called inversion of the pair. 

The size of the pairing elements has no effect on the relative 
motion of the bodies connected by the pair. Thus, if a rod, Fig. 
12, has a hole which fits over a stationary pin, the path of any 
point on the rod is a circle whose center is the center of the pin, 
regardless of the diameter of the pin. Sometimes by simply expand- 
ing pairing elements the appearance of a machine may be changed 
beyond recognition without altering in the slightest degree the 
character of the motions. For example, a simple machine is 
shown in Fig. 13 consisting of four bars connected by turning pairs. 
By simply expanding the pairing elements this machine may be 
made to assume the different forms shown in Figs. 14 to 19 1 
without changing in any way the relative motions. The machine 
1 See Kennedy's Mechanics of Machinery, pp. 293 to 402. 



INVERSION OF PAIRS 



shown in Fig. 14 is precisely the same as that in Fig. 13, except 
that the pair A has been expanded till the bar A B has been changed 
to a disk whose circumference is one of the pairing elements. 




FIG. 12. 

In Fig. 15 the same bar AB has been changed to a circular disk, 
this time by the expansion of the pair B. In this case the bar AB 
has been changed to an eccentric; hence comparing Figs. 13 and 

15 it is evident that an eccentric 
is equivalent to a crank whose 
length is equal to the eccentricity. 
In Fig. 16 the pair D is enlarged. 
Evidently the motion of DC 
will be equally constrained if 
the sides of the circular disk 
are cut away, leaving the zone 
shown by dotted lines. An- 
other modification is shown in Fig. 17, where the link CD is 
given the form of an annular ring instead of a disk. The next step 
is to cut away all of the ring except a small block containing the 
pair C. Finally since by the rotation of the crank A B the pair C 




FIG. 13. 




8 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

is made to reciprocate in a limited circular arc, we may as well 

shorten the slot in 
which the block 
moves to the length 
actually required, 
as shown in Fig. 
18. Comparing Fig. 
18 with Fig. 13 we 
see that thelower pair 
shown in Fig 18 is 

FlG 14 equivalent to a turn- 

ing pair whose center 

is the center of curvature of the slot. The block containing the 

point C is equivalent to a bar of length equal to the radius of 

curvature of the slot and 

pivoted at the center of 

curvature. By increasing 

the radius of the slot the 

equivalent of very long 

links may be obtained 

without increasing the 

dimensions of the machine. 

A sliding pair is thus equi- Fio. 15. 

valent to a turning pair 

whose center is at an infinite distance. By a series of modifica- 
tions similar to those described, the mechanism shown in Fig. 19 





FIG. 16. 



INVERSION OF PAIRS, MULTIPLE CONTACT 





FIG. 17. 

is obtained. All of the arrangements shown in Figs. 13 to 19 

are kinematically identical. 

6. Lower Pairs with Multiple Contact. It frequently occurs 

in machine con- 
struction that a 

single lower pair 

may have several 

contact surfaces. 

Thus the shaft 

of an engine runs 

in two bearings, FIG lg 

and both the 

crosshead and piston are provided with sliding pairs. A single 

sliding pair may have any number of contact surfaces provided 

these surfaces are 
parallel, and a single 
turning pair may 
have any number of 
journals and bearings 
provided that they 
all have the same 
axis. In the analy- 
sis of mechanisms, 
it is important that 
the student guard 
FlG - 19- against the common 

error of counting each contact surface a pair. 




10 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

7. Links. A body provided with two or more pairing elements, 
so that it may be connected to at least two other bodies, is called a 
kinematic link, or simply a link. A body with two pairing ele- 
ments is called a binary link, one with three pairing elements a 
ternary link, one with four pairing elements a quaternary link. 

In machines having more than two members each link must 
have at least two pairing elements in order that the motions of all 
links may be constrained. It should be noted that the term link 
applies to all parts of a machine which are rigidly fastened together 
so that there can be no relative motion. For example, in the 
steam engine, Fig. 1, we may identify the following links: 

1. Cylinder and cylinder heads, base plate, foundation, 

main bearing, crosshead guides, etc. 

2. Piston and rings, piston rod, crosshead, wedges, bolts, etc. 

3. Connecting rod, brasses, straps, wedges, etc. 

4. Crank shaft, crank arm, crank pin, counterweight, fly- 

wheel, etc. 

The motions in the machine depend entirely on the relative 
positions of the pairs, and not at all on the size and shape of the 
links. In particular the stationary link may be very large and 
heavy, and extra material may be added indefinitely, without in 
any way affecting the motions, so long as the positions of the pairing 
elements are unchanged. 

8. Chains. If a number of links are provided with suitable 
pairing elements they may be connected by joining the elements. 
If this is done so that each element is provided with a mate and 
none left unpaired, the resulting structure is called a kinematic 
chain, or simply a chain. Such chains are divided into three 
classes as follows: 

1. Constrained chains, in which relative motions of the 

links are possible, and where such motions are com- 
pletely predetermined by the character of the pairs. 

2. Locked chains, in which no relative motion is possible. 

3. Unconstrained chains, in which the relative motions are 

indeterminate. 

In Figs. 20 to 22 are shown chains consisting of three, four, and 
five binary links respectively. In Fig. 20 evidently no relative 



CHAINS, LOCKED, CONSTRAINED, UNCONSTRAINED 11 



motion is possible and the chain is therefore locked. Thus if 
link 1 is held fast, say to the paper, the axis of the pair B consid- 
ered as part of link 2 must move, if at all, in an arc about A as a 
center. Likewise considering the axis of B as a point on link 3 
it must move, if at all, in an arc about C as a center. But, since 
the point cannot move simultaneously in two different paths, it 
cannot move at all, and the chain is therefore locked. 




FIG. 20. 



FIG. 21. 



In Fig. 21 if link 1 is held fast and link 2 is moved, the only 
possible motion of link 2 is a rotation about A as a center, and 
point B must therefore move in a circular arc. Also point C 
must move in a circular arc about D as a center. But, since B 
and C must remain at a constant distance apart, for each position 
of B there is in general only one position which C can take. The 
motions of all 
points on any of 
the links are there- 
fore completely 
constrained. 

The five-link 
chain, Fig. 22, is 
evidently uncon- 
strained. As in 
Fig. 21, if link 1 
is held stationary 

two points B and D must move in circular arcs about A and E as 
centers. But, since points B and D need no longer remain a con- 
stant distance apart, they may be moved independently of each 
other, and the point C may be made to travel in an infinite 
variety of paths. The motions are therefore indeterminate, and 
the chain is unconstrained. 




FIG. 22. 



12 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

Each of the three types of chains is useful for particular pur- 
poses. Locked chains are not used in machines, since the parts 
of a machine are supposed to have relative motions. They are, 
however, extensively employed in bridges, roofs and other struc- 
tures. 

In the great majority of machines constrained chains are 
employed. Unconstrained chains are used only in cases where a 
greater degree of freedom is desired than is afforded by constrained 
chains. The following are examples: 

1. In a steam engine with shaft governor the motions of 

each point on the piston, crosshead and connecting 
rod can be accurately determined from those of the 
crank; but the governor mechanism and the slide 
valve may be given motions independent of that of the 
crank. The engine mechanism as a whole is there- 
fore unconstrained. 

2. Reversing gears for steam engines may be given 

motions independent of the motions of the rest of the 
engine, and may thereby be made to influence the 
motions of the valves. 

3. In the Corliss engine the movement of the valve, when 

the hook is released, is entirely independent of the rest 
of the mechanism. 

4. In many machines springs or other flexible members 

are introduced to allow extra movements when the 
forces in the machine become excessive. 

It will be noted that in practically all these cases the machines 
run as if constrained. In the examples cited, in case: 

1. The engine runs as if constrained except when the 

governor is actually shifting: 

2. Except when the engineer is moving the reverse lever; 

3. Part of the time the motion of the lever is constrained, 

the atmospheric pressure supplying the required 
force to move the valve after the hook releases. 

4. Except when unusually heavy forces occur; 

In most cases the unconstrained chain is entirely useless. 
Practically all the discussions in the following pages will apply to 
constrained chains. 



MECHANISMS, SKELETON LINKS, 13 

9. Mechanisms. The step from the constrained chain to the 
mechanism is a simple one. Suppose one of the links of a con- 
strained chain to be held fixed relative to the ground or some other 
standard. Then since each of the links has a constrained motion 
relative to the fixed link, it also has a constrained motion relative 
to this standard. The chain with one of its links fixed is called a 
mechanism. Since any one of the links may be chosen as the 
fixed link, a chain has as many mechanisms as it has links. 

10. Skeleton Links. As the motions of a constrained chain 
depend only on the location of the pairing elements and not on 
the size and shape of the links, we ^ 
may represent the links by sim- Binary Links 

pie geometrical figures. Thus a 
binary link may be represented 
by a straight line having pairing 
elements at its ends, a ternary- 
link by a triangle, a quaternary 
link by a quadrilateral, etc., having 
the pairing elements at the verti- 
ces. Such skeleton links are rep- 
resented in Fig. 23. In this fig- FlG 23 

ure an element of a sliding pair is 

represented by a small block, an element of a turning pair by a 
small circle, and an element of a higher pan- either by a small 
triangle or by the profile of the pairing element. In this way 
mechanisms may be represented by simple figures, all unnecessary 
lines being eliminated. 

11. Formation of Constrained Chains. As machines are 
usually constrained chains, two problems now arise: (1) To study 
the methods of connecting a series of links so as to form con- 
strained chains; (2) to determine the conditions which must be 
satisfied in order that a chain shall be constrained. 1 

To build up a constrained chain from binary and ternary links, 
Fig. 24, start with the links 1, 2, 3, and 4, giving a four-link chain 
with pairs A, B, C, and D. Add ap element E to link 3 and an 
element F to link 4, thus making links 3 and 4 ternary; join E 
and F by means of two binary links 5 and 6 having a common pair 

1 In this and the following discussions only mechanisms in which all points 
move in parallel planes are considered. 




14 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 




G. The result is a six-link chain with two ternary and four binary 

links. The chain is con- 
strained, for points E and F 
have different motions in 
known paths, and this fact 
imposes upon G a deter- 
minate motion. If E and 
F are connected by one 
link, the chain is locked, 
and if more than two links 
are used the chain is uncon- 
FlG - 24 - strained. This chain is 

known as the Watt chain, 
because with link 3 fixed 
it gives the mechanism 
of the Watt Beam En- 
gine. 

If the element E is 
added to link 2, making 
the non-adjacent links 2 
and 4 ternary links, the 
result is a second six- 
link constrained chain as 
shown in Fig. 25. This FIG. 25. 

arrangement is known 

as the Stephenson chain. 
In a ternary link two 
of the elements may coin- 
cide. Suppose that the 
elements B and E of link 
3, Fig. 24, thus fall to- 
gether, the resulting chain 
is shown in Fig. 26. This 
chain is also obtained by 
making B and E of link 
2, Fig. 25, coincident. 
Furthermore, elements F 
and D of link 4 may be 
chain shown in Fig. 27, in 





FIG. 26. 
made coincident, thus obtaining 



the 



CONSTRAINED CHAINS 



15 



BE 



which all the links are binary. It is to be noted that neither F 

nor G may be made to coincide with C, for in that case three of 

the links would form 

a locked triangle and 

the chain would in 

reality have only four 

links. 

A joint such as 
B, E, or D, F, Fig. 27, 
which is made up of 
three elements, is called 
a ternary joint, one FIG. 27. 

made up of four ele- 
ments is called a quaternary joint, and so on. The ordinary joint 
of two elements is a binary joint. Evidently a ternary joint is 
equivalent to two binary joints; a quaternary joint to three binary 
joints; and a joint of i elements to i-\ binary joints. 

Starting again with the 4-link chain, let links 2, 3, and 4 be 
made ternary by the addition of elements E, F, and G, Fig. 28. 





FIG. 28. 

At these points connect three binary links 5, 6, and 7, and pair 
these with a fourth ternary link 8. The result is an 8-link chain 
with 4 binary and 4 ternary links. That this chain is constrained 
is established by the following reasoning. Consider link 1 fixed 
and let the elements of the pair E be separated. Denote the 
element of pair E that belongs to link 2 by #2, and the element 
that belongs to link 5 by EQ. If link 2 be moved, EI must move in 



16 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

a curve m and the joints F and G will likewise move in definite 
paths. Suppose E 2) F and G to be moved to some position; then 
keeping F and G at rest, links 6, 7, and 8 may be given a con- 
strained motion and consequently, H, K, and L will move in 
definite paths as shown. At the same time 5 may be made to 
follow the path m and therefore may be brought into coincidence 
with E 2 . These two operations may be repeated indefinitely; 
E 2 , F, and G are first moved; then held fixed while K y L, 
and H are moved and E 5 is made to coincide with E 2 . But these 
successive motions may be made as small as we please and may 




FIG. 29. 

take place at the same time instead of in succession. In this 
case E 2 and E 5 will remain together and may be joined to form a 
pair. Now since for every position of E, F, and G, there are 
definite positions for H y K y and L in which E$ can be made coin- 
cident with E 2) it follows that if E& and E 2 remain together H, K, 
and L will move in definite paths. This fact proves the constrain- 
ment of the chain. 

Other 8-link constrained chains are shown in Figs. 29 and 30. 
In the chain Fig. 29 two 4-link chains with a common link are 
joined by the link 8. If link 1 is held fixed the points E and F 
are evidently constrained to move in definite paths; hence if E 
and F are joined by the link 8, that link must have a definite 
motion relative to link 1. 



CONSTRAINED CHAINS 



17 




Figs. 31 and 32 show chains with 10 and 12 links, respectively. 
The constrainment of the 12-link chain is easily proved, there 

being no less than 
three 4-link cells. 
It is not easy, how- 
ever, to determine 
whether the chains 
shown in Figs. 30 
and 31 are con- 
strained, as all the 
cells are bounded by 
five or more links. 
In general, it is dif- 
ficult or impossible 
to determine by in- 
spection whether a 
FIG. 30. chain of more than 

six links is con- 
strained. There is, however, a simple relation between the 
number of links and the number of joints in a constrained chain 
that may be used as a 
criterion of constraint. 
This relation is reduced 
in the next paragraph. 

12. Criterion of Con- 
straint. In order that 
several points A, B, C, 
D, etc., Fig. 33, may 
occupy always the same 
relative position a cer- 
tain number of con- 
ditions are necessary. 
With two points A and 
B tliere is one condition 
required, namely that FlG 31 

the points shall always 

be the same distance apart or that the length of the line 
joining A and B shall be constant. If a third point C is 
added to the system, two more conditions are required to 




18 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 



fix C relatively to A and B. These may be the lengths of the 
lines AC and BC or the angles ' and a", which these lines make 




FIG. 32. 

with AB. Three points are therefore fixed by three conditions. 
Similar reasoning shows that a fourth point D requires two more 
conditions, and each additional point two additional conditions. 

In general, to fix the rela- 
tive positions of n points 
2n-3 independent condi- 
tions are necessary. 

In a chain the joints 
are the characteristic 
points whose relative po- 
sitions are to be investi- 
T gated; and the necessary 
conditions are furnished 
by the links. Consider 
the chain shown in Fig. 
25. If the joints B and 

D are connected by a rigid link, as shown by the dash line, the 
chain is locked. Thus adding one condition to a constrained 
chain locks it, and conversely, the removal of one condition in the 
case of a locked chain, permits constrained motion, that is, gives 
the chain one degree of freedom. 




FIG. 33. 



CRITERION OF CONSTRAINT, LOWER PAIRS 19 

Let / = number of joints in the chain; 

C = number of conditions furnished by the links. 

We have seen that 2J-3 conditions just lock the chain; hence 
2J-4 conditions leave the chain constrained. Expressed in alge- 
braic language: 

If C = 2J-3, the chain is locked; 

C = 2J-4, the chain is constrained; 

C < 2 J-5 the chain is unconstrained. 

Each link of the chain furnishes a number of conditions depend- 
ing on the number of its pairing elements. 1 A binary link gives 
one condition, the distance between the joints connected by it; 
a ternary link gives three conditions, the three required to fix the 
relative positions of its three elements; a quaternary link con- 
tributes the five independent conditions that fix its four elements. 
In general, a link with k pairing elements contributes 2k-3 inde- 
pendent conditions. 

Let N = total number of links in the chain; 

J total number of joints in the chain; 
E = total number of pairing elements; 
712, tta, ft4, etc. = respectively, the number of links of 2, 3, 4, etc., 

elements; 
mz, ws, W4, etc. = respectively, number of binary, ternary, etc., 

joints; 
C = number of conditions furnished by the links. 

Then 

#=2n 2 +3tt 3 +4tt4 ........ (1) 

Also 

(2) 



= n 2 +n 3 +n 4 ......... (3) 

(4) 
....... (5) 



1 In this paragraph it is assumed that the chain contains only lower pairs. 
The case of higher pairing is considered in the next article. 



20 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

If this chain is constrained, 

C = 2J-4. 
Hence 

7i2+37i3+5rc4+7rc 5 . . . = 2J-4 ..... (I) 

This equation is the criterion of the constrainment of the chain. 
Adding equations (3) and (I), 



. ... (6) 
Subtracting (3) from (1), 

tt2+2ri3+3tt4+4tt 5 =E-N. .... 1 (7) 
Comparing (6) with (7). 

2(E- N) =2J- 4+ N 
Or 

E-J = %N-2 ..... ' . . (8) 

If the chain has only binary joints, 

E = 2n 2 = 2J 
and (8) reduces to 



(II) 



Equation (II) is the form in which the criterion is generally used. 
Even when a chain has ternary and quaternary joints, we may 
consider each ternary joint replaced by two binary joints, each 
quaternary joint by three binary joints, and so on; and use the 
form (II). From the form of (II) it appears that a chain with 
only lower pairs cannot in general have an odd number of links. 

As an exercise, let the student apply both criterions to each of 
the chains shown in Fig. 20 to 32. 

13. Criterion of Constraint with Unclosed Pairs. The term 
joint is restricted to the connection of two links by a complete or 
closed pair. The mere point of contact which ordinarily occurs in 
higher pairing is not a joint in this sense. In chains with higher 
pairs, therefore, there may be links with only one joint. Gear 
wheels and cams are nearly always links of this character. In 
conformity with the notation of the preceding paragraph, the 



CRITERION OF CONSTRAINT, HIGHER PAIRS 21 

number of such links is denoted by n\. It has been shown that a 
link with k joints gives 2k 3 conditions; following this rule each 
link with one joint gives 2X1 3= 1 conditions. The inter- 
pretation of this statement is simple: such a link with its one joint 
furnishes no conditions at all, but instead requires one condition, 
namely that it shall be in contact with or have higher pairing 
with some other link. 

In addition to the conditions furnished by the links carrying 
two or more joints, each unclosed pair furnished one condition, 
namely that some pair of links have contact. If H denotes the 
number of unclosed pairs, then these higher pairs furnish H con- 
ditions. The total number of conditions furnished is therefore: 

C=-rn+n2+3n 3 +5n 4 + ...+#. . . (1) 

As before, the number of conditions necessary to render the chain 
constrained is C = 2J 4; hence 

-ni+tt2+3n3+5rc 4 . . . +H = 2J-4. . . (Ill) 

Equation (III) is the most general form of the criterion for 
chains with unclosed pairs. A form analogous to Equation (II) 
may be derived as follows: 



(2) 
(3) 

2J-4- #=-7ii+tt2+3tt3+5n 4 +7tt5. . . (4) 

Combining Equations (2) and (3), and Equations (3) and (4): 

E- 

2J-4- 
Therefore 

2(E-N)=2J-4-H+N, .... (5) 
or 

E-J = (3N-4-H) ...... (6) 

If all the joints are binary, E = 2J', and therefore 

or J+ifl r =fAT-2 1 . . . (IV) 



1 The criterion . here developed is for plane mechanisms only; that is, 
mechanisms in which all points move in parallel planes. 



22 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 

In the rarer cases in which higher pairs are closed pairs, they 
should be counted among the joints J rather than among the 
unclosed pairs H. 

14. Application of Criterion. If the criterion in this form be 
applied to any mechanism the left-hand member may be equal to, 
larger than, or smaller than the right-hand member. If the two 
are equal, the mechanism is constrained. If the left-hand member 
is the larger, the mechanism is locked; if the smaller, the mechan- 
ism is unconstrained. The left-hand member may be taken as 
representing the number of restrictions imposed on the motions 
of the links, and the right-hand member as the number of restric- 
tions necessary to constrain their motion. If the restrictions are 
too numerous no motion is possible, if too few there is no con- 
straint. It will be noted from the form of the criterion that an 
unclosed pair imposes on the motion of the pairing links only 
half the constraining effect of a closed pair. (Compare Art. 6.) 

15. Exceptional Cases. In some cases the test for constraint 
indicates that a mechanism is locked, while we find that it is 
actually free to move. In every such case a study of the motions 
reveals that some point is constrained to move in the same path 
by two separate devices, either of which would alone be sufficient. 
The Davis engine. Fig. 34, is an example. Two cylinders are 
placed at right angles to each other as shown. The long piston 
rods 2 and 3 are guided at both ends and are joined by turning 
pairs to the connecting rod 4. The crank 5 is joined to link 4 at A, 
the middle point of the connecting rod. We have here five links 
and six joints, namely 12, 13, 15, 24, 34 and 45. 

The criterion of constraint gives 6 = fX5 2 = 5J and the 
machine appears to be locked. 

A study of the mechanism shows that if the crank 5 were 
omitted the point A would still be obliged to travel in a circle 
about as a center, and that therefore the crank really adds 
nothing to the constraint. If the crank were attached to the con- 
necting rod at any other than the middle point, the mechanism 
would be locked as every point of the rod such as B is constrained 
to move in an ellipse. Since a crank attached at B would con- 
strain point B to move in a circle the mechanism would be locked. 

In a few cases machines which are indicated as locked, are 
enabled to run by allowing a small amount of lost motion or by 



EXCEPTIONAL CASES 



23 



slightly bending some member. An example is shown in Fig. 35. 
In this mechanism A as a point on link 5 must move in a horizontal 




FIG. 34. 




FIG. 35. 



line. As a point on 6 it must move in an arc of a circle. The 
machine is therefore apparently locked. However, a small amount 



24 MACHINE MOTIONS, PAIRS, LINKS, CHAINS MECHANISMS 

of lost motion in the pair between links 1 and 5, or a slight bending 

of link 5 will allow the machine to 
move. 

In one very simple mechan- 
ism the criterion as stated is at 
fault. In the wedge shown in 







Fig. 36, we have three links and 
1 three sliding pairs. The criterion 

I gives: 



3+0 = fX3-2 = 2i, 

F IGt 36 , and the machine is apparently 

locked, although it is evident that 

movement is possible. A chain containing only sliding pairs 
possesses an extra degree of freedom of motion as compared with 
other chains having the same number of links and joints. 1 

As examples in the use of the criterion let the student test for 
constraint the skeleton mechanisms shown in these pages, and any 
other plane mechanisms which present themselves. In every 
case where the criterion is not satisfied, study the motions care- 
fully and try to discover the cause. 

16. Analysis of Mechanisms. It has been seen that the 
motions of the links depend entirely upon the character and loca- 
tion of the pairing elements, and that consequently a binary link 
may be conveniently represented by a straight line connecting the 
elements, a ternary link by a triangle, etc. The analysis of a 
mechanism consists of identifying the links, and noting with what 
other links each one pairs. Then each link may be represented by 
its skeleton form straight line, triangle, etc. and the pairing 
elements properly marked. As an example take the Stephenson 
link motion Fig. 37. Here we can identify eight links: 

(1) Frame of engine. 

(2) Shaft and two eccentrics. 

(3) Eccentric rod. 

1 The complete criterion of constraint is given by 



where is the number of complete cells containing only sliding pairs. See 
Kinematics of Machinery, by A. W. Klein. 



NALYSIS OF MECHANISMS, SKELETON MECHANISM 25 



V 4) Eccentric rod. 

(5) Slotted link. 

(6) Hanger. 

(7) Block moving in slotted link. 

(8) Slide valve and stem. 




Stephenson Link 



FIG. 37. 

These links are shown in skeleton form in Fig. 38. Taking the 
links in order we see that link 1 is a ternary link pairing with links 
2, 6 and 8. It is therefore repre- 
sented by a triangle. The joints 
with links 2 and 6 are turning while 
that with link 8 is sliding. Link 2 
has three turning pairs with links 
1, 3, and 4. Link 3 has turning 
joints with links 2 and 5, and the 
same is true of link 4. Link 5 has 
four turning joints with links 3, 4, 
6, and 7. Link 6 has turning joints 
with links 5 and 1. Link 7 has 
turning joints with links 5 and 8. 
Link 8 has a turning joint with link 
7 and a sliding joint with link 1. 

Now, if desired, the links may be 
connected to form a chain, which 
then represents the machine in its 

simplest or skeleton form, as shown in Fig. 39. This chain may 
readily be shown to be constrained by applying the criterion. 



IZ O 



120 



230 



35 




160- 

570- 
700 



035 
045 



056 
078 



FIG. 38. 



26 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 




FIG. 39. 



Often when analyzed in this way, machines which externally 
are very different in appearance are found to be built on the same 

skeleton chains. 1 For ex- 
ample, if we omit links 7 and 
8, the chain just described is 
identical with that of the 
Wanzer needle bar, Fig. 106. 
If link 2 or link 5 is held sta- 
tionary, it becomes the chain 
for the Atkinson gas engine, 
Fig. 165; the Blake stone 
crusher, Fig. 147; or the 
shaper, Fig. 54. 

17. Inversions of Mechan- 
isms. It has been seen that 
by holding different links of a chain fixed in turn, various mechan- 
isms are formed. This process is called inversion of mechanisms. 
For example, in Fig. 
40 if link 
stationary 



1 is held 

and gear 

2 made to revolve, 
the compound gear 3 
will revolve and drive 
gear 4 as in the ordi- 
nary back gear of a 
lathe. If, however, 
one of the gears, say 
gear 2, is held station- 

ary and link 1 allowed to revolve, the result is an epicyclic or 
planetary gear train, which is an inversion of the ordinary gear 
train. Similarly the rotary engine, the oscillating engine, the 

1 The number of constrained chains which can be formed from links 
having only lower pairs is given by the following table : 




FIG. 40 



Number of links. 

6 

8 

10 

12 

See Klein's Kinematics of Machinery. 



Number of Chains. 

2 

14 

228 

4000 



INVERSION OF MECHANISMS, EXERCISES 



27 



Whitworth quick return and the crank-shaper quick-return 
motions are all inversions of the ordinary slider crank or steam 
engine mechanism. 

EXERCISES 

1. The slider crank chain shown in Fig. 1 is the mechanism of the ordinary 
steam engine when the link 1 is fixed. Transform this mechanism into three 




r i 


i 5 



FIG. 41. 

other steam engines by successively fixing links 3, 2, and 4. Make the length 
of link 2 = 1 inch and that of link 3 = 3| inches hi each case. Be sure that in 
each case: 

Link 1 joins sliding pair D to turning pair P; 

Link 2 joins turning pair P to turning pah* B; 

Link 3 joins turning pair B to turning pair C; 

Link 4 joins turning pair C to sliding pair D. 

In the solution of this problem it will be desirable in one case at least to inter- 
change the elements of one or more pairs. The exchange of the solid and 
hollow elements of a lower pair does not affect the relative motion of the links 



28 MACHINE MOTIONS, PAIRS, LINKS, CHAINS, MECHANISMS 



connected by the pair, and is always permissible. Take, for example, the 
pair B, Fig. 1; usually the crank 2 carries a pin which fits into an eye in the 
connecting rod 3; sometimes, however, the rod 3 carries the pin and the crank 
2 the eye, this construction being common in the cheaper classes of machinery. 
In the case of the sliding pair D, link 1 usually carries the hollow enclosing 
element, i.e., the cylinder and guides; this pair may likewise be inverted by 
making 4 carry the enclosing element, as shown in Fig. 11. The inversion is 

inexpedient when 1 is the fixed link, but 
is useful when 4 is taken as the fixed link, 
as it makes the lighter element the mov- 
ing part. 

Conditions to be observed: 

(1) Provide each engine with inlet 
and outlet for steam, but do not 
design valve gear. 

(2) Provide each engine with means 
of communicating motion to a 
shop shaft, but do not add extra 
links (an extra belt may be used 
when 4 is the fixed link). 

(3) See that no link interferes with 
the proper motion of any other 
link. 

(4) See that the parts are given 
such dimensions that the re- 
quired motion is possible. 

(5) Avoid arrangements that give 
heavy masses reciprocating mo- 
tions, and balance rotating parts 
as far as possible. First make 

FIG. 42. free-hand sketches in note books 

and submit before beginning 

finished drawing. Make the drawing neat and workmanlike, but 
do not waste time drawing small details such as keys, bolts, etc. 

2. Invert the hypocyclic straight line motion Fig. 41 so that the smaller 
gear 2 revolves eccentrically around a fixed center, and the larger gear recip- 
rocates in a straight line without turning. 

3. Invert the shaper quick-return mechanism Fig. 42 so as to obtain 
a quick-return mechanism in which there are two cranks, one of which may 
rotate at constant speed and the other at a variable speed. 

NOTE. This inversion may be accomplished by simply changing dimen- 
sions, retaining link 1 as the stationary link. 

Work independently in solving these problems. 

REFERENCES 

Klein, Kinematics of Machinery; Weisbach, Machinery of Transmission, 
Introduction; Rankine, Machinery and Mill Work; Kennedy, Mechanics of 
Machinery. 




CHAPTER II 
MOTION OF RIGID BODIES 

18. Plane Motion of Rigid Bodies. Plane motion of a body is 
defined as motion such that all points of the body move in parallel 
planes. For example, a shaft rotating in a bearing or a slide 
moving between parallel guides has plane motion. In many 
important machines all the moving points have motion of this 
kind. 1 In general the motions of any such link are completely 
determined by the motions of any two points, or of any line, 
lying in the plane section under consideration. In Fig. 43 let the 
irregular figure represent any 

rigid body having plane mo- 

tion and suppose that the 

points A and B are constrained 

to follow the paths m and n 

respectively. For any chosen 

position of A there is a per- 

fectly determinate position for 

B, and from these two the 

position of any third point C 

of the system is completely 

determined. It is therefore 

sufficient in studying the plane motion of any rigid body to confine 

the discussion to the motion of any line such as AB, 

19. Rotation about a Fixed Center. The simplest case of 
plane motion is that of rotation about a fixed center. In Fig. 44 
let PAiBi be the rigid system rotating about the center P. After 
an interval of time the body has turned through an angle 6 and 
has reached the position PA 2 B 2 . Evidently the angle AiPA 2 = 

1 In such machines the links may be represented by plane sections or pro- 
jections on a plane parallel to the planes of motion. 

29 




. 43. 



30 



MOTION OF RIGID BODIES 



BiPB 2 = 0, and the line A 2 B 2 also makes the angle 6 with A\B\. 
The displacements of AI and BI are proportional to the radii 
PAi and PBi. Measured along the arcs these displacements 
are PA\ 6 and PB\ respectively. Measured along the lines 

f) fi 

AiA 2 and BiB 2 the displacements are 2PAi sin - and 2PBi sin -, 

2 i 



therefore 



displacement of A i PA i 



displacement of BI PB\ 
If the center P is removed to an infinite distance from AI and 

PA\ 

the ratio ^=- approaches 1 as a limit and the angle 



_ - f 




0=2 sin -if - ) approaches 0. If the distance PA\ becomes 

infinite the motion becomes a translation. In this case 6 equals 

and A\A 2 = BiB 2 , or in 
other words the displace- 
ments of all points of the 
system are equal and 
parallel. 

20. Any Displacement 
Equivalent to a Rotation. 
Suppose the line A\B\, 
Fig. 44, to be moved to 
the position A 2 B 2 by any 
path whatever. Join A iA 2 
and BiB 2 and on these 
lines erect perpendicular bisectors which intersect at P. Then 
the displacement from A\B\ to ^2-62 could be effected by a single 
rotation about P. The truth of this statement is evident from 
the equality of the triangles PA\B\ and PA 2 B 2} the three sides 
of one being equal to three sides of the other. 

21. Instantaneous Center. If the line AiBi, Fig. 44, is part 
of a machine it is generally impossible for the displacement to 
take place in the form of a simple rotation, for the points AI and 
BI will usually not be able to travel in circular arcs about P. 
Take for example the connecting rod of a steam engine, Fig. 45. 
If this were a free body it could be moved from the position AB 
to A'B' by a simple rotation about point P. But the constraints 
imposed upon the motions of the rod by the crosshead and crank 



FIG. 44. 



INSTANTANEOUS CENTER 



31 




FIG. 45. 



make such a motion impossible. If, however, the displacement 
be made extremely small as in Fig. 46, then a very small rotation 
about P may become possible if there is any lost motion in the 
joints. In the limit, if 
the displacement be re- 
duced indefinitely, the 
point P approaches PO, 
the intersection of the 
normals erected at A 
and B to the paths of 
these points. For this 
case an infinitesimal 
rotation around PO is 

possible. The motion of the connecting rod may then be con- 
sidered as composed of a series of infinitesimal rotations about 
a series of centers PO. As each point PO is the center of rotation 

only for an instant 
it is known as an 
instantaneous center. 
What has been 
said about this par- 
ticular case applies 
equally well to any 
link in any mech- 
anism; the instan- 
taneous motion may 
be regarded as a 
rotation about an 
instantaneous cen- 
ter, and the total 
motion as a series 
of infinitesimal ro- 
tations about a series of instantaneous centers. In the case of 
a link rotating about a fixed point the instantaneous center always 
coincides with the fixed point. 

22. Properties of the Instantaneous Center. In the preceding 
article the instantaneous center PO was determined by drawing 
normals to the directions of the motion of two points on the link, 
Fig. 45. Now this center may be used in turn to determine the 




FIG. 46. 



32 



MOTION OF RIGID BODIES 



direction of the instantaneous motion of any other point C on the 
link. This direction is normal to the instantaneous radius PoC. 
It has been shown that the displacements of points on the link are 
proportional to the instantaneous radii. As the displacements 
take place in the same element of time dt it follows that the veloci- 
ties are also proportional to the radii. Expressed mathematically: 
Vel. A; Vel. B; Vel. C = P A; P B', P C. 

The size and shape of the link has no effect on this motion so 
long as the relative positions of the pairing elements are unchanged. 
We may, therefore, imagine a plane attached to the link and 




y/////////////////////////, 



FIG. 47. 



moving with it, that is, we may imagine the link expanded into a 
plane of indefinite extent, but still constrained by the pairs A and 
B. If this plane is extended to include the instantaneous center 
Po (see Fig. 47) then PQ is a point on the plane which has no motion. 
A pin might be stuck through the plane at this point into the fixed 
link without interfering with the instantaneous motion. The 
instantaneous center might then be defined as the point on the 
link (extended if necessary) which for the instant is at rest. 

The only possible instantaneous motion of a link in a mechan- 
ism is a rotation about the instantaneous center. Hence, any 
force applied to the link tends to produce such rotation. The 



COMBINED ROTATION AND TRANSLATION 



33 



direction of motion is determined by the sense of the moment of 
the force about the instantaneous center. 

23. Combined Motions. In many cases it is more convenient 
to regard the motion of any link as the resultant of several motions 
rather than as a simple rotation about an instantaneous center. 
For example, the 
motion of the 
wheels of a lo- 
comotive may be 
regarded as a 
translation with \|\ % ^ 

the engine to- 
gether with a ro- 
tation about the 
axles; the motion 
of the connect- 
ing rod of an 
engine may be 
regarded as made 

up of a translation with the crosshead together with a rotation 
around the wrist pin. Many other examples will occur to the 
reader. 

Referring again to Fig. 44 it was seen that the displacement 
from AiBi to ^2-62 was effected by a simple rotation about P. 
This displacement, however, could be effected in many other ways. 
For example, in Fig. 48 the triangle A\B\P is first rotated about 
AI through an angle 6. This brings A\B\ to the position A\B', 
which is parallel to A 2 B 2 , and shifts P to P' a distance of 




FIG. 48. 



If now the whole triangle AiB'P' is given a transla- 



2AiP sin -. 
2 



tion AiA 2 = B'B 2 = P'P = 2AiP sin -. AI will fall at A 2} B r at B 2 

and P' at P. Thus the original rotation around P has been 
replaced by an equal rotation about A \ together with a translation 

2AiP sin -. From this result the following principle is deduced : 

A rotation about any center is equivalent to an equal rotation 
about another center together with a translation equal to twice 

a 

the distance between the centers times the sine of -. In the case 



34 



MOTION OF RIGID BODIES 



of an infinitesimal rotation dd about an instantaneous center, 

2 sin , becomes equal to d8, and therefore an infinitesimal 
& 

rotation dd about an instantaneous center is equivalent to an 
equal rotation about another center together with a translation 
pd6, where p is the distance between the centers. The direction 
of the translation is normal to the instantaneous radius p and is 

readily seen to be the 
displacement of the 
second center. 

As an example con- 
sider the motion of 
the sprocket wheel 6, 
of a bicycle, Fig. 49. 
The bicycle is moving 
forward with velocity 
V=RV. The sprock- 
et wheel therefore has 
a combined motion of 
translation with velo- 
FIG. 49. city V, together with 

a rotation about its 

center with angular velocity co. Since the chain runs without 
slip 

V 




r=Qn= 






(1) 



The total motion of the sprocket wheel is a rotation with angular 
velocity co about some instantaneous center P. The velocity of 
the center is given by the equation 



yn 

Rr' 



and therefore 



OP = R-. 
n 



(2) 
(3) 



Equation (3) gives the distance of the instantaneous center P 
from the center 0. Since the instantaneous radius OP must be 
perpendicular to the motion of 0, P lies vertically below 0. 



INSTANTANEOUS CENTER OF RELATIVE MOTION 



35 




The motion of the sprocket wheel can be exactly replaced by 
the rolling of a wheel of radius OP along a horizontal line MN. 

EXERCISES 

1. A wheel rolls along the ground as in Fig. 50. Compare the results 
obtained for the velocity of the point A, (1) regarding the wheel as rotating 
about P with angular velocity <a. (2) re- 
garding the wheel as having a combined 

motion of translation with the center 
and rotation about O. 

2. In Fig. 46 compare the results 
found for the velocities of B and C (1) 
regarding the rod as rotating with angu- 
lar velocity co about P, (2) regarding 
the rod as having a combined motion 
of rotation about A and translation 
with A . 

3. The wheel Fig. 50 is 6 feet in 
diameter and rotates at 200 r.p.m. If 
the center advances at 60 feet per second 

find the speed of slipping; find the instantaneous center. Compare the 
velocity of slipping found by the method of instantaneous centers with that 
found by the method of combined rotation and translation. 

24. Instantaneous Center of Relative Motion. Let the two 

irregular bodies 2 and 3, Fig. 51, represent two links of a mechan- 
ism. These links may be regarded as two planes of indefinite 
extent rotating about the instantaneous centers P% and PS. 
Any point M may be considered as covering two coincident points 
one on each link. These points may be distinguished as M?. and 
MS. As M2 is rotating about P% with angular velocity o>2 and 
MS about PS with angular velocity 003, the velocities of these 
points will in general be different in both magnitude and direction. 
If the point is chosen somewhere on the line joining P% and PS 
as at A the directions of the velocities will be the same but the 
magnitudes will in general, be different. There exists, however, 
one point where the velocities will coincide in both magnitude 
and direction. This point may be found as follows: From any 
point A on the line joining P2Ps draw vectors representing the 
velocities of At and A%. Join the ends of these vectors to P% 
and PS respectively. From the intersection of these lines drop a 
perpendicular on P^Pz. The foot of this perpendicular B, is a 
point whose velocity is the same whether considered as a point 



36 



MOTION OF RIGID BODIES 



on link 2 or 3. The point B is called the instantaneous center 
of relative motion. If the mechanism is inverted so that link 2 
becomes stationary, B becomes the instantaneous center for link 3. 
It is the only point where a pin could be placed connecting links 
2 and 3 without interfering with their instantaneous relative 
motion. 

25. Law of Three Centers. The centers P 2 and P 3 , Fig. 51, 
are the instantaneous centers of motion of the links 2 and 3 rela- 
tive to the fixed link say the paper. If we call this link 1, we 
see that the centers of relative motion between 1 and 2, 2 and 3, 
and 1 and 3 lie in a straight line. If some other link 4 (not shown) 
were taken as the stationary link, links 2 and 3 would rotate 
about two other centers Q 2 and Qs- These are the centers of 




FIG. 51. 

relative motion between links 2 and 4, and 3 and 4. Reasoning 
similar to that in the last paragraph will show that the center of 
relative motion between 2 and 3 will lie on the line Q 2 Qs. 
In general, if three links r, s, and t are chosen, the centers of rela- 
tive motion between r and s, r and t, and s and t, lie in a straight line. 
This law is called the law of three centers. For the sake of brevity 
instantaneous centers of relative motion will be denoted by the 
symbols 12, 16, 28, etc., where 12 represents the center of relative 
motion between links 1 and 2, etc. These symbols should be read 
one-two, one-six, two-eight, etc., not twelve, sixteen, twenty- 
eight. The law of three centers is of great importance in the 
theory of machines and is used extensively in studying machine 
motions. 



LAW OF THREE CENTERS, APPLICATION 



37 




26. Application of the Law of Three Centers. It is usually 
possible to find all of the instantaneous centers of relative motion 
in a mechanism by the law of three centers. As the simplest case, 
consider the 4-link chain, Fig. 52. Here are four systems all 
having motions relative 
to one another. In all, 
there are six relative 
motions and conse- 
quently six centers of 
relative motion, name- 
ly: 12, 13, 14, 23, 24, 
34- Of these six rela- 
tive motions, four are 
rotations about the axes 
of the joints namely: 
12, 23, 34, 14; and of FIG. 52. 

the six centers, four are 

located at the axes of the joints as shown. The remaining centers, 
13 and 24 are found by the law of three centers. Thus, consider- 
ing the links 1, 2 and 3, the centers 12, 23 and 13 must lie in a 
straight line. As the centers 12 and 23 are known, this determines 
a locus for 13. Again considering links 1, 3 and 4> the centers 

14, 34 and 13 must 
lie in a straight line. 
This determines a sec- 
ond locus for 1 3, which 
must, therefore, lie at 
M, the intersection of 
1 the loci. Similarly, 24 

must lie on the lines 
12-14 and 23-34 and 
is consequently locat- 
ed at N, the intersec- 
F IG 53 tion of these lines. It 

is evident that if 2 is 

held stationary, N is the instantaneous center of rotation of 4, and 

that if 3 is held stationary, 13 is the center around which 1 rotates. 

In Fig. 53 is shown the slider crank or steam engine mechanism. 

In this case, since link 4 has a motion of translation, the center 14 




38 



MOTION OF RIGID BODIES 



lies at an infinite distance. To connect any point to 14, simply 
draw a vertical line through that point. The centers 12, 23 and 

14 are located at the axes of 
the turning joints. To find the 
unknown centers, 13 and 24, 
use the law of three centers. 13 
lies on the lines 12-23, and 14- 
43. 24 lies on the lines 12-14 
and 23-34- This may be con- 
veniently represented as follows: 




13 



12-23 
14-34 



24 



12-14 
23-34 



As an example of the applica- 
tion to more complicated mechan- 
isms, consider the shaper mech- 
anism, Fig. 54. In general, if a 
mechanism has n links, there are 

n(n 1) , ,. ,. 

^- -centers of relative motion. 
2 

Thus, in the six-link mechanism there are - - -=15 centers. 
These may be conveniently tabulated as shown below: 



FIG. 54. 



(24) 




(25) 

35 
45 



(26) 
36 
46 
(56) 



In this table all known centers may be indicated by small full 
circles. An examination of the mechanism shows the following 
known permanent centers: 12, 14, 16, 23, 25, 34, 56, and these 
are accordingly enclosed in full circles in the table. The remain- 
ing 8 centers are to be determined by the law of three centers. 
Further examination shows two quadric or four-link chains in the 



INSTANTANEOUS CENTERS, LOWER PAIRS 



39 



mechanism, namely 1234 and 1256. The centers 13, 24, 15 and 26 
are then readily found by the same method as that employed in 
Fig. 53. These are now marked in the table by dotted circles. 
The remaining centers are found as follows: 



35 



36 



(13) -(15) 

()-() 
(16) -(13) 



(23) -(26) 



@-() 

(l4)-(43) 



!25)-(56) 




12-C14 



46 



These lines coincide, so a third locus 
must be found. 



As checks on the accuracy of the work, it should be noted that 23, 
26, 36, must lie in a straight line; 34, 35, 45 must lie in a straight 
line; 35, 36, 56, must lie in a straight line; 45, 4, 56 must lie in a 
straight line. 

27. General Directions. The following general directions will 
be found useful: 

1. Construct a table such as shown in the preceding article 

including all the centers. 

2. Indicate all known centers by drawing full circles around 

them. Note that all axes of turning joints are centers, 
and that the centers of sliding pairs are at an infinite 
distance in a direction perpendicular to the direction 
of relative motion. 

3. Examine the mechanism and pick out any quadric 



40 



MOTION OF RIGID BODIES 



chains. If necessary, draw the skeleton in its typical 
form to assist in picking out these chains. As in Figs. 
52 and 53 intersections of opposite sides of quadric 
chains give additional centers which may be marked 
by dotted circles. 

4. Apply the law of three centers to find any center which 

is still unknown. 

5. Test the accuracy of the work by drawing all loci which 

pass through each center. If there are n links, then 
there are n 2 loci for each center, e.g., the center 12 

13-23 

14-24 

15-25 



lies on the lines 



ln-2n 



If one of these lines, say 15- 
it passes through 12. The total number of loci l is 



, has not been drawn, see whether 

n(n-l)(n-2) 
1.2.3 " 



28. Higher Pairs. To determine all the instantaneous centers 
of mechanisms with higher pairs, it is usually necessary to draw 
normals to paths of points or to contact surfaces. One or two 
examples will illustrate this statement. 

In Fig. 55 two members, 2 and 3, rotate about fixed points at 
12 and 13, respectively. Link 3 drives link 2 by direct contact, 
thus giving a case of higher pairing. It is required to determine 
the center 23 of the relative motion of links 2 and 3. Since links 
2 and 3 cannot cut into each other, and since they do not separate, 
the relative motion of the points in contact must be along the 
common tangent t at the point of contact P; for if the relative 

1 This is not as long and tedious a process as it may appear. Thus, 
for the six-link mechanism, Fig. 54, there are 20 loci as follows: Those not 
drawn on the figure are indicated by stars. Such loci are useful as checks 
on the accuracy of the work. 

25-45 
26-46 



12 



13-32 
14-42 
15-52 
16-62 



13 



14-43 
15-53 
16-63 



14 



15-54 
16-64 



15 1 16-65 



23 



24-43 
25-53 
26-63* 



24 



25 26-56 



35-54* 
36-64 



35(36-56* 45 1 46-56' 



INSTANTANEOUS CENTERS, HIGHER PAIRS 



41 



motion has a component along the normal, the members must 
either cut into each other or separate. But the motion of 2 relative 
to 8 is a rotation about the center 23 and evidently, therefore, this 
center must lie on a line perpendicular to the direction of the rela- 




-7~-* 25 




FIG. 56. 



tive motion, that is, perpendicular to the tangent t] hence 23 
lies on the common normal drawn through P. By the law of 
three centers, 23 also lies on the line 12-13] hence it lies at the 
intersection of the latter line with n. 

In the case of parts rolling upon each other as the pitch lines 
of circular or elliptical gears, Fig. 56, the point of contact must 
lie on the line of centers and is 
itself the instantaneous center of 
the two systems. 

In Fig. 57 there is a higher pair 
between the pin on the block 2 and 
the slot in the swinging arm 3. 
The relative motion of the pin and 
slot is a sliding along the slot, and 
the center 23 lies therefore in a line 
through the pin perpendicular to 
the direction of the center line of 
the slot. The center 12 is at in- 
finity. The locus 12-13 is there- 
fore a line through 13 perpendicular 
to the motion of link 2. By the 

law of three centers 23 lies also on the line 12-18 and is there- 
fore fully determined. 




FIG. 57. 



42 



MOTION OF RIGID BODIES 



29. Special Cases. In some cases the law of three centers alone 
is insufficient to determine all the instantaneous centers. Thus 
in the skeleton mechanism, Fig. 58 (which is the same as Fig. 28), 
the centers 13 and 24 are readily found; but it may be shown by 
trial that none of the remaining centers can be determined by the 
law of three centers. Another center may be found by a sort of 
trial and error method as follows: 



26, 



\ 



Locus a+ 




FIG. 58. 

The center 26 is known to lie on the line 28-68. Assume this 
center to be at some point on this line as 26 \. Then other centers 
may be determined as follows : 



15, 



12)- 



gfl- 16! 



- 65 




18- 86! 



23)-^6j 
- 



A trial position of center 15 is thus located. But 15 must lie on 
the line 1J+-1+5. Hence the assumption as to the location of center 
26 at 26 \ is incorrect. Assume center 26 to lie at 26 2 and repeat 



SPECIAL CASES, CENTRODES 



43 



the process as shown. By a series of such assumptions a number 
of trial positions of 15 are found. These determine a locus for 15, 
and the point where this locus cuts the line 1^~45 is the true posi- 
tion of 15. After 15 has been located correctly the remaining 
centers can be found by the law of three centers. 1 

30. Centrodes If any link, say link o, of a machine has 
motion relative to the fixed link 1, then this motion may be regarded 
as a series of rotations around successive instantaneous centers. 
If a curve be drawn connecting all the positions of the instanta- 
neous center this curve is called the fixed centrode. This locus may 
be regarded as a curve attached to link 1 and always containing 
the center 15. Similarly, if a plane be attached so as to move 




with link 5, a second curve which always contains the center 15 
can be drawn on this plane. The second curve is called the 
moving centrode. For example, if a wheel rolls along a rail, the 
instantaneous center of the wheel is always at the point of contact 
with the rail. The top of the rail is, therefore, the locus of this 
center, that is, the fixed centrode. The run of the wheel also 
always contains the instantaneous center and is, therefore, the 
moving centrode. A second example is shown in Fig. 59. Two 
points A and B of a rod are caused to move along two grooves 

1 When this method is followed the locus is usually found to be a straight 
line. In some cases, however, conies or even higher degree curves are involved. 



44 MOTION OF RIGID BODIES 

which intersect at right angles. The instantaneous center 13 is 
readily located at P. As the distance OP is always equal to AB, 
it follows that the fixed centrode is a circle about with radius OP 
equals AB. The moving centrode is easily found to be a circle 
whose diameter is AB. For the angle APB is always a right 
angle, and the locus of a point at which a line AB subtends a right 
angle is a circle whose diameter is AB. Thus if the bar moves 
(carrying the circle APB with it) to the position A'B f t the new 
center P' lies on this circle. 

In most cases the geometrical character of the centrodes is 
not easily determined. They must then be constructed point by 
point. A convenient way to construct the moving centrode is as 
follows : On a sheet of tracing paper, make a drawing of the moving 
link, as AB. Place this tracing in the successive positions of the 
link as AB, A'B' , etc., and at each position mark on the tracing 
paper the corresponding instantaneous center. The points thus 
located on the tracing paper determine the moving centrode. 

31. Centrodes of Relative Motion. The relative motion of any 
two links may be considered as a series of rotations about suc- 
cessive instantaneous centers of relative motion. A pair of 
curves, one attached to and moving with each link, and which 
always contain this center, constitute the centrodes of relative 
motion of these links. Each of these centrodes is readily found 
by means of a sheet of tracing paper as previously described. 
There is, therefore, a pair of centrodes for each pair of links in a 

chain, or if there are n links, l - pairs in all. In case two links 

l 

are connected by a turning joint, both centrodes reduce to a point 
the axis of the joint. 

EXERCISES 

4. In Fig. 60 let 1 be the fixed link and 2 and 4 the cranks rotating about 
P 2 and P 4 respectively. Taking link 3 as a 
moving system S, construct the fixed and 
moving centrodes. 

Make P 2 A =P 4 B = 3" and P 2 P* = AB=2". 

Construct centrodes of relative motion 
between links 2 and 4. 

5. The moving system S, Fig. 61, consists 
of two lines rigidly connected and making an 
FIG. 60. angle of 60 with each other. The motion of 




INSTANTANEOUS CENTERS. EXERCISES 



45 



the system is such that one of these lines always passes through a fixed point 

M, the other through a fixed point N. Con- 
struct the centrodes. Make the distance 
MN = 4". 




FIG. 63. 

6. In the slider crank mechanism, Fig. 62, take 1 as the fixed link, and 3 
as the moving system S, and construct the centrodes attached to 1 and 5, 
respectively. Let AB = l" and BC = 3%". 

7. In the same mechanism take 4 as the fixed link and construct the cen- 
trodes attached to 4 and 2, respectively. 

8. In the six-link mechansim, Fig. 63, construct the centrodes attached to 
links 3 and 6, respectively. 



46 MOTION OF RIGID BODIES 

32. Motion Produced by Rolling of Centrodes. At the point 
of contact between two centrodes, that is, the instantaneous center 
of relative motion, there can be no relative motion between the 
two links to which the centrodes belong. Therefore the centrodes 
roll on each other without slip. Consequently the motion of any 
links can be reproduced by simply rolling the two centrodes 
together.- For example, in Fig. 59 if the bar AB be detached from 
the guides, links 2 and 4) its motion can be reproduced by simply 
rolling the smaller circle on the inside of the larger. This is accom- 
plished in the mechanism shown in Fig. 41. Here the large internal 
gear 1 is held stationary and the smaller gear link 3 is caused to 
roll inside 1 by means of the crank link 2. It is easily proved that 
two points, A and B, on the pitch circle at opposite ends of a 
diameter travel in straight lines which are perpendicular to each 
other. The bar 4 adds nothing to the constraint. It simply 
emphasizes the fact that B must travel along a vertical diameter 
of the larger gear. 

33. Equivalent Mechanisms. Since the motions of the links 
may be completely replaced by the rolling of the centrodes on 
each other, it follows that mechanisms which give the same cen- 
trodes are kinematically equivalent. This equivalency can be 
shown to exist in machines bearing no external resemblance to 
each other. Thus the elliptograph, Fig. 59, the Davis engine, 
Fig. 34, the hypocyclic parallel motion, Fig. 41, and the slider 
crank or steam-engine mechanism where the crank and connecting 
rod are of the same length, all give the same centrodes between 
links 1 and 8. These mechanisms therefore are all equivalent to 
each other, and suitably chosen points on each may be shown to 
follow identical paths. 

34. Gears as Centrodes. In geared mechanisms, the pitch 
lines of the spur gears (whether circular or not) roll on each other 
without slip. These pitch lines are therefore centrodes. In the 
crossed four-bar chain, Fig. 60, for example, the centrodes between 
links 1 and 3 are found to be ellipses. The whole mechanism 
might then be replaced by a pair of elliptical gears each rotating 
about a focus of the ellipse. In any mechanism the motions of 
any two links may be reproduced by substituting for these links a 
pair of gears whose pitch lines are the centrodes of relative motion 
of these links. Of course in many cases this is practically impos- 



ROLLING CENTRODES, GEARS AS CENTRODES 



47 



sible; the centrodes may be of infinite length, or of shapes quite 
unsuited for gears; but theoretically this substitution is always 
possible. 

35. Pitch Lines. The determination of pitch lines of 
gears may be approached from two points of view. If the 
gears are to replace some linkage whose properties are known 
then the pitch lines must be the centrodes as shown in the 
preceding paragraph. If on the other hand the pitch line of 
one gear is arbitrarily chosen 
a second gear can be designed 
so as to insure proper mesh- 
ing with the first. In Fig. 
64 let the closed curve 2 rep- 
resent the arbitrarily chosen 
pitch line of one gear, which 
is to rotate about the center 
P2. It is required to find 
the pitch line of a second 
gear which will rotate about 
Pa and mesh with the first. 
As the pitch curves are to 
roll on each other, the in- 
stantaneous center of relative 
motion is at the point of 
contact, which therefore by 
the law of three centers must 
lie on the line PzPs. It 
follows that the sum of the 
radii to the point of contact must be constant. Divide the 
perimeter of gear 2 into parts so small that the length of each 
chord may be considered equal to its arc. From A strike an 
arc of radius AB, and from PS an arc of radius P2Ps-P2#. 
The intersection of these arcs gives the point B' of the second 
gear, which is to come in contact with B. From B f strike an 
arc of radius BC and from Pa an arc of radius PzP^-P2C. 
The intersection of these gives C". This process can be con- 
tinued until the profile of the second gear is completed. 1 

Certain special forms of pitch lines such as ellipses and log- 
1 Ordinarily two non-circular gears will have the same perimeter, or 




FIG. 64. 



48 MOTION OF RIGID BODIES 

arithmic spirals are found to be suitable for non-circular gears. 1 
Of course in the case of spirals the curves must have points of dis- 
continuity. 

36. Gear Teeth. The actual contact between gear teeth is in 
general not at the pitch point but at one side. Consequently the 
actual pairing between two toothed wheels is of the same nature 
as that described for cams. In order that the motion of the wheels 
shall be actually equivalent to the rolling of the two pitch curves 
on one another, the instantaneous center of relative motion must 
lie at the pitch point. But this center lies on the common normal 
at the point of contact. Consequently for correct action the teeth 
must be cut in such form as to satisfy the condition that the 
common normal always passes through the pitch point. The- 
oretically if the profile of the teeth of one gear is arbitrarily chosen, 
a profile can be found for the second so as to give correct motion. 2 
In practice only two forms are used extensively, namely the 
involute and cycloidal. The discussion of the form and construc- 
tion of tooth profiles will be found in Chapter X. 

37. Space Motion. Axodes. In all the preceding work it 
was assumed that every point of the mechanism had plane motion 
only. In the more general case, the points may travel in any paths 
in space whatever. The most general instantaneous motion of a 
rigid body in space may be shown to be equivalent to an infin- 
itesimal rotation about some axis together with a translation along 
that axis. For example, a nut moving on a stationary bolt rotates 
about the axis and at the same time travels along the axis. The 
most general motion possible may be resolved into a series of such 
infinitesimal rotations and translations. 

If a surface be passed through all the positions of the instan- 

their perimeters will have a simple ratio such as 2 : 1, 3 : 1, etc. It should 
be noted that in general the process described will not give closed pitch curves 
which satisfy this condition. That is, the gears cannot make complete revo- 
lutions. The conditions which must be satisfied in order that the second gear 
shall make one revolution while the first makes 1,2, . . . .n, revolutions, 
are discussed in note (A). The application of these principles is a matter 
of considerable difficulty. The converse problem of making the second gear 
revolve 1,2, . . . .n, times for each revolution of the first, is in general not 
capable of solution. 

1 Dunkerly, Mechanism, pp. 339-346. 

2 Dunkerly, Mechanism, pp. 275-280 and 353-356. 



GEAR TEETH AXODES 49 

taneous axis, this surface is called an axode. Exactly as in the 
case of centrodes, there exists for any constrained motion of a rigid 
body a fixed axode and a moving axode. The moving axode is a 
second ruled surface attached to the body and moving with it 
and which always contains the instantaneous axis of rotation. 
The motion of the body may be exactly reproduced by rolling the 
moving axode on the fixed, and at the same time sliding it along 
the element of contact. Also the relative motions of two moving 
bodies may be replaced by the rolling and sliding of two axodes. 1 
The only important applications of axodes are in conical and 
hyperboloidal (skew bevel) gears. In conical gears the axes of 
rotation intersect, and the motion is exactly equivalent to the 
rolling of the two pitch cones on each other. The axodes are these 
pitch cones. In this case there is no slip in the direction, of the 
line of contact. In skew bevel gears the axodes are the hyper- 
boloidal pitch surfaces and here both rolling and sliding occur. 

1 For further discussion of the question of axodes the reader is referred to 
Weisbach's Mechanics of Engineering and Machinery, Vol. 2, introduction. 



CHAPTER III 
VELOCITIES IN MECHANISMS 

38. Introductory. In the preceding chapters the motions of 
machine parts have been considered without particular reference 
to the velocities or rates of motion. In the present chapter 
methods will be developed by means of which the velocities in 
all parts of a machine can be completely determined. These 
methods may be classified as Analytical and Graphical. Of the 
latter there are two general classes namely, the method of Instan- 
taneous Centers, and the method of Relative Velocities. In some 
cases a combination of these two may be profitably employed. 




FIG. 65. 

39. Analytical Methods. Analytical methods are used exten- 
sively in geared mechanisms and in some very simple linkages. 
For more complex linkages the mathematical difficulties become 
so great and the equations so cumbersome that graphical solu- 
tions are much to be preferred. 

(a) The most useful field for the employment of analytical 
methods in finding velocities is in the case of geared mechanisms. 
Consider the ordinary gear train, Fig. 65. Let a, b, c, d } e, and / 

50 



VELOCITIES, ANALYTICAL METHOD, GEARS 



51 



represent the numbers of the teeth in the gears A, B, C, D, E, 
and F respectively. It is required to find the ratio of the angular 
velocity of gear F to that of gear A. The numbers of teeth in 
two gears which are in mesh are proportional to the circumfer- 
ences of the pitch circles. As the pitch circles roll on each other 
without slip, evidently the number of revolutions of the gears 
are inversely proportional to the circumferences of the pitch 
circles, or to the numbers of teeth. 
Therefore: 

co( = angular velocity of B) _ _o 
co a ( = angular velocity of A) b' 
or 

a 
a 6* 

If B and C are rigidly connected so that they must rotate together 
co c = co 6 ; similarly w<i= e = co c (-ij ==w a(r'j)> 



and 



hence 



a c e 



^ = etc 
eo a bdf 



From this analysis we deduce this important law of gearing: 
The angular velocity of the driven wheel F is to the angular 
velocity of the driver A as the continued product of the numbers 
of teeth in the driving wheels (A, C, E) is to the continued prod- 
uct of the numbers of teeth on the driven wheels (B, D, F). 
Attention must be paid to the direction of rotation. Thus if A 
rotate in the counter-clockwise (positive) direction B and C will 
rotate in the clockwise (negative) direction, D and E counter- 
clockwise and F clockwise. The angular velocities are completely 
described therefore in the following table: 



Gear 


A 


B 


C 


D 


E 


F 


Angular veolcity. 


u a 


/a\ 
-"(l) 


-~(S 


+wa (^) 


+- 


lace\ 
~ aa \bdf) 



52 



VELOCITIES IN MECHANISMS 



(6) In many cases the same gear appears both as driver and 
driven. Such a gear is called an idler, and is shown in B, Fig. 
66. For this case 




FIG. 66. 



It follows that an idle gear does not change the magnitude of the 

velocity ratio, but does 
alter the direction or 
sign. 

(c) In many cases 
the pitch circles of two 
gears are tangent intern- 
ally instead of externally. 
In these cases the larger 
gears are called internal 
gears. Thus in Fig. 67, 

A is an internal gear. The velocity ratio is found as before: 

COb _OL 
^a = b' 

It should be noted, however, that in this case the gears rotate 
in the same direction. 

(d) In the foregoing examples it has been assumed that each 
gear rotates about a fixed axis passing through its center. In 
many mechanisms one gear is held stationary and a second one 
rolls round its circumference. Such an arrangement is shown 
in Fig. 68. The gear A link 1 is held fixed and gear C, link 3, 
rolls on A, the arm B } link 2, serving to hold the two in mesh. 
In this arrangement C is called an epicyclic or planetary gear. 
The motion of C may be regarded as composed of a rotation with 
B about the center 12 together with a rotation relative to B about 
the center 23. Taking first the motions relative to B, if A be 

given + 1 revolution. C will make - revolutions relative to 

c 

B. Now the whole mechanism may be given such a motion with 
B that A is brought back to its original position. That is, the 
whole mechanism must be given one revolution in the negative 



VELOCITIES, ANALYTICAL METHOD, GEARS 



53 



direction. The total motion of each link is then given by the 
following table: 



A 


c 


B 




a 1 a c 


01 i 


-f-1 1 U 


c c 





It follows that C must rotate in the same direction as B and that: 

Wc = a-j-c 
co& C 





FIG 67. 



FIG. 68. 



(e) The same scheme. can be followed if we add other gears 
such as D and E shown dotted in Fig. 68. If D be made to revolve 
with C, and E be free to revolve about 12, the motions of each 
are shovvn in the following table: 





A 


B 


C 


D 


E 


Motions relative to B 


1 


-0 


a 
c 


a 
c 


^ 


Motions with B 


-1 


-1 


-1 


-1 


-1 


Total motion 





-1 


-M) 


-M 


-1+2* 








\ c/ 


\ c/ 


^ ce 



54 



VELOCITIES IN MECHANISMS 



ad 



The gear E can, therefore, rotate either in the same direction 
as B or the opposite direction, according as <1 or >1. 

Ce 

Instead of the numbers of teeth the diameters might be used. 
If the diameter of A> the diameter of E, then ->1; also ->1. 

> C 

Therefore 1 <0 and therefore, the rotation of E is in the 

ce 

opposite direction to that of B, and vice versa. 

Another conclusion to be drawn is that, since >0 in all 

ce 

cases, <1, when the rotations are in the same direction. 

0) 6 

Since there is no limit to the ratio , it follows that the ratio 

ce co& 

can be made as large as desired when the rotations are in opposite 
directions. It also follows that if A and E are of nearly the same 

diameter the ratio becomes exceedingly small. For example, 

CO& 

if the numbers of teeth are chosen a = 51, c = 50, d=49, e = 50, 
then, 

we ad 2499^ 1 
co 6 ~ ce 2500 2500* 

This is a useful principle when 
very large reductions of speed 
are desired without the use of 
an excessive number of gears 
or very large gears. 

(/) The same principles may 
be applied to hypocyclic gear 
trains, that is, gear trains hav- 
ing a stationary internal gear. 1 
The development of the de- 
tails is left to the reader. 

EXERCISE 

In Fig. 69, gear A has 72 teeth, 
and B has 32 teeth. Find the 
number of revolutions of gears B 
FIG. 69. and C for 1 revolution of arm D. 

1 All of the results obtained thus far can be found by the graphical methods 
described later in the chapter. 




VELOCITIES, ANALYTICAL METHOD, BEVEL GEARS 55 

40. Bevel-gear Trains. (a) Bevel-gear trains may be treated 
in a manner similar to that used for spur gears. As before, the 
ratio of the angular velocities is given by : 

Continued product of numbers of teeth in driving wheels 
Continued product of numbers of teeth in driven wheels' 

The question of direction must, however, be treated differently. 
Fig. 70 shows an ordinary bevel gear train. As before: 




FIG. 70. 

But as B and A do not rotate in parallel planes we cannot say 
that o)& is either in the same direction as that of A, or in the 
opposite direction. Consequently, neither the + nor sign is 
appropriate to describe the direction. 

0)d_ _OC 

o^~ 53' 

o) _ . ac 
co a be 

Formal rules can be drawn up for determining the signs, but it 
is better to consider each case as it arises. 

(6) Mechanisms involving the use of stationary bevel gears 
are sometimes employed. The velocity ratios are determined in 
a manner analogous to that employed with epicyclic spur gear 
trains. As an example, consider Humpage's gear, Fig. 71. Gear 
A is stationary. Gears B and C rotate together about a shaft 
H which is carried by a bracket G, the latter in turn being free 



56 



VELOCITIES IN MECHANISMS 



GO rotate about the central shaft F. Gear B meshes with the 
driving gear E and C meshes with D. To determine the velocity 






ratio , consider the motions of all gears as composed of rotations 




FIG. 71. 

relative to H and a rotation with H around the central axis. 
The following table shows the results : 





A 


B 


C 


D 


E 


H 
















Motion relative to H . . 


+ 1 


b 


b 


_i_^i 
^bd 


e 





Motion with H 


-1 


-1 


-1 


-1 


-I 


-1 


Total motion .... 





-1 * 


i a 
1, T 


-(i-o) 


-6+-) 


-1 






' b 


' b 


\ M/ 


V e) 





1 The motion of B and C consists of two rotations about different axes: 
__ 7 1 about the central axis F, and a/6 

about H. These cannot be added alge- 
braically. They can, however, be com- 
bined vectorially into a single rotation 
about XX, whose magnitude can be found 
by the ordinary method of combining vec- 
tors. The pitch cones of A and B are in 
short the fixed and moving axodes for the 
motions of B and C. Fig. 72. 




FIG. 72. 



VELOCITIES, ANALYTICAL METHOD, SIMPLE LINKAGE 57 
Therefore: 



41. Velocity of Piston of Engine. In some very simple link- 
ages analytical methods may be used for determining velocities. 
As an example, consider the slider crank or steam engine mechan- 
ism. Here, the crank usually rotates at a nearly uniform speed 
and the problem is to find the speed of the piston. Referring to 

jrr 

Fig. 73, the velocity of the piston is expressed by . 




FIG. 73. 

From the geometry of the figure 

S = R cosa+L cos/8. 



Also 
Hence: 



L sin /3 = R sin a. 



I T? 2 

L cos /8 = LA/! -j~2 sin 2 a. 

('R\* 
Y ) sin 2 a is a small term we may write with sufficient 



*f-<fi-%mifa-l-gl*fa. ' 



accuracy, 



r>4 

This is equivalent to adding the small quantity sin 4 a under the 
radical. 



58 VELOCITIES IN MECHANISMS 

The same result can be obtained by expanding 



according to the binomial theroem and neglecting all except the 
first two terms. Substituting this value for cos/3 a new expres- 
sion for S is found : 

S = R cos a pry sin 2 a+L. 

iLt 

Differentiating with respect to the time, 

dS I R 2 \da / . . R _ \ 

= ( R sm . a ___2 sin a cos a }-rr= R[ sm a +777- sin 2a a>. 
at \ ZL /at \ ZL I 

It may be remarked that this velocity is much more easily found 
by graphic methods. 

42. Methods of Instantaneous Centers. Angular Velocity 
Ratios. Consider any two moving links S and T, Fig. 74. They 




FIG. 74. 

are rotating about some instantaneous centers P s and P t , with 
angular velocities w s and o>i respectively. The instantaneous 
center of relative motion st lies on the line P s Pt- The center st 
is defined as a point which has the same velocity whether con- 
sidered on link S or link T. Considered as a point on link S the 
velocity of st is given by the equation: 



Considered as a point on link T, 



VELOCITIES, INSTANTANEOUS CENTER METHOD 59 
Therefore 

c^ = P,-sf 

The angular velocities are measured relative to a third link R 
(say the paper) which is regarded as fixed. The centers P s and 
PI may, therefore, be denoted by rs and rt respectively. In 
general, therefore: 

o) tr rsst 



This equation is to be read as follows: 

" The angular velocity of link T relative to link R is to the 
angular velocity of link S relative to link R as the distance between 
the centers rs and st is to the distance between the centers rt 
and st." 

If the center st lies between rs and rt, then && and co* r are in 
opposite directions. If st lies beyond either of the centers rs or 
rt } then the two angular velocities are in the same direction. 

43. Examples. (a) In two gears revolving about fixed axes 
the instantaneous center of relative motion is the pitch point. 
The rule just derived then states that the angular velocities of 
the gears are inversely proportional to the radii of the pitch 
circles. 

(6) In the epicyclic gear train, Fig. 68, 



Compare this result with 
that obtained analytically in 
Art. 39. 

(c) In Fig. 75 link 2 
drives link 3 by direct con- 
tact. The instantaneous 
center 23 is at the point 
where the common normal 
intersects the line of centers 
12 -13. 




co 2 i~~13-23' 



FIG. 75. 



60 



VELOCITIES IN MECHANISMS 



That is, the angular velocities of the two links are inversely pro- 
portional to the segments into which the common normal divides 
the line of centers 12-13. 

(d) In the steam engine mechanism, Fig. 76, the angular 
velocity of the crank being known, it is required to find the angular 
velocity of the connecting rod. Again: 

12-23 

13-23* 




FIG. 76. 



44. Special Cases. (a) one of the centers involved may fall 
at infinity. For example, in the slider crank mechanism, Fig. 
76, the center 14 lies at an infinite distance. Therefore: 




C041 
0)21 



= 
14-24" oo 



In other words the cross- 
head, link 4, has no angu- 
lar velocity a result which 
is self-evident. 

(6) The instantaneous 
center of relative motion 
may fall at infinity. In 
the Whitworth quick-return 
motion shown in skeleton 
form in Fig. 77, there is a 
sliding pair between links 
3 and 4. 

cosi 14-34 oo 



FIG. 77. 



ANGULAR VELOCITY RATIOS, EXAMPLES 61 

The angular velocity ratio is therefore apparently indeterminate. 
The distance 14-34 may be written (14-13) + (18-84). 
Hence 

am = 14-13 13-34 = 
a>4i 13-34^13-34 

That is, 3 and 4 have the same angular velocity again a self- 
evident conclusion. 

(c) The ratio - - may be indeterminate. In the reverted 

Tt SI 

epicyclic gear train, Fig. 68: 



= _ 

co2i~14-24~0' 

To evaluate this indeterminate result write: 



0041 _ ^3i co4i _ 12 23 13 
~~ * 



13-2314-34' 

/ rA/^A 
e \ r b ]\r e /' 



r e 
or since the radii are proportional to the numbers of teeth, 



co ea ,< i a 

~ 



a(a+b) ae 1 _a(d+e) ae ad 

- : -- I , J. JL |"~7 , 

be be be 

which agrees with the value found by the analytical methods. 
It should be noted that this method does not give the direction 
of rotation. 

45. Relative Angular Velocity. If it is desired to find the 
relative angular velocity between two moving links $ and T the 
same line of reasoning may be employed. 

w st _rs rt 
co rs st rt' 



62 



VELOCITIES IN MECHANISMS 



For example, consider two gears, links 2 and 5, in mesh, 
Fig. 78. 

C02312 13 



T2 
C023= W21 + C021. 




FIG. 78. 



But 

Therefore 
For example, if 

co; 

and 

Wj 

then 



T2 

C021= C031- 

/ 3 



W23 = W21 W31. 

= +4 radians per second, 
= 5 radians per second, 
23 = +4 (5) = +9 radians per second. 



This equation may be generalized so as to cover any case, as 
follows: 

46. Linear Velocity Ratios. In many cases it is required to 
find the ratio of the linear velocities of points on different links. 
Let A and B, Fig. 79, be points on links S and T respectively. 

V a = velocity of A = P S A co s 
Vj,= velocity of B = 

p,- 



LINEAR VELOCITY RATIOS, SPECIAL CASES 



63 




Or 

P t B P s -st 

& P 4 * P Q/" 

rV* /j Sf 

This equation is perfectly 
general and may be ap- 
plied to any plane mech- 
anism. 

47. Special Cases. 
(a) The two points con- 
sidered may lie on the 
same link. In this case the equation reduces to 

v-v PB 

Va PA> 

where P is the instantaneous center. 

(6) One of the centers P s or P t may lie at infinity. In this 
case if P s is at infinity : 

TT- rr PtB 



FIG. 79. 



a P t -st' 

For example, in the slider crank mechanism, Fig. 80, the 
center 14 is at infinity. Suppose the velocity of point A, say on 

the flywheel, to be 
known. To find the 
velocity of a point D 
on the piston write: 

v _ OA 

Va ~ Vd oc 

since C is the center 24- 



We may interpret this 
result by noting that 
the center C (24) is a 
point which has the same 
velocity whether con- 
sidered on link 2 or 




FIG. 80. 



link 4- Since every point on link 4 has the same velocity, 
the center C might be briefly described as a point on link 2 



64 



VELOCITIES IN MECHANISMS 



which has the same velocity as link 4- Considering C as a point 
on link 2: 



Therefore, 



V a ~OA 
V^-V 

c - Va. 



(c) The center between the two moving links may lie at 
infinity. In this case the general equation reduces to : 



" 3vT 

For example, in the four-link mechansim, Fig. 81, the center 
24 lies at infinity. Given the velocity of A on link 2 to find the 
velocity of B on link 4- The 
center 14 is located at 0. 

OB 
Va P 2 A' 

This result may be interpre- 
ted as follows: Since links 2 and 
4 have the same angular velocity 
(see Art. 44) the velocities of 
points on these links are directly 




FIG. 81. 




proportional to the distances of the points from their respective 
instantaneous centers. 

(d) Sometimes the center of relative motion cannot be found 
by ordinary methods. In such cases special devices must be used. 



RELATIVE VELOCITIES 



65 



For example, in the shaper mechanism shown in Fig. 42 there 
are two positions where the center Jfi cannot be determined by 
the law of three centers. These positions are those where the 
center line of the oscillating lever is vertical, Fig. 82. In this 
case the centers are located as follows: 



13 



26 



36 



46 



12-23 
14-43 

12-161 
25-56 J 

13-16 
23-26 
35-56 
34-46 



14-16 
24-26 
54-56 
34-36 



at infinity 15 



at A 



35 



12-25 1 
16-65] 

13-15 1 
23-25 



at 



at B 



24 



12-14} 
23-34 J 



at B 



Of these loci, the first three coincide with the vertical 
center line, and therefore have no intersection. The 
fourth contains the unknown center J$. 



Of these loci again three coincide with the vertical 
center line and the fourth is unknown. Therefore, 
the centers 86 and J$ are indeterminate. 



The velocity of C can be determined from the following con- 
siderations: Since the center 13 falls at infinity all points on 
link 3 have- the same velocity. Hence F C =F 6 . Considering 
B now as a point on link 2 and A as a point on link 5. 

T _ Tr P%B 



according to the result obtained in special case (c). 

If desired, the location of 36 and 46 may now be found by a 
simple calculation. 

48. Method of Relative Velocities. While the method of 
instantaneous centers is a general one and may be applied to all 
plane mechanisms, many cases arise where its application is 
laborious and inconvenient. In mechanisms having a large 
number of links, the number of centers increases very rapidly, 
arid many of them may fall at considerable distances. On these 
accounts the constructions become tedious, and the chances of 
inaccuracies become serious. For complex linkages, a more 



66 VELOCITIES IN MECHANISMS 

direct and less troublesome method of determining velocities may 
be employed. 

The relative velocity of a point A with respect to a second 
point B is defined as the vector difference between the velocities 
of B and A. In other words, it is that velocity, which added 
vectorially to that of B will give the velocity of A. In Fig. 83, 
V a is the velocity of A, Vt> that of B, and Vab the relative velocity 
of A with respect to B. 

The relative velocity is unaffected by any motion which is 
given simultaneously to both points. Thus the relative motion 
of the piston and cylinder, or the wrist pin and crank pin of a 





FIG. 83. FIG. 84. 

locomotive is independent of the travel of the locomotive along 
the track. In particular, the entire system may be given a ve- 
locity equal and opposite to that of B, as illustrated in Fig. 84. 
B is simultaneously given the velocities Vi> and F&, and con- 
sequently remains at rest. A is given the velocities V a and 
7 6 . The resultant of these is the relative velocity F a6 . This 
is readily seen to be the same result as that obtained in the pre- 
vious figure. Therefore a second definition of relative velocity 
may be given as follows: 

The relative velocity of a point A with respect to a second 
point B is the velocity which A would have if B were brought 
to rest, by imparting to both points a velocity equal and opposite 
to that of B. 



RELATIVE VELOCITIES, SPECIAL CASES 



67 



49. Special Cases. Two special cases of importance in kine- 
matics require separate treatment. 

(a) In the case where one link drives another by direct con- 
tact, the relative velocity of the two points which are in contact 
is along the common tangent. For if the relative velocity 
Vba had any component along the common normal the bodies 
would cut into one another, or would separate. See Fig. 85. 

(6) If the points A and B are connected by a rigid link the 
only motion A can have if B is brought to rest is a rotation about 
B as a center. Hence the relative velocity of A with respect to 
B must be in a direction at right angles to the line AB. See 
Fig. 86. 




FIG. 85. 



FIG. 86. 



These principles are very important in the study of velocities 
in mechanisms. 

60. Velocity Images. Consider any link, Fig. 87, rotating 
about an instantaneous center P with angular velocity o>. The 
velocities of any points A, B, C, etc., of this link are proportional 
to the instantaneous radii, PA, PB, PC, etc., and are perpendicular 
to these radii. 

From a common pole lay off vectors Oa, Ob, Oc representing 
V a , V&, V c , etc., in direction and magnitude. Then 



, etc. 



Since 
and therefore 



Z.aob = Z.APB 
AAPB is similar to A0o&, 



68 . VELOCITIES IN MECHANISMS 

Similarly 



ac=coAC, 



and 



Hence the triangles ABC and abc are similar, whence follows the 
important theorem : 

" If from a common pole vectors are drawn representing the 
velocities of the points of a rigid link, the ends of these vectors 

determine a figure simi- 
lar to the link." 

This figure is called 
the velocity image of the 
link. Evidently, the 
sides ab, be, ac, are per- 
pendicular to AB, BC, 
AC. This is plain from 
the geometry of the fig- 
ure, or it may be proved 
independently ab repre- 
sents the velocity of A 
relative to B } since it is 
the vector which must 
be added to F& in order 
to obtain V a . This re- 
lative velocity must be 
perpendicular to AB. 
FlG - 87 - Similarly, be is the ve- 

locity of C relative to 

A, and ca the velocity of A relative to C. It has been shown 
that ab = uAB. Hence follows another important theorem: 

" The relative velocity between two points connected by a 
rigid link is perpendicular to the line joining the points, and is 
equal to the distance between the points multiplied by the angular 
velocity of the link. In case a link has a motion of translation 
its angular velocity is zero. In this case there is no relative 
velocity between the points of the link, and the velocity image 
reduces to a point. 




VELOCITY IMAGES, REVOLVED VELOCITIES 



69 



51. Revolved Velocities. In most cases it will be found more 
convenient to revolve the vectors representing the velocities 
through a right angle. If this is done, the sides of the image 
become parallel to the corresponding sides of the link. See 
Fig. 88. 

52. Velocity Polygons. If vectors are drawn from a common 
pole representing the velocities of all the points of a mechanism, 
then the resulting fig- 
ure contains a velocity 

image of each link. 
The complete figure is 
called the velocity poly- 
gon for the mechanism. 
As stated in the pre- 
ceding paragraph, it 
is usually more con- 
venient to revolve all 
the vectors through 
90. Then the sides 
of each image are par- 
allel to the corre- 
sponding sides of the 
links. Since the size 
of the image is propor- 
tional to the angular 
velocity of the link, if 
follows that the larger 
link may be represented in the polygon by the smaller image. 

For finding the velocities of different points the method of 
relative velocities is employed. For this purpose an equation 
of the form 

v v =v z +v yx , 

is used. The plus sign in this equation represents vectorial 
addition. 

53. Examples. (a) As a simple example, consider the pair 
of cams, Fig. 89. Given the velocity of A on link 2 represented 
by the vector V a . It is required to find the velocity of B on link 3. 
From a pole draw a vector Oa to represent the velocity of A 




FIG. 88. 



70 



VELOCITIES IN MECHANISMS 



revolved through 90. The velocity of C is made up of the ve- 
locity of A and the velocity of C relative to A, thus 

v c =v a +v ca . 

The relative velocity is perpendicular to AC, or when revolved 
through 90 is parallel to AC. Through a draw a line parallel 
to AC. The end of the vector representing the velocity of C 

lies on this line. The 
direction of V c is perpen- 
dicular to P 2 C. There- 
fore draw a line through 
parallel to P 2 C. The 
intersection c of this line 
with the line drawn 
through a determines Oc 
as the velocity of C. 

The velocities V d and 
V c are connected by the 
relation : 




The direction of V d is 
perpendicular to PsD, and 
the direction of V dc is 
along the common tan- 
gent. Therefore, if a line 
FIG. 89. be drawn through paral- 

lel to PsD, and a line 

through c parallel to the common normal, the intersection of 
these lines determines the velocity of D. Again 

v*=v d +v bd . 

The directions of V b and V bd are known. Therefore, if lines be 
drawn through parallel to PsB, and through d parallel to DB, 
the intersection b of these lines determines the velocity of B as 
Ob. Note that all the velocities in the polar diagram are drawn 
perpendicular to their real directions. To find the relative ve- 
locities of any two points as A and B, simply draw a vector connect- 
ing the corresponding points a and b of the polar diagram. The 



VELOCITY IMAGES, VELOCITY POLYGONS, EXAMPLES 71 




triangles Oac and Odb are images of P%AC and PsDB respectively. 
The image of the sta- 
tionary link is simply 

2 

second 



the pole 0. 

(b) As a second ex- 
ample take the pair of 
gears 2 and 3, Fig. 90. 
From a pole draw a 
vector Oa parallel to P^A 
to represent the velocity 
of A. The image of 
gear 2 is a circle about 
as center with radius 
Oa. The velocity of C 
is found by drawing the 
diameter aOc. As C has 
the same velocity as D, 

the image of D coincides F IG . 99. 

with that of C. The 

image of gear 3 is a circle about having a radius Oc that is, 

the images of gears 2 and 3 coincide. The image of B coincides 

with that of A. It will be noted that the image of gear 2 is 

erect while that of gear 3 is inverted. The significance of this 

fact is simply that the gears rotate in opposite directions. 

(c) Another example is shown in the epicyclic gear train, 
Fig. 91. If the angular velocity of the arm 2 is known, the velocity 
of A is readily found. From a pole draw a vector Oa equal 
to V a . This vector is the image of P-zA, that is link 2. The 
image of link 3 is a pair of circles having a common center at a. 
As point B has no velocity, its image lies at 0. This determines 
the radius of the circle representing the smaller of gears 3. The 
radius of the image of the larger gear is found by proportion and 
thus c is located. The image of the gear 4 is a circle whose center 
is and whose radius is Oc. The fact that this image is inverted 
shows that 4 rotates in the direction opposite to that of 2. It 
should be noted that 

Oa = o>2 P2 A W3 BA , 
and 



72 VELOCITIES IN MECHANISMS 

The reader may profitably study this construction, and see 
whether he can prove that the values of the ratios and , 

C02 C02 

found by this method, are the same as those found in Arts. 39 
and 43. 




FIG. 91. 

54. Velocity Polygons for Linkages. 1 Consider the four-link 
chain, Fig. 92. Given the velocity of point A, it is required to 
find the velocities of points B, C, and D. From a pole lay 
off a vector Oa = V a . 



Vt> and VM are known in direction. Therefore draw through 
a line parallel to PB, and through a a line parallel to AB. 
The intersection of these lines locates the image of B. To find 
the image of C simply locate on ab a point c so that 

AC = ac 
AB~aV 

1 It is well at least in the first few polygons constructed to draw vectors 
from the different points of the machine representing the velocities of these 
points. This will give the student a clear idea of what is actually taking 
place in the machine. 



VELOCITY POLYGONS. COMPLEX LINKAGES 



73 



To locate the image of D draw through b a line parallel to BD, 
and through a a line parallel to AD. Then Oa is the image of 
link 2, Ob of link 4, and abed of link 3. The velocities of C and 
D are 0" and Od respectively. 




55. Applications to More Complex Linkages. A great many 
mechanisms are built up from the simple arrangements shown 
in the preceding examples by merely adding extra gears or extra 
links two at a time. In such mechanisms all the velocities can 
be found by applying the preceding principles. 

(a) Joy Locomotive Valve Gear. In the steam engine mechan- 
ism, Fig. 93, let the velocity of the crank pin A be known. From 




FIG. 93. 

a pole 0, Fig. 95, draw a vector Oa equal Va revolved through 
90 as shown. 



To determine V* draw a vertical line through and a line through 
a parallel to AB. Then Oa is the image of link 2, ab that of link 
3 and the point b that of link 4- 



74 



VELOCITIES IN MECHANISMS 



Now suppose two additional links 5 and 6 to be added to the 
mechanism as shown in Fig. 94. The image of C is found by 



CLC AC 

locating on ab a point c, such that -T=TTD- 

Co L/JJ 



locate the image 




FIG. 94. 

of D draw cd and Od parallel to CD and Ps-D, as shown in Fig. 95. 
Now add two more links, 7 and 8 in Fig. 94. To locate the image 



simply divide the line cd in the ratio ^^ = -5. To locate 

eu, 



-=. Then locate the image of H by 



of 

the image of F draw ef and Of parallel to EF 
and PsF respectively. Finally, suppose two 
links 9 and 10 are added. To locate the 
image of G simply extend ef to g, so that 
fy. FG 
^9 

drawing gh parallel to GH, intersecting a 
vertical through at h. Then Oh repre- 
sents the velocity of the valve, link 10. 
Each line of the velocity polygon, Fig. 95, 
is the image of the similarly lettered line of 
the mechanism. 

(6) Pilgrim Step Motion. In this mechanism, there is a 
four-link chain on which is superimposed a gear train, see Fig. 96. 




VELOCITY POLYGONS. COMPLEX LINKAGES 



75 



The gear train consists of a small gear whose center is A, rigidly 
attached to the crank 2, and rotating with 2, an idle gear 5, whose 




FIG. 96. 

center is B on link 3, a compound gear 6, whose center is the 
moving pin C, and a driven gear 7, meshing with the smaller 




FIG. 97. 



gear 6 and rotating about the fixed center P 4 . To construct the 
velocity polygon, Fig. 97, assume the value of V a to be represented 
by the vector Oa. Considering the four-link chain the velocity 



76 



VELOCITIES IN MECHANISMS 



of C is readily found. Then Oa is the image of P\A and Oc the 
image of P2C. To find the image of the different gears, first 
locate d by drawing Od parallel to P\D. The image of the small 
gear 2 is a circle, having point a as a center and with a radius ad. 
Next, locate B, the center of gear 5, by dividing the line ab so that 

= -^=. The image of gear 5 is a circle about 6 as a center and 

CO O.D 

with radius bd. The image of the larger gear 6 is found by draw- 
ing a circle of radius ce 
about c as a center. 
The image of the smaller 
gear may be found by 
proportion or by drawing 
the line ef parallel to 
EF. The image of gear 
7 is a circle about 
with radius Of. 





FIG. 98. 



FIG. 99. 



(c) Shaper Mechanism. In the shaper mechanism, Fig. 98, 
the velocity of A, the center of the crank pin is known. Since 
the sliding block, link 3, is equivalent to a binary link of infinite 
length it is impossible to find directly the velocity of a second 
point on link 3, and the method of the preceding example 
therefore fails. It is possible, however, to determine the velocity of 
a point on link 4 which is located directly under A. Let the latter 
point be known as A 4. From the definition of relative velocity 



From the pole 0, Fig. 99, draw Oa equal V u revolved through 



SLIDING PAIRS. SPECIAL CONSTRUCTIONS 



77 



90. The direction of V ata is along the- center line of link 4, that 
is, parallel to P 2 A. Through a draw a line perpendicular to 
P 2 A. This line is a locus of the image of A. Through draw 
a line parallel to P 2 A. This line is also a locus for the image 
of A 4. The velocity of A 4 being thus determined the velocity 

. . Ob P 2 B 
of B is found by making ^ = ^ r . 



56. Special Constructions. The methods developed in the 
preceding paragraphs can be employed in determining the veloc- 
ities of all points in all mechanisms, in which the paths of the 
various points can be found by means of the straightedge 
and compass alone. Where this is not the case more complex 
methods must be employed. 

The special method most frequently employed is known as 
the Three-line construction. Suppose that in a revolved velocity 
\ 

\ 




FIG. 100. 

polygon the images of three points of a link ABC, Fig. 100, must 
lie on the three lines a' a', b'b', and c'c' respectively. Let mi be the 
intersection of two of these lines, say b'b' and c'c'. From B and 
C draw lines parallel to b'b' and c'c' respectively and let these 
lines intersect at M\. From MI draw M\A, and through m\ draw a 
line m\a parallel to M\A . The intersection of m\a with a' a' locates 
the velocity image of A at a. From a draw ab and ac parallel 
to A B and AC respectively. Then b and c are the images of B 
and C. If the work is correctly done, be will be parallel to BC. 
Proof. Imagine the link ABC expanded so as to include MI. 
Then write* 



78 VELOCITIES IN MECHANISMS 

Vmc (revolved) has the direction MiC, that is, the direction c'c'. 
Hence, as the image of C lies on c'c' the image of MI also lies on 
c'c'. Similarly, the image of M i must lie on b'b'. Therefore mi 
is the image of MI. This determines the velocity of one point 
MI on the link. To find the velocity of a second point A note 
that 

Va=V m +V am . 

V m is known and V am (revolved) has the direction MiA. There- 
fore drawing m\a parallel to M\A the image of A is located at a. 
The images of B and C are located as usual by drawing ab and 
ac parallel to AB and AC. 1 It should be noted that either of 
the intersections mz or mz can be used as well as mi. Correspond- 
ing points MI and MS on the link would then be located. The 
choice of the intersection is simply a matter of convenience. An 
example will make the use of this construction more clear. 

47. Example. Stephenson Link Motion. The Stephenson 
link mechanism, which is shown fully drawn in Fig. 37, is redrawn 
in its skeleton form in Fig. 101. In this mechanism there is a 
four-link chain, 2345, but the stationary or reference link is not 
one of its sides. 

Fig. 102 shows the velocity polygon which is arrived at as 
follows: If the velocity of the eccentric center A is known, the 
velocity of point B is readily found. It is required to find the 
velocities of points C, D and E. 

Note that 

V e =Va+V ca , 

where V C a revolved has the direction AC. This gives one line ac 
on which the image of C must be located; also since, 



where V e b revolved has the direction BE, the image of E must 
lie on the line be. The direction of Vd must be parallel to PoD. 
This gives a locus Od for the image of D. 

1 The validity of this construction can also be proved by showing that 
the triangle abc has its sides parallel to those of ABC. abc is therefore the 
image of ABC. As ab and ac are parallel to AB and AC by construction, 
it is necessary only to prove be parallel to BC. 



VELOCITY POLYGONS THREE-LINE CONSTRUCTION 79 

Choose the intersection of two of the lines ac, be and Od, say 
m 2 . Prolong AC till it intersects P 6 D at M 2 , Fig. 101. Now 
imagine link 5 expanded so as to include the point M 2 . Then, 



It is readily seen that V d and V m *i are in the same direction, 




FIG. 102. 



FIG. 101. 



hence their resultant is also in this direction, and the locus Od 
is therefore a locus for the image of M 2 . 
Now 



and 
Therefore 



Both V ca and Vm^ are in the same direction, and therefore 
ac is also a locus for the image of M 2 . The intersection of ac 
and Od is consequently the image of M 2 , and thus we have found 
the velocity of one point of link 5. 

To get the velocity of a second point E, it is only necessary 
to draw m 2 e parallel to M 2 E. The intersection of this line with 
be determines the velocity image of E. C and D are now readily 
located by drawing ec and ed parallel to EC and ED respectively. 



80 



VELOCITIES IN MECHANISMS 



The accuracy of the construction may be tested by noting that 
cd must be parallel to CD. 

EXERCISES 

1. Construct the velocity polygon for the Stephenson link motion, making 
use of point Mi or M s instead of point M 2 . 




FIG. 103. 



2. Construct the velocity polygon for the Wanzer needle bar mechanism, 
Fig. 103. The speed of rotation of the crank, link 6, is known. 

3. Construct the velocity polygon for the Watt engine, Fig. 104. 

The velocity of the crank pin A 
is known. 



In some cases a mechan- 
ism is built on a four-link 
chain and the extra links are 
added in groups larger than 
two. In such cases the 
three-line construction is> 
often useful. 

4. Construct the velocity poly- 
gon for the skeleton mechanism, 
Fig. 105. Link 1 is fixed and the 
velocity of point A is known. 




FIG. 104. 



VELOCITY POLYGONS COMBINED METHODS 



81 



58. Combined Method of Instantaneous Centers and Relative 
Velocities. In some cases a velocity polygon can be readily con- 




FIG. 105. 

structed by the determination of certain centers, when it would 
be difficult or impossible to construct it by the method of relative 

velocities alone. As an example, con- 
sider the Wanzer needle-bar mechan- 
ism, Fig. 106. Suppose the velocity of 
25 





FIG. 106. 



FIG. 107. 



C is known and that it is required to construct the polygon. Let Oc, 
Fig. 107, represent the revolved velocity of C. b'cb' is then a locus 
for the image of B, and a'ca' is a locus for the image of A. Also 



82 VELOCITIES IN MECHANISMS 

the direction of motion of any point on link 2 is known. With 
such data it is impossible to draw the polygon by any of the 
methods given previously. The construction is, however, easily 
completed by locating the center 25, and remembering that 
the velocity of 25 is the same whether it be regarded as on link 
2 or link 5. Considering 25 as a point on link 2 its revolved 
velocity is parallel to P<$5. A line through in this direction 
is, therefore, a locus of 25, that is, the image of 25. Considering 
25 as a point on link 5, write 



The revolved velocity V% 5c is parallel to C25. This gives a second 
locus for eg as shown. Having thus determined the velocity of 
one point on link 2, the velocity of any other point on the same 
link is readily found. The completion of the polygon is left to 
the reader. 

EXERCISE 

5. In the Stephenson link mechanism, Fig. 101, the velocity of point D 
is known. Construct the velocity polygon. 

59. Four-line Construction. -In some complex mechanisms a 
method involving both relative velocities and instantaneous 
centers can be employed. Suppose that one locus is known for 
the velocity image of each of four points, two of which are on a 
link $ and two on a link T. If the instantaneous center ST is 
known the following construction will give the velocity images 
of the links: Let AB, Fig. 108, represent the link S and CD the 
link T and let Z be the instantaneous center ST. The velocity 
images of A, B, C and D are known to lie on the lines a' a', b'b', 
c'c' and d'd' respectively. Through A, B, C and D draw lines 
parallel to a'a', b'b', c'c' and d'd' respectively, Fig. 108. The 
lines through A and B intersect at X and those through C and 
D at Y. The lines a'a' and b'b' intersect at x and c'c' and d'd' 
at y. Connect X and Y to Z. Through x and y draw lines parallel 
to XZ and YZ respectively. These lines intersect at z. Lines 
drawn from z parallel to ZA, ZB, ZC and ZD locate the images 
of A, B, C and D on the lines a'a', b'b', c'c' and d'd'. 

Proof. Consider links S and T extended so as to include the 
points X and Y respectively, and so that each includes the instan- 



VELOCITY POLYGONS FOUR-LINE CONSTRUCTION 83 




FIG. 108. 



taneous center Z. Then the velocity of X is made up of the veloc- 
ity of A and the velocity 
of X relative to A. The 
velocity of X relative to 
A is drawn parallel to 
XA, since X and A are on 
the same link. The ve- 
locity image of A lies on 
the line a' a', and to find 
that of X we must add a 
component along the same 
line. Hence the velocity 
image of X lies also on 
line a'o! . Similarly the 
velocity of X must lie on 
the line b'b', and there- 
fore x is the velocity image of point X on link S. Similarly, y 
is the velocity image of point Y on link T. Hence the velocity 

of one point on each link is 
found. The instantaneous 
center Z has the same veloc- 
ity whether considered on 
link S or link T. Consid- 
ered as a point on link S the 
velocity of Z is made up of 
the velocity of X, which is 
known, and a velocity in the 
direction XZ. Considered as 
a point on link T the velocity 
of Z is made up of the velocity 
of Y and a component in the 
direction YZ. Hence z is the 
velocity image of the instan- 
taneous center Z. Now the 
velocities of two points X and 
FIG. 109. ZoflmkandFandZoflmk!F 

are known. To find the com- 
plete images of the links it is necessary only to complete the figures 
zabx and zcdy, with sides parallel to ZABX and ZCDY respectively. 



a 1 




84 



VELOCITIES IN MECHANISMS 



EXERCISE 

6. Construct a velocity polygon for the Walschaert valve gear, Fig. 110. 
The velocity of the crank pin is known. 

60. Velocity Curves. In many cases it is desirable to know 
the velocities of different parts of a machine, not only for one 
position, but for every possible position. Usually, some one link 
can be assumed to run at constant speed. For example, in the 
slotter mechanism, Fig. Ill, the crank, link 2, may be assumed 
to run uniformly. It is desired to determine the variation in 
the velocity of the slide, link 6, during one revolution of the crank. 




FIG. 110. 

The mechanism can be drawn in skeleton form for each 15 
through which the crank revolves. For each such configuration 
the velocity of the ram can be found either by instantaneous 
centers or by velocity polygons. Then at each position of the 
ram a line perpendicular to the stroke can be drawn representing 
the instantaneous velocity for that position. Velocities upward 
may be drawn toward the right, and those downward toward 
the left. In this way a curve may be laid down, showing the 
velocity for each position of the ram. Such a curve is shown 
in Fig. 111. 

A second curve can be drawn to show the variation of the 
velocities during one revolution, by drawing a base line and divid- 
ing it into twenty-four equal parts, each space representing 15 
of revolution of the crank. As the crank is supposed to rotate 



VELOCITY CURVES 



85 



6 



at constant speed these spaces represent equal intervals of time. 

Such a curve is shown in Fig. 112. Some interesting conclusions 

may be drawn from this 

diagram. As the ordinate 

is velocity and the abscissa 

time, the slope of the curve 

represents the acceleration 

of the moving part. Also 

the area enclosed between 

the curve and the base line 

gives the distance traveled 

in any interval of time. 

The area above the base 

line represents the length 

of the cutting stroke, and 

that below the base line 

the length of the return 

stroke. Evidently these 

areas should be equal. 

If the tool cuts uniformly 

at maximum speed the 

amount of cutting, in the 

interval of time required for one revolution would be repre- 

sented by the area of the rectangle EFKM. The actual cutting 

is represented by the fig- 
ure EGHDE. The ratio 

EGHDE 

- wm may be regard- 

ed as a sort of " Time 
Efficiency " of the ma- 
chine. This furnishes a 
useful criterion for com- 
paring the effectiveness 
of various types of quick- 
return motions. 




Fie. ill. 




FIG. 112. 



EXERCISE 

7. Construct velocity curves and determine time efficiency of the quick- 
return motion shown in Fig. 98. The velocity of gear 2 is constant. 



CHAPTER IV 
ACCELERATION IN MECHANISMS 

63. Introductory. The study of accelerations in mechanisms 
will be carried out by methods somewhat analogous to those 
used in finding velocities. Analytical methods will be used in 
some simple cases, but for complex linkages graphical construc- 
tions are employed. The constructions for accelerations are 
more difficult than those for velocities and must be studied with 
great care. 

64. General Principles. Acceleration is defined as the rate 
of change of velocity. The general method of finding the acceler- 
ation of a point is as follows: Let V\, Fig. 113, denote the velocity 
of a point at any instant, and 2 the velocity after an interval 
of time AZ. Then V 2 >V'i denotes the vector difference between 
2 and Vi, that is, the velocity which must be combined with 

Vi in order to give V 2 as a resultant. The quotient 2 ~* - is 

the mean rate of change of velocity, or in other words, the mean 
acceleration, and the limit of this quotient as AZ approaches zero 
is the instantaneous acceleration. Denoting the acceleration by 
A we have 

limit F 2 ->Fi 
~ ~~ 



If, instead of the vector difference, we take the scalar differ- 
ence of velocities we get quite a different quantity, the rate of 
change oj speed. Denoting this by / 

limit V 2 -Vi , . 

J ~A* = A* ....... (2) 



GENERAL PRINCIPLES 



87 



In Fig. 114 let OA represent the initial velocity Vi and OB 
the velocity V 2 that the point has attained after an interval of 
time A. Then the closing side AB is the velocity component 
that must be added to OA to give OB. Let OC be laid off equal 





FIG. 113. 



FIG. 114. 



to OA: then CB = OB-OC = V 2 -Vi. The limit of f^ 

gives the acceleration or rate of change of velocity, while the 

(C* /?\ 
J gives merely the rate of change of speed. 

Now vector AB has the components AC and CB; that is, by 
vector addition AB = AC+CB. 

Whence 

AB_AC CB 
A A A 
and 



In the limit, the angle COA=0, and angle ACS = 90, thus the 
components are at right angles. 

r i dV 

hm 



rCl r 

m { A?] =hm - 

FACl 
im lATj =lim - 



in which is the angle between the directions of the velocities 
Vi and Vzy and o> is the rate of change of direction of motion equal 



88 ACCELERATION IN MECHANISMS 

-r:. If r is the radius of curvature of the path at the first instant, 
at 

Fi = rco and 



Since the discussion applies to every point in the path, the 

V 2 
subscripts may be dropped. The component is in the limit 

perpendicular to the direction of the velocity of the moving point, 
in other words, normal to the path. Hence, it is called the 
normal or radial acceleration of the point. The other component 

-TT-T72 h as the direction of the tangent and is therefore called 
at at 

the tangential acceleration. Denoting these components by A n 
and A 1 respectively, the resultant or total acceleration A is 
given by the equation 

. ..... (4) 



and the angle 4> which A makes with the normal to the path is 
given by the relation 1 

A t 

tan <t>=^ ......... (5) 

65. Translation. A plane system has a motion of translation 
when the paths of all points are equal and parallel. These paths 

may be either straight lines or 
curves. Evidently, since all points 
have precisely the same motion, 
they have the same velocity and 
also the same acceleration. 

66. Rotation about a Fixed 
Axis. Suppose a rigid system to 
rotate about a fixed axis, and 
let co denote the angular velocity 
FIG. 115. at a given instant. In Fig. 115 

let be the center of rotation 
and P and Q be points of the system. The point P describes a 

1 Unless otherwise stated, all additions and subtractions of velocities and 
accelerations will be considered as vectorial, and the usual symbol of vectorial 
addition and subtraction will be omitted. 




GENERAL PRINCIPLES 89 

circle with OP = r as a radius. The linear velocity of P (not 
shown in Fig. 115) is 



= rco ......... (1) 

The tangential acceleration represented by PA is 

dV rdaj . . . 



where co' denotes the angular acceleration. The normal com- 
ponent represented by PB has the magnitude 



r r 
For the angle </> we have 



(3) 



Equation (4) shows that the angle is independent of r and is 
the same for every point. 

The total acceleration has the magnitude 



(5) 



Equation (5) shows that the acceleration of a point is proportional 
to the distance r between the point and axis of rotation. 

The existence of the normal component of the acceleration 
occasions the fundamental difference in the treatment of acceler- 
ations and velocities. As there is no component of the velocity 
normal to the path, we can easily determine the direction of the 
velocities of different points. This is not true of accelerations. 
It is this fact which makes the determination of accelerations 
more difficult than that of velocities. It should be noted that 
the tangential acceleration is due to change of speed, normal 
acceleration to change of direction. 

67. Combined Motions. Coriolis' Law. Frequently it is con- 
venient to consider the motion of a system relative to a fixed body 
as composed of two motions: (1) The motion of a first system 
relative to a second moving system ; (2) The motion of the second 



90 



ACCELERATION IN MECHANISMS 



system relative to the fixed body. As an example, take the motion 
of the shaft governor. Fig. 116. Consider first the motion of the 
governor relative to the flywheel as though the wheel were at rest, 
and then consider the motion of the wheel relative to the fixed bed. 
The combination of the two motions gives the absolute motion 
of the governor weight or the motion relative to the bed. Any 
point of the given system, Si, may therefore be considered as 
moving in a curve mx in a second system 82, while the curve 

at the same time moves in the 
fixed system 83. The velocity 
of a point of system S\ is the 
resultant of its velocity along 
the curve and the velocity of 
the coincident point of the 
curve. Thus, in Fig. 116, the 
point P moves along the curve 
m with the velocity u, while the 
point of the curve coincident 
with P has in the fixed system 
a velocity w. That is, if the 
curve were at rest, the point P 
would have the velocity u alone, 
while if P were fixed on the 

curve it would have the velocity w alone, due to the motion of 
the curve. It has been shown that the actual velocity of the point 
P relative to the fixed system is the resultant of u and w. 

Consider the actual acceleration of the moving point. Sup- 
pose the vector PM to represent the acceleration of P due to its 
motion along the curve m, that is, the acceleration of P if the 
curve m were at rest, and suppose the vector PN to represent 
the acceleration of the point m coincident with P due to the motion 
of the curve in the fixed system. If the same law holds for acceler- 
ations as for velocities, the actual acceleration at P is the resultant 
of the relative acceleration PM and the acceleration PN of the 
coincident point. The problem now before us is to determine 
if such is the case. Fig. 117 shows the general case where the 
curve has any motion. In Fig. 117 mi and m^ represent con- 
secutive positions of the curve m, and P is the instantaneous 
center of the motion of m at the instant. A point at S' moves 




FIG. 116. 



CORIOLIS' LAW 



91 



in the curve with a velocity ui and the coincident point A\ of 
the curve moves with a velocity Va\ perpendicular to the instan- 
taneous radius. 

The point S moves along the curve with variable velocity 
so that if the curve remained stationary S would have a relative 
velocity uz after an interval of time AZ. But in this interval 
of time the curve has revolved around P through an angle A0 
and the point A\ of the curve has reached A 2, and has attained 
a velocity Vaz. The point BI has reached 82 and its velocity 







FIG. 118. 



FIG. 117. 



is 7& 2 . If Vi be the total velocity of the point S at the beginning 
of the interval of time M and 2 its velocity at the end of that 
time. Then 



U\ 



Now Fa, V a , = A Fa = change of velocity of point A on curve m. 
Hence 



92 ACCELERATION IN MECHANISMS 

u 2 is numerically equal to U2 but differs in direction by the 
angle A0. (See Fig. 118.) Hence, 



sin . 



Hence, 



Now 112' u\= Au = change of velocity considering the curve 
stationary. 
Therefore 

AV=V 2 -Vi=AV a +Au+Vi, 2 a 2 +2u 2 sin , 



(see Art. 50). 
Therefore 



At At At A* 

In the limit, 

dF a . du 



since A 2 B 2 = udt. 

Now the component F& 202 is at right angles to A 2 B 2 (that is, 
normal to the curve in the limit) and in the direction given by 
revolving the curve about A 2 in the sense w. In the limit the 

dV 

component uzuz' is in the same direction and sense, ~ 

dt 

is by definition A m , -r- is A r . 
Therefore 



The law thus expressed mathematically is known as Coriolis* 
Law. The third component 2itco is called the compound supple- 




CORIOLIS' LAW 93 

mentary acceleration. The direction of this component, as we 
have seen, is perpendicular to that of the relative velocity y,. 
The following rule will always give the proper sense of 2uu. The 
sense of 2ua> is such that, considered as a force, applied at the 
end of the vector representing u, it would rotate the vector, 
in the sense of , the angular 
velocity of the path m about 
its instantaneous center. Thus 
in Fig. 119 suppose the curve 
m to be rotating clockwise 
about P while the point S has 
at the same time the velocity 
u = SA in the curve. A force 
applied at the end A of this JP IG> u$ 

vector must have the sense 

AB to turn the vector clockwise about A; the acceleration 
2uu has the same sense as AB and passes through S perpen- 
dicular to SA. 

The compound supplementary acceleration is the result of 
two factors. 

(1) The velocity U2 is measured relative to a new point 
B on the curve which has a velocity different from 
that of A . Therefore the difference in velocity between 
B and A enters into the change of velocity of the 
point S. 

(2) The direction of velocity u is along the tangent to the 
curve m. If this tangent turns through an angle 
A0, the direction of u is changed by the same angle. 
This change of direction implies anothei component 
of the acceleration. 

Special Case. If the motion of the system carrying the curve 
m is a translation, then the points A and B have the same velocity. 
Also the angle 6 through which the curve turns is zero, and there- 
fore the compound supplementary acceleration disappears, as is 
also evident from the fact that co = 0. This is equally true whether 
the translation is rectilinear or curvilinear. 

As an example consider the motion of a wheel or circular disk 
rolling on a plane surface. A point P of the disk actually de- 



94 



ACCELERATION IN MECHANISMS 



scribes the arc of a cycloid (Fig. 120), but we may consider that 
P moves in a circle m which at the same time has a motion of 

translation parallel to AB. 
Suppose the center moves 
with a constant velocity w; 
then P moves with a con- 
stant speed u in the circle 
m and the circle has a trans- 
lation with constant velocity. 
Under these conditions, the 




B 



relative acceleration A r = 



PO 



FIG. 120. 

and is directed toward the 

center 0, and the acceleration of the point of the curve m coin- 
cident with P, that is A m is zero. Hence, the actual acceleration 
of Pis 



OP 



OP' 



which is the same as if the disk were rotating with the same angu- 
lar velocity about as a fixed center. 

As a second example, let a point move with constant speed 
in a helix, say the mean helix of a screw thread. This motion 
is equivalent to the motion of the point in a circle with a constant 
speed u, while the circle at the same time has a motion of trans- 
lation perpendicular to its plane with a uniform velocity w. The 
relations between u, w, and V are as follows : 

V = Vu 2 -\-w 2 , u=V cos a; and w = u tan a, 

where a is the helix angle, and V is the actual velocity of the point 
in the helix. 

If r is the helix radius the relative acceleration is 

. _ u 2 _ V 2 cos 2 a 
Ar ~~7~ ~T~ 

The acceleration A m of the coincident point on the circle is 
zero, hence, 

V 2 cos2 a 



APPLICATION OF CORIOLIS' LAW 



95 



68. Application of Coriolis* Law. (a) The following numerical 
example will serve to illustrate the use of Coriolis' Law. In Fig. 
121 the system 82 rotates about the fixed point and a second 
system Si rotates about a point C of the first system. The angu- 
lar velocity of 82 about or co is 5 radians per second, that of Si 
about C or ft, is 8 radians per second. Both co and ft are constants. 
OC = 13 feet. Find the acceleration of the point P of Si knowing 
that PC = 5 feet and PO = 12 feet. The component A r , the acceler- 
ation of P relative to system Si is 

= 320 feet per second per second. 



This component has the direction PC. 





FIQ. 121. 



FIG. 122. 



The component A, the acceleration of the point P of the 

system 2, is 

A m = OP<J = 12X5 2 = 300 feet per second per second. 

This component has the direction PO. 

To get the component 2wco, we have u = CPft = 5X8=40 feet 
per second, and co = 5; hence 

2u co = 2 X 40 X 5 = 400 feet per second per second. 

This component is perpendicular to u in the direction CP. 

The three components are shown in Fig. 122. Since A r and 
2wco are opposite but in the same line they may be added algebra- 
ically, giving as their sum 80 feet per second per second, repre- 
sented by OE. OE combined with A m ( = OF) gives a resultant 
represented by the vector OG. 



A = V80 2 +300 2 = 310 feet per second per second. 



96 



ACCELERATION IN MECHANISMS 



The angle between the direction of A and the radius OP, 
Fig. 121, is 



EXERCISES 

1. In Fig. 123, gear 1 is stationary and gear 3 rolls around the circum- 
ference of gear 1, the two gears being held in mesh by the arm link 2. 

Let gear 1 be 8 inches in diameter, 
gear 3 be 4 inches .in diameter, 
and o> = angular velocity of link 2 = 10 radians per second. 




FIG. 123. 



FIG. 124. 



Find the accelerations of points A, J9, C, and D on gear 3 by means of 
Coriolis' Law. 

2. The vane A of a centrifugal pump, Fig. 124, rotates with uniform 
angular velocity 30 radians per second. Find the acceleration of a particle 
of water P, which moves along the vane with a relative velocity of 20 feet 
per second and relative acceleration 100 feet per second per second. 

Radius of curvature of vane 8 inches, center of curvature 6 inches from 
center of rotation. Calculate each component of the acceleration and find the 
resultant graphically. 



RELATIVE ACCELERATION 



97 



3. The flywheel of an engine, Fig. 125, rotates with angular velocity 
o) = 30 radians per second, ami angular acceleration a = 20 radians per second 
per second. The governor weight 

is moving outward relative to the 
wheel with a uniform velocity of 
10 feet per second. Using the 
dimensions given in Fig. 125, find 
the acceleration of G, the center 
of gravity of the governor. 

4. In Fig. 126 the bar link 2 
has an angular velocity o> = 10 
radians per second and an un- 
known angular acceleration a. A 
particle P moves outward along 
the bar with relative velocity u 10 
feet per second, and unknown 



acceleration -=-. 



The total accel- 



eration of the point P is 200 feet 
per second per second directed 
horizontally to the left. Find the 




FIG. 125. 



du 



angular acceleration a of the bar and the acceleration of P along the bar = -7-. 




'O _v 



The acceleration of P is 



made up of the following components : 
(a) Normal acceleration of P 2 = rw 2 ; 
(6) Tangential acceleration of P 2 = r; 

~ 



(c) Acceleration along bar 



aa 



Of these components (a) and (d) can 
readily be found, and the directions of 
(6) and (c) are known. Lay out the 
known components (a) and (d) as two 
sides of a polygon. Add two com- 
ponents in f,he direction of 'ra and 

-37- so that the closing side of the poly- 
gon will be the known resultant A. 

69. Relative Accelerations. 

The acceleration of a point A 
relative to a second point B is 
FIG. 126. defined as the acceleration which 

added vectorially to the acceleration of B will give the acceler- 
ation of A. If the two points A and J5, Fig. 127, are on the same 



98 



ACCELERATION IN MECHANISMS 



link, the total motion of A is composed of a translation with 
velocity and acceleration equal to the velocity and acceleration 
of B, together with a rotation about B with angular velocity o> 
and angular acceleration a. The acceleration due to the rotation 
is composed of two parts: 



(a) A component ABu 2 



AB 



in the direction AB, 



(b) A component ABa at right angles to AB. 




FIG. 127. 

The component (a) is known as the normal acceleration of A rela- 
tive to B, and the component (b) is known as the tangential 
acceleration of A relative to B. 

70. Notation. The following notation will be used to denote 
the various acceleration components: 

A a = Total acceleration of the point A. 
Ai, a = Acceleration of point B relative to point A. 
A a n = Acceleration of A normal to the path of A . 
Ad = Acceleration of A tangential to the path of A . 
A& a n = Acceleration of B relative to A in the direction nor- 
mal to Ffta. 

A&o'= Acceleration of B relative to A in the direction 
parallel to F&Q. 

In studying velocities a single equation of the form 



NOTATION, ACCELERATION COMPONENTS 



99 



was often used. In studying accelerations a similar equation 
may be used, but it is usually necessary to resolve the acceler- 
ations into two components as follows: 



V 2 

71. Graphical Constructions. Terms of the form are used 

r 

so frequently in determining accelerations that graphical con- 
structions for finding such terms are very useful. Usually there 
is given a drawing of 
the link to a suitable 
scale, that is, r is repre- 
sented by a fixed length 
on the paper. Two 
problems now arise: 
(1) A vector represent- 
ing the velocity 
may be given and 
the length of the 
acceleration vector 



required. 

(2) The acceleration 
vector may be giv- 
en and the length 
of the velocity vec- 
tor required. 




FIG. 128. 



1. Let AB, Fig. 128, represent the radius r, and V^ give the 
length of the velocity vector. On AB as a diameter draw a circle. 

1 Students are often puzzled in trying to determine whether to include 
the acceleration component 2wu> in any given case, and whether to use the 
absolute or relative angular velocity of the link. A general rule covering 
these questions may be stated as follows: 

" When the acceleration of a point on one link is to be determined by 
comparison with the acceleration of the coincident point on another link, 
include 2wo> and use the relative angular velocity and acceleration of the two 
links in determining u and A r . When the acceleration of a point is to be 
determined by comparison with the acceleration of another point on the same 
link, omit 2uto. and use the absolute angular velocity and acceleration of the 
link in determining the relative acceleration of the two points." 



100 



ACCELERATION IN MECHANISMS 



With B as a center strike an arc CFD whose radius is F& a . Draw 

y 2 
CD intersecting AB at E. Then BE = -. 

Proof. Triangles ABC and BCE are similar. 
Therefore 



^ 
CB AB 



or 



BE = 



AB 



If BC>AB this construction fails. A more general construction 
is shown in Fig. 129. 



\ 



\ 



\ 



\ 



\ 



X 



X 



X 



X 



^ 



X 
X 
X 

FIG. 129. 



As before let AB = r and the vector F 6a give the velocity of 
5 relative to ^4 . 

At 5 erect a perpendicular 5C of length F 6a . Join AC and 
through C draw CD perpendicular to AC intersecting AB extended 
at D. Then 

17 2 

' ba 

r 



BD 



Proof. Triangles ABC and CBD are similar. 
Therefore 

AB BC 



BC BD' 



or 



BD = 



AB 



ACCELERATION VECTORS VfiLOC?TY : VECTORS 



101 



2. Let AB, Fig. 130, represent the radius r and let AC be the 
acceleration vector. From C drop a perpendicular CD on A.B. 
Then 



On A B as a diameter draw a semicircle intersecting CD at E. 
Then A#=7 &a . 

Proof. Triangles A ED and A EB are similar. 




Therefore 



Therefore 



Hence 



FIG. 130. 

AE = AD 
AB AE' 



AD = 



AB r 



If AD>A this construction fails. In this case extend ABD 
to E so that DE = AB } Fig. 131. 

With AE as a diameter describe a semicircle intersecting Z)(7 
at F. 

ThenDF=F 6a . 

Proof. Triangles AFD and DF# are similar. 



102 ; i .* ; 

Therefore 
or 



Therefore 



ACCELERATION IN MECHANISMS 



AD = 



DF AD 1 

DF 2 DF 2 DF 2 




FIG. 131. 

It should be observed that these constructions give the magni- 
tude but not necessarily the directions of the vectors, V ba or 

V 2 

&g . The directions should be determined from the conditions 
r 

of the problem. 

72. Projection of Acceleration Components. Let A a and A, 
Fig. 132, be the accelerations of two points A and B connected 




1C 



FIG. 132. 




by a rigid link. These accelerations may be resolved into com- 
ponents along AB and at right angles to AB. Since: 



ACCELERATION COMPONENTS ACCELERATION IMAGE 103 
the components of A a and A b along the link must differ by Aba n , 



(, a 

that is, by -r4r- The component 



must be in the direction 



BA. Therefore for the case shown A a has a larger component 

along the link than AD. If 

these components were re- 

versed in direction A b would 

have the larger component 

along the link. 

73. Acceleration Image. 
The acceleration of any point 
B, Fig. 133, on a moving link 
may be regarded as composed 
of the acceleration of any 
other point A, together with 
the acceleration due to the 
rotation of the link about 
A. This latter relative ap- 




celeration consists of two 
components: 

(1) The normal A b 
toward A. 

(2) The tangential 



Fro. 133. 



= ABw 2 , which is directed from B 



where >'=, at right 



angles to AB. 
The resultant of these two is given by the equation: 



This resultant makes an angle <f> with BA, where 



Similarly for another point C 



and this resultant makes the angle </> with CA 
Therefore 



BA' 



104 



ACCELERATION IN MECHANISMS 



Now from a pole 0, Fig. 134, lay off Oa = A a , and lay off from a; 
ac=A ca and ab=Ai> a . Then the triangle abc is similar to ABC 



because -=-= -r-= and the angle cab = angle CAB. 
ao A.JJ 



Hence 



and 



= A c . 



It follows, therefore, that if form a common pole vectors 
be drawn representing the accelerations of the points of a rigid 
link, the ends of these vectors form an image of the original link. 
The sides of this image form 
an angle </> with the sides of 
the original link, where 

w' 

tan 4> = 3. 

CO 

The acceleration image is 
of much importance in deter- 
mining accelerations in mech- 
anisms. If the accelerations 





FIG. 134. 



FIG. 135. 



of two points of a link are known, the accelerations of all other 
points can be found by simply constructing a figure similar to 
the link. 

74. Acceleration Center. Each point of the acceleration 
diagram gives the acceleration of a similar point on the link. The 
pole is therefore the image of a point of zero acceleration on the 
original link. This point is called the center of acceleration. It 
is found by constructing the quadrilateral ABCP, Fig. 135, similar 
to abcO. The lines PA, PB, etc., make the angle <f> with A a , A&, 
etc. The accelerations A a , A 6 , etc., are also proportional to PA, 



ACCELERATION CENTER ACCELERATION POLYGON 105 




FIG. 136. 



PB, etc. That is, the link behaves, as far as acceleration is con- 
cerned, as if it were rotating about a fixed center at P. 

The center of acceleration and the center of velocity are quite 
distinct points. 

The center of acceleration is not practically of much impor- 
tance. In the first place there is no simple manner of determin- 
ing this center, such as was furnished by the law of three centers 
in determining centers of velocity. Secondly, the location of 
the center of acceleration 
depends not only on the 
configuration of the mech- 
anism, but also on the 
accelerations themselves. 
A simple illustration will 
make this clear. Consider 
a wheel, Fig. 136, rolling 
at uniform speed along a 
rail. The center of the 
wheel has no acceleration 

and is consequently the center of acceleration. On the other 
hand, if the speed of the wheel is changing the center is being 
accelerated. Therefore, the center of acceleration must now be 
at some other point. The locus of the center of acceleration is 
discussed in note C. 

75. Graphical Methods. The accelerations in mechanisms 
may be determined by means of the method of relative acceler- 
ations, by Coriolis' law or by a combination of these two. Where 
possible, the method of relative accelerations is usually preferred. 
In some cases this method fails and CoriohY law must be applied. 

76. The Acceleration Polygon. If from a common pole vectors 
are drawn, representing the accelerations of different points of 
a mechanism, each link will be represented by an image. The 
complete figure showing the accelerations of all points is called 
an acceleration polygon for the mechanism. In general, in the 
mechanisms for which the velocity polygons are most easily con- 
structed, the acceleration polygons are also relatively simple; 
and those in which special constructions must be used to form the 
velocity polygon require more elaborate devices to determine the 
accelerations. 



106 ACCELERATION IN MECHANISMS 

In determining the accelerations, components of the form 

V 2 

occur so frequently that it is best to start, in most cases, by 

constructing a velocity polygon. This gives at once, all the rela- 
tive velocities which may be required. 

77. Selection of Scale. In determining the proper scale for 
constructing this polygon, the principles of Art. 71 should be 
followed. Thus if the acceleration of any point is given, its normal 

V 2 
component is . When this is laid off, the vector representing 

the velocity is first found, and then the velocity polygon is con- 
structed to the scale thus determined. 

One case deserves special attention. Suppose some link to 
be rotating at constant speed about a fixed center. The acceler- 

V 2 
ation of any point is normal and is equal to . Now if the scale 

of accelerations be so chosen that the radius itself represents the 
acceleration. 

Then 

= r 
r 

Therefore V = r. 

That is, the radius also represents the velocity to scale. This 
scale is often used in such cases. 

78. Ordinary Gear Train. In Fig. 137, let gear 2 rotate with 
a known angular velocity and known acceleration. 

It is required to find the acceleration of gears 3 and 4. From 
a pole 0, Fig. 138, lay off Of= normal acceleration of A=R2&2 2 j 
and from / lay off fa = tangential acceleration of A =#20/2. 

Then the acceleration image of gear 2 is a circle about as a 
center with radius Oa. Locate the image of B by constructing 
angle aOb^AP^B. The acceleration of B can be resolved into 
normal and tangential components. Since the speed of C must 
always be equal to that of B the tangential component of A c is 
the same as that of A&. The normal component, however, is 

quite different. 

y 2 T72 y2 ro ro 

A c n = = = X = A b n . 

7'3 J*3 7*2 7*3 7*3 



ACCELERATION IN GEARS 



107 



That is, the normal accelerations are inversely proportional to 
the radii. Also A c n is opposite in direction to A& n . Lay off 
Og=A c n parallel to CPs. From g draw gc=fa perpendicular to 
Og. Then Oc = acceleration of C. The image of the compound 
gear 3 is a pair of circles with centers at 0. The larger has a 
radius Oc, and the smaller radius is found by proportion. The 
image of D is found by constructing angle cOd = CP^D. The 
image of E is found in exactly the same manner as that of C. The 
polygon is thus a system of concentric circles. 




FIG. 137. 



FIG. 138. 



79. Epicyclic Gear Train. In the reverted train, Fig. 139, 
wheel 1 is stationary. Arm 2 revolves about with angular 

velocity to and acceleration . The two wheels 3 are carried 

at 

by the pin at A and revolve together. The gear 4 is free to turn 
about 0. 

The acceleration of A is composed of two parts: 



since 



C03=- CO. 



108 



ACCELERATION IN MECHANISMS 



As B always moves twice as fast as A , 
. t _ . t _ , dw 
dt' 

From the pole 0, Fig. 140, lay off Om = lu 2 and from m draw 
ma = I -jr. ThenOa = A a . 

From a lay off an = - (Om) and from n draw nb = I -7-. Then 

/ CLL 

The acceleration image of the pair of gears 3 is a pair of circles 
having a as a center. The radius of the inner circle is ab. The 

radius of the outer one is found by 
proportion. The accelerations of C 
and D are found by simply drawing 
the diameter bacd. As may be readily 





FIG. 139. 

shown the acceleration of C is entirely along the radius CA. D 
has a tangential acceleration dp which is readily seen to be equal to 

cd ida> CD 



The acceleration of E has two components: (1) a tangential = dp 

y 2 y 2 
and (2) a normal = -^ = ^= f since D and E have the same velocity. 

Since the wheel 3 revolves about C, 



FOUR-LINK CHAIN 



109 



From these data -^= can be found either by calculation or by 

a graphical method, the details of which are left to the student. 
The same results may be obtained by purely analytical methods 
or by use of Coriolis' Law. 

80. The Four-link Chain. In the four-link chain, Fig. 141, 
let AC represent the acceleration of A. It is required to find 
the acceleration of B. From C drop a perpendicular CD on A P. 




FIG. 141. 



FIG. 142. 



Then 



Vj? 
r 



To find V a prolong AP 2 D to E so that DE = AP 2 . On AE 
construct a semicircle intersecting CD at F. 



Then 



DF=V a . (Art 71.) 



Complete the velocity polygon by drawing DG perpendicular 
to PB and FG perpendicular to AE. 



Then 



DG=V b and FG=V ba . 



110 ACCELERATION IN MECHANISMS 

From a pole 0, Fig. 142, lay off Oa = AC = acceleration of A. 

A b =; _ 

Vbl 
AB' 



. A ba n = 



The normal component is found by drawing a circle on A B as a 
diameter, and striking an arc with B as center and radius FG = V& a . 
Then 



This acceleration is directed from B toward A. Lay off 
aj = BJ in the direction BA. The third component A ba l is known 
to be perpendicular to AB. Through j draw a line jm in this 
direction. Then the acceleration image of B lies on jm. It is 
also known that since B rotates about P\ its acceleration is made 
up of two components as given by the equation 



where 






Ab n is directed from B toward P and is represented by BK. 
From lay off Ok = BK. The other component A b is perpen- 
dicular to PB. Through k draw a line kn perpendicular to PB. 
Then the image of B lies on kn. This locates the image of B at 
b. Connecting ab and 06 it is readily seen that Oa is the image 
of link 2j Ob the image of link 4 and ab the image of link 3. The 
angular accelerations of the links are easily found from this figure. 
Thus: 



EXERCISE 

5. Construct the acceleration polygon for the crossed quadric chain, 
Fig. 60. Crank 2 rotates uniformly. Find the acceleration of the center 24: 

(a) Considered as a point on link 2. 
(6) Considered as a point on link 4- 



STEAM ENGINE MECHANISM 



111 



81. The Steam Engine Mechanism. In the steam engine 
mechanism, Fig. 143, let AC represent the acceleration of the 
crank pin A. It is required to find the acceleration of the wrist 
pin B. The velocities of the crank pin, of the wrist pin and the 




FIG. 143. 



relative velocity of the two are found precisely as in the preceding 
case. 



= V ba . 



BJ found precisely as in the preceding paragraph represents the 

Vba 2 

component . From a pole lay out Oa = AC = A a . From 



a draw aj = BJ = A ba n parallel to BA. The third component 
Aba 1 is perpendicular to AB. Draw a line jb in this direction. 
Then the image of B lies on this line. The acceleration of B is 



112 



ACCELERATION IN MECHANISMS 



in a horizontal direction. Therefore a horizontal line through 
also contains the image of B. This locates b. The acceleration 
of B is Ob. ab is the image of the rod. 

This construction differs from that for the four-link chain 
only in the fact that B has no acceleration normal to its path. 

82. Klein's Construction. The case where the crank rotates 
at uniform speed occurs so frequently that it deserves special 
consideration. 




FIG. 144. 

In this case it has been shown (Art. 77) that the crank radius 
itself can be used to represent both the acceleration and the 
revolved velocity of the crank pin A . Let P 2 , Fig. 144, be regarded 
as the pole for both accelerations and velocities. Then 

V a = P 2 A, 
V b = P 2 C, 



P 2 A = A a reversed in direction. 

On BA as a diameter draw a circle. Around A as a center 
draw a circle of radius AC, intersecting the first circle at D and 
E. Draw DE, cutting BA at F and BP 2 at G. Then 



AF 



AB 



ba ' 



KLEIN'S CONSTRUCTION SLIDING PAIRS 
reversed in direction 



113 



The first two of these components are known, and the third 
must be in a direction perpendicular to AB, that is, in the direc- 
tion FE. The resultant acceleration of B must be in a horizontal 
direction. Therefore PzG represents the acceleration of B reversed 
in direction. AG is the image of the rod AB. This is known as 
Klein's Construction. 

83. Sliding Pairs. In the four-link mechanism shown in 
Fig. 145, the block, link 4, moves in a circular slot in link 3. It 
was shown in Art. 
5 that such a block 
is equivalent to a 
link BA , whose 
length is the ra- 
dius of curvature 
of the slot, and 
which is joined to 
link 3 at B, the 
center of curva- 
ture. The mech- 
anism is therefore 
equivalent to the 
ordinary four-link 
chain P 3 BAP 2 . If 
the acceleration of 
A is known, the 
acceleration of B 
is readily found, 
and therefore the JT IG 145. 

acceleration of any 

other point on link 3 is easily determined. If, however, the 
slot is made straight, the center B recedes to infinity, and the 
construction fails. To determine accelerations in this case 
Coriolis' Law must be applied. 

In the shaper mechanism, Fig. 146, the acceleration of A is 
known. It is required to find the acceleration of B. Let A4 
be a point on link 4 directly behind A. Then by Coriolis' Law, 




114 



ACCELERATION IN MECHANISMS 



where A r is the acceleration due to the sliding of link 3 along 4, 
u is the velocity of this sliding and co is the angular velocity of 




FIG. 146. 

link 4 about P 4 . In the usual way find the velocity of A=EF, 
and complete the velocity polygon as shown. Then, 

Vai and a2d m = u. 

A -A n I A t 

*1<U -rt-CU \-t*-O4 ) 

V 2 

A n_ v Q * _ p, 7 
Aat "" 4 ' 



From A lay off AK = Oa m perpendicular to P 4 A. Draw a line 
from H, the middle point of P 4 A to K. Then 

Oa m 

= 2u. 



SLIDING PAIRS CORIOLIS' LAW 115 



From H lay off HL = a 2 am = u. Through L draw a line 
parallel to AK intersecting HK at N. Then LN=HL tan AHK 
= 2uw. If the rotation is clockwise, u is directed upward along 
PA. The usual rule, therefore, shows that 2uo> is perpendicular 
to PA, and is directed to the right. Of the four components 
of A a (Aa l n ) Aaf, A r and 2wco) two are now completely known. A r is 
along PA and A <? t is perpendicular to PA. This gives the 
directions of the remaining two components. Since the resultant 
A a is known, this suffices to construct the acceleration polygon. 

From a pole lay off Oa,2=A a and Oj=A at n . From a^ lay 
off a2S = 2wco reversed in direction. From j draw a line perpendic- 
ular to P^ A , and from s draw a line parallel to PA . The inter- 
section of these lines is the image of A 4. Oa4=Aa 4 . To find 
the acceleration of 6 prolong Oa* to b so that 



Study this construction carefully. Note that Oa2 = A a is the 
resultant of four components: 



(1) Qj-A 

(2) jat = A 

(3) as=A 

(4) s2 






EXERCISE 

6. An 8X12 inch engine runs uniformly at 300 r.p.m. Length of connect- 
ing rod = 42 inches. Find the acceleration of the piston when the crank is 
30 from the head end dead center. 

Use Klein's Construction. Check the results by Coriolis' Law. 

84. Blake Stone Crusher. The principles developed in the 
preceding paragraphs are sufficient to find accelerations in all 
mechanisms where the velocity polygon can be constructed with- 
out resort to special devices such as the three-line or four-line 
constructions. As an example, consider the Blake stone crusher, 
Fig. 147. 

Let AL = acceleration of A. It is required to find the acceler- 
ations of B, C, D and E. Determine the scale of velocities and 
construct the velocity polygon as shown. 



116 



ACCELERATION IN MECHANISMS 



The acceleration polygon is constructed as follows: From a 
pole 0, lay out Oa = AL = acceleration of A. The acceleration 

of B is made up of three components A t> =A a +A t) a n +A b a. Of 

y 2 
these A a is known, A ba n = - and is in the direction BA, and 





FIG. 147. 

A tot is at right angles to BA. To find A ba n lay off from B the 
velocity Vba = BN (taken from the velocity polygon) at right 
angles to BA } join A N and draw NH at right angles to AN. Then 

BN 2 

=Ab a n . Lay off ah = BH from a in the direction BA 



BH = 



AB 



and draw hz perpendicular to AB. Now the image of B lies on 



EXAMPLE 117 

hz. The acceleration of B may also be regarded as composed of 
two parts, Ab = Ai, n -\-Ab t , along BP and perpendicular to BP 
respectively. 

A n = J^- 

From P lay off PR=Vb (taken from the velocity polygon) 
at right angles to PB. Join BR and draw RT perpendicular 
to BR. Then PT = A b n . From the pole lay off Ot = PT and 
draw tb perpendicular to Ot. The point b where this line inter- 
sects hz is the image of B, and Ob = Ai>. Join ab. Then ab is 
the acceleration image of AB. The acceleration of C is readily 
found by constructing triangle abc similar to ABC. Then Oc = A c . 

A similar process gives the acceleration of D and E. A d 

V dc 2 

= A c -+-Adc n -\-Adc. Ac is known. A dc n = =-^ r and is the in 

DC 

direction DC. A d c is at right angles to DC. 

To find A dc n draw from C the line CS=V dc (taken from the 
velocity polygon) at right angles to DC. Join DS and draw SV 
perpendicular to DS. Then : 



Lay off cv= VC and draw vy perpendicular to cv. Then the image 
of D lies on vy. 

The acceleration of D may also be regarded as composed of 
two parts, 



along DQ and perpendicular to DQ respectively. 

V? 



From Q lay off QW=V d (taken from the velocity polygon) at 
right angles to QD. Join DW and draw WX perpendicular to 
DW. Then: 



From lay off Ox = QX and draw xd perpendicular to Ox. The 
point d where this line intersects vy gives the acceleration of 
D = Od. Join cd. The acceleration of E is found by constructing 



118 



ACCELERATION IN MECHANISMS 



a triangle Ode similar to QDE and the polygon is now complete. 
Note that Oa is the image of link 2, abc the image of link 3, Ob 
the image of link 4) cd the image of link 5 and Ode the image of 
linktf. 

EXERCISES 

7. Construct acceleration polygon 
for the quick-return motion, Fig. 112. 
Gear 2 rotates uniformly. 

8. Construct acceleration polygon 
for the Joy valve gear, Fig. 94. The 
crank, link 2, rotates uniformly. 

9. Construct acceleration polygon 
for the shaper mechanism, Fig. 148. 
The crank rotates uniformly. 

10. Construct acceleration polygon 
for the Pilgrim step motion, Fig. 96. 

Hint. The accelerations of the 
pitch points of two gears in mesh 
have the same components along 
the tangents to the pitch circles, but 
different components along the line 
of centers. 

After completing these polygons draw vectors from each point 
whose acceleration has been determined, showing these acceler- 
ations in direction and magnitude. This will serve to give the 
student a better physical conception of the meaning of the polygon. 

85. Three-line Construction. For mechanisms in which the 
velocity polygon is found by the three-line construction, Art. 56, 
the acceleration polygon is found by an analogous method. 
Example: The Stephenson link motion, Fig. 149. Let link 2 
revolve at constant speed and let Oa of the velocity polygon 
represent the revolved velocity of A. The velocity polygon is 
constructed by the method of Art. 56. The accelerations of A 
and B are represented by Oa and 06 of the acceleration polygon. 
The ordinary methods of finding accelerations of other points 
fail and it is necessary to adopt a new method. Consider the 
link 5 extended so as to include the point M. Then: 




me > 



THREE-LINE CONSTRUCTION 



119 



Then: 



Of these A a is known, 



v 2 y 2 

n* A _ * 1 

1 ~ AC 1 "" ~MC' 




FIG. 149. 



The velocities V ca and Vmc can be taken from the velocity 
polygon. The components Aca n and A me" are thus found. They 
lie in the direction CA and AfC, repsectively, and therefore can 
be added directly. The components Acd and Ami are both at 
right angles to ACM. The acceleration of M is therefore composed 



120 ACCELERATION IN MECHANISMS 

of three known quantities and a component whose direction is 
known. 
Also: 



~PD' 

in the direction of DP, 



MD 



in the direction MD, and the other components are both at right 
angles to PMD. 

We can now find the acceleration of M as shown in the acceler- 
ation polygon. From the pole lay off Oa=A a . From a lay 
off ah=A ca n and from /i.lay off hk = A mc n . Draw kx perpendicular 
to ok. Then the acceleration image of M lies on kx. 

From lay off On = A a 1 and from n lay off nl Amd 1 - Through 
I draw ly perpendicular to On. Then the acceleration image of 
M lies at m, the intersection of ly and kx. 

To find the acceleration of a second point E on link 5 con- 
sider the following equations: 



V 2 
A m is known =0ra. A em n = -^-j- in the direction EM, and 



is at right angles to EM. 

V 2 

From m lay off mp = ~ and draw pz perpendicular to mp. 



Then the acceleration image of E lies on pz. A* is also known = Ob. 
2 
m tne direction EB. A e b f is perpendicular to EB. 



y 2 
From b lay off br = -^ and draw rs perpendicular to 6r. 



Then the acceleration image of E is at e, the intersection of pz 
and rs. 

Join me. Then to find the accelerations of C and D it is 
necessary only to construct a figure meed similar to MCED. To 
check the accuracy of the work it should be noted that d should 
lie on a line through n perpendicular to PD, and c on a line through 
h perpendicular to AC. 



FOUR-LINE CONSTRUCTION 



121 




86. Four-line Construction. In Art. 59 it was seen that the 
velocity images of two links can be constructed if loci are known 
for the images of two 
points on each link, and 
provided that the in- 
stantaneous center of 
relative motion can be 
found. In constructing 
the acceleration polygon 
for such a mechanism 
two cases may arise: 

(1) There may be a joint 
between the two 
links. 

(2) There may be no 
joint between the 
two links. 

As an example of the p IG 159 

first case, consider the 

eight-link mechanism shown in Fig. 150. Let link 1 be the 
stationary link, and let the velocity and acceleration of A on 
link 2 be known. Let Oa, Fig. 151 represent the velocity of A. 

The velocity of B is first found, 
and a velocity image of link 2 is 
constructed as Oab. Then c'ac' 
parallel to AC is a locus of the 
image of C on link 7, and e'Oe' 
parallel to P$E is a locus of the 
image of E. Similarly d'bd' and 
f'Of are loci for the images of D 
and F on link 8. The instantane- 
ous center 78 is the point G. To 
construct the velocity polygon 
imagine link 7 extended to include 
Y the intersection of AC and P S E 
prolonged. Then y the intersec- 
tion of c'ac' and e'Oe' is the image of Y on link 7. Similarly X 
is located on link 8 by the intersection of BD and PF. Then 
the intersection of d'bd' and f'Of' is the image of X. Join XG 




FIG. 151. 



122 ACCELERATION IN MECHANISMS 

and YG. Draw xg and yg parallel to XG and YG respectively. 
Then the intersection of xg and yg is the image of G. The images 
of E, C, D, and F are now found by drawing lines from g parallel 
to GE, GC, GD and GF. The polygon is now complete. 

To construct the acceleration polygon start with Oab, Fig. 
152, as the acceleration image of link 2. Next find the acceler- 




ation images of X and Y. To get the acceleration image of X 
write the equation 



Of these four components A f n and 4*/ n are readily found. 
From lay off Oh = A f n = ^p, parallel to P 4 F. Through h draw 

f'hf perpendicular to PF. Then f'hf is a locus for the image 

V 2 
of F. From h lay off hi = A xf n = parallel to XF. Oi then 



represents A f n +A fx n . The remaining components Af and AJ 
are both perpendicular to P*F. Therefore a line x'x' through i 
perpendicular to PF gives a locus for the image of X. 
Next consider the equation 



FOUR-LINE CONSTRUCTION 123 

Of these five components, three are known: 



A "- 
* ~DB* 



A - 

xd ~ 



XD' 



From b lay off V-jjg parallel to DB. Through j draw d'jd' 

perpendicular to BD, giving a locus for the image of D. From j 

V 2 
lay off jk = -~ parallel to XD. The two components Aaf and 



are both perpendicular to BD. Through k draw a line perpen- 
dicular to BD, giving a second locus for the image of X. The inter- 
section of the two loci gives x as the image of X. In the same 
manner the image of Y is located at y. Next to find the image 
of G consider the equations 



= A v -\-A gv 

V z 2 
Of these components A x and A v are known, A ffx n = ~f-r t and 



Y 2 Y 2 

-?.. From x lay off xn = rF parallel to Gx and through 



ft draw a line perpendicular to GX. This line is a locus 

V 2 
of the image of G. Similarly from y lay off ym = -~ parallel 



to GYj and through m draw a line perpendicular to GY. This 
line is a second locus for the image of G. The intersection of 
the two loci gives g, the image of (7. 

To complete the figure gxfd similar to GXFD, and gyec similar 
to GYEC. Note that there was already one locus for each of 
the points /, d, e, and c. If the work is correctly done these 
points should fall on the proper loci. 

EXERCISE 

11. Construct an acceleration polygon for the Walschaert valve gear, 
Fig, 110. The crank, link 8, rotates uniformly. 



124 



ACCELERATION IN MECHANISMS 



CASE 2. The acceleration polygon for this case cannot be 
constructed directly by any of the methods so far considered. 




FIG. 153. 

The accelerations may be-determined in some instances by means 

of the trial and error method described below. 

Consider the eight-link mechanism shown in Fig. 153. The 

velocity polygon, Fig. 154, is constructed in the usual manner. 

Starting with the acceleration of A, 
Fig. 155, the accelerations of X and 
Y are found precisely as in the pre- 
ceding article. In finding the images 
of X and Y one locus is determined 
for the image of each of the points 
B, C, D and G. Since X is considered 
as a point on link 6 we may write 

i ' t 



FIG. 154. 




which gives a locus for the image of F. 
Also since F is a point on link 7 
V 2 

I y g V I A t 

v EY ' 



which gives a locus for the image of E. 



SPECIAL CONSTRUCTION 



125 



As there is only one locus for the image of each point, it is 
impossible to find the image of any point directly. Choose some 
point 61 on the locus of b and assume temporarily that this is the 
image of B. Then since link 2 is represented by a similar image 
abc, the image of C is readily found at a. Also since link 6 is 
represented by a similar image, f\ the image of F is found by con- 




<*' 



FIG. 155. 



structing triangle xbifi similar to XBF. Now e\, the image of 
E is determined by the equations 



-I- A t 
r-"-eei 



V 2 

ec 



I *& 
ft i TTcT 



Of these six components all except the last one in each equation 
are known, and the two unknown components are perpendicular 



126 ACCELERATION IN MECHANISMS 

to CE and FE respectively. Thus the image of E is located at 
e\. But the image of E must lie on the line e'e', and therefore 
the original assumption that the image of B falls at 61 is wrong. 

Assume a second position 62 for the image of B, and repeat 
the process. This gives a second position 62 for the image of E. 
By repeated trials a series of points 6%, e, etc., is found. A curve 
drawn through e\, 2, e%, etc., gives a second locus for the image 
of E. The intersection of this locus with e'e' gives e, the true 
image of E. The remainder of the polygon is easily completed. 

To avoid complication of the figure, only two trial positions 
are shown worked out. The locus e\e^e^ is found to be a straight 
line. The completion of the figure is left to the reader. 

87. Accelerations in Cams. In Fig. 156 let the motion of 
cam 2 be completely known, and let it be required to find the 
accelerations in cam 3. 




FIG. 156. 

The relative motion between the cams at A and B is either, 

(1) A rolling of one cam on the other; 

(2) A sliding of one cam over the other; 

(3) A combination of (1) and (2). 

Let M and N be the centers of curvature of the parts of the 
two cams which are in contact. Suppose M and N were con- 
nected by a rigid link M N. Then for a very small displacement 
the faces of the two cams would be free to have any of the three 
kinds of relative motion mentioned. As the acceleration of M 



ACCELERATION IN CAMS 127 

is known, that of N is readily found by treating OMNP as a 
quadric chain. When the acceleration of N has been determined 
that of any other point B on link 3 is found by the proportion 

A b : A n = PB : PN. 

Of course a moment later, when different points are in con- 
tact, the centers of curvature will be different, and the process 
must be repeated. 



CHAPTER V 
INERTIA FORCES OF MACHINE PARTS 

88. General Statement. The importance of the study of 
accelerations in mechanisms is due to the fact that this study 
makes possible the determination of the forces which cause the 
accelerations. These forces tend to produce vibrations or shak- 
ing of the machine, and often cause considerable stresses in machine 
elements which should be taken into account in design of these 
parts. 

All moving parts of a machine are subject to accelerations, 
and the forces necessary to produce the accelerations must gener- 
ally be applied by other members of the machine. Conversely 
the moving parts resist the accelerations, and thereby produce 
forces, known as " inertia forces," which are transmitted to the 
constraining elements. 

In many machines it is desirable to eliminate or neutralize the 
inertia forces as far as practicable. In most mechanisms this 
would require the addition of moving bodies and would con- 
sequently complicate the machine. In the majority of such cases 
the benefits to be derived from reducing the inertia forces would 
not compensate for the extra cost and complexity of the machine. 
In such cases the only available remedy for excessive inertia 
forces lies in proper design, by means of which the mass and accel- 
erations of the moving parts are kept within reasonable limits. 
In one important class of machines, however multi-cylinder 
steam and internal combustion engines it is often possible to 
minimize the inertia forces without adding greatly to the expense 
or complexity of the mechanism. This problem has received 
much attention, and is treated fully in the following chapter. 

89. Acceleration Produced by a Single Force. Consider any 
rigid body M , Fig. 157, which is constrained by outside mechan- 

128 



RESULTANT ACCELERATING FORCE 



129 



ism (not shown) to have any desired plane motion. This motion 
must be produced by forces FI, %, Fs, etc., acting at points A, 
B, C, etc., where the body M is in contact with other mem- 
bers of the machine, together with the weight and other external 
forces acting on the link. These forces can be combined into a 
single resultant F having the line of action shown in Fig. 157. 
In other words the total acceleration of the mass M can be pro- 
duced by a single force F whose magnitude, direction and line of 
action are indicated. 

The motion of the body is not affected by the introduction 
of any system of balanced forces. It is, therefore, permissible to 
introduce at G, the center of gravity of M, two opposed forces 




FIG. 157 

F' and F" which are each equal and parallel to F. The force F" 
produces an acceleration of the center of gravity such that 

F" = F = mA g , 

where ra = the mass of M, and A g is the acceleration of G. The 
forces F and F r form a couple of moment Fh which produces an 
angular acceleration of M such that 



where /= moment of inertia, and k = radius of gyration of M 
about G as a pole, and a is the angular acceleration. It is evident 
that by a proper choice of F and h any desired linear and angular 
acceleration of M can be produced by means of the single force F. 
Three special cases may arise: 



130 INERTIA FORCES OF MACHINE PARTS 

(1) The forces F\, F<z, F% . . . , may form a balanced system. 
In this case the body has momentarily no angular acceleration 
and no acceleration of the center of gravity. An example of this 
condition is a flywheel rotating at uniform speed. 

(2) The resultant of the forces FI, %, F% . . . , may be a 
couple. Here the center of gravity has momentarily no acceler- 
ation. The body M, however, undergoes angular acceleration. 

(3) The resultant F may pass through G. In this case there 
is no angular acceleration. 

These special cases present no difficulties. The laws derived 
for the general case apply without modification. 

90. Kinetically Equivalent Systems. In the case of an actual 
machine part the forces FI, F^ F% . . ., are unknown. The 
accelerations, however, may be found by the methods of Chapter 
IV, and the problem to be solved consists in determining the 
forces which cause these accelerations. 

To find the resultant F completely, three things must be 
known : 

(a) The magnitude of F, 
(6) The direction of F, 
(c) The line of action of F. 

The first two of these are readily found, since 

(a) F = mA g . 

(b) The direction of F is the same as that of A g . 

The most convenient method of finding the line of action of 
F is to substitute for the link M what is known as a kinetically 
equivalent system, which may be defined as a group of bodies, 
rigidly connected together, which will be given the same acceler- 
ations as the actual link under the action of the same forces. To 
meet this requirement three conditions must be fulfilled : 

(a) The two systems must have the same mass. 

(b) The two systems must have the same center of gravity. 

(c) The two systems must have the same moment of inertia. 

The simplest kinetically equivalent system which can be sub- 
stituted for the actual link is shown in Fig. 158. It consists of 



KINETICALLY EQUIVALENT SYSTEM 



131 



two heavy particles mi and m^ connected by a weightless link. 
Then to satisfy the conditions of equivalence 



m\hi 2 H-W2/Z2 2 = / = mk 2 . 



(a) 
(6) 
(c) 



It will be noted that there are three equations to be satisfied and 
four unknowns mi, m2, h\ y fo. Therefore one of these unknowns 




FIG. 158 

can be assumed to have any convenient value, and the equations 
can be solved to find the other three. This principle will be used 
later. 

Eliminating the masses mi, ni2, and m these equations reduce 
to 



Assuming any convenient value for hi, /i2 can be found, thus 
locating the masses mi and m^. In Fig. 159 let the masses mi 




FIG. 159. 

and m2 be located as shown. Let AI, A^ and A g be the acceler- 
ations of mi, m2, and G respectively as found from the acceler- 
ation diagram. Then the accelerations of mi and m2 may be 



132 INERTIA FORCES OF MACHINE PARTS 

regarded as produced by two forces Fi = miAi, and F^ 
respectively, passing through m\ and m 2 and lying in the direc- 
tions of A i and A 2 . Their resultant F must pass through the 
intersection X of the lines of action of these two forces. Since 
F = mA & the resultant is now known in magnitude, direction, and 
line of action. 

It is customary to choose the location of mi at some point 
whose acceleration is already known, and to determine the loca- 
tion of m 2 by means of the relation 



For example, in studying the inertia forces of the connecting rod 
of a steam engine it will be found convenient to locate mi at the 
wrist pin. The line of action of FI is then the line of the stroke. 

It should be noted that in the preceding construction it is 
not necessary to determine the values of the masses mi and. m2 
or of the forces FI and F 2 . The only requirement is to find X, 
the intersection of the lines of action of these forces. This point 
fixes the line of action of F. 

91. Calculation of the Line of Action of the Resultant. In 
some cases it is more convenient to calculate the distance h, Fig. 
157, rather than to locate the line of action of F by means of a 
kinetically equivalent system. 

Let ABC, Fig. 160, represent the link M and let abc be its 
acceleration image drawn from the pole 0. ba represents the 





a 

FIG. 160. 

acceleration of A relative to B. This may be resolved into two 
components: 

xa = ABu 2 parallel to AB, 
and 

bx = ABa at right angles to AB, 



STEAM ENGINE MECHANISM 133 

where co and a are the angular velocity and angular acceleration 
of link M. According to Equation (2), Art. 89 

Fh = la, 

or 



Therefore 



F must, of course, be parallel to Og. Therefore its line of action 
is fixed. To determine on which side of G the distance h must 
be laid off consider that the tangential acceleration of A relative 
to B is in the direction bx, and the angular acceleration a is there- 
fore in the clockwise sense. 

In problems where the accelerating forces are to be found 
only for a single position of the mechanism this method of finding 
the line of action is more convenient than the use of a kinetically 
equivalent system. Where the forces are to be found for a number 
of configurations of the mechanism the latter method is usually 
preferable. 

92. Components of the Resultant Force. The force F is the 
resultant of all the forces acting on the link. These are of two 
kinds: 

(1) Pressures applied by other members of the mechanism. 

(2) External forces, such as weight, pressure of steam or 

gas on pistons, etc. 

The forces of class (2) are usually given by the conditions of 
the problem. They can then be subtracted vectorially from the 
resultant F } leaving as a remainder the resultant of the forces 
of class (1). The methods of splitting up this remainder into 
its components and thus determining the pressures at the pairing 
elements are illustrated in the following articles. The methods 
there developed will suffice to determine forces in practically 
any plane mechanism. 

93. The Steam Engine Mechanism. 1 Fig 161 represents the 
steam engine mechanism. The pressure P on the piston, link 4, 

1 The results of a complete investigation of a standard six-cylinder gasoline 
motor are given in Note E. 



134 



INERTIA FORCES OF MACHINE PARTS 



is known. The crank, link 2, rotates at constant angular velocity 
co. The weights, centers of gravity, and moments of inertia of 
all links are known. It is required to find the tangential com- 
ponent T of the force acting at A in other words the turning 
effort. 

Since the crank OA rotates at uniform angular velocity the 
length OA can be taken to represent both the velocity and acceler- 
ation of the pin A, the velocity being revolved through 90 and 
the acceleration through 180. Then OAD is the velocity polygon, 
OD representing the velocity of the piston, and AD the relative 
velocity between crank and piston. By Klein's construction OC 




FIG. 161. 



represents the acceleration of the piston, reversed in direction, 
and AC is the acceleration image of the rod AB. 

Consider first the forces acting on the piston, link 4- These 
are: 

(a) The weight W, acting at G* the center of gravity 

of the piston. 

(6) The steam pressure P acting horizontally along the 
center line of the rod. 

(c) The reactions of the cylinder wall and crosshead guide. 

These are vertical if friction is neglected, but their 
point of application is not known. 

(d) The pressure between the wrist pin and the connect- 

ing rod. The point of application of this force is at 
B, but its direction and magnitude are unknown. 



INERTIA FORCE OF CONNECTING ROD 135 

This force may be regarded as made up of two com- 
ponents: 

(1) A thrust in the direction AB, 

(2) A force acting across the rod, and due to the weight 

and inertia of the rod. This force can be com- 
pletely determined as shown below. 
(e) The resultant of the forces (a), (6), (c) and (d) is known. 

This resultant acts horizontally through B and has 

the magnitude 



Where 

mi = mass of link 4, 

Ab = acceleration of the piston = 0c, Fig. 161. 

The determination of the forces on the piston now involves 
the following steps: 

(1) Determination of the inertia forces of the rod. 

(2) Combination of this inertia force with the weight of 

the rod. 

(3) Resolution of this resultant into two parallel compo- 

nents acting at A and B. 

(4) Combination of the known forces acting on the piston. 

(5) Determination of the rod thrust and the guide reaction. 

The inertia force of the rod is given by the equation 



where ma is the mass of the rod and A is the acceleration of G, 
the center of gravity of the rod. Its direction is the same as that 
of A a . To determine the line of action of this force replace the 
rod by a kinetically equivalent system consisting of two heavy 
particles located at B and E, Fig. 161. According to Art. 91 



where k is the radius of gyration of the rod about G. Then the 
inertia force may be regarded as the resultant of the inertia forces 
of the particles D and E. The latter act in the directions of the 
accelerations of these two points. Draw Ee parallel to OB. 
Then Oe is the acceleration of E, since AC is the image of the rod 



136 



INERTIA FORCES OF MACHINE PARTS 



AB. Draw EX parallel to Oe. This is the line of action of the 
inertia force of E. The acceleration of B is horizontal. Hence 
BX is the line of action of the inertia force of B. The intersection 
X lies on the line of action of the resultant of these forces, in other 
words of the inertia force of the rod. Draw Gg parallel to OB. 
Then Og is the acceleration of G. Through X draw XFparallel to 
Og. This is the line of action of the inertia force of the rod. The 
magnitude of this force is given by the relation 



Next this force must be combined with the weight of the rod. 
The weight Ws acts vertically at Gs. To combine these two draw 
Fig. 162, vertically to intersect XY at S. Let -SY represent 




the inertia force, or the accelerating force reversed in direction. 
Let ST be the weight of the rod. Then SM is the resultant of 
these two forces. 

To resolve SM into two parallel components acting at A and 
B draw BN equal and parallel to SM t and join AN. Prolong 
SM to cut AB at R and AN at K. Through K draw KL parallel 
to AB. Then the resultant SM can be considered as formed of 
two components, LB acting at B and NL acting at A. For tri- 
angles ABN and ARK are similar. 
Therefore 

LN_ = LN = KR = BL 
LK~ BR AR AR' 
Or 

LNXAR=BLXBR 
Also 

BL+LN = BN = SM. 



FORCES ON PISTON 



137 



Therefore LN and BR are the components of SM acting at A 
and B respectively. 

Now returning to the piston, link 4, the following forces are 
known : 

(a) Pressure P=--Op, Fig. 163. 
(6) Weight W 4 = pw, Fig. 163. 

(c) Component of inertia force and weight of rod acting at 
B = BL (Fig. 162) =wl (Fig. 163). 




FIG. 163. 



The following forces are known in direction: 

(d) Thrust in rod AB. 

(e) Side thrust, which is vertical neglecting friction. 



The resultant of all these forces = m4A4, where m.\ is the mass 
of the piston and A 4 is the acceleration of the piston and is com- 
pletely known. Let Ox, Fig. 163, represent this resultant. 
Through x draw xk parallel to AB, and through I draw Ik vertical. 
Then xk = rod thrust and Ik = side thrust. 

The determination of the forces acting on the piston is now 
complete except for the location of the line of action of the side 
thrust. To find this consider that the pressure P and the weight 
1^4 both act through G 4 . Their resultant Ow, Fig. 163, there- 



138 INERTIA FORCES OF MACHINE PARTS 

fore also acts through this point. Through Gt draw GY parallel 
to Ow intersecting the line of action of the forces BL, Fig. 162, 
at Y. Through Y draw YZ parallel to 01. YZ is then the line 
of action of the resultant of the pressure, the weight of the piston 
and that component of the inertia force and the weight of the rod 
which acts at B. This resultant is represented in Fig. 163 by 
01. Prolong AB to intersect YZ at R. Then R is a point on the 
line of action of the side thrust, since the side thrust, rod thrust 
and the resultant 01 must meet in a point. The forces acting on 
the piston are now completely known. 

The force acting on the rod at A is easily found. Its two 
components are the rod thrust = kx, Fig. 163, and the second part 
of the inertia force and weight of the rod = LN, Fig. 162. 

Let AK (Fig. 164) =xk (Fig. 163), 

and 

AN (Fig. 164) =NL (Fig. 162). 




FIG. 164. 

Then the resultant AL is the pin pressure at A. Resolving this 
into two components, 

(a) AT perpendicular to the crank OA, 

(b) LT parallel to the crank. 

It is evident that AT represents the turning effort, and LT the 
thrust in the crank. If the thrust LT is combined with the weight 
and inertia force of the crank the resultant gives the bearing 
pressure, or the force which the bearing must exert in order to 
hold the crank in place. 

NOTE. All of the constructions in Figs. 162, 163, 164, could have been 
placed on Fig. 161. It was thought best, however, for the sake of clearness 
in showing the successive steps in the solution, to separate the constructions. 
In applying the principles set forth in the preceding paragraphs it is usually 
more convenient to do all the work on a single figure. 



ATKINSON GAS ENGINE 



139 



94. The Atkinson Gas Engine. In the Atkinson gas engine, 
Fig. 165, the crank, link 2, revolves at constant angular velocity 
co. The pressure P on the piston, link 6, is known, as well as the 
weights, centers of gravity and moments of inertia of all the links. 
It is required to find the turning effort exerted at the crank pin A . 

The velocity polygon, Fig. 166, and the acceleration polygon, 




?/$}' 




FIG. 165. 



FIG. 166. 





FIG. 167. 



FIG. 168. 



Fig. 167, are drawn by the methods of the preceding chapters. 
The first part of the solution follows exactly the same procedure 
as in the ordinary steam engine. The inertia force of the rod 
is found, combinid with the weight, and resolved in o parallel 
components acting at the ends of the rod. The construction 
is shown in Fig. 168, the method followed being exactly the same 
as that employed in Fig. 162. The forces on the piston are then 



140 



INERTIA FORCES OF MACHINE PARTS 



shown in Fig. 169 in the same way as in Fig. 163, the only differ- 
ence being that since the center of gravity is taken at D the side 
thrust acts through this point, so that it is not necessary to deter- 
mine its line of action. In Fig. 170 the pressure CL on pin C is 
found in the same manner as the crank pin pressure in Fig. 164. 

The next step is the determination of the forces acting on 
links 3 and 4- The forces acting on link 3 are as follows: 

(a) The weight TF 3 . 

(6) The pin pressure at C. 

(c) The pin pressure at B. 

(d) The pin pressure at A. 




FIG. 169. 



FIG. 170. 



Of these (a) and (6) are completely known. The points of 
application of the forces (c) and (d) are at the pins B and A. The 
resultant 

F 3 



where ma is the mass of the link, and A is the acceleration of the 
center of gravity G$. Fs is thus known in magnitude and direc- 
tion, but its line of action is still to be found. The pin pressure 
at B can be resolved into a component in the direction PB and 
two components due to the acceleration and weight of link 4- The 
latter components can be comple ely determined. Let P^B, Fig. 
171, represent link 4 and Ob its acceleration image. Substitute 
for the link a kinetically equivalent system consisting of two 
particles located at P and M. Then P 4 (r4XCr4M = A; 2 4 where 
4 is the radius of gyration of link 4 about G. Locate on Ob 
the image m of the point M . Since P has no acceleration the 
inertia force of link 4 passes through M . As usual its direction 
and magnitude are given by the equation 



FORCES ON TRIANGULAR LINK 



141 



where A g is the acceleration of the center of gravity <2 4 . A g is 
given by Og, Fig. 171. The force F is therefore completely 
known and is represented by MS, Fig. 171. F 4 is the resultant 
of the weight of the link and the pin pressures exerted at P 4 and 
B. Through G 4 d aw a vertical in ersecting the line MS at X. 
From X lay off XT = MS and from T lay off TY=W. Then 
XY represents completely the vector difference between F 4 and 
W in other words, the resultant of the pin pressures at P 4 and 
B. Resolve XY into two par- 
allel components BZ and PW 
acting at B and P 4 respective- 
ly. BZ and PW are components 



a 


M 


& 


\ 


\ 

\ V J 

\ 

^ 






FIG. 171. 



FIG. 172. 



of the pin pressures at B and P 4 , the other components being equal 
and opposite forces acting along the line PB. Thus another 
of the forces acting on link 3 has been resolved into a known com- 
ponent and a second force whose line of action is known. 

Returning now to link 3 the next step is to locate the line 
of action of the inertia force FS. In Fig. 172 le ABC represent 
link 3 and abc its acceleration image, taken from Fig. 167. Let 
<% be the center of gravity of link 3 and g% its accele ation image. 
Draw an parallel to AB and bn perpendicular to AB. Then 
bn = ABa. It was shown in Art. 91 that F2 > h = Ia = m^ 2 a t where 
h is the perpendicular distance from (73 to the line of action of ^3. 
Therefore 



003 -AB' 

Since nb represents the tangential component of B's acceleration 
relative to A it follows that a is in a clockwise direction, and that 



142 



INERTIA FORCES OF MACHINE PARTS 



therefore ^3 must be located as shown in Fig. 172 and not at F'%. 
If an additional force equal and opposite to F 3 were applied to 
the link the whole system would be in equilibrium. The forces 
then acting on the link would be: 

(a) ^3 reversed in direction. (Completely known.) 
(6) The weight Ws- (Completely known.) 

(c) Pin pressure at C. (Completely known.) 

(d) Pin pressure at B. 

(1) Component due to weight and inertia of link 
4 = Fb. (Completely known.) 

(2) Thrust along P 4 B. (Direction known.) 

(e) Pin pressure at A. (Point of application known.) 

All the known forces can be combined into a single resultant 

R as shown in Fig. 173. 
The link may then be con- 
sidered in equilibrium under 
the action of three forces : 

(1) The resultant R. 

(2) The thrust along 
P 4 B=T 4 . 

(3) The pin pressure 
Sit A = Pa. 

These three forces must 
meet in a point. Let X, 
Fig. 173, be the intersection 
of the lines of action of 
the first two. Then AX is 
the line of action of the 
pin pressure at A. The 
value of the unknown forces 

can now be found from the triangle of forces, Fig. 173. The pin 
pressure at A may be resolved into a radial component B and 
a tangential component T. The problem is thus completely 
solved. 

95. The Wanzer Needle-bar Mechanism. In the mechanism 
shown in Fig. 174 the disk link 2 revolves at uniform angular 
velocity co. The pins A and B on the triangular link 8 move in the 




FIG. 173. 



WANZER NEEDLE BAR 



143 



slots in link 2, and the pin C is guided in a vertical line. Given 
the angular velocity co and the weight, center of gravity and 
moment of inertia of link 3, it is required to find the pressures 
on pins A, B, and C. 

The velocity polygon, Fig. 175, is constructed in the usual 
manner, and the acceleration polygon, Fig. 176, is found by the 

use of the auxiliary point M, and 
CoriohV law as explained in Arts. 85 
and 67. 

The magnitude, direction, and line 
of action of the resultant accelerating 
force acting on link 3 is found pre- 
cisely as in the Atkinson Gas Engine. 





FIG. 174. 



FIG. 175. 



Let XY, Fig. 174, represent this force. If a force equal and 
opposite to this resultant is applied to the system link 3 will be 
in equilibrium under the action of 5 forces as follows: 

(1) F Z = XY. (Completely known.) 

(2) Wz = YV. (Completely known.) 

(3) Pin pressure at A acting in line AM. 

(4) Pin pressure at B acting in line BM. 

(5) Pin pressure at C acting in line CN. 

The first two can be combined into a single force YZ, thus 
reducing the number of forces to four. These forces can be com- 
bined in pairs, the resultant of one pair being equal and opposite 
to that of the other pair. The resultant of YZ and the pin pres- 
sure at C must pass through N, thp intersection of their lines of 
action. Similarly the resultant of the pin pressures at A and 
B must pass through M . 

Therefore MN is the line of action of both resultants. From 
N lay off NL = YZ and draw LK parallel to CN. Then KL is 



144 



INERTIA FORCES OF MACHINE PARTS 



the pin pressure at C and KN is the resultant of this pressure 
and YZ. NK is then resolved into components NP and PK 




. FIG. 176. 

parallel to BM and AM respectively. These components repre- 
sent the pin pressures at A and B. 

If a known force is to be overcome at C, it is only necessary 
to combine this known force with the resultant YZ and proceed 
exactly as before. 



CHAPTER VI 



BALANCING OF ENGINES 

96. Introductory. One of the most important applications 
of the study of inertia forces is in the balancing of engines. The 
inertia forces in high-speed engines are of great magnitude and 
the balancing of them is a very important consideration. This 
subject has been given extensive study, the most complete treatise 
in the English language being by W. E. Dalby. The following 
discussion is based in general on the methods of Dalby. 

97. Kinetic Load to an Unbalanced Mass. Suppose a shaft 
S, Fig. 177, to carry a mass M whose center of gravity is at a dis- 





FIG. 177. 



FIG. 178. 



tance r from the shaft axis. If the shaft rotates with an angular 
speed co, the connection between the mass and the shaft is sub- 
jected to a stress of magnitude Mrco 2 . The bearings must there- 
fore take up a kinetic load whose magnitude is Mrco 2 pounds. As 
the direction of this load is continually changing, vibrations is 
the frame or foundation which carries the bearings may be set 
up when the shaft rotates at high speed. 

98. Centrifugal Couple. If the shaft S, Fig. 178, carries two 
masses MI and MZ in different planes of revolution, but in the same 
axial plane, and if further the centrifugal forces Minco 2 and 

1 The Balancing of Engines, by W. E. Dalby. Longmans, Green & Co. 

145 



146 



BALANCING OF ENGINES 



M 2 r 2 o) 2 are equal, then the shaft is subjected to the action of a 
centrifugal couple. This couple tends to turn the shaft in an 
axial plane and must be resisted by an equal couple applied by 
the bearings. 

99. Masses in a Single Plane of Revolution. If several 
masses, MI, MI, etc., Fig. 179, lie in the same transverse plane, 
the shaft is subjected to concurrent forces, Miriu 2 , M 2 r 2 u> 2 , etc., 
acting in the plane of revolution of the masses. The condition 
that the load on the shaft shall be zero is given by the equation 



(1) 



or since w 2 is a common factor, 



In other words, if the products Mtf\, M 2 r 2 , etc., are laid off in 
succession as vectors, each in its proper direction, these vectors 
should form a closed polygon. 





FIG. 180. 

Returning again to Fig. 179, when the products Miri, M 2 r 2) 
etc., are laid off as in Fig. 180, the polygon will not in general be 
closed. The vector EA gives the magnitude and direction of 
the kinetic load MQTQ required to close the polygon. Hence if 
any convenient value of r$ is assumed, the mass MQ is definitely 
located. If desired, the gap EA may be closed by means of two 
or more vectors chosen at will, and the given system may thus 
be balanced by means of two or more masses instead of one. 

Since the masses are proportional to the weights, Equation (1) 
may be written 



Wiri+W 2 r 2 + . . . 



(2) 



ROTATING MASSES 



147 



100. Masses in Different Transverse Planes. Suppose masses 
MI and M 2} Fig. 181, to be connected to the shaft S at A and B 
respectively. Through some point on the shaft axis pass a 
transverse plane (called the plane of reference, or briefly, the R.P.). 
The mass M\ gives rise to a kinetic load Fi=M\riu 2 . At 
introduce two equal and opposite 
forces FI parallel to this kinetic load. 
The load FI at A and the opposite 
force FI at form a centrifugal couple 
whose moment is M\r\u 2 a\. Hence 
the single force FI acting at A may 
be replaced by an equal and parallel 
force acting at and a couple whose 
moment is FI<ZI. Likewise the force 
F2 acting at B may be replaced by an 

equal and parallel force F acting in the R.P. and a couple whose 
moment is F '2,0,2 




Therefore an R.P. can be chosen at will and the system of 
kinetic loads may be reduced to a system of concurrent forces 
acting in the R.P. and a system of couples acting in various axial 
planes. The forces have a single resultant, and the couples can 
be reduced to a single couple. Hence, in general, the system of 
kinetic loads may be reduced to a single force and a single couple. 
The magnitude of the couple will depend on the position chosen 
for the R.P. 

The conditions to be satisfied in order that the shaft shall be 
kinetically balanced are evidently the following: 

(1) The resultant centrifugal couple shall vanish; that is, 

u?(Miriai+M2r2a2+ . . . +!/>#) =0; 

(2) The resultant centrifugal force shall vanish; that is, 

. . . +M n r n )=0. 



To balance a shaft with given revolving masses, at least two 
additional masses in different transverse planes are required. 
To determine these balancing masses proceed as follows: let the 
given masses be denoted by M\, M^ etc., Fig. 182, and the bal- 
ancing masses by MO and M'Q. Chose the transverse planes in 



148 



BALANCING OF ENGINES 



which the balancing masses are to lie, and take the plane of 
M'Q as the R.P. Denote by oo, a\ t 02, etc., the distances of the 
planes of MO, MI, MI, etc., from the R.P. The couples M\r\a\, 
, etc., may now be calculated. The only unknown couple is 



A/, 




/% 



Jt> 




FIG. 182. 



M r ao, since by choosing the R.P. as the plane of the mass M'Q 
the couple produced by this mass is zero. Laying off vectors 
representing the known couples as the sides of a polygon, Fig. 
183, the closing side gives the unknown couple MQTQOQ. The 
arm oo is known, and therefore the product MQTQ can be found 
immediately. The direction of the closing couple vector gives 
the plane of the couple and therefore the direction of the mass MQ 
from the axis. 





FIG. 183. 



FIG. 184. 



By the addition of the couple MQTQOQ condition (1) is satisfied. 
It is now necessary to satisfy condition (2). Lay off the prod- 
ucts Min, M2T2, etc., Fig. 184, including MQTQ as just found, 
as the sides of a polygon. The closing side gives the product 
Af'o/o. Choosing a convenient value for r'o, the mass M'Q (which 
must be placed in the R.P.) is readily determined. The direc- 
tion of the closing side gives the direction of M'Q from the axis. 



BALANCING OF ROTATING MASSES 



149 



The following are rules for drawing the sides of the polygons: 

(1) The force vectors (Afiri, Mtfz, etc.) are drawn from the 
axis outward toward the masses and parallel to the cranks. 

(2) The couple vectors are likewise drawn parallel to the 
respective crank directions, outward for masses on one side of 
the R.P., and inward (toward the axis) for masses on the other 
side of the R.P. 




tJ 



M t 



FIG. IQ5 

Schedule I 



Plane. 


Mass 


Rbdius 


a 


e 


Mr 


Mra 


1 





II 


4.4 


o 


tt 


?66 


Z 


3 


e 


Z.G 


105 


24 


GZ.4 


3 


1.5 


& 


1.4 


225 


9 


12.6 


O 


(Z.OI) 


10 


4.4 


(ZI6~) 


(20. 1) 


(3Q.4) 


O 1 


(1.70) 


5 





025*} 


(8.5) 


O 



FIG. 185. 

101. Example. Figs. 185, 186, 187 show the application of 
the method to an example. The masses MI, M^ and MS in the 
planes 1, 2, and 3, Fig. 185, are to be balanced by masses in two 
planes. Let one of the masses be located in plane 1. The plane 
of the other mass is chosen at random and is taken as the R.P. 
In Schedule 7 the masses, their directions, their radii, and their 
distances from the R.P. are entered in the proper columns. Then 
the products Mr and Mra are calculated and entered as shown. 



150 



BALANCING OF ENGINES 



The couple polygon is now drawn as shown in Fig. 186. AB is 
laid off in the direction of r\ outward from the shaft, its length 
representing the value of the product Minai = 96.8. Then EC 

and CD are laid off in 
order in the directions of 
7-2 and r 3 , their lengths 
representing the values 
of the products Mtfzdi = 
62.4, and M 3 r 3 a 3 = 12.6 
respectively. The closing 
side DA represents the 
couple Moroao, due to 
the balancing mass MO 
Z in plane 1. This couple 
is found by measurement 
to be 88.4; hence 

M ro= 88.4 =20.1. 




FIG. 186. 



The product M r may 
now be entered in its 
proper place in the 

column headed Mr. The addition of the mass M at the radius 

TO in the plane 1 has made the resultant couple vanish. There 

are left, however, the forces in the 

reference plane, including the force 

Moro, just found, and these forces 

will in general not be balanced. 

Laying off the products Mr in 

order, as shown in Fig. 187, the 

closing vector EA gives the prod- 
uct M'or'o, which measures 8.5. If 

now the products M r and MVo 

are divided by the assumed radii 10 

and 5 M = 2.01, and M' = 1.70. 

The directions of the balancing 

masses are of course given by the 

directions of the closing lines DA, Fig. 186, and EA, Fig. 187. 
102. General Relations. So far it has been assumed that 

the masses have had unequal radii. Evidently the radius of a 




CONDITIONS OF BALANCE 151 

mass can be varied at will provided that at the same time the mass 
itself varies so that the product Mr remains constant. It is there- 
fore possible to choose some convenient radius, say that of an 
engine crank, and all masses that have different radii can be 
reduced to masses having this common radius. In this case the 
common factor r can be dropped, the reduced masses can be used 
for the sides of the force polygon, and the reduced mass moments 
Afiai, M2d2 become the sides of the couple polygon. 

In the balancing of a given system of masses, it is essential 
to know how many quantities may be determined beforehand, 
and how many must be left undetermined. For example suppose 
that five masses are to be placed along a shaft so as to form a 
balance system. If all the masses, their angles, and their dis- 
tances from some R.P. are fixed at random beforehand, the system 
will undoubtedly be unbalanced. Some of these variables must 
be left for subsequent determination. 

With n masses carried on a shaft, the first question to be decided 
is, how many variables enter into consideration. Suppose that 
one of the masses M\ is fixed, that its plane of revolution is also 
fixed, and that its direction from the axis at some instant is 
assumed. The other masses may be placed in n 1 planes of 
revolution, which may be chosen arbitrarily; they may be placed 
in n 1 axial planes which make nl different angles with the 
plane of MI. Evidently it is not the absolute values of the masses 
which count, but their ratios to each other, and therefore there 

will be nl mass ratios -^3, -^, etc., which can be chosen arbi- 

M2 MS 

trarily. Hence as independent conditions there are 

nl planes of revolution, 

nl angles between axial planes, 

nl mass ratios. 

The total number of independent conditions is therefore 
3(n-l)=3n-3. 

To form a balanced system the force polygon and the couple 
polygon must both close. The closing vector in each case defines 
two quantities, one given by its magnitude and the other by its 
direction. Hence four variables can be determined by the poly- 



152 BALANCING OF ENGINES 

gons. Of the 3n 3 variables, therefore (3n 3) 4 = 3n 7 may 
be assumed at will, but four must be left for determination by 
the polygons. Thus for five masses 3X5 7 = 8 of the variables 
may be assumed. For example, if the five masses are chosen, 
four mass ratios are fixed; if, further, three planes of revolution 
are assumed, one of these may be taken as the reference plane, 
and the distances from this plane to the other two fixes two more 
of the variables; finally, two angles between axial planes can be 
assumed. The remaining four variables two planes of revolu- 
tion and two angles can now be found from the force and couple 
polygons. 1 

The fact that the conditions of balance require two closed 
polygons with sides parallel, but usually not in the same ratio, 
leads directly to some interesting and obvious conclusions. The 
student may verify the following statements and give reasons: 

(1) Two masses cannot form a balanced system unless 

they are in the same plane of revolution. 

(2) Three masses, to form a balanced system, must be 

either in the same plane of revolution or in the same 
axial plane. 

103. Analytical Methods. In many cases it is convenient 
to use analytical instead of graphical methods in determining 
the shaking forces and the size and position of balancing masses. 
For this purpose the forces are resolved into horizontal and ver- 
tical components as shown in Fig. 188. Each of the groups of 
forces and couples formed by this resolution must be balanced 
separately. The following equations then express the conditions 
for balance: 

Horizontal forces: 

rco 2 (Mi cos di+M 2 cos 2 + . . . M r . cos 0) =0. 
Vertical forces: 

sin di+M 2 sin 2 + . . . Msin0 n )=0. 



1 Freedom of choice of the 3w 7 variables is not absolutely unrestricted. 
For example, if one mass is chosen greater than the sum of all the others, 
or if the angles are so chosen that all the masses lie on the same side of one 
axial plane, the system cannot be balanced. In such cases it will be found 
impossible to construct polygons which satisfy the assumed conditions. . 



ANALYTICAL METHOD. ROTATING MASSES 

Horizontal couples: 

ru> 2 (Miai cos 0i+Af 2 a 2 cos 2 + . . . M n On cos 0)=0. 
Vertical couples: 

sin 6 ] \-\-M "2^2 sin 02+ . . Af n a ra sin n )=0. 



153 



Evidently the common factor ro> 2 can be omitted from all these 
equations. For convenience and brevity write cos0i=i, 




M, 



FIG. 188. 

sin 0i =2/1, etc. Then the four equations can be conveniently 
written: 

0, ........ (I) 

0, ........ (II) 



SMaz=0, ...... (Ill) 

2Mai/ = ........ (IV) 

Equations (I) and (II) simply state that the center of gravity 
of the system lies in the center of the shaft. A system which 
satisfies Equations (I) and (II) but not Equations (III) and (IV) 
is in static balance but not in running balance. 



154 



BALANCING OF ENGINES 



104. Inertia Effects of Reciprocating Masses. Harmonic 
Motion. Suppose a mass MI, Fig. 189, to be given a recipro- 
cating harmonic motion by means of a crank C\. If co is the angu- 
lar speed of the crank, and 0i is the angle which the crank makes 
with the line of motion of M i then the radial acceleration of the 
crank pin is rco 2 , and the horizontal component of this acceler- 
ation is rco 2 cos &i. This horizontal component is also the acceler- 
ation of MI. The truth of this statement is almost self-evident, 




FIG. 189. 

but it may be shown analytically as follows: the displacement 
of M i from the middle point of its motion is 

x = r cos 0i. 
Hence 



dx 



M* . n aui 

-rr = r sin 0i -7- = rco sin 0i, 



and 



dt 

d 2 x 



dt 



= r cos 



-7- 
at 



-r,.,2 



rco cos 



The direction of the acceleration of MI is always toward A, the 
mid-point of its stroke. The accelerating force required is 

MiTut 2 COS 0i, 

and this force must likewise be directed toward A . 



RECIPROCATING MASSES. HARMONIC MOTION 155 

If the reciprocating masses MI, MI, etc., are the piston and 
crossheads of an engine, then forces equal and opposite to the 
accelerating forces act on the frame or bed of the engine. Suppose 
that the engine has four cranks and four reciprocating masses 
as shown in Fig. 189. Then the engine frame will be subject to 
action of four inertia forces, 

FI = Mirco 2 cos 0i, F2 = M2ru> 2 cos 62, etc., 

all of which will be directed from the mid plane outward. Of 
course all these forces lie in one plane, the plane of reciprocation. 

The system of parallel forces may be treated in the same 
manner as the system of forces along a shaft, Art. 100. A refer- 
ence plane is chosen perpendicular to the plane of the recipro- 
cating masses. Then the force FI may be replaced by an equal 
force in the reference plane and a couple whose moment is Fidi, 
where ai is the distance of F\ from the R.P. Similarly the forces 
F2, FS, etc., can be replaced by equal forces in the R.P. and suit- 
able couples. The system is therefore reduced to a set of collinear 
forces in the R.P. and a system of couples in the plane of reciproca- 
tion. The forces can be reduced to a single resultant, the scalar 
sum 



which causes a backward and forward movement of the frame 
as a whole. The system of couples reduces to a single couple 
whose moment is the scalar sum 



The effect of this couple is to rock the frame in the plane of recipro- 
cation. 

105. Balancing Conditions. The system of forces acting on 
the engine frame will be balanced when the single resultant and 
the resultant couple are both equal to zero ror every position 
which the masses can take. These conditions are expressed by 
the equations 



Or 

ru?(Mi cos 0i+ M 2 cos 62+ . . . M n cos n )=0 

cos 0i+M 2 a 2 cos 2 + . . . M n On cos n ) =0. 



156 BALANCING OF ENGINES 

Evidently the factor rco 2 may be omitted from each equation. 
In order that the system may be balanced for every position these 
equations must hold when the cranks are turned through any 
angle <j>. In this case the angles Oi, 62, 63 . . . , become 0i+<, 
^2+0 .... Substituting these values in the equations of bal- 
ance we get 



M i cos (0+0) +M 2 cos (0 2 +0)+ . . . M n cos(d n +<t>) = 
MI cos di cos 0+7kf 2 cos 02 cos 0+ . . . M n cos B n cos <f> 
(Mi sin 0i sin 0+71^2 sin 02 sin 0+ . . . M n sin 6 n sin 0) = 
cos <j>(M i cos 0i+Af 2 cos 2 + . . . M n cos n ) 
sin (f>(Mi sin Bi+Mz sin 02+ . . . M n sin n ) =0. 



MI cos 0i+M2 cos 2 + . . . M n cos9 n = 0, 
MI sin 0i+Af2 sin 02+ . . . M n sin n = 0. 



In order that this equation may hold for every value of <j> each 
of the quantities in parentheses must vanish separately. That is: 



Or 



Similarly from the equation for the couples 



In other words the conditions for balancing a system of recipro- 
cating masses having harmonic motion are precisely the same as 
those for balancing the same masses concentrated at the cranks 
from which their motion is derived. 

106. Engines with Finite Rods. In the preceding discussion 
the masses were assumed to have harmonic motion. In an actual 
engine with connecting rods of finite length the motion is not 
strictly harmonic, but may be assumed to be to if the rod is not 
too short relative to the crank. 

It is customary to assume part of the mass of the rod itself 
as concentrated at the crosshead pin and the remainder at the 



FINITE CONNECTING ROD 



157 



crank pin. Thus if L is the length of the rod, and h the distance 
from the crosshead pin to the center of gravity, the fraction h/L 
will be considered concentrated at the crank pin and the remainder 
at the wrist pin. 

Two approximations are thus introduced, (1) in considering 
the motion as harmonic, and (2) in dividing the mass of the rod 
between the two pins. It was shown in Art. 93 that if the rod 
is replaced by two equivalent masses, one of which is placed at 
the crosshead pin, the other cannot in general be at the crank 
pin. Later it will be shown how the first of these approximations 
may in some cases be corrected. A correction of the second is 
not possible. 

107. Acceleration of Reciprocating Masses. Finite Rod. 
The deviation of the 'actual inertia forces from those which would 
exist in the case of true harmonic motion introduces errors in 
the balancing which become serious when the connecting rod is 
short relative to the crank. Under certain conditions the masses 
may be so arranged that these errors practically disappear. 




FIG. 190. 

In Fig. 190 the crosshead is driven by a connecting rod of 
finite length L. For a given crank angle 6, measured from the 
inner dead center, the travel of the wrist pin B is 

s = L+r L cos </> r cos 0. 
From the geometry of the figure 

L sin <j> = r sin 0, 
and therefore 



L cos < = VL 2 -L 2 sin 2 < = \/L 2 -r 2 sin 2 = LJl- sin 2 0. 



158 BALANCING OF ENGINES 



Since, however, is usually less than J, 



L+ll jz sm2 ^ = ^~oT sm2 approximately. 
Hence 

s = r r cos 0+757" sm2 0- 
Differentiating twice 

ds a dd . r 2 . Q d0 / . . r A 

di = r Sm ^ ^ + L Sm ^ COS ^ = rc ( Sm 6 +2L Sm 20 j' 
and 



s 0+ cos 20 = ra> 2 cos 0+ cos 20. 
The inertia force due to a mass M having this acceleration is 



This inertia force, it will be noted, can be divided into two 
parts, 

(1) Mru 2 cos 0, 

which is precisely the inertia force of the same mass having 
purely harmonic motion, and 

r 2 

(2) M j cos 20w 2 , 

which is the error introduced by the finite rod. The second term 
may be written 

2w 2 cos 20. 



This is evidently the inertia force which would be developed 
by a mass M which is given harmonic motion by a crank of 

r 2 
length JY rotating at a speed 2co. 

The actual inertia force caused by the reciprocation of the 
mass M may thus be separated into a primary and a secondary 
part. The primary part is the projection on the line of stroke 
of the centrifugal force of the mass M transferred to the crank 



f SECONDARY BALANCE 159 

pin, while the secondary part is the projection on the line of the 
stroke of the centrifugal force which would result if the mass 
M were transferred to the crank pin 2 of an imaginary crank 

r 2 

of length rotating in the same plane as the main crank, but 

at double the speed. 

108. Secondary Balance. In order that complete secondary 
balance may be secured two conditions must be satisfied for 
every value of the angles B\, 62, . . . 0. 

SM(2co) 2 (r 2 /4L) cos 20 = or SM cos 20 = . . (a) 
SaM(2a>) 2 (r 2 /4L) cos 20 = or 2aM cos 26 = . . (6) 



Since these equations must hold for every value of 0, that is for 
every possible position of the masses, it is necessary as before to 
satisfy four equations giving the conditions for balance as follows: 



cos 20 = .or ZAf(z 2 -2/ 2 )=0, 

sin 20 = or 2Mxy = Q, 
2Ma cos 20 = or 2Ma(x? - y 2 ) = 0, 
2Af a sin 20 = or ZMaxy = 0. 

For complete balance of primary and secondary forces and 
couples eight equations must therefore be satisfied. 

SMz=0 ...... (I) 1 

\ Primary forces balanced. 

(II) J 



(Ill) 

\ Primary couples balanced. 

(IV) 



^M(x 2 -y 2 )=Q. ... (V) 1 

\ Secondary forces balanced. 
...... (VI) J 



2Ma(x 2 -y 2 )=Q . . . (VII) 1 

\ Secondary couples balanced. 
..... (VIII) J 



109. Partial Balance. In order that an engine may be in 
complete primary and secondary balance, the eight equations of 



160 BALANCING OF ENGINES 

the preceding article must be satisfied that is, eight variables 
must be left undetermined. As the number of variables at our 
disposal is 3 (n 1) it follows that no engine of less than four cranks 
can be completely balanced. Practically five cranks is the mini- 
mum, the solution for the four-crank engine involving an impos- 
sible arrangement from the standpoint of construction. 

For the smaller numbers of cranks partial balance may be 
secured by satisfying part of the equations of Art. 108. These 
equations must always be taken in pairs. It is impossible, for 
example, to satisfy Equation (I) for all positions of the cranks 
unless Equation (II) is likewise satisfied. 

The conditions for balancing the forces must always be ful- 
filled before attempting .to balance the couples. Any solution 
which apparently balances the couples leaving the forces unbal- 
anced is illusory. If the forces are not balanced the couples may dis- 
appear with respect to one particular reference plane, but not with 
respect to other planes. Such a solution simply shows that the 
unbalanced resultant force lies in the chosen reference plane. 
It is of course useless to attempt to get rid of the secondary forces 
if the much larger primary forces are left unbalanced. There- 
fore, in attempting to balance an engine partially it is always 
necessary to satisfy first Equations (I) and (II) of the preceding 
article. Either the primary couples or the secondary forces may 
then be balanced. In other words after Equations (I) and (II) 
are satisfied either Equations (III) and (IV), or Equations (V) 
and (VI) may be taken next. The secondary couples must be 
left to the last. 

110. The Single-crank Engine. In this case 

n-1, 

3(n-l)=0. 

Hence none of the eight equations of balance can be satisfied. A 
modification of the shaking forces in such an engine can, however, 
be effected by counterbalancing. The primary force in such an 
engine is TWVco 2 cos 6. If a counterweight M is placed upon the 
shaft at a distance from the center TO such that M ro Mr and 
directly opposite the crank pin, this weight will exert a centrif- 
ugal force Mru> 2 in a direction opposite to the crank radius. 



f DUPLEX ENGINE 161 

Resolving this force into horizontal and vertical components 
there results, 

Horizontal force = Mrco 2 cos 6. 

Vertical force = Mru 2 sin 6. 

The primary shaking forces are thus balanced and vertical shak- 
ing forces substituted. 

If a smaller counterweight is used such that MQTQ = kMr where 
k is less than 1 the primary horizontal force becomes 

1/rco 2 cos B-kMru 2 cos B = Mr<f(l-k) cos 6 
and the vertical shaking force 

kMru 2 sin 9. 

EXERCISE 

1. A 10X12 inch single-cylinder engine runs at 250 r.p.m. Length of 
connecting rod = 36 inches. Weight of piston, crosshead, etc., including 
portion of connecting rod which is considered concentrated at wrist pin = 100 
pounds. Determine shaking forces for each 30 of the revolution, using 
fc = 0, |, I, f, and 1. Draw a 3-inch circle representing the crank circle, and 
mark the 30 points. From each of these points draw vectors outward, 
representing the shaking forces reversed in direction. Join the ends of these 
vectors by a smooth curve. Repeat for each value of k. Scale of forces 200 
pounds = 1 inch. The resulting curves show the effect of various counter- 
weights. 

111. The Two-crank Engine. Here 



3(n-l)=3. 

Since the equations of Art. 108 must be taken in pairs only (I) 
and (II) can be satisfied. These equations reduce to 

M 1X1 +M 2X2 0. 



These equations can be satisfied by making 



-Xi=X 2 . 

-2/1 = 2/2. 



162 BALANCING OF ENGINES 

In other words the reciprocating masses should be equal and the 
cranks opposite to one another. This arrangement is usually 
undersirable on account of the uneven turning effort, and the 
cranks are ordinarily placed at right angles. 

EXERCISE 

2. A duplex engine has two cranks at right angles. The sizes, weights, 
speed, etc., are the same as for the single cylinder engine of Exercise 1. Con- 
struct diagram similar to that in Exercise 1, showing the combined shaking 
forces for the two cylinders, each crank being counter-weighted as before. 
The center lines of the two cylinders are 4 feet apart. Taking a reference 
plane midway between the two cylinders construct a similar diagram for the 
shaking couples. Scale for couples, 1 inch = 400 foot-pounds. 

112. Three-crank Engine. In this case 



3(n-l)=6. 

Hence the first six equations of Art. 108 can be satisfied. Choose 
the position of crank 1 as horizontal so that 



and take the reference plane through crank 1 so that 

ai = 0. 
The six equations then reduce to 

Mi+M 2x2 +Af 3x3 = 0, ....... (I) 

M 2 2/2+Af 32/3 = 0, ....... (II) 

M 2 a2Z2 + Af 33X3 = 0, ..... (Ill) 
^2022/2 +^3^32/3 = 0, ..... (IV) 

M 1 +M 2 (x 2 2-y 2 2)+M 3 (x 2 3-y 2 3 )=Q ) . (V) 
A/2X22/2+M 3 x 3 2/3 = ...... (VI) 

From Equations (II) and (IV) either 

Z 2 = x 3 or 2 = 3 = 0. 



THREE-CRANK ENGINE 163 

The second solution is untenable since Equation (V) cannot be 
satisfied in this case. 
Therefore 



2/2 = 2/3. 

If 2/2 = 2/3 Equation (II) cannot be satisfied. 
Hence 

2/2= 2/3- 
Now from Equation (II), 



The six equations are now reduced to four as follows: 

M 1 +2M 2 x 2 = 0, ....... (I)' 

M 2 z 2 (a 2 +a 3 )=0, ...... (Ill)' 

^22/2(02-03) =0, ...... (IV)' 

Mi+2M 2 (3?2-y 2 2)=0 ..... (V)' 

Since #2 and 2/2 cannot be zero it follows from (III)' and (IV)' that 

2 = CL3 = 0. 

In other words all the cranks must lie in the same radial plane. 
Subtracting (I)' from (V)' and substituting 2/2 2 = l X2 2 , 



Therefore 

z 2 = l or . 



If X2= 1, M 2 = JAfi, which is impossible*. 
Hence 

z 2 = z 3 =-i 
and 

2/2 =-2/3 = 5 V3. 
Therefore the cranks are at 120, and the masses are all equal. 



164 BALANCING OF ENGINES 

Since all the cranks are in the same plane this solution is of 
no value. We must therefore be content with satisfying only four 
of the conditions of balance instead of six. For this purpose 
either the primary forces and primary couples may be balanced, 
or all the forces may be balanced without regard to the couples. 
In the first case the first four equations must be satisfied. These 
are as before: 



..... (I)' 

..... (II)' 

Q, .... (Til)' 

M 2 ci2y 2+ M 30,31/3 = 0. .... (IV)' 

From (II)' and (IV)' either a2 = a 3 , or y 2 = 2/3 = 0. 

The first of these alternatives reduces to the same solution as 
found for the case where it was attempted to satisfy six equations. 

If 2/2 = 2/3 = 0. 2 = =bl and #3 = 1. 

Then 



These equations admit of an infinite number of solutions, all 
of which have the following characteristics : 

(a) The cranks are all in one axial plane, two being parallel 

and the third at 180 to these two. 
(6) One of the masses is equal to the sum of the other two, 

and is attached to the crank which lies opposite the 

other two. 
(c) Considering the masses concentrated at their respect- 

ive crank pins, the center of gravity of the two 

masses on the same side of the shaft lies directly 

opposite the third mass. 

Since cranks at 180 are undesirable from the standpoint of 
uniformity of turning effort this arrangement is seldom used. 



FOUR-CRANK ENGINE 165 

It is also possible to balance the primary and secondaiy forces 
disregarding the couples. For this purpose Equations (I), (II), 
(V), and (VI) must be satisfied. These equations become: 

(I)' 

(II)' 

2 -y 3 2 )=0, . . (V)' 
(VI)' 



From (II)' and (VI)' 

x 2 = x 3 , or 2/2 = 2/3 = 0. 

If 7/ 2 = 2/ 3 = o, Equation (V)' becomes MI +Af 2+^3 = 0, which 
is impossible. 
Therefore 

x 2 = x 3 and -2/2 = 2/3. 
From (II)' 



Substituting these values in (I)' and (V)' and combining, there 
results 

2x 2 2 -x 2 -l = Q. 
Hence 



2/3=-iV3, 



That is, the masses are all equal and the cranks are at 120. The 
distances along the crank were not taken into account, and the 
couples are unbalanced. This arrangement is much used in 
practice. 

113. Engines Having More than Three Cranks. For detailed 
discussion of the balancing of engines with more than three cranks 
the reader is referred to Dalby's " Balancing of Engines," where 
solutions for engines of four, five, and six cylinders are given in 
detail. Only those types which are commonly used in automotive 
engineering will be discussed here. 



166 



BALANCING OF ENGINES 



114. The Four-cylinder Automobile-type Engine. In this 
type the cranks are arranged in pairs, the outer cranks being 
parallel and opposite to the inner cranks. The arrangement is 

shown in Fig. 191. Each 
of the two pairs of cranks 
I is symmetrical wi h respect 

t to the central reference 




4 plane. Accordingly 



FIG. 191. 



= 24= 2= xs, 
= 2/4 =~2/2 =-2/3, 



0,2= as. 



Substituting these values in the eight equations of Art. 108, all 
the equations are satisfied except (V) and (VI), which become, 



That is to say that the primary forces and couples are balanced, 
but the secondary forces are not. The secondary couples are 
balanced with respect to the chosen reference plane, but not with 
respect to any other 
plane. 

115. The Six- 
cylinder Automobile- 
type Engine. In 
this engine the cranks 
are arranged in three 
pairs symmetrically 
disposed with respect 
to the central refer- 
ence plane. The 
cranks form angles 
of 120 with each 

other. The scheme is illustrated in Fig. 192. Since all the 
masses are equal, and since corresponding cranks are at equal 
distances on opposite sides of the reference plane all eight of 




^ 

\ 


J 3! 


T } 



FIG. 192. 



SIX, EIGHT, AND TWELVE-CYLINDER ENGINES 167 

the equations of Art. 108 are satisfied. This is readily verified 
by substituting in these equations the values of the re's, y's 
and a's. For convenience in making this substitution it is best 
to assume one of the pairs of cranks in either a horizontal or 
a vertical position. Then 



2/1=2/6 = . 2/2 = 2/5 

ai=dQ. Cl2=a5. 03= #4. 

With these values all the equations are satisfied, and the engine 
is therefore completely balanced. 

116. The Eight-cylinder Automobile-type Engine. This 
engine is usually built in the form of two units each composed 
of four cylinders arranged as described in Art. 114. The axes 
of the cylinders in one group are inclined at some angle to those 
of the other group. In each of these units the primary forces 
and couples are balanced, but the secondary forces are not. Since 
the two sets of unbalanced secondary forces are inclined to each 
other at an angle </>, they cannot neutralize each other, and the 
engine is therefore not balanced as far as the secondary forces 
are concerned. 1 

117. The Twelve-cylinder Automobile-type Engine. This 
engine is usually built up of two groups of six cylinders, each 
group being arranged as described in Art. 115. Since each of 
the groups is completely balanced in itself, the whole engine is 
in perfect balance with regard to both primary and secondary 
forces and couples. 

118. The Radial Engine. The radial or star engine is a 
multiple-cylinder engine having a single crank. The cylinders 
are all in the same radial plane, and all the connecting rods are 
attached by special devices to a common crank pin. The size 
of all the cylinders and the weights of all the reciprocating parts 
are equal. The general arrangement is shown in Fig. 193. 

1 It can be readily shown that if the angle </> is a right angle, the resultant 
shaking forces are in a plane at right angles to the plane bisecting the angle 
</>. The value of the unbalanced force is the sum of the components of the 
two individual sets of secondary forces in a direction normal to the bisecting 
plane. 



168 



BALANCING OF ENGINES 



If the engine has n cylinders, the angle between the lines of 
stroke of two adjacent cylinders is 2ir/n = 0. Then when the com- 
mon crank makes an angle a with the line of stroke of cylinder n, 
it makes an angle 

a with the line of stroke of cylinder number 1. 
a 20 with the line of stroke of cylinder number 2. 
a 30 with the line of stroke of cylinder number 3. 

a &0 with the line of stroke of cylinder number k. 




FIG. 193. 

The inertia forces of the different sets of reciprocating parts 
then become 

Mr co 2 (cos a+r/L cos 2a) for cylinder n, 

Mrco 2 [cos (a 0)+r/L cos 2(a-0)] for cylinder 1, 

Mrco 2 [cos (a-20)+r/L cos 2(a-20)] for cylinder 2, 

Mru?[cos (a k<j>)-\-r/L cos 2(a k<f>)] for cylinder k. 



RADIAL ENGINES 169 

Taking cylinder n as horizontal, the horizontal and vertical com- 
ponents of the inertia forces are found to be as follows: 
Horizontal forces: 

Mru?(cos a+r/L cos 2a) ......................... for cylinder n, 

Mrco 2 [cos (a 0)+r/L cos 2( 0)] cos ........... for cylinder 1, 

Mrco 2 [cos (-20)+r/L cos 2(a-20)] cos 20 ........ for cylinder 2, 

A/>co 2 [cos (a fc0)+r/L cos 2(a fc0)] cos &0] ....... for cylinder k. 

Vertical forces: 
Afrco^O) ....................................... for cylinder n, 

Mrco 2 [cos (a 0)+r/L cos 2(a 0)] sin ........... for cylinder 1, 

Mrco 2 [(cos (a-20)+r/L cos 2(a-20)] sin 20 ....... for cylinder 2, 

Afrco 2 [cos (a fc0)+r/L cos 2 (a 0)] sin ........ for cylinder fc. 

The first term in each of these expressions represents the 
primary force, and the second term gives the secondary force. 

Let X represent the sum of the horizontal components and 
Y the sum of the vertical components of the primary forces. 

Then 

X = Afrco2[cosa+cos (a 0) cos <+cos (a 20) cos 20 . . . ] 
= Mro) 2 cosa[l+cos 2 0+cos 2 20 . . . cos 2 /b0 . . .] 
H-Mrco 2 sin a[0+sin cos 0+ sin 2 cos 20 . . . sin fc</> cos k$ . . .]. 

It is readily proved that the second series reduces to zero, 
and the first to n/2. 1 . 
Therefore 

X = JnMrco 2 cos a. 

F=Mrco 2 [cosa sin 0+cos (a 0) sin 0+cos (a 20) sin 20 . . . . ] 
= Mrco 2 cos a[sin cos 0+sin cos 0+sin 20 cos 20 . . . ] 
+Mrco 2 sin a[sin 2 0+sin 2 0+sin 2 20 ... ] = 0+|nAfrco 2 sin a. 



Combining X and F, the resultant is found to be JnMrco 2 , and the 

direction of this resultant is along the crank radius. The shak- 

1 See any standard work on trigonometry. 



170 BALANCING OF ENGINES 

ing force due to the reciprocating masses can therefore be com- 
pletely neutralized by a counterweight of mass \nM placed 
directly opposite the crank and at a distance from the center equal 
to the crank radius. 

Let X' and Y r represent the sums of the horizontal and vertical 
components of the secondary forces. Then 



X'= j [cos 2a+cos 2(a-0) cos 0+cos 2(a-20) cos 20 . . . ] 

= ( j ) cos 2a [cos 20 cos 0+cos 40 cos 20 ... cos 2fc0 cos &0] 
\ L I 

/Mr 2 o> 2 \ . 

+ ( 7 Ism 2a[sm 20 sin 0+sm 40 sm 20 ... sin 2&0 sin fc0]. 
\ L I 

Both these series vanish, and X' is therefore zero. In a pre- 
cisely similar way it may be shown that the vertical force Y' 
vanishes. 

This type of engine is therefore in complete secondary balance, 
and can be put in primary balance by means of a single counter- 
weight. As all the cylinders are in the same plane, there are of 
course no couples to be taken care of. 

It may be noted that this solution holds only when there are 
more than three cylinders. With two cylinders the primary 
forces are unbalanced. With three cylinders the secondary forces 
do not vanish. 

119. The Opposed Engine. In this type of engine two cranks 
are placed at an angle of 180, and driven by two similar cylinders 
which are on opposite sides of the shaft, as shown in Fig. 194. The 




FIG. 194. 

center lines of the opposed cylinders are in the same straight 
line. 

From the symmetry of the arrangement it is evident that the 
motions of the two pistons are exactly equal and opposite. There- 



OPPOSED ENGINE. ROTARY ENGINE 171 

fore the accelerations are equal, and the inertia forces exactly 
balance each other. Since the cylinders are in the same straight 
line there are no couples to be taken into account. The engine 
is therefore completely balanced. 

The engine may have two or more pairs of opposed cylinders, 
each of which is completely balanced, and therefore there can be 
no shaking forces. It can be shown that the inertia forces of 
the connecting rods are also equal and opposite. Since these 
forces are not exactly in the same line they have a slight effect 
on the turning effort. 

120. The Rotary Engine. This engine is an inversion of the 
ordinary slider-crank mechanism, the crank becoming the station- 
ary member, and the cylinder and connecting rod making com- 
plete revolutions about fixed centers. Usually rotary engines 
are made with seven or nine cylinders, all of which revolve in the 
same plane. By special devices all the connecting rods are 
attached to a common pin. The general arrangement is shown 
in Fig. 195. 

If all the cylinders are alike, and if all the angles between 
adjacent cylinders are equal, evidently the revolving cylinders 
form a balanced system. Since all the cylinders are in the same 
plane there are no inertia couples. Therefore only the inertia 
forces of the pistons and connecting rods require investigation. 

Let the number of cylinders be n. 

Let <t> = 2ir/n be the angle between the center lines of adjacent 
cylinders, and let a be the angle between the center line of the 
nth cylinder and the center line of the stationary crank. 

Then the angles between the axes of cylinders 1, 2, ... k and 
the center line of the crank are +<, a+2</>, . . . a-\-k<f>, as shown 
in Fig. 195. 

Let |8 be the angle between the connecting rod and the line 

of stroke for the nth cylinder; 
r, the length of the stationary crank; 
Lj the length of the connecting rods; 
s, the distance from the center of rotation to the center of 

the wrist pin of the nth cylinder; 
u, the velocity of sliding of the piston in the cylinder; 
and 

oj, the angular velocity of rotation of the cylinders. 



172 
Then 



BALANCING OF ENGINES 

s = r cosa+L cos (3 = r cos a+LVl r' 2 /L 2 sin 2 a 

= r cosa+L r 2 /2L sin 2 a approximately. 
u ds/dt = ru sin a+r 2 /2Lco sin 2a. 




FIG. 195. 

The acceleration of the wrist pin has three components, namely: 

(1) Acceleration of the point on the cylinder coinciding with 
the center of the wrist pin. This is directed along the line of 
stroke and its value is sco 2 . 

\2) Acceleration due to sliding of piston in cylinder = -j- 
directed along the line of stroke. 

(3) 2uu at right angles to the line of stroke. 



RADIAL ENGINE 173 

Substituting the values of s and u these accelerations become: 



(1) 

(2) dujdt (rco 2 cos a+r 2 /Lco 2 cos 2a). 

(3) 




These accelerations are shown in Fig. 
196. The inertia forces are found by 
multiplying the accelerations by M , the 
mass of the piston. Evidently the com- 
mon factors M and co 2 may be omitted 
in discussing the balance of the engine. F IG< 195. 

Omitting the common factors and resolving the forces into 
horizontal and vertical components there results: 

Horizontal component. 

, du 1 2woo sin a 

X n = scosa+- -T.2 cos a --- -5 
dt co 2 or 

r 2 

= r COS 2 a.-\-L cos 01 7:7 sin 2 a cos a+r cos 2 a 
iLt 

r 2 r 2 

-\~r cos 2a cos a 2r sin 2 a -7- sin 2a sin a 
Li LI 

r 2 

= 2r(cos 2 a sin 2 a) +L cos a. -~j sin 2 a cos a 

r 2 2r 2 

+y (cos 2 a sin 2 a) cos a -- j- sin 2 a cos a 

Li ' Li 

r 2 7 r 2 

= 2r cos 2a+L cos a+- 7- cos 3 a TJT sm2 a cos a - 
L^ ^L/ 

Vertical component. 
1 du . 



j 5 ^^ cos a 

or at co 

r 2 
r cos a sin a+L sin a jr=- sin 3 a+r sin a cos a 



^r cos 2a sina+2r sin a cos a+y sin 2a sin a 
L Li 



174 BALANCING OF ENGINES 



r 2 
4r sin a cos a+L sin a ^r sin 3 a 



+T cos2 sin a 7- sin 3 a +-7- 2 sin 2 a cos a 
Li LJ Li 

3r 2 . r 2 2r 2 

= 2r sin 2a+L sin a 7^7- sin 3 a+y cos 2 a sin cH f sin 2 a cos a. 
ZLi Li LJ 

For the other cylinders substitute for a the values 



Summing up the horizontal forces for all the cylinders, 

= 2rS cos 2(a+/b0)+L2 cos(a 



-^ 2 sin 2 (a+/c</>) cos (<*+/b0)+ T 2 cos 3 (a 
ZL/ Li 

Each of these summations is separately zero if n is greater 
than 3. 1 Therefore all the horizontal forces vanish. 
Similarly for the vertical forces 

o2 

SF = 2r S sin 2a+fc+L S sin 



r 2 2r 2 

T S sin (a 



Again each of these summations is equal to zero, and the vertical 
forces therefore also vanish. 

It can also be shown that the inertia forces of the connecting 
rods form a balanced system. The engine is therefore completely 
balanced. 

NOTE. An investigation by the authors shows that the inertia forces 
exert no influence on the turning effort. The sum of the moments of these 
forces about the center vanishes for all positions of the engine. 

121. The Offset Engine. In this type of engine the line of 
stroke of the wrist pin does not pass through the center of the 
main bearing. These engines are used in many makes of auto- 
mobiles, and are built with four, six, eight, or twelve cylinders. 

1 For the case where n = 2, the term S cos 2(a+fc0) does not vanish, and 
for the case where n = 3, the terms S sin 2 (<*+&<) cos (a-\-k<t>) and 2 cos 3 
(a+/c0) do not vanish. 



OFFSET ENGINE 



175 



The arrangements of cylinders are the same as those described 
in Arts. 114, 115, 116, and 117. 

This engine gives a quick-return motion, the working stroke 
taking place while the crank revolves through the angle ACB, 




FIG. 197. 

Fig. 197, and the compression and exhaust strokes lasting through 
the arc BDA. In Fig. 198 are shown the variation of velocity 
and acceleration for an engine 
having a stroke of 6 inches, a con- 
necting rod of 15 inches and an 
offset of 1 inch. 

Let L- length of connecting 

rod; 

r = crank radius; 
b = offset; 
S = horizontal distance 

from crank center to 

wrist pin. 

Then from Fig. 197, 

S = r cosa+Lcos0, 
L sin /3 = r sin a 6, 





FIG. 198. 



COS 



176 BALANCING OF ENGINES 

Expanding by the binomial theorem, 

(rsina b) 2 (rsina 6) 4 



cos/3 = l- 



2L 2 8L 4 

(r sin a b) 2 (r sin a 6) 4 



Differentiating twice 



Q 7 

r cos a j- cos 2a -= sin a 



[ 


3 (r sin a b} 2 r cos 2 a (r sin a b) 3 r sin a 

2L 3 '.' ' '}' 

The first term of this series may be called the primary acceleration, 
the second and third terms the secondary acceleration, and the 
remaining terms may be called the tertiary and higher acceler- 
ations. It is evident that the primary and secondary inertia forces 
and couples can be balanced under exactly the same conditions 
as those of ordinary engines. In other words, if the eight equa- 
tions of Art. 108 are satisfied the engine is in complete primary 
and secondary balance. 

122. The tertiary acceleration was investigated for an engine 
of the dimensions given above. The maximum value was found 
to occur when a was nearly 270 and to be about one-tenth as 
great as the secondary acceleration. For most purposes, this effect 
is too small to be considered, and the engine may therefore be 
regarded as balanced if the eight equations of Art. 108 are sat- 
isfied. 



CHAPTER VII 



GOVERNORS 

123. Purpose. The purpose of the governor is to control 
the supply of steam furnished to an engine, so that the machine 
may run at approximately constant speed under all loads. In 
the ideal case the supply of steam is so regulated that the speed 
is the same at all loads. Practically such operation is impossible, 
and all governors are so designed that under heavy loads the speed 
is less than under light loads. 1 

124. Classification. For the purposes of analysis it is con- 
venient to divide governors into two classes: 

(1) Flyball governors. 

(2) Shaft governors. 

A governor of the first class has two or more weights W, Fig. 
199, carried by arms A, which in turn are pivoted on a revolving 
shaft S. The shaft is. driven at a 
speed proportional to the speed of 
the crank. As the weights revolve 
with the shaft they tend to move 
outward under the action of centrif- 
ugal force. If this force is great 
enough the weights W will assume a 
new position, and by means of the 
links B will move the slide w along 
the shaft S. The supply of steam 
is controlled by the position of w, 
decreasing as w rises. 

1 Stodola (The Siemens Governing Principle, Zeitschrift des Vereines 
deutscher Ingenieure, 1899) has shown that under certain conditions a gover- 
nor may be so designed that the engine will run at higher speed under heavy 
loads than under light loads. In practice no such governors are built. 

177 




178 



GOVERNORS 



A governor of the second class has a weight w, Fig. 200, placed 
inside the flywheel and revolving with it. The mass W is usually 
pivoted to one of the arms so that it can assume various positions 
with respect to the flywheel. As the c ank revolves the mass w 
tends to move outward under the action of centrifugal force. 
This motion is opposed by the tension of the spring S. If the 
speed becomes sufficiently high w will move to a new^ position. 
By suitable linkage (not shown) the eccentric is shifted* relative 
to the shaft, thus regulating the supply of steam. In addition 

to the centrifugal 
force there is an- 
other force tending 
to move the weight 
W. Suppose the 
flywheel and gov- 
ernor to be rotating 
together about the 
shaft center P with 
angular velocity oo. 
Now if the load is 
lightened the engine 
tends to speed up 
and the flywheel 
receives an angular 
acceleration^. Due 
to its inertia the 
governor weight re- 
sists this accelera- 
tion and thus tends to rotate about the pivot in a sense 
opposite to that of a. This action is known as the inertia 
effect. Governors in which the inertia is relied upon to furnish 
the greater part of the regulating force are called inertia 
governors. Those where the centrifugal force predominates 
are called centrifugal governors. As wil be shown later, it is 
possible to construct governors operating purely by centrifugal 
force, but it is not possible to build a governor which works by 
inertia only. 

125. Force Reduction. Suppose any body M, Fig. 201, to 
have constrained motion and to be acted upon by a force F applied 




FIG. 200. 



FORCE AND MASS REDUCTION 



179 



at the point A. Let the velocity of the point A be V a , and let 
6 denote the angle between the force vector F and the velocity 
vector V a . Then in the interval of time dt the force performs 
work FV a cos ddt. Let a second point B of the body have a 
velocity V b . Then if a force F' is applied at B, in the interval 
of time dt it will do work F'Vb cos <f>dt, / 

where < is the angle between the vec- 
tors Vb and F'. 
If 



FV a cos edt = 



cos 



(1) 



then the force /'"would produce exactly 
the same effect on the motion of the 
body as the force F. Since the velocities 
V a and Vb are proportional to the dis- 
tances from the instantaneous center of 
the link to the points A and B, Equa- 
tion (1) may be written 




FIG. 201. 



(2) 



where R a and Rb are the instantaneous radii of the points A and 
B. The force F acting at A may therefore be replaced by the 
force F' acting at B without in any way altering the motion of 
the body. 

In particular if the point B is the instantaneous center of 
relative motion between the link M and some other link N t then 
the force F acting at A on link TV may be replaced by the force 
F' acting on link N at B, the instantaneous center of relative 
motion between the links M and N. 

In this way all the forces acting on the various links of a mech- 
anism may be replaced by forces acting on a single chosen link. 
These new forces may finally, if desired, be replaced by still other 
forces acting at any chosen point of the given link, and all the 
forces acting on the mechanism can thus be combined into a single 
resultant acting at any desired point. The process of replacing 
a given force by another acting at a different point is called reducing 
the given force to the given point. The sole condition to be 
observed in making such a reduction is that for any given move- 
ment of the mechanism the new force must do the same amount of 



180 GOVERNORS 

work as the one replaced. It should be clearly understood that 
the equivalence of the two forces extends only to their effect on 
the motion of the mechanism, and that this method is of no 
value in computing stresses in machine parts or shaking forces 
in the mechanism. 

126. Mass Reduction. Referring again to Fig. 201, suppose 
a mass m to be concentrated at A. This mass will have kinetic 
energy \mv. Suppose a second mass m' concentrated at B. 
Its kinetic energy will be JmW. Then the mass m concentrated 
at A may be replaced by the mass m' concentrated at B provided 
that: 



Or since V a /Vb=R a /Rb Equation (3) may be written 

m'Ri?. . ...... (4) 



Thus the entire mass of the link M may be reduced to a single 
mass concentrated at one point B. The magnitude of this mass 
is found as follows: 

Let Am be the mass at any point, and let R be the distance 
of this point from the instantaneous center of rotation of the link 
M. Then, if mo is the reduced mass of the entire link, 



(5) 

The right-hand member of Equation (5) is the moment of inertia 
of the link M about its instantaneous center. Therefore Equation 
(5) may be written: 

m R b 2 = m(R Q 2 +k 2 ), ...... (6) 

where m is the mass of link M , RQ is the distance from the center 
of gravity of the link to the instantaneous center, and k is the 
principal radius of gyration of the link. 

Thus the entire link M may be replaced by a single mass 
mo concentrated at any given point B. In particular, if the point 
B be chosen as the instantaneous center of relative motion between 
the links M and N, the link M may be replaced by the mass mo 
concentrated at the point B of the link N. In this way all the 
masses of all the links of a mechanism may be reduced to a system 
of masses attached to a single link, and finally, all these new 
masses may be reduced to a single mass concentrated at a single 



MOMENT OF CENTRIFUGAL FORCE 



181 



point. If all the forces acting on the mechanism have previously 
been reduced to the same point, the study of the motions in the 
mechanism is reduced to the problem of the effect of a single force 
acting on a single concentrated mass. 

For example, in the steam engine, all the forces and masses 
might be reduced to the crank pin, and the accelerations through- 
out the mechanism could be determined from the consideration 
of the action of the single resultant force on the single reduced 
mass. The principles of force and mass reduction will be much 
used in the study of the behavior of 
governors. 1 

127. Moment of the Centrifugal 
Force. Let the mass M, Fig. 202, 
revolve about the axis OY with angu- 
lar velocity o>, and also be free to 
turn about some point P. It is 
required to find the moment about 
the point P due to the centrifugal 
force of the mass M . Consider any 
elementary mass dM whose distance 
from OY is x t . and whose distance 
from OPX is y. Then the centrifugal force of this elementary 



<fa* 



T 

> 



Y 



FIG. 202. 



mass is 



(1) 



and the moment of the centrigugal force about the point P is : 



(2) 



Let r;o, 2/o be the coordinates of the center of gravity G, and let 



Then 



and 



(3) 



T = co 2 Cxy dM = rf(xoyo M+x fh dM+y Cj dM+fjh dM). (4) 



1 For formal proof of the principles of force and mass reduction see any 
standard work on mechanics. 



182 GOVERNORS 

But since G is the center of gravity, 



Therefore 



(5) 

(6) 



The term xoyoMu 2 is the moment of the centrifugal force, 
assuming the entire mass concentrated at the center of gravity. 

In most flyball governors the second term co 2 I jh dM is relatively 

small. In some cases, how- 
ever, it must be taken into 
account. 

128. The Elementary Fry- 
ball Governor. In Fig. 203 is 
shown the simplest form of 
flyball governor. The weight 
W is supposed to be concen- 
trated at the point P. The 
arms A and B and the slide 
w are supposed to be weight- 
less. Then if the shaft 
S revolves with an angu- 
lar velocity w, the cen- 




FIG. 203. 

trif ugal force of the weight W is : 

W 



(1) 



and the moment of this force about is: 

T c = 

9 



The moment of the weight W about is: 

T w =Wr. 



(2) 



(3) 



1 The integral J jh dM is called the product of inertia. For most sym- 
metrical figures the product of inertia is zero. 



ELEMENTARY FLYBALL GOVERNOR 183 

If T w = T c the governor is in equilibrium and therefore, 

h = g/u> 2 . . (4) 

The same result can be obtained by the principle of reduced 
forces as follows: the point Q is the instantaneous center between 
the arm A and the slide w. Therefore 

V P : V, = OP :OQ (5) 

Hence the force C can be replaced by a vertical force Co acting 
upward at Q, and of such magnitude that 

W 

g 

The force W can be replaced by a force WQ acting downward at 
Q where 

(7) 



If the governor is in equilibrium C = WQ and therefore 

........ (8) 



9 

or 

h = g/u? (9) 

The mass M = W/g can be replaced by a mass M o located at 
Q provided that the relation 



is satisfied. Therefore 

MOP 2 



Since the velocity of the point Q is the same as that of the slide 
w the forces and mass might equally well have been reduced to 
any point of the slide w. If the governor is not in equilibrium 
the slide w will be given an upward acceleration according to 
the equation 

, Co-Wo 



184 



GOVERNORS 



It will be noted that in this type of governor the height h 
depends solely on the speed co. The curve Fig. 204 shows the rela- 
tion between h and co. In this curve h is given in inches and co 
in r.p.m. From the figure it is evident that at high speeds a large 
change in the speed produces only a very small movement of the 
governor. 

Example. An elementary flyball governor runs normally at 
300 r.p.m. The speed of the engine is now changed to 400 r.p.m. 

Find the movement of the 



governor. Solution : 
Equation (4) 

32.2-12 



from 



0.38 
in. 




360 400 



FIG. 204. 



V 60 / 

2 32.2-12 ^0.21 
2 g/< * 2 ~/40Q-27r\ 2 ~ in. 

In other words a change of 
speed of 100 r.p.m. causes 
the governor to rise only about inch. 

On the other hand, if Equation (4) gives a value of h greater 
than the length OP the governor will not move at all. In this 
case the weight W hangs vertically against the shaft S. If W be 
moved out a distance dr from the shaft a centrifugal force is set 
up such that: 

dC- 

~ g 

The moment of this force about is: 

W 

dT c =<JOPdr. 

g 

The weight W has a moment about given by: 

dT w =Wdr. 
Since 

0/co 2 >OP, dT c <dT w . 



WEIGHTED GOVERNOR 



185 




Therefore the weight W will simply fall back against the shaft. 
For example if OP = 12" the minimum speed at which the governor 
will act is given by the equation, 



5.68 radians per sec. = 54.3 r.p.m. 



129. Weighted Flyball Governor. Let the slide w, Fig. 205, 
have a weight 2b W. As in the preceding article each of the masses 
w can be replaced by a mass 

M = W/g I Q 2 j moving 

with the slide w. In this equa- 
tion k denotes the principal 
radius of gyration of the arm 
and weight W . The forces C 
and W can be replaced by 
forces C and WQ acting on 
w, these forces being given by 
the equations : 



and 




FIG. 205. 



For equilibrium 
or 

Therefore 



In some governors the dimensions are so chosen that OQ 
Then Equation (3) reduces to the simple form : 



CD 

(2) 

(3) 
= r. 

(4) 



It is evident that increasing the weight of the slide increases the 
speed at which the governor will operate in a given position. 



186 



GOVERNORS 



Many engines are equipped with a device for varying the weight 
of the slide and thus controlling the speed of the engine. 

A weighted flyball governor will not operate if the speed falls 
below the value which makes h equal to the length OP. For 
example if OP =12" and 6 = 2 Equation (4) gives: 

12=^(1+2). 

Therefore co = 9.85 radians per second = 94.1 r.p.m. is the mini- 
mum speed at which the governor will act. 

130. Weighted Flyball Governor, Second Type. In order 
to make the flyball governor act at low speeds the construc- 




FIG. 206. 

tion is slightly changed as indicated in Fig. 206, the points and 
the joints F being set at a distance c from the center line of the 
shaft. 

As before, the centrifugal force C and the weight W can be 
reduced to forces Co and Wo acting on the slide according to the 
relations: 

* w - \, CD 



and 



r c 

-QQ- 



(2) 



HORIZONTAL GOVERNOR 187 

Then for equilibrium 



....... (3) 

or 



Therefore 

h = g/o> 2 (l-c/r+b'OQ/r) ...... (5) 

or 

...... (6) 



When the weights hang vertically downward r = c, and OQ = 0. 
Therefore co = and the governor operates at all speeds. 

131. Horizontal Spring-controlled Governor. In some gov- 
ernors, particularly for use on small gas engines, the governor 
spindle is horizontal, and the centrifugal force is opposed by springs 
instead of by weights. There are two possible arrangements: 

(1) The spring may act on the revolving weights, thus 

directly opposing the centrifugal force. 

(2) The spring may act on the slide. 

In the first case the spring tension increases uniformly with 
the distance of the weights from the shaft. That is* 

S = A+Br, ........ (1) 

where S is the spring tension, r the radius, and A and B are con- 
stants. The centrifugal force varies directly as the radius. Thus : 



(2) 
y 

If the governor is in equilibrium in a given position TQ at a given 
speed WQ, then 

W 
A+r = r o 2 ........ (3) 

g 

Suppose the weight is now moved an additional distance Ar from 
the shaft, the speed remaining constant. The centrifugal force 
now becomes: 

, ..... (4) 



188 GOVERNORS 

whence 

W 
AC = co 2 Ar ........ (5) 

[/ 

The spring tension is also increased by an amount: 

AS = Ar ......... (6) 

If AS>AC the weight will return to its original position and the 
governor is in stable equilibrium. Comparing (5) and (6) 

W 
>-"o 2 ........ (7) 

W 

But A+Br = r co 2 , and therefore 

A<0 ......... (8) 

If AS<AC the weight does not return to its original position, 
but moves on out as far as the construction of the governor allows. 
Such a governor is said to be unstable, and is quite useless for 
practical purposes. 

If A*S=AC, A =0. In this case the weight is in neutral equi- 
librium in all positions as long as the speed remains constant at 
COQ. If he speed increases above coo the weights move out to 
their extreme positions, and if the speed falls below coo the weights 
move inward until stopped by the spindle. Such a governor is 
said to be isochronous. Although this represents the ideal gover- 
nor, which maintains absolutely constant speed under all condi- 
tions, it is too sensitive for practical use. 

In all practical spring-controlled governors of this type A < 0. 
The proper values for the constants A and B are found as follows: 
let r\ and r<2, be the maximum and minimum radii permitted by 
the construction of the governor, and let coi and co be the maxi- 
mum and minimum allowable speeds. Then 



......... (9) 



and 



SPRING-CONTROLLED GOVERNOR 



189 



Solving Equations (9) and (10) 

W nr 2 



A = 



g ri 



(o^-co! 2 ) (11) 



D = 



r 2 co 2 2 



. . . . (12) 



g nr 2 

The relations between A, B, co, S, and C are shown in Fig. 
207. Here the spring tension S and the centrifugal force C are 




FIG. 207. 

plotted as ordinates against the radius r as abscissa. The ordi- 
nates of the lines radiating from the origin 0, represent the centrif- 
ugal forces at different speeds. The dotted parallel lines show 
the spring tension for a spring of known stiffness B, but with 
varying initial tension A. The intercepts of the dotted lines on 
the vertical axis give the initial tensions A. Choosing any one 
of the dotted lines, the point where it crosses any of the lines of 
constant speed gives the position in which the governor will be 
in equilibrium at that speed. Evidently from the figure, if A 
is positive the speed must be reduced as the governor weight 
moves out from the center, or in other words the equilibrium is 
unstable. If A is negative, the higher speed corresponds to the 
larger radius, and the governor is stable. If A is zero, the line 



190 



GOVERNORS 





representing the spring tension passes through the origin, and 
therefore coincides with one of the lines of constant speed; in 
other words, the equilibrium is neutral and the governor is iso- 
chronous. 

Fig. 207 gives values only for a spring having a given stiff- 
ness B. If a stiffer spring is used the dotted lines become steeper. 

Evidently for a spring of a 
given stiffness and initial ten- 
sion there is a definite speed 
above which the governor 
will not function. This maxi- 
mum speed corresponds to 

i_:ijuuuinnjiiiiiAAn ^ Q ^ as ^ ^ ^ e constant- 

5 __1_ WjrVmWW ~ speed lines which is inter- 

^""^ T^& ^n sected by the spring tension 

J/~-%i/'*' *~~~ "H ^ ne within the limits of move- 
/ . \ ment of the governor. 

132. Horizontal Spring-con- 
trolled Governor, Second Type. 
In Fig. 208 is shown a hori- 
zontal flyball governor con- 
trolled by a spring which acts 
on the slide w in the direction of 
the center line of the spindle. 

= 2S=(A-Bx) (1) 

~ h 




2>.^.. 

tmtm 



FIG. 208. 



The spring tension 

The reduced centrifugal force 

The reduced mass 

For equilibrium 2S 



OQ 2 



or 



Therefore 



h = 



Q 



(A-Bx) 



OQ 



(2) 



(3) 



r 

In most governors of this type a = b ^L. Then h = x and 
OQ = rc. Let ho be the yalue of x for which the spring tension 
is zero. Then 

S = B(h -h) (4) 



SPRING-CONTROLLED GOVERNOR 191 



The reduced centrifugal force becomes: 



. , y h W 9 h 

o = C^.= r co 2 - ....... (5) 

OQ g rc 



For equilibrium Co = S, or 
W 



(6) 



To test the stability of the equilibrium suppose the weights 
to be moved out a distance dr, the speed co remaining constant. 
The slide will be moved a distance dh, and the spring tension will 
be increased by an amount Bdh. The height h is reduced by an 
amount dh. The reduced centrifugal force now becomes: 

W 2 ( r +dr)(h-dh) m 

Co+dC Q = -* (r+dr _ c) ..... (7) 

For stability S+dS> C +dCo or, 

B( ho 

Clearing of fractions : 

B(ho-h+dh)(r-c+dr)>(,?(r+dr)(h-dh) . (9) 

g 

or 

B(ho-h)(r-c)+B(r-c)dh 

W 
+B(ho-h)dr>(**(rh+hdr-rdh). . (10) 

Combining (6) and (10) and reducing: 

(r-c)dh+(h -h)dr>(ho-h)(r-c)(dr/r-dh/h). . (11) 
Collecting terms: 

. (12) 



This inequality is evidently satisfied in all cases where ho h>Q. 
In other words the governor is in stable equilibrium provided 
that the spring tension opposes the outwaid movement of the balls. 
133. Vertical Spring-controlled Governor. In some vertical 
governors a spring is arranged to oppose the centrifugal force of 



192 



GOVERNORS 



the balls directly, as shown in Fig. 209. There are four forces 
acting on the mechanism as follows: 

W 

(1) Centrifugal force of weights W = C = ro> 2 , 

(2) Weight of W = W, 

(3) Tension of spring = S= (A +Br), 

(4) Weight of slide = 2bw. 



4 




FIG. 209. 

All these forces can be reduced to the slide. Considering only 
one weight and one-half the weight of the slide, the reduced forces 
are: 

h W h 



_W(r-c) 



OQ 

} OQ = 
) = bW. 






For equilibrium, 
or 



(i) 

. . (2) 



STABILITY 193 

All the variables in Equation (2) can be expressed as func- 
tions of the angle a. In the simplest case where x = y = %L Equa- 
tion (2) reduces to: 

Leo 2 . /i IA A cos a BL sin a cos a Be cos a /ox 

sinacosa=(l+6)sina+ ^ +~ -pjT~ ~ + ^T~- ( 3 ) 

Lc^ = 1+b A ,BL Be 
g cos a W sin a W W sin a 

To investigate the stability of this governor assume that there 
are two angles, a\ and 2, at which the governor is in equilibrium 
at the same speed w, Then 

1+6 . __ A __ Be = 1+6 A Be 

cos ai TF sin ai W sin i cos az W sin 2 W sin #2' 

or 

(l+6)(secai seca2) = ( ^ j(coseca2 cosecai). . (6) 

sec2>0, and cosec2 coseci>0. 



Therefore to satisfy Equation (6) A+Bc>0. But A+Bc is the 
spring tension when the weights hang vertically. Consequently 
the governor will be unstable over some part of its range if the 
spring is in tension when the weights hang vertically. Example: 

Let ai = 30, a 2 = 10, and 6=1. 

From Equation (4) 

/r.iA ,A+Bc .BL 

= (1+6) secaH -- == coseca+-. 



yy 

The left-hand member of this equation represents the reduced 
force tending to raise the slide, and the right-hand member the 
forces opposing this motion. If the left-hand member is larger 
than the right the governor will rise, and vice versa. 
From Equation (6) 

A+Bc 2(sec 30 -sec 10) 



W coseclO-cosec30 



194 GOVERNORS 

Substituting this value in Equation (4), 

sec 10+0.0742 cosec 10 

= (1+6) sec 30+0.0742 cosec 30 = 2.458. 

The governor is therefore in equilibrium at the same speed o> in 
both the 10 and the 30 position. For any intermediate position, 
say a = 20 the right-hand member becomes : 

(1+6) sec 20+0.0742 cosec 20 = 2.345. 

The left-hand member is now larger than the right-hand, which 
means that the governor will not remain in position at 20, but 
will move on out until it comes to rest in its new equilibrium 
position at 30. 

Spring-controlled vertical governors must therefore be designed 
with great care so that the region of instability is outside the range 
of possible positions of the governor. From the inequality (7) 
it follows that the governor will be stable if the spring is in com- 
pression when the weights W hang vertically. It is not usually 
feasible to use a spring in compression acting directly on the gover- 
nor weights, but it is possible to design the governor so that in 
its lowest position the spring will be in tension, although the spring 
would be compressed if the weights could move downward so 
far as to hang vertical. 

134. Oscillations of Flyball Governors. In the preceding 
articles the governors have been considered unstable if there exists 
more than one equilibrium position for any speed, or if the equi- 
librium is unstable. In this article a second form of instability 
is considered. If, on account of a change of speed or for some 
other reason, the governor is not in its equilibrium position, it 
will move toward that position. On account of the kinetic energy 
gained in this motion the governor may go past its equilibrium 
position and will later return toward it. In this way oscillations 
may be set up which may prevent the governor from properly 
controlling the speed of the engine. As the governor oscillates 
on either side of the proper position, the engine is given alter- 
nately too much steam and too little, thus causing fluctuations 
of speed. In some cases such actions may cause serious trouble 
in the operation of the engine. 



f OSCILLATIONS 195 

As an example consider the weighted governor of Art. 130. 
The reduced force tending to raise the slide, according to Equa- 
tion (1), Art. 130, is 



The reduced forces resisting this motion are 

2W(r-c) 



and 



OQ 



(2) 



2w = 2bW 
The reduced mass of the mechanism is 



If CQ>W+WQ the slide will be given an upward acceleration. 
Let y be the distance of the slide from its equilibrium position. 
Then 






The oscillations may be studied by means of Equation (4). In 
the simplest case, where OG = FG = \L the geometry of the figure 
gives the following relations: 



h = L cosa=x, 
rc = OQ = L sin a, 



Then 



g rc 
W = W, 
W 



Then Equation (4) becomes: 

ru 2 (1+% 

d? r-c 



196 GOVERNORS 

Let ho be the value of h for the position of equilibrium. Then 

y=x-h = h-h , 

and 

cPy^cPh 

dt 2 dt 2 ' 

At the position of equilibrium there is no acceleration, or in other 
words 



c 



dt 2 



Adding Equations (5) and (6), and substituting L 2 h 2 = (r c) 2 , 
-Q+fyg 



For small oscillations we may write 
L 2 -h 2 = L 2 -h 
Equation (7) then reduces to 
d 2 h ho 



VL 2 -ho 2 

Since 

,, , , , d 2 h d 2 y 

(h-ho)=y, and ^2 = ^2" 

Equation (8) may be written 



The general solution of Equation (9) is 

pt, ..... (10) 



where A and B are arbitrary constants. Equation (10) signifies 
that small oscillations of the governor are harmonic, provided (a) 
that there is no friction, and (6) that the change of speed during 
the period of oscillation is negligible. Such a governor, there- 



OSCILLATIONS 197 

fore, if once started oscillating would continue to do so indefinitely. 
In practical governors there is always friction in the joints which 
tends to damp out the oscillations, and in many cases an oil dash- 
pot is used to help bring the governor to rest. The resistance 
of such a dashpot is usually assumed to be proportional to the 
velocity. If such a device is used Equation (10) becomes: 



where / 2 is the resistance of the dashpot at unit speed. 

Equation (11) is a linear differential equation whose general 
solution is 

y = Ae m *+Be m *, (12) 

where A and B are arbitrary constants, and mi and m% are the 
roots of the quadratic equation 

m 2 +/ 2 w+/3 2 w = (13) 

That is 

-/-V/*-4P 



in/i n ~) ano. iivz fj 

Substituting the values of mi and m<z in Equation (12) 

y = e~* f * t (Ae* f *~^-\-BeT^ /4 ~ 4 ^ 2 ). . . . (14) 

If / 2 >2j3 the roots m\ and mi are real. It is then evident that 
y approaches as a limit as t indefinitely increased, and that y 
does not change sign. In other words, the governor moves 
steadily toward its equilibrium position but does not pass it. 

If J 2 < 2/3 the roots mi and W2 are imaginary. Then the gen- 
eral solution becomes 

y=Ae~^ sin (+i\/4/3 2 -/ 4 (15) 

Therefore the governor will oscillate back and forth on either 
side of the equilibrium position. The oscillations rapidly decrease 
in magnitude and practically disappear within a very short period 
of time. 



198 GOVERNORS 

The arbitrary constants in Equations (12), (14) and (15) are 
found by assuming any set of initial conditions such as, t = 0, 

y = yo, and -5^= VQ. Then from Equation (12) 

yo=A+B, 

VQ = m\A -|-ra2/3. 
From Equation (15) 

yo=A sin/3, 

F =A(\/40 2 -/ 4 cos jS-j 2 sin /3). 

NOTE. In the preceding articles two factors have been neglected: (a) 
the effect of the centrifugal force of the link B, Fig. 199, (6) the term J jh dM 

in the moment of the centrifugal force (see Article 127). Both these items 
are of comparatively little importance in most practical governors. For 
discussion of the effects of these items, with numerical data taken from actual 
governors, the reader is referred to R. C. H. Heck's " The Steam Engine," 
Vol. II, page 379 et seq. 

135. Effect of Changing Speed of Engine. Suppose the 
engine to be running at uniform speed coo, and the governor to be 
in equilibrium with the slide at a height AO- If the load is now 
lightened the engine tends to speed up and the governor to rise. 
Let coi and hi represent the new speed of steady running and 
the new position of equilibrium of the slide. In general it may 
be said that as long as the governor remains in a position lower 
than hi the speed tends to increase, and that when the governor 
is above the position hi the speed tends to decrease. This tend- 
ency is, however, modified by the fact that after cut-off the gover- 
nor can exert no further influence until the beginning of the next 
stroke. 

When the slide reaches the position hi it has a certain velocity 
V, and will therefore not stop in this position, but will move on 
to a new point hz. Thence it will again descend, passing the 
position hi with a speed V. As the engine decreases in speed 
during the time which elapses while the slide is moving from hi 
to /i2 and back, the centrifugal force of the weights during the 
downward movement is less than on the upward movement. In 




f OSCILLATIONS 199 

Curve I, Fig. 210, is shown approximately the manner of vari- 
ation of this force. In this curve the abscissa is the height h, 
and the ordinate is the centrifugal force reduced to the slide. 
Then the area under the curve represents the work performed 
by the centrifugal force. In the same manner Curve II shows 
the action of the forces oppos- 
ing the upward movement of the 
governor the weight of the balls 
and slide, the spring tension, etc. 
As these forces depend only on the 
position and not on the speed, 
Curve II is the same for both 
the upward and downward move- 
ment of the slide. Therefore 
the work expended in overcom- 
ing these forces on the upward FI G- 210- 
movement is returned during 

the return downward. But the work represented by the area 
enclosed in the loop of Curve I is not restored, but goes to increase 
the kinetic energy of the governor. Therefore the velocity V 
on the downward stroke will be greater than the velocity V on 
the upward movement. Consequently the slide will continue to 
move downward past its original position ho to some other point 
hsj whence it returns upwards with greater velocity than before. 
It follows that a frictionless governor will oscillate back and 
forth with ever-increasing violence until the limits imposed by 
the construction of the mechanism prevent further increase in 
the amplitude of oscillation. 

If the friction in the mechanism is just sufficient to absorb 
the work represented by the area enclosed in the loop of Curve I, 
the amplitude of the oscillations remains constant, and the slide 
will have an approximately harmonic motion. If the friction is 
still further increased the oscillations will die out and the governor 
will come to rest in the new position of equilibrium hi. 

Since excessive mechanical friction is undesirable, such gov- 
ernors are often provided with oil brakes, which quickly damp 
out the oscillations without impairing the sensitiveness of the 
governor. 

As stated previously these conclusions must be modified on 



200 



GOVERNORS 



account of the fact that the governor can act only while steam is 
being admitted to the cylinder. After cut-off the governor can 
do nothing until the beginning of the next stroke. This results 
in greater irregularity of the oscillations, but does not impair the 
validity of the general conclusions. Heck 1 has attempted to 
analyze this effect. For further discussion of this point the reader 
is referred to Heck's treatise. 

136. Elementary Centifugal Shaft Governor. In Fig. 211 is 

shown an elementary 
centrifugal shaft gov- 
ernor. The weight W 
is free to move back 
and forth in a groove in 
the flywheel F. This 
motion is used to adjust 
the position of the ec- 
centric on the shaft, 
thus regulating the quan- 
tity of steam furnished 
to the engine. The out- 
ward motion of the 
weight is opposed by 
the tension of the spring 




FIG. 211. 



S. If the flywheel rotates with 
trifugal force of the weight is 



angular velocity co, the cen- 



n W 2 

C = rco 2 . 



(1) 



The spring tension increases uniformly with the distance of the 
weight from the shaft center. That is, 

S = A+Br. 

It will be noted that the forces in this governor are precisely the 
same as those in the governor described in Art. 131. Also the 
conditions for stability and the method of calculating the proper 
strength of the spring are the same as those developed in Art. 131. 
While a few governors have been made to operate on this 

1 The Steam Engine. Volume II. 



INERTIA GOVERNOR 



201 




plan, a different construction is usually employed and other forces 
besides the spring tension and centrifugal force are brought into 
play. 

137. Elementary Inertia Governor. In Fig. 212 is shown 
an elementary inertia governor. The weight W is pivoted at 
the center of the shaft 

and revolves with the fly- 
wheel. As long as the 
flywheel rotates at uni- 
form speed co there is no 
relative motion between 
the weight and the wheel. 
If, however, the wheel is 
given an angular acceler- 
ation a, the weight, due 
to its inertia, resists this 
acceleration, and a mo- 
ment la is set up which 
can be used to shift the 
eccentric relative to the FIG. 212. 

shaft. 

This type of governor is more sensitive than the centrifugal 
governor. It begins to act as soon as the flywheel begins to change 
its speed, while the centrifugal governor does not move until the 

W 

increase in centrifugal force r(co2 2 o>i 2 ) is great enough to over- 

9 
come the friction of the mechanism. 

The elementary inertia governor cannot be used in practice, 
since it does not fix the speed at which the engine will run. If 
the engine is running at any speed to this governor will regulate 
at that speed. If, by any means, the speed of the engine is changed 
to some other value, the governor will regulate equally well at 
the new speed. 

Nearly all practical shaft governors operate by combination 
of inertia and centifugal force. Governors are classified accord- 
ing to which force predominates. 

138. The Inertia Forces. Consider any mass M, Fig. 213, 
pivoted at P to an arm of the flywheel. The flywheel rotates 
about its center with angular velocity co and angular acceler- 



202 



GOVERNORS 



ation a. The mass M has a rotation about P with angular veloc- 
ity 12 and angular acceleration relative to the wheel. Then 
by Coriolis' law any point Q of the mass has the following acceler- 



ations: 



(1) Sco 2 in the direction QO, 

(2) Sa at right angles to QO, 

(3) z!2 2 in the direction QP, 

(4) x at right angles to QP, 



(5) 2wco = 2xcol2 along QP. 

If an elementary mass dM is concentrated at Q, this mass 

must be subject to forces 
equal to dM multiplied by 
these various accelerations. 

The force SoPdM may 
be broken up into two com- 
ponents ru 2 dM and cu> 2 dM 
parallel to GO and QG, re- 
spectively, where G is the 
center of gravity of .the 
FlG 213 mass M. The sum of all 

the elementary forces ru 2 dM 

is T(j?M and passes through G. The sum of all the elementary 
forces cuPdM is 

X* 

0. . . . . (1) 




Both these results follow directly from the definition of the center 
of gravity. The entire system of forces of the form Su> 2 dM can 
then be replaced by a single force JVfrco 2 acting through G in the 
direction GO. 

Similarly all the elementary force of the form SadM can be 
resolved into components radM and cadM at right angles to OG 
and GQ respectively. This resolution is shown in Fig. 214. 

The resultant of all the forces radM is a single force raM 
acting through G at right angles to OG. The summation of the 
forces cadM is given by the expression 

Q (2) 



INERTIA FORCES 



203 



It should be noted, however, that, though the sum of these forces 
is zero, their moments about G do not vanish. The moment of 
cadM about G is c?adM. The sum 
of the moments of all these forces 



s 

C 
J 



(3) 



where / is the moment of inertia of 
the mass M about G. 

The forces xQ 2 dM can be com- 
bined into a single resultant att 2 M 
acting through G in the direction GP. 
The proof is exactly the same as for 
the forces of the form Su> 2 dM. 

The forces of the form x dM 
can be combined into a single force 
Ma passing through G in a direc- 







tion perpendicular to PG, and a couple I -7-. The proof is the 



same as for the forces of the form SadM. The forces 
can be combined into a single resultant 2Mo>Qa acting through 
G along the line PG. The proof is the same as for the forces 
of the forms Su 2 dM and x&dM. 

The entire system of forces is then reduced to the following: 

(a) A single force Mru 2 in the direction GO. 
(6) A single force MaQ 2 in the direction GP. 

(c) A single force Mr -a at right angles to GO. 

(d) A single force Ma -^ at right angles to GP. 

(e) A single force 2Macoft along the line GP. 
(f ) A couple la. 

(g) A couple 7-^. 

(h) The spring tension A+Br reduced to the center of 
gravity. 



204 



GOVERNORS 



The six forces all act through the center of gravity G. Forces 
(6) and (e) pass through P and consequently have no tendency 
to turn the mass about P. They can therefore be dropped from 



further consideration. 



As is usually small the force (d) and 

Civ 



the couple (g) are often ignored in governor calculations. 

The inertia forces are of course equal and opposite to the 
accelerating forces. 




The two forces Afrco 2 and Mra, and the couple la may all 
tend to turn the mass M in the same sense about the pivot P or 
the moment of one may oppose the other two. The relative 
positions of the points P, 0, and G determine the sense in which 
the different moments act. With the line OP, Fig. 215, as the 
diameter construct a circle, and extend the diameter OP indefi- 
nitely. The plane is thus divided into four regions I, II, III, and 
IV as shown. Suppose the rotation co and the acceleration a 
to be clockwise. The sense of the moments of the forces Mrco 2 
and. Mra about P can be seen by placing the center of gravity G 
successively at Gi, G^ G$, and G in the regions I, II, III, and IV 



ACTION OF INERTIA GOVERNOR 



205 



respectively. Calling counter-clockwise moments positive the 
results may be shown in compact form in the following table : 



Region 


I 


II 


Ill 


IV 


la 


-i 


+ 


_|_ 


+ 


Mru- 


+ 


+ 


- 


- 


Mra 


+ 


- 


+ 


- 



It is evident that the governing action is most powerful when 
point G lies in region I. Regions III and IV can be used only 
when I is small, as in these regions the centrifugal force opposes 
the inertia couple la. If the couple la is large the immediate 
effect of increasing the speed might be to increase the supply of 
steam furnished. 1 

139. Action of the Governor. When the engine runs at uni- 
from speed co the forces Mra and the couple la disappear, and 
there remains only the centrifugal force Mrou 2 . This must be 
balanced by the spring tension A-\-Br<). Then if the load on the 
engine changes the speed tends to change also and an angular 
acceleration a is given to the flywheel. On account of the increased 
speed the centrifugal force becomes greater than the spring ten- 
sion and the governor therefore tends to rotate about the pivot 
P. On account of the angular acceleration a the force Mra and 
the couple la are also set up and help to cause rotation of the 
governor weight about P. As soon as the governor starts to move, 
dtt 



the force Ma and the couple / come into play. 
dt ctt 



These 



usually tend to retard the motion. Under the influence of all 
these forces the governor tends to move to a new position. 

There is in general only one position at which the governor 
will be in equilibrium under the new load, and one speed at which 
the engine will run steadily. Let r\ be the distance of the center 
of gravity from the shaft center in this position, and let coi be the 
new speed of steady running. Then when the governor again 
reaches equilibrium, 



Zeitschrift des Vereines 



^todola. The Siemens Governing Principle. 
deutscher Ingenieure, 1899. 



206 



GOVERNORS 



Before reaching this new condition of equilibrium the governor 
may, however, oscillate back and forth on either side of the posi- 
tion. Friction in the mechanism tends to damp out these oscilla- 
tions, but if the friction is small, the governor may under certain 
conditions continue to vibrate back and forth for a considerable 
time, thus causing very irregular running. 

140. Mathematical Analysis of Governor Action. The mathe- 
matical study of the ac- 

J & 



tion of the inertia gover- 
nor involves great diffi- 
culties, and it may be 
doubted whether a general 
solution is possible. Sup- 
pose that after a time t 
the engine is running at 
a speed co and that the 
governor has reached a 
position where the center 

of gravity is at a distance r from the center of the shaft. Then 

taking moments about P there results: 




FIG. 216. 



Moment of 



Mroo 2 a sin </> (Fig. 216) ..... (1) 



Moment of M ra = Mraa cos 



(2) 



TVT n/r ,, o 

Moment of Ma-j-= Ma 2 -j-. 
at at 



Moment of Ia = Ia. 



(3) 
(4) 



,, , T T 

Moment of / -j7 = I -77 
at at 



Moment of spring tension = (A-\-Br) a sin 



(5) 



(6) 



In general the acceleration a is approximately proportional 
to the distance of the governor from its equilibrium position. 
Or 



(7) 



OSCILLATIONS OF SHAFT GOVERNOR 207 

Substituting this value and writing & = , we get 
Total moment = 7lfro> 2 a sin <j>+MarC(ri r) cos </> Ma 2 -^ 

+IC(ri-r)-I~ 2 -(A+Br)asm<t> = T. ... (8) 

This moment tends to cause rotation about P. 
Then 



T=(Ma*+I) 2 ....... (9) 

Combining (8) and (9), the resulting equation is 

J2a 

2(Ma?+I) ^ 2 ==Mra 2 a sin 4>+MarC(n-r) cos 4> 

+7C(n-r)-(A+r)asin</> ....... (10) 

Also 

co = coo + ( arf = wo+C | (ri-r)rfr. ... 11) 
Jo Jo 

Since < and r can be expressed as functions of 6, Equation (10) 
reduces to a single differential equation between 6 and t. Unfortun- 
ately, this equation does not admit of any general solution. It 
is possible, however, to draw some general conclusions from the 
investigation. Suppose the engine to be running steadily at a 
speed o>o, and the governor to be in equilibrium at a position 
denned by the angle . Now let the load be suddenly lightened. 
Let the new speed of steady running be o>i and the new position 
of equilibrium of the governor be denned by the angle B\. The 
governor at once begins to move towards its new position of 
equilibrium, and the engine begins to speed up. By the time 
the governor reaches the position B\ the speed of the engine will 
have increased to some value co which may be greater or less than 
i. The governor will have acquired a certain angular velocity 
ft about the pivot P, and will therefore continue to move outward 
beyond 0i to some new position 62- After the governor passes 
0i the speed of the engine begins to diminish and the centrifugal 
force consequently tends to grow less. After the governor has 
reached its extreme position 02 it starts to return toward its original 
position. As the speed of the engine decreases after the governor 



208 



GOVERNORS 



passes 0i, the centrifugal force is less during the return movement 
of the governor than during its outward travel. In Curve I, Fig. 
217, is shown the moment of the centrifugal force about the pivot 
P. Since the ordinate of this diagram represents moment about 
P and the abscissa represents angular movement about the same 
point, the area under the curve represents the work done by the 
centrifugal force during the motion of the governor. In like 
manner Curves II, III, and IV represent the moments of the 




FIG. 217. 

inertia force Mra, the inertia couple la and the spring tension 
A+Br, respectively. Since these forces are independent of the 
speed and are functions of only, the curves are the same whether 
the governor is traveling from do toward 62 or vice versa. There- 
fore the work done by these forces in accelerating the motion of 
the governor during -its outward movement is absorbed in retard- 
ing the governor during its return over the same path. But the 
work represented by the area enclosed in the loop of Curve I is 
not thus reabsorbed, and this work serves to increase the kinetic 
energy of the governor. Therefore the governor returns at a 



OSCILLATIONS OF SHAFT GOVERNOR 209 

higher speed than it had on its outward travel. Consequently 
it will pass its original position and go on to another position 
63, whence it again moves outward with greater velocity than 
before. In other words, a frictionless governor will move back 
and forth with increasing amplitude of oscillations until a limit 
is reached which is fixed by the construction of the mechanism. 

Friction in the mechanism tends to reduce the speed of the 
governor's motion and to diminish the amplitude of the oscilla- 
tions. Static friction in the joints, etc., opposes the motion with 
a nearly constant resistance. If this resistance is sufficient to 
absorb the work represented by the area enclosed in the loop of 
Curve I, the amplitude of the oscillations remains constant, and 
the motion of the governor is approximately harmonic. If the 
friction is further increased, the oscillations gradually diminish 
until the governor comes to rest in its new position of equilibrium. 

Since excessive friction is objectionable on account of its 
tendency to hinder the prompt adjustment of the governor to 
small changes of load, it is customary to provide this type of 
governor with an oil brake. The resistance of this brake is 
approximately proportional to the velocity. With this device 
the sensitiveness of the governor is not impaired, while large and 
violent oscillations are quickly damped out. For more complete 
discussion of the oscillations in shaft governors, the reader is 
referred to the article by Stodola in the Zeitschrift des Vereines 
deutscher Ingenieure, May, 1899. 

The friction and inertia of the eccentric, valve rod, etc., also 
deserve consideration. For a discussion of these points the 
reader is referred to Lanza's " Dynamics of Machinery." 



CHAPTER VIII 
THE MECHANICS OF THE GYROSCOPE 

141. Introductory. One of the most interesting examples 
of inertia forces and kinetic reactions is found in the gyroscope. 
Inertia forces in general and the kinetic reactions produced by 
them may be very undesirable, and under extreme conditions 
even dangerous, but in some cases they may be made very useful 
in accomplishing desired results. This is especially true in gyro- 
scopic motion. 

Recent applications of the gyroscope to engineering problems, 
especially its use in the stabilizing of ships, has given fresh interest 
to what for years has been considered mainly as a toy for children 
or as a problem to test the ability of mathematicians. 

142. Gyroscopic Couple. The gyroscope here considered 
will be a rigid body, symmetrical with respect to three intersecting 
axes mutually at right angles, the body rotating or spinning about 
one of the axes with freedom to turn about one or both of the other 
two axes as the case may be. 

Referring to Fig. 218, let OZ, OX, and OY be three intersecting 
axes mutually at right angles, ABCD being a horizontal circular 
disk. Let the disk rotate or spin with a high constant angular 
velocity, co, about the Z axis and at the same time let the disk 
turn about the Y axis with an angular velocity 2. A particle, m, 
at the distance p (assumed at the circumference for convenience) 
from the center, 0, will have a constant velocity, pco in the plane 
of the disk. It will also have a velocity perpendicular to the plane 
of the disk (parallel to Z axis) due to the angular velocity 12. 
This component of the total velocity, in one revolution of the 
disk about the Z axis, will change in magnitude from zero at A 
increasing downward to a maximum at B, then decreasing to zero 
at C, then increasing (in opposite direction) to a maximum at 

210 



GYROSCOPIC COUPLE 



211 



D and finally decreasing to zero at A again. The magnitude of 
this velocity is 12p cos 6. 

As stated above, a force is always required to change the mag- 
nitude of a velocity, the direction of the force being in the direc- 
tion of the change in velocity. Hence a force must be acting 
downward on the particles which are traveling from D to B and 
upward on particles traveling from B to D. The velocity cop 
may be replaced by two components, one perpendicular to OF, 
cop sin 6, and one parallel to OF, cop cos B. The component, 
cop cos 6, may be neglected as far as rotation about OF is con- 




FIG. 218. 



FIG. 219. 



cerned since the force causing a change in this component has 
no moment with respect to OF. The component cop sin 6 suffers 
a change in direction due to the angular velocity 12. A force is 
required to change the direction of the velocity of a particle, the 
direction of this force being the same as the direction of the change 
in direction of the velocity. It is evident from the figure that 
this force acts downward on particles which are traveling from 
D to B and upward on particles traveling from B to D. The disk, 
therefore, due to these two forces acting on each particle, would 
start to rotate about the X axis unless a couple acted to prevent 
the turning. This couple is called the Gyroscopic Couple. 

One of the simplest physical illustrations of this gyroscopic 
reaction is obtained by holding a bicycle wheel with one hand 



212 THE MECHANICS OF THE GYROSCOPE 

on either end of the projecting axle (dismounted from frame). 
If the wheel is spinning at a high velocity in a vertical plane, an 
attempt to turn the axle (and hands) in the horizontal plane will 
meet with resistance and the wheel will start to turn in the vertical 
plane unless the hands exert a couple to prevent this turning. 

In order to determine the magnitude of the gyroscopic couple 
proceed as follows: The component of the acceleration of any 
particle, m, parallel to OZ is a = 0,1+0,2 where a\ represents the 
rate of change of the magnitude of the velocity ftp cos 6 and 02 
represents the rate of change of direction of the velocity cop sin 8. 



_^ v ..^cos0)_ . fi dd_ 

dt dt p "" y ' 

EF, Fig. 219, represents the change in direction of the velocity 
cop sin during an interval of time dt while the disk turns through 
an angle dd about OF. Hence, remembering lhat an arc equals 
the subtended angle times the radius and that the sine of a small 
angle equals the angle itself, we have : 

EF cop sin d<t> 
0,2= 77- = ~T. - = cop sin 0ft = coft?/, . . (2) 

therefore 

........ (3) 



the force then acting on the particle of mass m, perpendicular to 
the plane of the disk producing the acceleration a, is 



p sin. 6 = m2utty ...... (4) 

The moment of this force with respect to the X axis is 

dT = Fy = 2muQy 2 , ..... . . (5) 

and the total gyroscopic couple is 

.... (6) 



= CdT = C 



but since the polar moment of inertia (with respect to Z axis) is 
I = 21 x we have 



1 It should be noted that this result could have been obtained with much 
less work by applying Coriolis' law. 



SURGING 213 

The following conclusion may be drawn: If a rotating or 
spinning body is turned about an axis perpendicular to the axis 
of spin a couple will be set up about an axis perpendicular to the 
other two axes. The magnitude of the couple equals the product 
of the angular momentum of spin times the angular velocity of 
turning. Or, in other words, a body spinning about a Z axis with 
angular velocity co, requires a couple C Y = /wl2 to maintain an 
angular velocity 12 about the Y axis. 12 is called the velocity of 
precession, or the body is said to precess with an angular velocity 12. 

143. Surging. Attention should be called to the fact that 
C = /col2 is the couple necessary to maintain a precession which 
has already been started. Referring to Fig. 220, during the time 
which elapses while the angular velocity 12 is being set up the wheel 
alternately dips and rises. While the wheel is descending the 
angular velocity 12 increases, and while the wheel is rising 12 
decreases. This action is known as surging. Usually friction 
soon causes the oscillations to die out, and the wheel continues 
to precess at nearly constant speed. 

Suppose the shaft of the wheel Fig. 220, to be supported in 
a horizontal position, and the wheel to be given an angular ve- 
locity a; about the axis O'Z. Now if the support is removed from 
the end of the shaft the wheel starts to drop, and thus a rotation 
about the axis O'X' is set up. If 6 is the angle through which 
the wheel turns about the axis O'X', the angular velocity of the 

rotation about O'X' is . As the wheel now has simultaneous 
at 

rotation about the axes OZ and O'X' a gyroscopic couple is induced 
which tends to produce rotation in a horizontal plane about the 

axis O'Y. The magnitude of this couple is 7co . As the end 

at 

of the shaft is free to move, the wheel is given an angular velocity 
12 about O'Y, and this rotation together with the rotation co pro- 
duces a gyroscopic couple of magnitude 7col2, which tends to cause 
the wheel to rise. 

After an interval of time t from the beginning of the surging 
action, let 

0= angle of dip, 
and 

Q = angular velocity about O'Y. 



214 



THE MECHANICS OF THE GYROSCOPE 



Then the moment of the gyroscopic couple tending to right the 
wheel is Icofi, and the moment tending to cause rotation about 
O'X' is 



For small values of 6, cos 6 is approximately 1 . Therefore the 
moment about O'X' is WlIu$l. This moment causes an angular 
acceleration about O'X' such that, neglecting friction, 



TTT 7 X Q I J | " Ml * 




FIG. 220. 
The gyroscopic couple Jco -^ produces an angular acceleration 

-TT about O'Fsuch that 
dt 



The surging action can be studied by means of Equations (1) 
and (2). Differentiating Equation (2), 



d?6 
dt 2 



dt 2 ' 



(3) 



SURGING 215 

Combining Equations (1) and (3), 



(4, 



Equation (4) may be written in the form 




or 

^720 

^ : +A2n=B2 ........ (6) 

The general solution of Equation (6) is 

...... (7) 



At the instant < = 0, = and ^=0. From Equation (2) 

if ^ = 0,^ = 0. Therefore 

dt ' dt 

Ci = 0, ......... (8) 

and 

C 2 =-f? ......... (9) 

Equation (7) then reduces to 

0=!^(l-cosAO ....... (10) 

From Equation (2) 

dO 1 dO B 2 . 



The interpretation of Equations (10) and (11) is that neglecting 
friction, the velocities of rotation about O'X and O'Y follow the 
law of harmonic oscillation. 

If friction is taken into account, Equation (6) must be modified 
by the addition of a term which takes account of the friction. 

Let it be assumed that the friction is proportional to the velocity 

jn 

-j-. From Equation (2) it follows that the friction is also pro- 
portional to -jr. Equation (6) then becomes 



216 THE MECHANICS OF THE GYROSCOPE 

The general solution of Equation (12) is 



where mi and m* are the roots of the auxiliary equation 

m 2 +F 2 m+A 2 = 
Therefore 

and 



(13) 
(14) 



rai = 



m2 = 






To determine the constants Ci and C 2 put 12 = 0, and -^7=0? 
when = 0. Then from Equation (13) 



and 
Therefore 

and 



r 



C 2 =- 



mi-m 2 A 2 ' 
mi B 2 



mz A 2 ' 



If F 2 >2A the roots mi and mz are real, In this case 12 increases 

B 2 Wl 
steadily to a maximum value of -r^ j, and thereafter remains 

constant. If F 2 <2A the roots mi and mi are imaginary. In 
this case the general solution of Equation (13) is 



(16) 



The interpretation of Equation (16) is that 12 approaches the 

B 2 

value -p as a limit, after oscillating back and forth on both sides 

of this value. In other words the phenomenon of surging is 
present, but the oscillations are soon damped out, and there- 
after the wheel continues to precess at constant speed. 



WHEEL ON CIRCULAR PATH 217 



144. Line of Action of Resultant. From the equation a = 
it is noted that all points at the same distance, y, from the X axis 
have the same acceleration perpendicular to the disk. The result- 
ant force acting (perpendicular to disk) on each half as divided 
by the X axis may be found by summing up the forces F = 2mcofly 
for each half, which leads to the equation, 



The action line of each of these two resultants, R 2 , passes through 
a point at a distance yo from the center along the Y axis as found 
from the moment equation 

4r 
Iwtt = MwQ 2y 

07T 

but since 7 = \M r 2 (for disk) 



145. Wheel on Circular Path. A disk or wheel rolling on a cir- 
cular track is an illustration of gyroscopic motion. Many other 
examples of such motion will occur to the reader, such as loco- 
motive drivers or automobile wheels and flywheels when rounding 
a curve, etc. A couple C = /col2 is required to keep the disk from 
overturning, or in other words the gyroscopic couple of the drivers 
helps the centrifugal couple to overturn the locomotive. It is 
interesting to note that the line of zero acceleration perpendicular 

to the disk is at y= = as shown as follows: Referring to Fig. 
f 

221, the velocity of any point (assumed at the circumference for 
convenience) is made up of three components as shown. The 
component of the acceleration of this point which is parallel to 
the Z axis is a = ai+a2-f-a3. i and #2 have the same meaning 
as in the case already considered ai = a 2 = utty. az = ti 2 R due to 
the rate of change of the direction of SIR. Hence, 



The velocity of the center of the disk is equal to the velocity of 
rim or SIR = cor. Equating a to zero and solving for y we get, 



2' 



218 



THE MECHANICS OF THE GYROSCOPE 



that is, all points on a line parallel to the X diameter midway 
between the track and the X diameter have no acceleration 
parallel to the Z axis. All points above this line have acceler- 
ations directed toward the inside of the track and all points below 
this line have accelerations directed outward, the acceleration in 
each case being proportional to the distance from this line. The 
resultant of the forces producing these accelerations may easily 
be found. This leads to the fact that two equal masses fixed 
along any diameter at the same distance from the center of the 



n 



top 



c/*w/0r petit 



FIG. 221. 

disk would cause no uneven pressure on the track when running 
on a straight track and would cause variation in pressure when 
running on a curved track. It is an interesting fact easily proved 
that three or more masses arranged symmetrically with respect 
to OZ will cause no variation in pressure on either track. The 
variation in pressure in the case of a locomotive driver is very 
small even when running at high speed around a curve of the 
smallest radius. 

146. Angular Momentum. As stated above, when an external 
couple acts on a rigid body the angular momentum of the body, 



ANGULAR MOMENTUM 219 

Jw, will change, the magnitude of the couple being measured by 
the rate at which the angular momentum changes. The three 
properties of an angular momentum which are involved in its 
physical meaning are (1) magnitude, (2) sense of rotation, (3) 
direction (not location) of the plane of rotation, as specified by 
the angle which the plane of rotation or its normal makes with 
some fixed plane (or line). A vector may fully represent these 
three factors by drawing it of a proper length to some scale per- 
pendicular to the plane of rotation with the arrow pointing in 
the direction in which a right-handed screw would advance if 
turned in the direction of rotation. Any change in the angular 
momentum is fully represented by the corresponding change in 
the vector. The couple required to produce this change is always 
in a plane perpendicular to the vector representing the rate of 
change in the angular momentum. 

It is evident that the rate of change of 7co may be (1) the 

rate of change of the magnitude only, as given by 7 -^- = 7a or (2) 

the rate of change of the direction of the 7o> vector only or (3) 
the rate of change of both magnitude and direction. 

The principle of conservation of angular momentum states 
that if no external couple acts in a given plane on a system of 
particles, the angular momentum of the system with respect to 
an axis perpendicular to the plane remains unchanged. 

147. Illustrations. A few familiar illustrations may serve 
to make clear the physical meaning of the principles mentioned 
above. Let two pulleys be mounted on a rotating shaft, one keyed 
to the shaft, and rotating with it, the other loose (not rotating). 
Let a clutch operate suddenly to throw in the loose pulley, thus 
starting it to rotate with the tight pulley and axle. The angular 
momentum of the whole system before the clutch was operated 
was 7 ion. By throwing in the clutch the moment of inertia of the 
rotating system was increased to (7i+72), while the angular 
velocity decreased to co 2 . But from the principle of conservation 
of angular momentum we get, 

7l60l = (7l + 72)c02 Or C02 = j , j 6)1. 

The angular velocity 002 is reached after a surging action 
dies out. While these vibrations are being extinguished kinetic 



220 THE MECHANICS OF THE GYROSCOPE 

energy is lost. The kinetic energy of the whole system before 
the clutch was operated was J/icoi 2 , and after the two pulleys 
have come to a constant velocity, a>2, the kinetic energy of the 
system is 



'TAT \2 ^ v-^i^i l T \ T I't 
,11+12)* ( /1 + /2 J 

hence 

i LL. 

JL T . -r 



represents the part of the original kinetic energy which is dis- 
sipated. This is analogous to the loss of energy in setting up 
precession in the gyroscope. 

A man if standing on a platform which is revolving with an 
angular velocity co can decrease the co by extending his arms thus 
increasing his moment of inertia. This develops a friction couple 
on the soles of his shoes but the product Ico for the man and plat- 
form must remain constant or if co is to be kept constant an 
external couple C must be applied to the platform such that 

dl 



A gymnast leaving the swinging trapeze at the top of a circus 
tent with angular velocity co, his body being extended, may make 
several complete turns in mid-air as he descends in the vertical 
plane by " doubling up." This decreases his moment of inertia 
and gives him a corresponding increase in angular velocity, the 
product 7co remaining constant, since no couple acts in the ver- 
tical plane. 

As is well known, a cat when held by the four feet will, if 
dropped, always turn through one-half a revolution and light on 
its four feet (providing it is held a reasonable distance from the 
floor). This is easily explained by the principle of conservation 
of angular momentum. The 7co of the cat with respect to an 
axis corresponding to its long dimension is zero when its feet are 
released, and must remain zero during its fall, since no couple 
acts to change this angular momentum. For small relative dis- 
placements the cat may be considered as made up of a fore and 
hind part with a swivel at the center. Let I\ and /2 represent 



GYROSCOPIC ACTION 221 

the moment of inertia of the fore and hind parts respectively. 
By extending the hind legs and contracting the fore legs the fore 
part is turned in one direction, throwing the hind part in the 
opposite direction. Since, however, /2>/i there is a net gain 
in favor of the fore part. Now by extending the fore legs and 
contracting the hind legs the hind part is turned, throwing the 
fore part back but to a less extent, since now I\ >/2. By a rapid 
succession of such actions and reactions the cat makes one-half 
a turn. The reader may, by the same method, turn through 
any desired angle, when sitting on a swiveled chair, by extending 
his legs and contracting his arms, etc., etc. The action is more 
positive if weights are held in the hands. 

In the above illustrations the change in the angular momentum 
was either zero or a change in magnitude only. That is, the plane 
of rotation remained the same, or in other words, ihe vector 
representing 7cu changed in length only. In gyroscopic motion 
the 1 oj vector changes direction. 

148. Analysis of Gyroscope by Method of Angular Momen- 
tum. As has been stated, a disk rolling around a curved path 
is revolving simultaneously about two axes mutually at right 
angles. This example of gyroscopic motion together with the 
case as shown in Fig. 220 will be used (for convenience) to show 
the applications of the foregoing principles concerning angular 
momentum. 

Referring to Fig. 222 let A\Bi represent the disk at a given 
instant when its angular velocity about its axle, OD, is o>i and its 
angular velocity around the curve (also its angular velocity about 
an axis through D perpendicular to plane of paper) is fli. Let 
A^Bz represent the disk after an interval of time dt, during which 
the angular velocities have increased to 002 and fe The plane 
of the disk during the time dt turns through an angle dQ. Let 
H denote angular momentum. The angular momentum of the 
disk at the first instant is represented by #i = (7co)i, and after 
the interval dt it is /^ 2 = (/co) 2 . The change in H is ab (exagger- 
ated), Fig. 222. This change is due to a change in direction of 
H, represented by ac, plus vectorially the change in magnitude 
represented by cb. A couple is always necessary to produce a 
change in H, the magnitude of the couple being the rate of change 
of H, the plane in which it acts being perpendicular to the vector 



222 



THE MECHANICS OF THE GYROSCOPE 



representing the rate of change of H and its sense of rotation 
being determined from the direction of the arrow. 
The rate of change in the magnitude of H is 

cb _dHi d(I u>) i T d&i T 
dt = 



dt 



dt 



I being constant. 

The direction of cb is perpendicular to the plane of the disk. 
Hence a couple Ci = lai acts in the plane of the disk causing the 
disk to roll faster. 




FIG. 222. 
The rate of change of the direction of H is 



dt dt 

cb and ac are perpendicular to each other, for the infinitesimal 
angle dB. Hence a couple 2 = Iwitti acts in the plane of the paper 
(perpendicular to plane of disk) in a clockwise direction. In 
other words, the disk would tend to overturn unless a couple 2 
acted to prevent the overturning. 

As already shown C2 = /o>12 is the gyroscopic couple. If o> 
is constant Ci = zero, the total change in H then being ac and 
the only couple required to maintain the velocity 12 being the 
gyroscopic couple. 



PRECESSION 223 

149. Precession. When the body is allowed to precess, the 
external couple acting on the body will be resisted by the gyro- 
scopic couple. This is readily seen by following the changes 
in the angular momentum in Fig. 220. When the couple Wl 
first acts, the change in the angular momentum produced is AB 
vertically upward. This requires a couple in the horizontal plane 
in an anti-clockwise direction, but since there are no bodies to 
develop or supply this couple, the disk turns (precesses) clock- 
wise in the horizontal plane, developing the necessary couple 
(Iy\l/) from the inertia of ;he disk. As soon as precession starts, 
however, the change AC is produced in the angular momentum. 
This requires a clockwise couple in the vertical plane to prevent 
the disk and axle from rotating anti-clockwise in the vertical 
plane. This couple is supplied by the two forces W having a 
moment Wl. Hence if the body is allowed to precess the external 
couple, Wl is resisted by the gyroscopic couple, or 

Wl=Iutt = k 2 o>U, 
from which the velocity of precession 12 = -, k being the radius 



of gyration of the disk and axle with respect to the y axis. 

If precession is prevented, a couple C = /col2 must be set up 
in the horizontal plane, ft here being he angular velocity at any 
instant produced by Wl, which will be the same whether the disk 
rotates or not, there being no resistance offered to the external 
couple unless precession is allowed. This explains why a heavy 
rotating flywheel or armature on board a ship, with axle hori- 
zontal and athwartship, will offer no more resistance to the rolling 
of the ship than if not rotating. The bearings of the axles, how- 
ever, must exert a large couple C = 7o>0 in a horizontal plane, 
which tends to " nose around " the ship. 12 here represents the 
angular velocity of roll. It is evident that the rotating bodies 
aboard ship will be subjected to less stress when their axes are 
parallel to the fore and aft direction, in which case ft would be 
due to the pitching of the ship. 

If an external couple is applied in a horizontal plane, Fig. 
220, to increase or hurry the precession the disk and axle will 
rise, since 7col2>TFZ. This is the principle employed in Brennen's 



224 THE MECHANICS OF THE GYROSCOPE 

monorail car, the precession being hurried by the rolling of the 
axle of the revolving flywheels, extended, on a shelf attached 
to the side of the car. This principle is also used in the " active 
type " of gyroscope for stabilizing ships. In this case the pre- 
cession is hurried by means of a precession engine, which acts 
after the ship has rolled a very small degree, thus producing a 
gyroscopic righting couple just sufficient to extinguish the roll. 
Since the roll is checked in its incipiency only a small amount 
of work is done. The stresses produced in the hull of the ship 
are also small for the same reason. The weight and displacement 
of the gyroscope may likewise be small. 

As already pointed out, energy is dissipated in establishing 
precession. In order to start the disk and axle, Fig. 220, pre- 
cessing the axle must dip due to the couple Wl through an angle 
d6. The work done during this displacement is Wldd. This 
work generates a certain amount of kinetic energy which appears as 
the kinetic energy of precession which is expressed by \I y l 2 . The 
kinetic work WldO, however, is not all expended in producing this 
energy, as may be shown as follows: When the axle dips an angle 
dd, the component of the angular momentum, 7o>, parallel to the 
y axis, is 7co sin dO. The total angular momentum with respect 
to the Y axis must be zero (conservation of angular momentum), 
hence the angular momentum of precession I u ti must neutralize 
this component, or 

7o> sin dd = lySl, 
multiplying by | ft we get, 



But 
Therefore 



that is, one-half of the work expended is transformed into kinetic 
energy of precession; the other half is dissipated through the 
friction which reduces the surging. 

For those interested in the various ways in which gyroscopic 
motion may arise and in the progress which has been made in 
the application of the gyroscope to the stabilizing of ships, mono- 



PRECESSION 225 

rail cars, and aeroplanes, to the guiding of torpedoes, to compasses, 
and to the Griffin grinding mill the following references are 
appended : 

1. Journal of The Franklin Institute, May, 1913. 

2. American Machinist, August 7 and 14, 1913. 

3. Iron Age, December 1, 8, 15, and 22, 1910. 

4. Scientific American Supplement, January 26, 1907. 

5. Popular Science Monthly, July, 1909. 



CHAPTER IX 



CRITICAL SPEEDS AND VIBRATIONS 

150. Introductory. In the preceding chapters the links of 
all the mechanisms discussed have been regarded as rigid, and the 
distortions of the members under the action of the forces applied 
have been neglected. In some cases, notably in revolving shafts, 
there may be appreciable deflections; and sometimes vibrations 
are set up which may have serious consequences. At certain 
" critical speeds " the deflections and vibrations become extremely 
severe. In design of steam turbines and other high-speed machines 
care must be taken to see that the speeds employed are far removed 
from the critical values. The purpose of this chapter is to dis- 
cuss in an elementary way the character and causes of such deflec- 
tions and vibrations. 

151. Revolving Shaft Loaded at Middle. Let the shaft, Fig. 
223, be supported at the ends and carry at the middle a disk or 

pulley of mass M . Let the 
center of gravity of this 
pulley be at a distance s 
from the center of the shaft. 
If the pulley revolves at 
an angular velocity co the 
shaft will bend under the 
action of the centrifugal 
force. Let y be the de- 
flection at the middle of 

the shaft. Then the distance of the center of gravity from the 

axis of rotation is 

R=y+s (1) 

The centrifugal force exerted by the pulley is 

!f(2/+s)co 2 (2) 

226 




FIG. 223. 



SHAFT LOADED AT MIDDLE 



227 



The centrifugal force is balanced by the elastic force tending to 
straighten the shaft. From the laws of mechanics of materials 



__4&EIy 

~~ 



(3) 



where E is the modulus of elasticity in tension (Young's modulus), 
7 is the moment of inertia of the cross-section of the shaft, and 
L is the length of the shaft. Then 



, .... (4) 



where 
Therefore 



ML*' 



(5) 



Evidently if co = a, y becomes infinite. The interpretation of 
this statement is that when co = a, the deflection will increase 
until stopped by the limitations of construction, or until the shaft 
receives a permanent set. 

The speed co = a is called the critical speed of the shaft. If 
co becomes greater than 
a, the value of y becomes 
negative. In other words, 
the deflection y is opposite 
in direction to the eccen- 
tricity s. The position 

assumed by the shaft at speeds above the critical value is 
illustrated in Fig. 224. 

As co increases above the critical value, y becomes less, until 
in the limit when co becomes infinite, y=s. That is to say, 
as the speed increases above the critical value the center of gravity 
moves inward toward the axis of rotation, reaching this axis when 
the speed becomes infinite. 

The centrifugal force is given by the equation 



x _j_ _ _[_ _ _ 




228 



CRITICAL SPEEDS AND VIBRATIONS 



f^ 

In Fig. 225 the values of , 2 are plotted as ordinates against 

Jyl o, s . a 

as abscissa. Fig. 225 shows 
that the centrifugal force 
increases rapidly up to the 
critical speed, and then de- 
creases, approaching Ma?s 
as a limit as w approaches 
infinity. 

If the shaft were held in 
rigid bearings instead of 
being simply supported at 
the ends, the same argument 
holds good, the only change 
being in the value of a. In 
this case 



(7) 



152. Natural Period of Vibration. Let the shaft shown in 
Fig. 223 be kept from revolving and be bent by a force P acting 
at the middle. Then from the laws of mechanics of materials 



Ma z 5 
1.8 
1.6 
1.4 
1.3 
1.0 
0.8 
0.6 
0.4 
0.2 


-0.4 
-0.6 
-0.8 
-1.0 
-1.2 
-1.4 
-1.6 
-1.8 
-2.0 ( 


















































- 










































































































































































































2 












































































































































































































/ 


























/ 


























/ 




























/ 
























1331567 

FIG. 225. 



L 3 



(1) 



If the shaft is now released it will spring back toward its original 
shape. The mass M will be given an acceleration 



where 



dt 2 



P 

"M" 

4SEI 
ML*' 



(2) 



The general solution of the differential Equation (2) is 

y = A sin at+B cosa, .... (3) 

where A and B are arbitrary constants. These constants can be 
evaluated by putting 



y = 2/0, and - = when t 



0. 



VIBRATION OF SHAFT LOADED AT MIDDLE 229 

From Equation (3) 

Bsmat) (4) 



When 

2 = 0, sin at = Q and ^7=0- 

Therefore 

A = (5) 

when 



Therefore 

B = yv (6) 

Equation (3) then reduces to 

y = y cosat (7) 

Evidently y = yo whenever cos at = 1 ; that is when 

at 2ir, 4ir, 6rr, etc. 
The time of a complete vibration is therefore 

27T 



and the number of complete vibrations per second is 
1 a 



r 2-ir' 



(9) 



From Equation (6) of Art. 151, the number of complete revolu- 
tions per second when the shaft is running at its critical speed is 



Since a has the same meaning in Arts. 151 and 152, it follows that 
the critical speed measured in revolutions per second is the same 
as the natural speed of vibration of the shaft. This statement 
holds equally good when the shaft is held in rigid bearings instead 
of being simply supported at the ends. 



230 



CRITICAL SPEEDS AND VIBRATIONS 




153. Critical Speed of Uniformly Loaded Shaft. Let the 
shaft in Fig. 226 carry a uniform load of w pounds mass per unit 

length, and let the shaft 
revolve at an angular 
velocity co. Consider a 
section of the shaft of 
length dx and situated at 
a distance x from the left 
FIG. 226. support. Let the mass 

wdx, of this section Have 

its center of gravity at a distance s from the center of the shaft, 
and let the deflection of the shaft at this point be y =f(x) . Then 
the centrifugal force of the mass wdx is given by the equation 

dC=(y+s)wdxa 2 (1) 

From the principles of mechanics of materials the bending 
moment at the section under consideration is 



M-- 



EI d2y 
BI & 



and the shear is 



(2) 



dx "dx*' 
The shear at a distance x-\-dx from the left support is 



J~3 I "* J~A d%' V*) 



The difference between the shears at the sections x and x-\-dx 
must balance the centrifugal force of the mass between these 
sections. Therefore 



or 



dx 4 



El 



(5) 



(6) 



Equation (6) is a linear differential equation of the fourth 
order. The complementary equation is 

d 4 y wu> 2 y 



W 



= 0. 



(7) 



UNIFORMLY LOADED SHAFT 231 

The complementary function is therefore 

yo = Ae kx +Be- kx +C coskx+Dsinkx, . . . (8) 
where 

4 



k = 

The general solution of Equation (6) is therefore 

y = Ae kx +Be~ kx +C cos kx+D sin kx+P(x), . (9) 

where P(x) is a particular solution of Equation (6). 

The arbitrary constants A, B, C and D are found from the 
conditions at the ends of the shaft. If the shaft is simply sup- 
ported at the ends, y = and j-^ = when x = and when x = L. 
Purring these values in Equation (9) we get 

(10) 

0, .... (11) 



Ae* L +Be-* L +C cos kL+D sin kL+P(L) = 0, . (12) 
Ae kL +Be~ kL -C cos kL-D sin fcL+p^ ( 2 L) =0 . (13) 
Solving Equations (10), (11), (12) and (13) 

f (Q,^- TO)+ i[gg.-.^l 

kL - kL 



" -". 

(16) 




2 sin kL 

From Equation (9) it follows that y will become infinite if any 
one of the arbitrary constants A, B, C, D becomes infinite. From 
Equations (14) to (17) it is readily seen that the only one of these 



232 CRITICAL SPEEDS AND VIBRATIONS 

constants which can have an infinite value is D. If sin &L = 0, 
D and consequently y increases without limit, unless the numer- 
ator of the right-hand member of Equation (17) vanishes when 
sin kL = 0. The critical speeds are therefore such that sin kL = 0, 
or 

/cL=7T, 27T, STT, (18) 

Substituting the value of k in Equation (18) 

Z/\W =7r> 27r, STT, etc., 
or 

_7T2 [El 47T 2 [El 97T 2 [El 

W ~Z2\1P T?\~^> L^ArST' ' ' 

For a uniformly loaded shaft there is therefore a series of critical 
speeds, these speeds being proportional to the squares of the natural 
numbers. That is 

eoi : co 2 : co 3 : 4 , etc. = I 2 : 2 2 : 3 2 : 4 2 , etc. . . (20) 

The investigation of the exceptional cases where D is not 
infinite even when sin kL vanishes, requires a knowledge of the 
form of the particular integral P(x), which in turn depends on 
the character of the function s. Since s is a perfectly arbitrary 
function of x, the numerator of the right-hand member of Equa- 
tion (17) is, in general, not equal to zero. It is easily possible, 
however, to find cases where this expression vanishes when 
sin kL = 0, and where therefore D may not become infinite even 
if sin kL may be zero. For example let the distance s, Fig. 226, 
be a constant s = a. Then Equation (6) may be written 



The general solution of this equation is 

y+a=Ae* x +Be- kx +C cos kx+D sin kx. 
It follows that the particular solution reduces to 

P(*)--a 
Substituting this value in Equation (17) 

r) _(l cosfcL) 
sin kL 



UNIFORMLY LOADED SHAFT- 



233 



When sin&Z/ = 0,^ 

cos-fcL = l. 

Hence the critical speeds will be such that cos kL= 1, or 

kL = ir, 3?r, STT, 77T, OTT, etc. 
Therefore 

_>* 2 [El QT^ [El 257T 2 /#/ 
~Z?\V L 2 \w> L 2 \w e 

The disappearance of the critical speeds corresponding to the 
values kL = 2ir, 4?r, GTT, etc., may be explained as follows. The 
deflection of the shaft as given by Equation (9) is made up of a 
series of terms such as 
shown in Fig. 227. In 
the case of a symmet- 
rically loaded shaft 
the deflections shown 
in Fig. 227 (6), (d), (/) 
etc., cannot take place. 
There remain there- 
fore only the deflec- 
tions shown in Fig. 
227 (a), (c), (e), etc. 
The critical speeds 
correspond to the con- 
ditions when any FIG. 227. 
one of the deflections 

increases without limit. When the shaft is so loaded that any 
of the deflections vanishes at all speeds the corresponding critical 
speed also vanishes. 1 

1 The condition under which one or more of the critical speeds will dis- 
appear is given by the equation, 

cos kx I sin kx s dx sin kx I cos kx s dx . 

In the example given above s=a, and P(x) = a. Substituting these values 
for s and P(x), 

k T a cos 2 kx-j- a sin 2 kx I = a. 
I k k J 

For the development of this criterion the writers are indebted to Dr. Cyril 
A. Nelson of the Department of Mathematics, Western Reserve University. 




'=AT 



234 CRITICAL SPEEDS AND VIBRATIONS 

154. Vibrations of Uniformly Loaded Shaft. Let the shaft 
in Fig. 226 be distorted by any set of external forces and then 
released. There will be set up a series of vibrations of varying 
amplitudes and periods according to the manner in which the 
shaft was originally bent. Let y be the deflection of any point 
of the shaft at a distance x from the left support, and at the time 
t. Then 

/=/(*, 0- 

At a distance x from the left support take a section of the shaft 
whose length is dx. The mass of this section is wdx, or 



(1) 

As in the preceding article the shear on the section is 



and the net force on the element under consideration is 

dV=EI^dx ....... (2) 

(The partial derivative is used to indicate that the time is con- 
sidered constant in this equation). Then the acceleration of the 
mass wdx is 



w x 
In order to find the solution of Equation (3) assume 



and 



Therefore 

m 4 =ra 2 , ......... (6) 

w 

and 



VIBRATIONS UNIFORMLY LOADED SHAFT 235 

An expression of the form 
y= (Ae mx +Be~ mx +C cos mx+D sin mx)(E cos nt+F sin nt}. (8) 

satisfies these conditions. 1 

At the ends of the shaft there is no deflection and no bending 

f?y 
moment. Therefore when x = 0, y = and -r-^ = 0. Substituting 

these values in Equation (8) and reducing, it follows that 

0, ....... (9) 



and 

A+B-C = 0, ....... (10) 

and therefore 

C=0, ....... (11) 

and 

A = -B ....... (12) 

If the time is measured from the instant of release it follows 
that when 



In other words, at the moment of release the particles of the shaft 
have no velocity. Substituting the values of B and C in Equa- 
tion (8) and differentiating with respect to t, 



= n(Ae mx -Ae~ mx +D sin mx)(-E sin nt+F cos nt) = 0. (13) 
ot 

when t = 0. As sin nt = if t = it follows that 

F = Q ............. (14) 

Equation (8) is now reduced to 

y=(Ae mx -Ae- mx +Dsiumx)E cosnt. . (15) 

1 The general solution of Equation (3) may be written in the form of 
Equation (8), substituting for A, B, C and D four arbitrary functions of t, 
and for E and F arbitrary functions of x. The validity of this solution may 

easily be shown by forming the derivatives ~ and \ and substituting 

the values thus found in Equation (3). The argument which follows can 
be adapted with slight modifications to the case where arbitrary functions 
are used instead of arbitrary constants. For the sake of simplicity A, B, 
C, D, E, and F are treated as constants in the text. The results expressed 
in Equations (26), (28) and (29) are identical with those obtained using the 
general solution. 



236 CRITICAL SPEEDS AND VIBRATIONS 

r o 

When x = L, y = and ^1 = 0- Substituting these values in 
Equation (15) it follows that 

A(e mL -e- mL )+DsmmL = Q, .... (16) 

and 

A(e mL -e- mL )-DsmmL = ..... (17) 

Hence 

A=0, ........ (18) 

and 

DsinraL = ......... (19) 

From Equations (15) and (18) it is evident that if D = 0, y = 
and there is no deflection. As this is contrary to the hypothesis 
that the shaft is bent, it follows from Equation (19) that 

sinraL = .......... (20) 

Therefore 

mL=7r, 27T, STT, 4ir, etc, .... (21) 

and 

o JEI 7T 2 lEI 47T 2 lEI 97T 2 lEI 

n = m z \ = T2\/ i ~J~2\ Ta-v/ > etc. . (22) 
\ w L 2 \ w ' L 2 \ w' L 2 \w' 

Equation (8) is now finally reduced to 

y = c sin mx cos nt, ..... (23) 
where c = DE. 

It has been shown that m and n can have a series of values 
mi, m2, ws, etc., and m, 712, ^3, etc., which are defined by Equa- 
tions (21) and (22). Any pair of these values satisfies Equation 
(23) . Furthermore the sum of the series of terms 

ci sin m\x cos nit+C2 sin m^x cos nrf+c^ sin m^x cos n^t-\- y etc., 
is also a solution of Equation (23). We can therefore write 



(24) 



where s can take all integral values from 1 to oo . 

At the time Z = the shaft has been distorted by external 
forces into any desired form. Let the original deflections be 
represented by the equation 

y = F(x). ... ......... (25) 



VIBRATIONS UNIFORMLY LOADED SHAFT 237 

Then when t = Q, 

y = F(x)=Ci sin -y-+C2 sin -= \-c% sin ~ . . . (26) 
Li Li Li 



This is a Fourier series. The coefficients ci, C2, ca, etc., are deter- 
mined by the equation 



sin 



STTX 



(27) 



The vibrations of the shaft are therefore made up of a set of 
harmonic oscillations of different amplitudes and different periods. 
The amplitudes are given by the values of the constants ci, C2, 
etc., and the periods are given by the equations 



. 
or *i = 



or t*= 



= 2ir or 



2L 2 Iw 

7T 

2L 2 
2L 2 

: 97T 



, . . . . (28) 



etc. The number of vibrations per second are the reciprocals 
of ti, k, etc. 



47T 



(29) 



etc. From Equation (19) Art. 153 the critical speeds of such a 
shaft are found to be: 

TT 2 [El TT lEI 

coi = -pxA / radians per second = jrrf>\ / revolutions per second. 
L 2 \w 2L 2 \ w 

4** [El 47r fEI 

0)2 = fj\ radians per second = or\/ revolutions per second. 



238 CRITICAL SPEEDS AND VIBRATIONS 

etc. Therefore the critical speeds expressed in revolutions per 

second are exactly the same as the speeds of the vibrations set 

up in the shaft. 

155. Shaft Rotating in Fixed Bearings. Uniformly Loaded. 

For the type of shaft illustrated in Fig. 228 Equations (1) 

to (9) of Art. 153 
hold good. The dif- 
ference in the solu- 
tion consists simply 
in the determina- 
FIG. 228. tion of the arbitrary 

constants A, B, C, 

D of Equation (9), Art. 153. When the shaft rotates in fixed 

bearings y = Q and -^ = at the two ends, that is, where z = and 

where x = L. 

From Equation (9), Art. 153, 

y = Ae**+Be-**+C cos kx+D sin kx+P(x). . (1) 
Differentiating, 

= k(Ae**-Be-* x -C sin kx+D cos kx)+&. (2) 




Putting y and -^- = at the ends, 

A+B+C+P(0)=0 ........ (3) 

....... (C 



Ae JcL +Be- lL +CcoskL-\-DsmkL+P(L)=0. . . (5) 



Ae kL -Be- kL -C sin kL+D cos kL+j =0. . (6) 

K 



Solving these equations: 
A = cg-bh 



bf-ag' 
ah-cf 
bf-ag 1 
_ (b-a)h+(f-g)c 

bf-ag J) > 

D= Q>+a)h-(f+g)c ldP(0) 
bfag k dx 



SHAFT IN FIXED BEARINGS 239 



where a = e kL cos kL sin kL ; 

J} = e~ kL ~ cos fcL+sin kL', 



c = P(0) cos kL+P(L) - - sin 

y = e kL -f sin &L cos kL ; 

/cL+cos &L; 



. 

k dx k dx 

If any of the arbitrary constants A, B, C, D are to become infinite, 
it must be when bfag = 0. 

Substituting the values of a, b, /, g, and expanding we get, 
after suitable reduction : 



4[l-J(e* z 4-e-* L ) cos kL]= 4(1 -cos kL cosh kL) = Q. (11) 
Hence 

cos kL cosh kL = 1 ........ (12) 

This gives for the critical speeds the values corresponding to 

/cL = 0, 4.729, 5-Ai, +A 2> -A 3 , .... (13) 



where AI, A2, AS, ... are very small quantities. The critical 

o 

speeds are such, therefore, that kL is slightly greater than -^, 

2 

?TT UTT 157T 5rr 9rr 137r 

~2~' ~2~' ~2~' i or slightly less than , , , .... Since 



4 foo> 2 
^nr, the critical speeds are approximately given by 

97T 2 lEI 257T 2 /#7 497T 2 



' 4L 2 \ w 

156. Vibrations of Uniformly Loaded Shaft Fixed at Both 
Ends. Let the shaft represented in Fig. 228 be bent into any 
desired form by external forces, and then released. A set of vibra- 
tions will be set up which follow the laws derived in Equations 
(1) to (8) of Art. 154. That is 



240 CRITICAL SPEEDS AND VIBRATIONS 

y=(Ae mx +Be~ mx +C cos mx+D sin mx)(E cos nt+F sin nt) , (2) 

=m y^ (3) 

^=- 2 (4) 

5t 2 " ^ ' 

and 

n 2 = m 4 (5) 

The end conditions of the shaft give: 

y = when x = 0, and when x=L and 

-^=0 when x = 0, and when x L. 

5x 

If the time is measured from the instant of release, 



Therefore F=0, exactly as in Art. 154. 
When z = 0, 

y= (A+B+C)E cos nt = Q, 
and therefore 

A+B+C = ........ (6) 

Also 



and therefore 

A-B+D = ........ (7) 

When x = L, 

y=(Ae mL +Be- mL +C cos mL+D sin mL)Ecos nt = 0, 

and therefore 

Ae mL +Be~ mL +C cos mL+Z) sin wL = 0. . . (8) 
Also 



- = m(Ae mL -Be- mL -C sin mL+D cos mL); cos n*=0, 

whence 

Ae mL -Be- mL -C sin mL+D cos wL = 0. . . (9) 



VIBRATIONS SHAFT IN FIXED BEARINGS 241 

Substituting in Equations (8) and (9) the values of C and D from 
Equations (6) and (7) 

A(e mL cosmL smmL)+B(e~ mL cos wL-fsin wL)=0, . (10) 

A(e mL +sm mL cos mL)+B( e~ mL +cos raL+sin mL) =0. (11) 
Equations (10) and (11) can be written in the form 

aA+bB = 0, ....... (12) 

cA+dB = Q ........ (13) 

Eliminating B from Equations (12) and (13) 

A(bc-ad)=Q ........ (14) 

Therefore A =0, or 

bc-ad=0 ........ (15) 

Eliminating A from Equations (12) and (13) 

B(bc-ad)=0, ....... (16) 

and therefore B = Q, or 

bc-ad = Q ....... (17) 

If both A and B are zero, C and D must also be zero, and there 
is no deflection of the shaft. This is contrary to hypothesis, and 
it is therefore necessary to adopt the other solution, namely 

bc-ad-0 ........ (18) 

Substituting in Equation (18) the values of a, 6, c and d from 
Equations (10) and (11), and reducing, 

4[1- J cos mL(e mL +e- mL )]=4:(l-cos mL cosh mL) =0. (19) 

Therefore 

cos mL cosh wL = l, ...... (20) 

and mL has exactly the same values as those found for kL in the 
preceding article, that is 

_ STT 5ir . 7ir 

= 0, - Ai, +A2, - As, etc., 



where AI, A2, As, etc., are small quantities. The value mL is 
inadmissable and may be discarded. The remaining values of 
mL are approximately 

T 3-7T 5?T 7-JT 

= ~~' ' ~~' ...... * ' ^ ^ 



242 CRITICAL SPEEDS AND VIBRATIONS 

Hence 



O7T O7T iTT /r\r\\ 

m = 2L'2L'2L' etc ' :- ( 22 ) 
l^_97r 2 ,'#/ 257T 2 [El 497T 2 /#/ 

~~ A T 2 \l , /IT2\] w > 42 ^ ^ > 6tC - (**V 



and 



The period of oscillation, r, is given by: 
so that 



2w 8L 2 fw 8L 2 /~^~ 8L 2 /w 

y.1 * l-r- X. li I '/-\-^- % IV / | V I-,- % ; /V f 



Oscillations per second 

/ 1 

= ; = 8f 2 \"^r'8Z^\"ST'8Z^V^ L (25) 

From Art. 155 the critical speeds are given by : 

Sir 2 [El 25** tEI 497T 2 /#? 
= 4L 2 \^' IZ^V^' 4L 2 \^~ ' ' ' radians P er sec ' 



9?r M 257T IEI 497T to 

= 8r 2 V^r' 8L 2 V^ szW^r rev - per second - 

In other words, the critical speeds, expressed in revolutions per 
second are equal to the natural periods of vibration of the shaft. 

The constants A, B, C, D, E. are determined in the same 
manner as in Art. 154. 

157. Inclination of Rotating Disk. In the preceding articles 
the unbalanced masses have been considered as concentrated at 
a point, or as concentrated along a line at a known distance s 
from the center line of the shaft. In other words, there is con- 
sidered to be only one mass in any plane drawn normal to the center 
line of the shaft, and this mass is regarded as concentrated at a 
point. In practical constructions, such as the shaft and wheel 
of a turbine, the load in any transverse plane is distributed over 
the area of the wheel or disk. This causes a modification of the 
laws of critical velocities and of vibrations of the shaft, on account 
of the angle through which the disk is turned when the shaft is 
bent. 



INCLINED DISK 



243 



As an example consider the shaft shown in Fig. 229. The 
shaft is supported in one bearing and carries a disk at its free end. 
If the mass is regarded as concentrated at its center of gravity 
G, the critical speed of the shaft is readily shown to be : 



3EI 
ML 3 ' 

If, however, the mass is distributed over the area of the disk a 
different value of the critical speed will be found. 




FIG. 229. 



Let 



distance of G from center line of the shaft; 
6 = angle of inclination of shaft at free end ; 
dM = mass of any element of the disk ; 

z = distance from G to dM, measured at right angles to 

the center line of the shaft; 
R = deflection at free end of shaft. 
Then, for small values of 6 the following relations hold : 
Distance of dM from center of rotation = R -\-s-\-z. 
Centrifugal force of 

(1) 
.. (2) 



Moment of dC about G=dT =(R+s+z)u 2 dMz sin 
And since for small angles, sin 0=0, 

dT = (R+s+z)Ow 2 zdM. 

Moment of centrifugal force about G=T = ( dTo 
= co 2 ((R+s+z)zdM = 



(3) 



244 CRITICAL SPEEDS AND VIBRATIONS 

From the definition of the center of gravity: 



Hence, 

(4) 



where J is the moment of inertia of the disk with respect to an 
axis passing through G and normal to the center line of the shaft. 
There is therefore a couple of moment u> 2 6J which is caused by 
the inclination of the disk, and which tends to straighten the shaft. 
The centrifugal force C = M(R-\-s)u> 2 , tends to bend the shaft. 
The moment of the centrifugal force about any section at a dis- 
tance x from the free end is: 



The net bending moment is therefore: 

T=Ti-To = M(R+s)u 2 x-rfeJ. . . . . (5) 
From the laws of mechanics of materials 



Integrating twice: 



ClX 

. , n 

EIy= ^ ' --- +Cix+C 2 . ... (7) 



When 
Hence 



(8) 



r , 

~2~ ~~3~ 

When a;=0, y = R, and -j-= tan 6= 6 for small deflections. 

CLX 



ROTATING DISK 



245 



Then from Equation (6) 



-Eie=Ci = u 2 ojL- 



M(R+s)w 2 L 2 



Hence 



When x = 0, y = R. Then from Equation (7) 
EIR = C2=-^^+ 



+ 



Therefore 



= (+)[- 



9 T T \ I 



72 = 



SH 



1 aj0/ * / J&/ 



+1 



Let 



And let 



i- 



z. 



i- 



Then Equation (13) reduces to: 

B 



3EIN 



(11) 



(12) 



. (13) 



(14) 



72 becomes infinite when JV = 0. The critical speed is therefore 
such that N vanishes. Putting N = Q and clearing of fractions: 

M L 4 Jco 4 + (4EIML* - 12 EIJL) o> 2 - 12 E 2 / 2 = 0. . (15) 
Hence, 

- ML 2 + 3 J 4- VM 2 L 4 3JML 2 + QJ 2 ! 

ML 3 J J' ' 

1 Stodola: The Steam Turbine. 



2 #7 



246 CRITICAL SPEEDS AND VIBRATIONS 

Example. A steel shaft 2 inches in diameter overhangs its 
bearing by 30 inches and carries at its free end a steel disk J inch 
thick and 30 inches in diameter. Neglecting the mass of the shaft 
find the critical speed of rotation. 

Weight of disk =7rXl5 2 XiX0.24 = 84.8 pounds. 

M = = 0.219, 



g being expressed in inches per sec. per sec. 

, MD 2 0.219X900 
16~ = 16" 

7= = 0.7854, 



12.32, 



E for steel = 30,000,000 pounds per square inch. 
Substituting these values in Equation (16) 



co = 115.1 radians per second = 18.34 rev. per sec. 

If the weight of 84.8 pounds were concentrated at the end of the 
shaft the critical speed would be given by 

I3EI _ 
co = urz v 11954 = 109.3 radians per second. 



= 17.41 revolutions per second. 

The critical speed is thus seen to be increased due to the inclina- 
rion of the disk. 

158. Vibrations of Shaft Loaded with Disk at Free End. 

Suppose the shaft shown in Fig. 229 to be held from rotating and 
to be bent by some external force. If the shaft is now released 
vibrations will be set up. It is required to find the speed of the 
vibrations of a shaft carrying a disk at the free end, as compared 
with the vibrations set up in a similar shaft carrying a concentrated 
load at the free end. 

When the shaft is set in vibration the disk will have an angular 

acceleration -^ , and a linear acceleration -^-. To give the mass 
its linear acceleration a force is required : 

'-*? 



VIBRATIONS OF DISK 247 

The moment of this force about any section of the shaft at a dis- 
tance x from the free end is : 



The moment necessary to give the disk its angular acceleration is: 

T = T (*\ 

The total moment tending to bend the shaft is therefore: 
Integrating twice : 



where Ci and 2 are functions of t. 

Whenz = J 

Therefore, 



When x = L, y = and = 0. 

oX 



Ci -^ *?+*** (7) 

2 (it 2 3 dt 2 

When z = 0, y = R and ~^-= tan0= for small deflections. 
Substituting these values in Equations (4) and (5) : 

_JLd 2 B ML 2 d 2 R 
El dt 2 2EI dt 2 ' 

JL 2 d 2 B . ML* d 2 R 

K == (Mi 

f)fj>T .J/2 ' Q J7 1 T *]-t2 * v* 7 / 
I . I CH OJjjL ut 

d 2 d 
Eliminating -^ between Equations (8) and (9) 

d 2 R 



248 CRITICAL SPEEDS AND VIBRATIONS 

and therefore, 

2/ML?d 2 R 



Eliminating -^ from Equations (8) and (9) 
2LB = JL 2 d 2 8 



From Equation (11) differentiating twice, 

d^ = ~L\l2Ei df~W)' ' ' ' (13) 

Substituting in Equation (12) the values of 6 and -p- from Equa- 
tions (11) and (13) and collecting terms 

MJL 4 d 4 R 3JL+ML 3 d 2 R 
12E 2 ! 2 dt* ~* 3EI ' dt 2 

This is a linear differential equation of the fourth order. The 
auxiliary equation is 

MJL 4 m4 3JL+ML* m2 - = 
The four roots of Equation (15) are given by 

n. (w) 



Two of these roots are real and two are imaginary. Let 



be the imaginary roots. Then the general solution of Equa- 
tion (16) is 

coskt+Dsiukt. . . . (17) 



The period of the vibrations is given by the value of k. If T 
represents the period of vibration then 

Tk = 2ir ......... (18) 

The number of vibrations per second is 



OTHER SYSTEMS OF LOADING 249 

Example. Find the number of vibrations per second of the 
shaft described in the example of the preceding article. 

As before 

M = 0.219; 

J= 12.32; 
7 = 0.7854; 
7 = 30,000,000. 

Substituting these values in Equation (16) 

A; = 98.77. 

Vibrations per second = ^ = 15.72. 

ZiTT 

If the load were concentrated in a point at the free end, the 
vibrations per second would be 17.41, corresponding to the critical 
speed under these conditions. It follows that when the obliquity 
of the disk is taken into account, the critical speed is no longer 
the same as the natural speed of vibration of the shaft. The 
effect of the obliquity of the disk is to increase the critical speed 
and to reduce the rapidity of elastic vibration. 

159. Other Systems of Loading. In the preceding articles 
the only cases considered have been those of a single concentrated 
load or a load uniformly distributed along the shaft. Many other 
arrangements might be investigated along similar lines. Some 
of the arrangements which may occur in practice are: 

Two or more concentrated loads. 

Two or more disks or wheels. 

Non-uniform distributed loads. 

A large number of disks, giving approximately the effect 
of a distributed load, in which the obliquity must 
be taken into account. 

Disks set at an oblique angle to the center line of the shaft. 

Shaft of non-uniform section with various loadings. 
Some of these cases have been investigated by Dunkerley, 
Stodola, Chree, Greenhill, and others. Limitations of space forbid 
a further consideration of such problems in this place. The reader 
who wishes to pursue the study of critical speeds further is there- 
fore referred to the works of the authors named. 



CHAPTER X 
TOOTHED WHEELS 

160. Introductory. Rolling Contact. In many mechanisms 
motion is transmitted from one link to another by direct rolling 
contact. Friction gears are examples of such operation and 
toothed wheels can be shown to be kinematically equivalent to 
links having pure rolling contact. 

Evidently the angular velocities of the links having rolling 
contact are inversely proportional to the radii from the centers 
of rotation to the point of contact. It follows that if the velocity 
ratio is constant the segments into which the line of centers is 
divided by the point of contact must be constant, and that there- 
fore the rolling curves must have a constant radius. 

All contact of this kind may be divided into three classes : 

(a) Axes parallel. 

(b) Axes intersecting. 

(c) Axes neither parallel nor intersecting. 

In cases (6) and (c) the rolling curves are not in the same 
plane. The angular velocity ratio in these cases is treated in 
Arts. 162 and 163. 

161. Axes Parallel. Rolling Cylinders. If two cylinders C 
and D, Figs. 230 and 231, are mounted on two parallel shafts A 
and B in such a manner that they are kept in contact in all posi- 
tions, the sum of their radii will equal the distance between the 
shafts in Fig. 230, while in Fig. 231, the distance between the 
shafts will equal the difference of the radii. In either case the 
line of contact is common to both cylinders, and if one cylinder 
be made to rotate with angular velocity co, it will drive the other 
by rolling contact, with angular velocity 12 such that 



250 



ROLLING CYLINDERS ROLLING CONES 



251 



Note that in Fig. 230 the wheels rotate in opposite directions, 
while in Fig. 231 they rotate in the same direction. 

162. Axes Intersecting. Rolling Cones. The derivation of 
rolling cones from rolling cylinders is probably best shown by the 
use of two right cylinders combined with two right cones as shown 
in Fig. 232. Each cylinder has one base in common with that of 
one of the cones, hence the axes of this cylinder and cone must 
coincide. While the bases of the cones need not be equal the 
slant heights must be the same. The bases of the two cones have a 
common tangent in their plane which passes through E and which 





FIG. 230.] 



FIG. 231. 



is perpendicular to the plane of the paper. If the axis AA be 
caused to rotate about this tangent, point a can be brought into 
coincidence with point b and the cones still have a common tan- 
gent and will be in contact along element bE, Fig. 233. Hence 
the cones can roll upon each other in their new position. Evidently 
the velocities of the two contact points are the same as for the 
original position. 

If from any point F of the line of contact planes are passed 
perpendicular to the axes AA and BB, these planes will cut circles 
from the cones which will roll together with the same velocity 
ratio as the bases. 

Hence, 

co FG R 

= rn = ~ ) a constant ratio. 



252 



TOOTHED WHEELS 




FIG. 232. 



FIG. 233. 



Having given the axes, and the velocity ratio for two rolling 

cones, it is re- 
quired to con- 
struct the cones. 

In Fig. 234 let 
OA be the driving 
axis and OB the 
following axis; let 
the velocity ratio 
of driver to fol- 
lower be 




In other words, 
OA is to make m 
revolutions while 
OB makes n revo- 
lutions. 

To any con- 
venient scale lay 
off OC = n divi- 
sions. Through C 
draw CD parallel 
FIG. 234. to OB and let CD 

be equal in length 
to m divisions on the same scale. Through D draw ODT, which 



ROLLING HYPERBOLOIDS 



253 



will be the line of contact. From any point T of ODT, let fall 
the perpendiculars AT and BT on the axes. Construct two 
right cones on OA and OB having AT and BT as radii of their 
respective bases. These cones will roll together with the required 
velocity ratio, for: 

sin e_AT . BT_AT 
~OT '' OT 



m 



sin 



BT' 



In other words the radii of the bases have the required constant 
ratio. 

163. Axes Neither Parallel nor Intersecting. Generation of 
the Hyperboloid of One Sheet. If any line AA } Fig. 235, be 
caused to rotate about any other 
line BB that does not lie in the 
same plane, the resulting surface is 
a hyperboloid of revolution. It is 
evident that a plane through line 
BB would cut from the surface a 
hyperbola of which line BB is the 
conjugate axis. 

The same hyperboloid of revolu- 
tion would evidently have been 
generated by the revolution of line 
CC about line BB as an axis. 
Hence the hyperboloid of revolu- 
tion is a double ruled surface. 

Sections cut from the surface 
by planes perpendicular to the axis 
are circles, the smallest of these 
being called the gorge circle. 

Referring to Fig. 236, line XX 
is a vertical axis and YY is another 
axis revolved from any position 

until parallel to a chosen vertical plane but inclined to the hori- 
zontal plane. KM is the common perpendicular to the two 
axes. OC is a straight line parallel to the vertical plane, 
o'c' makes an angle a with X'X' and an angle with Y'Y'. 

If an hyperboloid of revolution be described by the revolu- 
tion of OC about the vertical axis and another hyperboloid be 




254 



TOOTHED WHEELS 



described by the revolution of OC about YY as an axis, the two 
hyperboloids will be in contact along the line OC. If one hyper- 
boloid be made to rotate about its axis with angular velocity 




co it will drive the other by rolling contact if the frictional resist- 
ance between the hyperboloids is sufficient. 

The construction of the hyperboloids is clearly shown in Fig. 
236. The end view is a projection of the inclined hyperboloid 
upon a plane perpendicular to its axis. This view is of service 
in the layout of the inclined hyperboloid. 



ROLLING HYPERBOLOIDS 255 

To draw a normal to an hyperboloid of revolution from any 
point on its surface proceed as follows: Through the chosen point 
draw the two generatrices. The plane containing these two lines 
will be tangent to the hyperboloid at the chosen point and at this 
point only. The normal to the surface will be perpendicular to 
this plane and since the surface is one of revolution, the normal 
must intersect the axis. 

The common normal to the surfaces at C will then be perpen- 
dicular to the tangent plane at that point and this tangent plane 
will contain the line of contact OC. Since the line OC is parallel 
to the vertical plane the vertical projection a'b' of the common 
normal must be perpendicular to o'c' '. But the common normal 
must intersect the axis of each hyperboloid. Hence a'c'b' at right 
angles to o'c' is the vertical projection of the common normal. 
The horizontal projection acb is easily found. 

Denote the vertical hyperboloid by the symbol M and the 
inclined hyperboloid by the symbol N. Let co and 12 be the angular 
velocities of M and N respectively. Let Vi be the linear velocity 
of the point considered as a point on the gorge circle of M and 
let 2 be the linear velocity of the point considered as a point 
on the gorge circle of N. Let r\ and r<i be the radii of the gorge 
circles of M and N respectively. 

Draw o'd', o'f and o'e' at right angles to X'X', Y'Y' and o'c' 
respectively. Lay off o'd' equal to Vi and draw d'e'f perpen- 
dicular to o'e', meeting o'e' at e' and o'f at /'. Since the motion 
is transmitted by rolling contact the components of \ and 2 
perpendicular to OC must be equal. Resolving Vi and 2 
parallel and perpendicular to OC the components of V\ are o'e' 
and d'e', while those of 2 are o'e' and e'f. In addition to the 
rolling of the one hyperboloid on the other there is also a sliding 
motion along the line of contact and the velocity of this sliding 
is represented by d'f. 

Here we have 

u = o'd' . o'f' = o'd'r 2 
12 n ' r 2 o'f n' 

But the triangle d'o'f is similar to the triangle o'a'b'. 
Hence 

o^_oV 
o'f o'b'' 



256 



TOOTHED WHEELS 



Also from the top view, 

r2_c6 
ri~ca 
or 



cb c'V 

=-r~> 

ca ca 



_ ___ 

12 o'6' c'a' o'b'' o'a'~sma 

On x'o'x' make o'g' equal w and draw g'h' parallel to Y'Y' to 
meet o'c' at A'. Then angle o'&y=/3, and ?^'=?5, but this 

is equal to , or, since o'g' = co, g'h' must be equal to 12. 

Having given the two axes and the relative angular velocities, 
it is required to find the line of contact and to draw the hyper- 
boloids. 

On x'x' lay off o'g' = w and draw g'h' parallel to Y'Y' and 
equal to 12. Then o' being joined to h' } the elevation of the line 

of contact is found. To 
find the plan of this line 
draw a'c'b' perpendicular 
to o'c' to meet x'x' at a' 
and Y'Y' at V. Hence 
b' is located and by draw- 
ing ab we can locate c. 
Then oc parallel to YY 
is the horizontal projec- 
tion and the surfaces 
may be drawn. 

If in any hyperboloids 
the radii of the gorge 
circles be decreased the 
meridian curves will become flatter, and the axes in the limit 
will intersect and the surfaces will be rolling cones. 

If the axes are neither parallel nor intersecting, it is possible 
to transmit motion from one to the other by pure rolling contact 
if two pairs of rolling cones are used as shown in Fig. 237. 

164. Friction Gearing. Friction gearing is employed where a 
constant velocity ratio is not imperative. When working with- 
out slip friction gears transmit precisely the same motion as 
toothed wheels. The instantaneous center of relative motion is 




FRICTION GEARING 257 

at the point of contact if there is no slippage, but becomes inde- 
terminate in case slipping exists. In the latter case the motion 
is unconstrained, and the drive is not positive. Hence in dealing 
with toothless wheels from the standpoint of kinematics, it is 
usually assumed that there is no slippage. The material and 
construction of the wheels together with the pressure with which 
they are held in contact will determine the magnitude of the force 
that can be transmitted. 

In many applications the possibility of slip is desirable. For 
example, in hoisting machinery if the car or skip strikes an 
obstruction while being hoisted or if the drum is overwound the 
slippage of the friction drive lessens the danger of breakage of 
parts of the machine. 

From the standpoint of kinematics, any of the surfaces dis- 
cussed in Arts. 161 to 163 will roll together and may be used as 
friction gears. Practically, rolling cylinders and rolling cones or 
special cases of the latter are most common. 

165. Spur Frictions. Grooved Spur Frictions. The simplest 
form of friction gearing consists of two plain cylinders, rotating 
on parallel shafts and held together by properly constructed bear- 
ings, Figs. 230 or 231. 

Such gearing is used for light power hoists, coal screens, friction- 
board drop hammers, etc. 

If the amount of power to be transmitted is increased, the 
pressure holding the rolls together must be increased. This pres- 
sure may be excessive upon the shaft and cause a considerable 
loss of power due to journal friction. To decrease this loss, a 
form of gearing known as grooved spur friction wheels, Fig. 238, 
is used. 

166. Beveled Friction Gearing. Crown Friction Gearing. 
Beveled friction gearing, Fig. 239, is used where it is desired to 
transmit power between intersecting shafts. 

If the shafts intersect at 90, crown friction gearing is often 
used to transmit power, as shown in Fig. 240. As usually installed 
disk b acts as the driver and wheel a as the follower, although 
from the point of view of uniformity of wear the reverse arrange- 
ment is preferable. The wheel a is so mounted that it can be 
drawn across the face of the disk 6, thus varying the velocity 
ratio and also the rotational direction of the driven shaft. 



258 



TOOTHED WHEELS 



167. Spur Gearing. Where the velocity ratio between the 
driving and driven members must be absolutely positive, or when 
the power to be transmitted is large, it becomes necessary to 
provide the contact surfaces with grooves and projections, or teeth, 




FIG. 238. 




thus providing a positive means of rotation. The rolling surfaces 
are then called Pitch Surfaces, and sections of them perpendicular 
to their axes are called Pitch Circles. The point of tangency 
of these circles is called the pitch point. 




FIG. 240. 

The teeth must have such a form as to satisfy the following 
conditions : 

1. For satisfactory operation and for the transmission of 
a uniform velocity ratio the teeth must have a form 
such that the common normal will always pass 
through the pitch point. 



SPUR GEARING. DEFINITIONS 259 

2. Friction and wear attend sliding contact. Hence the 

relative motions of the teeth should be as much a 
rolling motion as possible. It is impossible to have 
a pure rolling contact with toothed gearing and at 
the same time a constant velocity ratio. 

3. The teeth should be symmetrical on both sides in order 

that the gear may run in either direction. 

4. The arc of action should be long enough to insure the 

meshing of more than one pair of teeth. 

A number of shapes of teeth will satisfy these conditions, but, 
the two commonly used are the involute and cycloidal teeth, so 
called, because of the curves on which they are based. 

168. Definitions. As a preliminary to the discussion of the 
forms of gear teeth it is necessary to define certain terms which 
are used in connection with gears. 

(a) The circular pitch is the distance measured along the 
pitch circle from a point on one tooth to the cor- 
responding point on the next tooth, or the circum- 
ference of the pitch circle divided by the number of 
teeth. 

(6) The chordal pitch is the distance measured on the chord 
of the pitch circle from a point on one tooth to the 
corresponding point on the next tooth. 1 

(c) The diametral pitch is the ratio of the number of teeth 

in the gear to the pitch diameter expressed in inches. 
It should be noted that this is not a dimension, but 
simply a convenient ratio. 

(d) The thickness of the tooth is the thickness measured on 

the arc of the pitch circle. See Fig. 241. 

(e) The tooth space is the width, measured on the arc of 

the pitch circle, of the space between two adjoin- 
ing teeth. 

(f) The backlash is the difference between the tooth space 

and the thickness of the tooth. 

(g) The addendum is the distance from the pitch circle 

1 The chordal pitch is used only in the layout of a drawing or by the 
pattern maker in forming the teeth on a wooden pattern. 



260 TOOTHED WHEELS 

to the outer ends of the teeth. It is shown as 

dimension a in Fig. 241. 
(h) The addendum circle is the circle bounding the outer 

ends of the teeth. 
(j) The dedendum is the distance from the pitch circle to 

the bottom of the tooth space. It is shown as 

dimension b in Fig. 241. 
(j) The dedendum circle is the circle bounding the bottom 

of the tooth spaces. 




FIG. 241. 

(k) Clearance is the amount of space measured on the line 

of centers between the addendum circle of one gear 

and the dedendum circle of the mating gear. 
(I) The face of the tooth is that part of the tooth profile 

lying between the pitch circle and the addendum 

circle. 
(m) The flank of the tooth is that part of the tooth profile 

lying between the pitch circle and the dedendum 

circle, 
(ft) The base circle is an imaginary circle used in involute 

gearing to generate the involutes which form the 

tooth profiles. It is drawn tangent to the line of 

action of the tooth thrust, 
(o) The describing circle is an imaginary circle used in 

laying out cycloidal gearing. There are two such 



SPUR GEARING DEFINITIONS 



261 



circles, usually of the same size, one to generate the 
epicycloid, which forms the face of the tooth, and 
the other to generate the hypocycloid which forms 
the flank of the tooth. 

(p) The angle of obliquity is the angle formed by the line 
of action of the pressure between a pair of mating 
teeth and a tangent to the pitch circle drawn 
through the pitch point. This angle is represented 
by a in Fig. 241. 

(q) The arc of approach is the arc measured on the pitch 
circle from the pitch point to the position of the 
tooth at which contact begins, as shown by AC or 
BC, Fig. 242. 




FIG. 242. 

(r) The arc of recess is the arc measured on the pitch circle 
from the pitch point to the position of the tooth at 
which contact ends, as shown by CK or CL, Fig. 242. 

(s) The arc of action is the sum of the arcs of approach and 
recess. 

(t) The velocity ratio is the ratio of the number of revolu- 
tions of the driver to the number of revolutions of 
the driven gear. 



262 TOOTHED WHEELS 

169. Relation between Circular and Diametral Pitch. By 
definition : 

^. , -p,.. Circumference of Pitch Circle 2?rr ird . 
Circular Pitch = - ^ - ^ -- r - r _ = == p'. 
Number of teeth t t 

Diametral Pitch = -, = p. 
Hence, 

irdt 



That is, the product of the circular pitch times the diametral 
pitch is equal to w. 

170. Rectification of Circular Arcs. In laying out gears it 
is frequently necessary to lay off a line equal in length to a given 
arc. The best geometrical constructions for this purpose are 
those due to Rankin (A Manual for Machinery and Mill Work), 
and are as follows: 

(a) To draw a straight line approximately equal to a given 
circular arc AB, Fig. 243. 





FIG. 243. FIG. 244. 



Join BA and produce it to D making AD = ^ AB. With 
center D and radius DB describe the arc BC cutting the tangent 
AC at C. AC is the straight line required. If the given arc 
subtends an angle greater than 60 it should be subdivided. 

The error varies as the fourth power of the angle A OB, where 
is the center of the circle of which AB is an arc. When the 
angle AOB is 30, AC is less than the arc AB by about i^ioo of 
the length of the arc. 

(6) To lay off on a given circle an arc AB approximately equal 
to a given length, AC, Fig. 244. 

Draw a tangent AC to the circle at A, and make AC equal 
to the given length. Make AD = \ AC. With center D and 



PITCH CIRCULAR ARCS CYCLOIDAL CURVES 263 

radius DC describe the arc CB to cut the circle at B. AB is the 
arc required. 

The error in (6) expressed as a fraction of the given length 
is the same as in (a) and follows the same law. 

171. Generation of Cycloidal Curves. A cycloid is the curve 
described by a point on the circumference of a circle which rolls 
on a straight line, the circle and straight line remaining in the 
same plane. 




FIG. 245. 

If the circle be rolled on the outside of another circle, the two 
circles remaining in the same plane, the curve described is called 
an Epicycloid; when rolled inside the curve described is called 
a Hypocycloid. 

(a) To draw an epicycloid, Fig. 245, let 

ri = the radius of the rolling circle; 
r% = the radius of the fixed circle. 



264 



TOOTHED WHEELS 



Draw through C a circle CD concentric with the fixed circle. 
Evidently, the radius of this circle will equal r \-\-r % and the center 
of the describing circle will always be on CD. Divide the describ- 
ing circle into a convenient number of parts and lay off these 
lengths (Art. 170) on the circumference of the fixed circle. Through 
the latter points of division draw radii and produce them to inter- 
sect the line of centers CD. The points thus located will be the 
successive positions of the center C of the describing circle. 

If the describing circle is drawn in any one of these positions, 
as 2, its intersection b with the circular arc through 2' will be 
a point on the required curve. Other points will be determined 
in the same way. 

A simpler and more accurate method for constructing the curve 
is to produce the arc /4' to intersect the radius through C4 at g 
and to lay off gd =/4'. Other points will be determined in the same 
way. 

This method gives the best results for points of the curve 
near T which is the part of the curve employed in the construc- 
tion of gear teeth. 

(6) The construction of the Hypocycloid is the same as that 

for the epicycloid, except 
that the radius for the line 
of centers is here 7*2 r\ . 

A given hypocycloid 
on a given base circle may 
be described by a point 
on the circumference of 
either of two rolling circles. 
In Fig. 246 APR is the 
hypocycloid described by 
point P of the rolling circle 
CP when rolled on the 
fixed circle ACBFDE. The 
same curve would be 
described by the point P 
considered as a point of 
the rolling circle DP. 

An important case of the hypocycloid is that where the diam- 
eter of the rolling circle is equal to the radius of the fixed circle. 




CYCLOIDAL CURVES CONSTANT VELOCITY RATIO 265 

In this case the hypocycloid becomes a straight line EOF, which 
is a diameter of the circle. 

(c) The construction of the Cycloid is the same as that for 
the epicycloid except that the center of the fixed circle is moved 
to infinity and the circumference becomes a straight line, Fig. 
247. 




FIG. 247. 




FIG. 248. 
In Fig. 248 a and 6 are two pitch circles with centers at and 

s~\l ' rn 

0' respectively. The velocity ratio is then - = 7- Let oT 



266 



TOOTHED WHEELS 



be the radius of any describing circle e such that oT is less than 
O'T. By rolling this circle on the outside of circle a describe 
the epicycloid Tc and by rolling it inside of circle b describe hypo- 
cycloid Td. Let the circles a, b, and c rotate about their respect- 
ive centers in rolling contact. 

At any instant the curves T'c' and T'd' will be in contact at 
the point T' in the circumference of the describing circle since 
by construction, 

arc 7T' = arc TV = arc Td 1 '. 

Referring to Fig. 249 it is evident that when a circle rolls on the 
outside of another circle in the same plane the instantaneous 



V a 




FIG. 249. 

center is the point of contact and hence the normal passes through 
this point. The same is true for the case of the hypocycloid. 
Therefore, the curves must have a common normal and a common 
tangent at the point T', Fig. 248. 

In Fig. 248 the instantaneous center is the point T and the 
common normal is the line T'T at that instant. The same proof 
can be repeated for any other position of the two curves. It 
will be noted that the common normal always cuts the line of 
centers at the point T, which is the pitch point. Hence these 
curves will transmit a motion which will exactly replace the 
rolling action of the pitch circles. 

If the diameter of the describing circle were made equal to 



LAY-OUT OF CYCLOIDAL TEETH 267 

the radius of the fixed circle, the hypocycloid would become a 
straight line, but the two curves would work together as before. 
The proof of this statement is left to the student. 

If the diameter of the describing circle is still further increased, 
the curve generated becomes convex in a direction opposite to 
that of the epicycloid, and it lies on the opposite side of the line 
of centers. In the limit where the diameter of the describing 
circle is equal to the diameter of the pitch circle the hypocycloid 
will have become a point. This point is the basis for pin gearing 
and may be driven by the epicycloid with the same velocity 
ratio as the pitch circles. 

Referring to Fig. 248, if the circles be rolled together, the 
point of contact T f must follow the arc T'T until points T 7 ', c' 
and d f coincide at T and, if the motion is continued, it must follow 
the arc TT", to the point where contact ceases. 

172. Teeth of Wheels. In Fig. 250, Oi and 2 are the centers, 
and a and b are the pitch circles of the driver and driven wheel; 
01 and 02 are the centers of the describing circles whose diameters 
are less than the respective radii of the pitch circles in which they 
roll. Let TV =7^' = the pitch and let T' be the point where 
the teeth quit contact and which is therefore a point on the 
addendum circle. 

If the circles are caused to turn about their respective centers 
in rolling contact the point T will trace the epicycloid Te, which 
will be the face of the driver's tooth, and the hypocycloid Tf, 
which will be the flank of the follower's tooth. 

To complete the addendum of the driver's tooth draw the 
addendum circle through T' ; bisect Td' at m and through m draw 
an epicycloid reversed in direction to intersect the addendum 
circle at n. 

Pn is the part of the flank of the follower that comes in con- 
tact with the epicycloid mn, but in order to make room for the 
point of the driver's tooth, as it revolves, Pn is continued by 
the amount of the clearance, until the root circle is met. The 
profile is usually joined to the root circle by a small circular arc 
or fillet. 

The face of the follower's teeth and the flank of the driver's 
teeth are determined by laying off T", the point where contact 
begins, and proceeding as before. 



268 



TOOTHED WHEELS 



The complete wheels can then be laid out by dividing both 
pitch circles into arcs equal to one-half the pitch, and by then 
drawing curves similar to those already found in their proper 
order through these points. 




(a) The arc of approach determines the addendum of the 

follower's tooth or vice versa. 
(6) The arc of recess determines the addendum of the 

driver's tooth or vice versa. 

173. Relation between Pitch and Arcs of Approach and 
Recess. Having given the diameters of the pitch and describing 
circles and the arcs of approach and recess, it is required to 
determine the limits between which the pitch may vary. 



RELATION BETWEEN ARCS OF APPROACH AND RECESS 269 

Let 7" and T", Fig. 251, be the points where contact between 
the mating teeth begins and ends. Professor Willis has pointed 
out that if the radial line 0\T' be drawn, the distance gc can 
nerver be greater than one-half the thickness of the tooth; in 
other words, the pitch must not be less than four times gc. If 
the pitch is made equal to four times gc the tooth will become 
pointed as indicated at rs. It is also evident that the pitch 
must not be greater than the arc of action, for in that case contin- 




FIG. 251. 

uous contact becomes impossible. The upper and lower limits 
of the pitch are therefore given by the inequality, 

40c p' 5 arc of action. 

174. Example. 

Let the radius of gear a, Fig. 250, be 2J inches; 
the radius of gear 6, Fig. 250, be 1| inches; 
the radius of the describing circles be f inch; 
the arc of approach = TT' = J inch ; 
the arc of recess = TT" = f inch. 

Constructing the teeth with these data, eg is found by measure- 
ment to be |J inch, and c'g' is found to be ff inch. Therefore 



270 TOOTHED WHEELS 

the pitch cannot be greater than f-f| = l| inches or less than 
4XjJ=ii i ncn - Both of these limits should be avoided. The 
pitch of, course, must divide both pitch circles evenly. 

175. Size of Describing Circle. It is evident that a describ- 
ing circle of any size, smaller than the pitch circle in which it 
rolls, may be used to produce a curve fulfilling the required con- 
ditions. 

In Fig. 246 it was shown that, when the diameter of the 
describing circle is equal to the radius of the pitch circle, the 
hypocycloid described becomes a radial line in the pitch circle. 
This condition is undesirable from the standpoint of the strength 
of the teeth. 

It follows that it is desirable to have the diameter of the 
describing circle less than the radius of the pitch circle. This 
condition gives the teeth spreading flanks which are much stronger. 
Where but two wheels are to mesh together, the diameter of the 
describing circle is usually made three-eighths the diameter of the 
pitch circle in which it rolls. 

It is evident that, for a given pitch, the length of teeth required 
for a given arc of action is less, as the describing circles are made 
greater in diameter. 

During the contact the face of the teeth on one wheel acts 
only on the flank of the teeth on the other. Hence the form of 
the face on one wheel does not effect that of its own flanks. There- 
fore there is no fixed relation as to size between the describing 
circles used in the two wheels. 

176. Interchangeable Wheels. In making a set of patterns 
or making cutters for cut teeth, it is desirable that any gear of a 
set shall work with any other gear of the same pitch. When this 
is the case the set is interchangeable. 

The conditions necessary in an interchangeable set of gears 
are that all of the wheels of the set shall have the same pitch and 
that all of the tooth curves shall be generated by the same describ- 
ing circle. The size of the describing circle usually used is such 
that the smallest gear of the set shall have radial flanks. This 
gear is generally taken as having from 12 to 15 teeth. The 

diameter of the describing circle would then be equal to -J-, where 

ZTT 

t = the number of teeth in the smallest gear. 



INTERCHANGEABLE WHEELS EXAMPLE 



271 



177. Example. Fig. 252, drawn half size, is the practical 
solution of the following problem, assuming cast teeth: 
Distance between shaft centers = 12 inches. 
Driving shaft turns at 200 r.p.m. 
Driven shaft turns at 400 r.p.m. 




Diametral pitch =4. 

Diameter of describing circle = fXdiameter of smaller pitch 

circle. 

400 



200+400 

200 
200+400 



X 12 = 8 inches rad. 
X12=4 inches rad. 



272 TOOTHED WHEELS 

Size of describing circle = |X8 = 3 inches diameter. 
Number of teeth in gear =4X16 = 64. 
Number of teeth in pinion =4X8 = 32. 

Circular pitch =^ = 0.7854 inch. 
From Art. 197: 

Addendum = 0.3p' = 0.3 X 0.7854 = 0.2356 inch. 
Dedendum = 0.4p' = 0.4 X 0.7854 = 0.3 1416 inch. 
Radius of addendum circle (gear) =8+0.2356 = 8.2356. 
Radius of addendum circle (pinion) =4+0.2356 =4.2356. 
Radius of dedendum circle (gear) =8- 0.3 14 = 7. 686. 
Radius of dedendum circle (pinion) =4 0.314 = 3.686. 

Lay out the pitch circles and describing circles making the 
lower wheel the pinion. 

Draw the addendum circle of the gear, thus locating T' where 
the teeth will quit contact. Draw the dedendum circle of the 
pinion. Through T' lay down the epicycloid T'c. On the pitch 
circle of the gear, lay off ce = one-half the pitch. Draw the radial 
line OiT', intersecting the pitch circle at g, then since gc is less 
than one-half ce, the case is a practical one. Draw through 
e an epicycloid reversed in direction, thus forming the addendum 
for one tooth. Lay down the hypocycloid T'd, and continue it 
to the dedendum circle, joining the two by a small fillet. On the 
pitch circle of the pinion, lay off dh = one-half the pitch, and draw 
through h a hypocycloid reversed in direction, joining it to the 
dedendum circle as before, thus forming the dedendum for one 
tooth. 

In the same way, the teeth of the driven wheel may be proven 
satisfactory. The addendum of its teeth, and the dedendum of 
the driver's teeth are next determined. 

The pitch points of the teeth are then laid off on the pitch 
circles, and proper curves, bounded by the addendum and deden- 
dum circles, are drawn in. 

178. Rack and Pinion. If we consider the center of one of 
the gears just described as being moved to infinity, the case of 
the rack and pinion, Fig. 253, would result. Thus it is evident 
that all of the deductions of the previous articles apply equally 
well, with obvious modifications, to this case. The faces and 



RACK AND PINION ANNULAR GEARS 



273 



flanks of the rack are cycloids; and in an interchangeable set of 
wheels they are alike. In this case, any gear of the set will work 
with the rack. The motion of the rack must be reciprocating. 




FIG. 253. 

179. Annular Gears. The theory of the forms and the method 
of drawing the tooth outlines for annular wheels, Fig. 254, are 




FIG. 254. 



the same as for spur gears. Here, however, the faces of the pinion 
teeth, and the flanks of the gear teeth are both epicycloids, while 



274 TOOTHED WHEELS 

the flanks of the pinion teeth and the faces of the gear teeth are 
both hypocycloids. 

180. Approximate Methods for Laying-out the Cycloidal 
Tooth. A number of simple methods for the approximate lay- 
out of tooth forms, represented by circular arcs, have been devised. 
Those due to Professor Willis and to Mr. George B. Grant are 
most generally used. These circular arcs, as will be seen, do not 
follow the exact tooth form, but gears with cast teeth, especially 
if the pitch is small, also depart from the ideal form. 

In the case of cut gears, the exact outline is usually used in 
making the cutters. The cutters are also made automatically. 

The method of layout devised by Professor Willis gives cir- 
cular arc tooth profiles which are correct for one point only, and 
which have radii equal to the mean radius of curvature of the exact 
curve. In this system the approximate faces lie entirely within 
the true epicycloid. 

In the method of layout devised by Mr. Grant, called by him 
the " Three-point Odontograph," the circular arcs pass through 
three important points of the exact outlines of the faces. These 
points are, at the pitch line; the addendum line and a point mid- 
way between. Thus the arc devised by Mr. Grant crosses the 
true curve twice, and, as the average error is much less than that 
of the arc devised by Professor Willis, it is to be preferred. The 
student should draw an exact tooth profile to a large scale, and 
superimpose an approximate profile on it for comparison. 

181. Grant's Odontograph for Cycloidal Gears. In Tables 
I and II, which follow, are given the radii for cycloidal teeth as 
worked out by Mr. G. B. Grant. The radii of the circular arcs 
and the radial distances, from their lines of centers to the pitch 
line, as given, are for One Diametral Pitch or for One-Inch Circular 
Pitch. The table corresponding to the kind of pitch adopted 
should be used. 

In laying out the profiles of cycloidal teeth, by circular arcs, 
Fig. 255, draw the pitch, addendum, and clearance circles and 
lay off the pitch of the teeth on the pitch circle, dividing the latter 
properly for the tooth thickness and tooth space. Draw the line 
of face centers, circle B, at a distance a, inside of the pitch line, 
and the line of flank centers, circle C, at a distance e outside of 
the pitch line. 



APPROXIMATE CYCLOIDAL TEETH 
TABLE I 



275 



Number of Teeth 


Divide by the Diametral Pitch 


Exact 


Approx. 


Rad. b 


Dist. a 


Rad. c 


Dist. e 


10 


10 


1.99 


0.02 


- 8.00 


4.00 


11 


11 


2.00 


0.04 


-11:05 


6.50 


12 


12 


2.01 


0.06 


00 


00 


13| 


13-14 


2.04 


0.07 


15.10 


9.43 


15| 


15-16 


2.10 


0.09 


7.86 


3.46 


17| 


17-18 


2.14 


0.11 


6.13 


2.20 


20 


19-21 


2.20 


0.13 


5.12 


1.57 


23 


22-24 


2.26 


0.15 


4.50 


1.13 


27 


25-29 


2.33 


0.16 


4.10 


0.96 


33 


30-36 


2.40 


0.19 


3.80 


0.72 


42 


37-48 


2.48 


0.22 


3.52 


0.63 


58 


49-72 


2.60 


0.25 


3.33 


0.54 


97 


73-144 


2.83 


0.28 


3.14 


0.44 


290 


145-300 


2.92 


0.31 


3.00 


0.38 


Rack 




2.96 


0.34 


2.96 


0.34 





TABLE II 



Number of Teeth 



Multiply by the Circular Pitch 



Exact 


Approx. 


Rad. 6 


Dist. a 


Rad. c 


Dist. e 


10 


10 


0.62 


0.01 


-2.55 


1.27 


11 


11 


0.63 


0.01 


-3.34 


2.07 


12 


12 


0.64 


0.02 


00 


00 


13 


13-14 


0.65 


0.02 


4.80 


3.00 


15| 


15-16 


0.67 


0.03 


2.50 


1.10 


171 


17-18 


0.68 


0.04 


.95 


0.70 


20 


19-21 


0.70 


0.04 


.63 


0.50 


23 


22-24 


0.72 


0.05 


.43 


0.36 


27 


25-29 


0.74 


0.05 


.30 


0.29 


33 


30-36 


0.76 


0.06 


.20 


0.23 


42 


37-48 


0.79 


0.07 


.12 


0.20 


58 


49-72 


0.83 


0.08 


.06 


0.17 


97 


73-144 


0.90 


0.09 


.00 


0.14 


290 


145-300 


0.93 


0.10 


0.95 


0.12 


Rack 




0.94 


0.11 


0.94 


0.11 















276 



TOOTHED WHEELS 



Fig. 255 drawn half size is the solution of the following prob- 
lem; assuming Brown & Sharpe cut teeth: 

Number of teeth =40. 

Diametral pitch = 2. 

Diameter of pitch circle = */ = 20 inches. 

From Table VI, Art. 198: 

Length of addendum = 0.3183p' = 0.3183 X 1.5708 = 0.501 in. 
Length of dedendum = 0.3683p' - 0.3683 X 1.5708 = 0.579 in. 
Radius of addendum circle = 10 in. +0.501= 10.501 in. 
Radius of dedendum circle =10 in. 0.579 = 9.421 in. 




FIG. 255. 

From Table I, opposite 40 teeth: 

0.22 0.22 



Distance a 



Distance c = 



Diametral pitch 2 
0.63 0.63 



= 0.11 inch. 



= 0.3 15 inch. 



Diametral pitch 2 
All tooth faces are circular arcs with centers on circle B. 
All tooth flanks are circular arcs with centers on circle C. 

2.48 2.48 



Radius b 



Radius c 



Diametral pitch 2 

3.'52 _3.52 

Diametral pitch" 2 



= 1.24 inches. 



= 1.76 inches. 



INVOLUTE CURVE 



277 



With these radii and centers on their respective lines of centers, 
draw circle arcs through the pitch points. Terminate the tooth 
profiles at the addendum and dedendum circles, and put in fillets 
at the bottoms of the spaces. 

If the circular pitch is used, the construction is similar to that 
just described, except that the values in Table II must be multi- 
plied by the circular pitch in inches. 

The smallest gear in the set as here given is one having ten 
teeth, while the smallest one for which standard cutters are manu- 
factured is one having 12 teeth. 

182. Involute System. The Involute Curve. The curve 
most commonly used for tooth profiles is the involute of a circle. 
This curve is the 
path traced by a 
point on a line as 
the line rolls on a 
fixed circle. It may 
also be described as 
the path traced by 
a point at the end 
of a string, as the 
string is unwound 
from a circle, the 
string being kept 
taut at all times. 
Thus in Fig. 256 
it will be seen that 
the string or line 
will be tangent to 
the circle in all po- 
sitions, and that the 
length to be laid 
off on the tangents 
must be equal to 
the length of the arc 
(rectified) between 
the point of tangency and the starting point P. 

It is evident that the involute is a special case of the epicycloid 
where the center of the describing circle is moved to infinity, and 




FIG. 256. 



278 TOOTHED WHEELS 

that, therefore, tne point of tangency is the instantaneous center 
of motion for the point P in any position. 
It is to be noted that: 

(a) Any tangent to a circle is a normal to any involute of 

that circle. 
(6) The direction of the velocity of any point P on the 

involute is perpendicular to the normal drawn from 

P tangent to the circle. 

183. Involutes in Sliding Contact. In Fig. 257 a and b are 
two pitch circles with centers at Oi and 62 respectively. The 

s\ rp 

velocity ratio is then -^ T^T' Draw the common tangent ft', 

and through T draw a straight line making any angle a with ft'. 
From the centers Oi and 0% drop the perpendicular OiE and 
2 F upon the inclined line ETF. With OiE and 2 F as radii 
describe the circles a' and b' which by construction must be tan- 
gent to the line ETF, and which will be the base circles for the 
involutes to be used. Two involutes of the circle a' and b r are 
tangent at T. 

An instant later the same involutes will be in contact at T f . 
In Art. 182 it was shown that the normal to one of the involutes 
at the point of contact will be a tangent to the circle a' drawn 
through the point of contact, and that the normal to the other 
involute will be a line drawn from the same point tangent to the 
circle b'. 

Since the curves are tangent at the point T f they have a com- 
mon tangent, and hence the normals must coincide in direction. 
This normal, however, must be tangent to both base circles a' 
and &', and must therefore be the line ETF. As this common 
normal cuts the line of centers in the pitch point T 7 , the velocity 
ratio is constant. 

184. Path of Point of Contact. Referring to Fig. 257, if the 
pitch circles be rolled together, the point of contact T' must follow 
the straight line ETF until points T 7 ', d, and / coincide at T, and 
if the motion is continued, it must follow the same line to some 
point T" where contact ceases. 

185. Teeth of Wheels. In Fig. 257, Oi and 2 are the centers, 
and a and b are the pitch circles of the driver and driven wheel. 



LAY-OUT OF INVOLUTE TEETH 



279 



tTt' is their common tangent and ETF is a straight line making 
an angle i'TE=a with the common tangent. By definition a is 
the angle of obliquity. It should be noted that in standard 
practice the angle of obliquity is made 14 J. This angle was 
probably chosen because the sine of the angle equals 0.25, a pro- 
portion that was easy for the millwright to lay out at the time 
when all gears were cast, a' and 6' are the base circles for the 




FIG. 257. 



profiles, drawn with radii 0\E and 6)2^, 0\E and OzF being per- 
pendicular to the line of obliquity of action ETF. Let Tc=Td' 
= the pitch. On the base circle of draw the involute cicT', inter- 
secting the line ETF at T', and on the base circle b f draw the 
involute did'T", intersecting the line ETF at T". Here T'CI 
is the profile of the driver's tooth and T"d\ is the profile of the 
driven tooth. 

It is evident that: 

(a) Clearance must be allowed for the points of the teeth 
as they revolve. 



280 



TOOTHED WHEELS 



O, 



(b) The involutes cannot extend within the base circle 
from which they are derived. Radial lines are used 
to complete the profiles inside of these circles, these 
lines being joined to the dedendum circles by small 
circular arcs. 

Draw the addendum circles through points T' and T" and 
draw the dedendum circles after making allowance for the clear- 
ance. 

Contact begins at T" and ends at T'. The complete wheels 
can then be laid out by laying off half the pitch around both 
pitch circles, and by then drawing involutes similar to those 

already found in their 
proper order through 
these points. 

In this problem 
the pitch was made 
equal to the arcs of 
approach and recess. 
However, this need 
not be the case. 

186. Relation Ex- 
isting between Height 
of Tooth and Arcs 
of Approach and Re- 
cess. In a pair of in- 
volute gears, Fig. 258, 
FIG. 258. let a and 6 be the pitch 

circles; a' and &' the 

base circles and ETF the line of obliquity of action. Lay down the 
involutes T'C and T'D] then if contact commences at the point T': 

TC=TD = Arc of approach. 
Draw DG normal to the involute T'D, then: 



or snce 




ET=GD, 



triangle 2 ET= triangle 2 GD 
therefore, arc HI = arc EG= TT'. 



ARCS OF APPROACH AND RECESS 



281 



The application is readily seen in Fig. 259. The pitch circles, 
base circles, and line of obliquity are the same as those used in 
Fig. 258. 

Lay off TD = arc of approach, and draw D02 to intersect the 
base circle at /. From 
T mark off 7T' = arc 
HI thus locating point 
T' where contact be- 
gins. A circle through 
T' having 02 as a cen- 
ter will be the adden- 
dum circle required. 

The addendum cir- 
cle of the driver may 
be located in a similar 
manner. 

Note that clear- 
ance must be allowed 
between the adden- 
dum circle of one 
gear and the root 
circle of the other. 

187. Example. Fig. 260, drawn half size, is the practical 
solution of the following problem, assuming cut teeth. 

Distance between shaft centers =12 inches. 
Driving shaft turns at 400 r.p.m. 
Driven shaft turns at 200 r.p. m. 
Diametral pitch = 4. 
Angle of obliquity = 14|. 




FIG. 259. 



n 



400 



200+400 

200 
: 200+400 



X 12 = 8 inches radius. 



X12=4 inches radius. 



Number of teeth in gear =4X16 = 64. 
Number of teeth in pinion = 4 X 8 = 32. 



Circular pitch = j = 0.7854 inch. 



282 



TOOTHED WHEELS 



From Art. 198: 

Addendum = 0.3183 p f = 0.3183 X 0.7854 = 0.25. 
Dedendum = 0.3683 p' = 0.3683 X 0.7854 = 0.289. 
Radius of addendum circle (gear) =8+0.25 = 8.25. 
Radius of addendum circle (pinion) =4+0.25=4.25. 
Radius of dedendum circle (gear) =8 0.289 = 7.711, 
Radius of dedendum circle (pinion) =4 0.289 = 3.711. 



o, 



VEt 




30T.-12PD. 



f 




FIG. 260. 

Lay down the pitch, addendum and dedendum circles for both 
gears, making the upper wheel the pinion, and draw the common 
tangent tTt'. Draw the line of obliquity through T, making an 
angle of 14J with tTt\. Lay down the perpendiculars 0\E and 
0%F and with these lengths as radii draw the base circles a' and 
6'. The addendum circle with center at Oi will cut the line 
of obliquity in the point T' which will, therefore, be the point 



INVOLUTE ANNULAR GEAR 



283 



where the teeth quit contact. Similarly the addendum circle 
with center at 0? will determine the point T" where contact 
begins. 

Step off divisions equal to one-half the pitch around both pitch 
circles a and 6, and through the points thus located draw the 
tooth profiles in their proper order. The tops of the teeth and the 
bottoms of the tooth spaces are evidently circular arcs. 

188. Annular or Internal Gears. As stated in " the Involute 
Gear/' 1 " the internal gear is commonly defined as being an 




Involute Annular Gear and Pinion. 
FIG. 261. 

external gear with the teeth turned inside out." Hence the 
addendum and dedendum are reversed in position and the tooth 
profile is concave instead of convex, see Fig. 261. This condition 
changes the tooth action to some extent, so that in special cases 
it is necessary that the conditions existing be carefully under- 
stood. Where the number of teeth on the pinion is small as 
compared with the number of teeth on the gear, the tooth shape 
1 Published by the Fellows Gear Shaper Co. 



284 



TOOTHED WHEELS 



requires no special consideration. The Fellows Gear Company 
propose as a general rule that the smallest difference between the 
number of teeth in the pinion and the number of teeth in the 
internal gear with which proper tooth action can be obtained 
should be 7 teeth for 20 stub-tooth form, and 12 teeth for full 
length 14| involute form. 

Several points should be noted relative to annular gears: 

(1) The center distance is much shorter than in external 

gears of the same ratio. 

(2) The pitch circles are inclined in the same direction 

resulting in a greater length of tooth contact. The 
sliding action between the teeth in contact is reduced 
on account of the smaller difference in angular 
velocity. 

Fig. 261 shows the layout of an annular gear. 




Pinion Teeth are limi+ed only by 

becoming poini~ccf. \ 

FIG. 262. 

189. Rack. When the pitch circle becomes of an infinite 
diameter, as in a rack, the base circle will also become of an infinite 
diameter, and the involute will become a straight line. Hence 
in a rack the profiles of both face and flank are straight lines and 
these lines are always perpendicular to the line of obliquity. 
While the addendum of a rack tooth is limited as explained in 



INTERFERENCE IN INVOLUTE GEARS 



285 



the following article, the addendum of a tooth on the mating 
gear is limited only by its becoming pointed. Fig. 262 shows 
the layout of a rack. 

190. Interference. Limit of Addendum. In Fig. 263 wheel 
A has 30 teeth and wheel B has 36 teeth, of 3-diametral pitch. 
The proportions follow the Brown & Sharpe standard for 14J 
involute teeth as listed in Art. 198. Let us assume that it is 
desired to lengthen the teeth of the wheels. ETF is the line of 
contact and it determines the base circles a' and b'. The teeth 
of wheel B can be lengthened until the addendum circle of the 
teeth passes through 
point E without 
causing interference. 
If carried beyond 
this line the adden- 
dum of each tooth 
will have to be al- 
tered in shape to 
prevent cutting into 
the root of the tooth 
on the other wheel. 
In the same way the 
teeth of wheel A may 

be lengthened until F IG> 263. 

the addendum circle 

passes through point- F, provided that the teeth do not become 
pointed before this length is reached. 

Points E and F are therefore called interference points and the 
addendum lines through these points are called interference lines. 

In practice it is found that where the difference in the size 
of the gears is great or in the case of a small pinion and a rack 
interference is almost certain to occur. The latter case is shown 
in Fig. 262. If we keep in mind the fact that contact is always 
along the line EF and that it cannot pass beyond these per ts 
in either direction it is evident that the teeth of the rack carr.Dt 
usefully be extended beyond the interference line, FF'. The 
portion between the line FF' and the top of the teeth may be 
altered to clear the teeth of the pinion. 

It should be noted that the interference line usually lies out- 




286 



TOOTHED WHEELS 



side of the addendum line in which case no interference 
occurs. 

191. Approximate Method for Laying-out the Involute 
Tooth. In making an approximate lay-out of the involute tooth, 
Fig. 264, according to the system devised by Mr. G. B. Grant, 
the pitch, addendum, root, and clearance circles should be laid 
down. The teeth should then be stepped off on the pitch circle 
starting at the pitch point. Draw the base line B one-sixtieth 
of the pitch diameter inside the pitch line A. Note that all 
centers are located on the base line B. Using the radius b, Table 
III or IV, draw that part of the tooth profile lying between the 




a. = 



FIG. 264. 

pitch circle and the addendum circle. Next, using the radius c, 
draw that part of the tooth profile lying between the pitch circle 
and the base circle. That part of the profile between the base 
circle and the fillet at the root of the tooth is a radial line. 

NOTE. 

(1) The table is for 14J involute teeth. 

(2) The values given are for 1-diametral pitch and 1 inch 

circular pitch. 

(3) The tables can be used for annular gears the same as 

for spur gears, but care must be taken that the tooth 
of the gear is cut or rounded off to avoid interfer- 
ence. 



APPROXIMATE INVOLUTE TEETH 
TABLE III 



287 





Divide by 




Divide by 




Divide by 


Number 


the Diametral 
Pitch 


Number 


the Diametral 
Pitch 


Number 


the Diametral 
Pitch 


of 




of 




of 

rri j.v 






Teeth 


Had. 6 


Had. c 


Teeth 


Rad.6 


Had. c 


Teeth 


Rad.6 


Rad.c 


10 


2.28 


0.69 


22 


3.49 


2.06 


34 


4.33 


3.09 


11 


2.40 


0.83 


23 


3.57 


2.15 


35 


4.39 


3.16 


12 


2.51 


0.96 


24 


3.64 


2.24 


36 


4.45 


3.23 


13 


2.62 


1.09 


25 


3.71 


2.33 


37-40 


4. 


20 


14 


2.72 


1.22 


26 


3.78 


2.42 


41-45 


4.63 


15 


2.82 


1.34 


27 


3.85 


2.50 


46-51 


5. 


06 


16 


2.92 


1.46 


28 


3.92 


2.59 


52-60 


5. 


74 


17 


3.00 


1.58 


29 


3.99 


2.67 


61-70 


6. 


52 


18 


3.12 


1.69 


30 


4.06 


2.76 


71-90 


7. 


72 


19 


3.22 


1.79 


31 


4.13 


2.85 


91-120 


9. 


78 


20 


3.32 


1.89 


32 


4.20 


2.93 


121-180 


13. 


38 


21 


3.41 


1.98 


33 


4.27 


3.01 


181-360 


21. 


62 



TABLE IV 





Multiply by 




Multiply by 




Multiply by 


Number 


the Circular 
Pitch 


Number 


the Circular 
Pitch 


Number 


the Circular 
Pitch 


of 

m J.'L 




of 

rr* xi_ 




of 

m .1 






Teeth 


Rad.6 


Rad.c 


Teeth 


Rad.6 


Rad.c 


Teeth 


Rad.6 


Rad.c 


10 


0.73 


0.23 


22 


.11 


0.66 


34 


1.38 


0.99 


11 


0.76 


0.27 


23 


.13 


0.69 


35 


1.39 


1.01 


12 


0.80 


0.31 


24 


.16 


0.71 


36 


1.41 


1.03 


13 


0.83 


0.34 


25 


.18 


0.74 


37-40 


1. 


34 


14 


0.87 


0.39 


26 


.20 


0.77 


41-45 


1. 


48 


15 


0.90 


0.43 


27 


.23 


0.80 


46-51 


1.61 


16 


0.93 


0.47 


28 


.25 


0.82 


52-60 


1. 


83 


17 


0.96 


0.50 


29 


.27 


0.85 


61-70 


2. 


07 


18 


0.99 


0.54 


30 


.29 


0.88 


71-90 


2. 


46 


19 


1.03 


0.57 


31 


.31 


0.91 


91-120 


3. 


11 


20 


1.06 


0.61 


32 


.34 


0.93 


121-180 


4. 


26 


21 


1.09 


0.63 


33 


.36 


0.96 


181-360 


6. 


88 



288 



TOOTHED WHEELS 



The profiles of the rack teeth are straight lines inclined at 
14| with the vertical. The outer half of the addendum should 
be drawn by means of a circular arc having its center on the pitch 
line of the rack and having a radius of 2.10 inches, divided by the 
diametral pitch or 0.67 times the circular pitch. 

192. Stub-tooth Gear. While the standard tooth of to-day 
is the involute having a 14J angle of obliquity, its use is not as 
universal, as might be supposed. A form of tooth known as the 
" Stub-tooth " has been successfully applied to automobile drives, 
machine tools, hoisting machinery, etc., and has become estab- 
lished to a degree that many do not realize. 

Stub teeth are made on the involute system, the features of 
this form of tooth being a 20 angle of obliquity (usually), and a 
shorter addendum and dedendum than used for ordinary gears. 

The minimum length of the tooth must be such that continuous 
action is obtained when using the smallest gear of the set, usually 
having twelve teeth. A tooth longer than required results in 
undue fraction and wear. 

The dimensions of stub-tooth gearing as made by the Fellows 
Gear Shaper Company are given in Table V. These dimensions 
are based on two diametral pitches. The clearance is made 
greater than in the ordinary gear tooth system and equals 0.25 
divided by the diametral pitch. 

TABLE V 



Pitch 


Thickness on the 
Pitch Line 


Addendum 


Dedendum 


4/5 


0.3925 


0.2000 


0.2500 


5/7 


0.3180 


0.1429 


0.1785 


6/8 


0.2617 


0.1250 


0.1562 


7/9 


0.2243 


0.1110 


0.1389 


8/10 


0.1962 


0.1000 


0.1250 


9/11 


0.1744 


0.0909 


0.1137 


10/12 


0.1570 


0.0833 


0.1042 


12/14 


0.1308 


0.0714 


0.0893 



The numerator of the fraction is used in obtaining the dimen- 
sions of the thickness of tooth, the number of teeth and the pitch 
diameter. The denominator is used in obtaining the dimensions 



STUB-TOOTH GEARS 



289 



for the addendum and dedendum. As an example, let us choose 
the 6/8 diametral pitch. Here, the numerator signifies that the 
gear is of 6-diametral pitch, while the denominator signifies that 
the addendum is one-eighth of an inch high. A 24-tooth gear 
of this pitch would have a pitch diameter of 4 inches and an out- 
side diameter of 4} inches. 

193. Stub-tooth Gears. Nuttall System. Mr. C. H. Logue 
of the R. D. Nuttall Company has designed a system of stub- 
gear teeth in which the dimensions are based directly upon the 
circular pitch. 

The dimensions are arrived at as follows: 

Addendum = 0. 250 X the circular pitch. 
Dedendum = 0. 300 X the circular pitch. 
Whole depth of tooth = 0.55 X the circular pitch. 
Working depth of too th = 0.50 X the circular pitch. 




40 Tooth Involute. 
Angle of Obliquity 14 



40 Tooth Stub. 
Angle of Obliquity 20. 




12 Tooth Involute. 
Angle of Obliquity 14 




12 Tooth Stub. 
Angle of Obliquity 20. 



FIG. 265. 



194. Comparison between 14J Involute and Stub-tooth 
Gears. The advantages of the stub-tooth gear may be stated 
as follows: 

(1) Greater strength. Fig. 265 gives a clear idea of the 

increase of strength with the increase in the angle 
of obliquity. 

(2) Extreme sliding avoided. At the pitch point there is 

pure rolling contact. At other positions the slid- 



290 TOOTHED WHEELS 

ing is proportional to the distance from the point 
of contact to the pitch point. 

(3) More even wearing contact. 

(4) Interference is a less serious problem in stub teeth. 

195. Interchangeable Wheels. For interchangeable wheels in 
the involute system the only conditions that have to be satisfied 
are a common pitch and the same angle of obliquity for all wheels. 

196. Comparison of Systems. In the involute system the 
distance between centers may be either greater or less than the 
theoretical distance without affecting the velocity ratio. This 
is an advantage not possessed by cycloidal gears. Therefore 
two wheels of different numbers of teeth, turning about one axis, 
can be made to gear correctly with one wheel. This feature can 
be utilized in making differential motions of various kinds. 

The profile of the involute tooth is a single curve, while the 
cycloidal profile is a double curve. Therefore the involute tooth 
is easier to form. 

In the involute system the angle of obliquity is constant and 
therefore the pressure tending to separate the gears is constant. 
Hence, the wear is more uniform. 

In the cycloidal system a convex surface is always in contact 
with one that is concave. Although theoretically there is line 
contact, practically there is surface contact, hence the wear is 
not so rapid as in involute teeth. 

Interference is greater in involute teeth. 

197. Proportions of Cast Te^eth. 1 The proportions of cast 
gear teeth are not standardized. The following proportions in 
terms of the circular pitch have proven satisfactory in actual 
practice: 

Pressure angle or angle of obliquity = 15. 

Length of the addendum =0.3p'. 

Length of dedendum = 0.4p'. 

Whole depth of the tooth =0.7p'. 

Working depth of the tooth =0.6p'. 

Clearance of the tooth = Q.lp'. 

Width of the tooth space = 0.525p'. 

Thickness of the tooth = 0.475p'. 

Backlash = 0.05p'. 

1 Leutwiler, Machine Design. 



PROPORTIONS OF GEAR TEETH 



291 



198. Proportions of Cut Teeth. The formulas proposed by 
the Brown & Sharpe Co. are used more extensively than any 
others and are commonly recognized as the standard for cut 
gear teeth. However, the other systems included in Table VI 
are used sufficiently to warrant their inclusion. The formulas 
due to Hunt apply to short teeth, while those due to Logue apply 
to the stub-tooth system. Proportions for the Fellows' stub- 
tooth system are to be found in Table V. 

TABLE VI 





Brown & Sharpe 


Hunt 


Logue 


Pr6SSUT6 angle 


- 14i 


14| 


20 


Length of addendum 


0.3183;/ 


O.SSy/ 


O'.25p' 


Length of dedendum 
Whole depth of tooth 
Working depth of tooth . . . 
Clearance 


0.3683// 
0.6866p' 
0.6366p' 
0.05p' 


0.30p' 
0.55p' 
O.SOp' 
0.05p' 


O.'30p' 
0.55p' 
0.50p' 
0.05' 


Width of tooth space 
Thickness of tooth 


0.50?' 
0.50p' 


0.50p' 
0.50p' 


0.50p' 
0.50p' 



199. Bevel Gears. It has been shown in Art. 162 that the 
contact surfaces of a pair of bevel friction wheels are frusta of a 
pair of cones whose vertices are at the point of intersection of the 
axes. In bevel gearing these surfaces are the pitch cones and 
teeth are formed on them in a manner analogous to the methods 
used in spur gearing. It should be noted that : 

1. The involute form of tooth is almost universally used 

for bevel gears. 

2. Bevel gears are made in pairs and are not interchange- 

able. 

3. Bevel gears are either cast or cut. The casting process 

is similar to the casting process for spur gears, but 
the cutting process is much more difficult because 
of the changing size and form of the tooth from 
one end to the other. 

4. Bevel gears are divided into three classes: 

(a) Plain right-angle bevel gears. Center angle 
= 90. In case the shafts are at right 



292 



TOOTHED WHEELS 



angles, and the gears have equal pitch diam- 
eters, they are known as Miter Gears. 

(b) Acute-angle bevel gears. Center angle less 

than 90. 

(c) Obtuse-angle bevel gears. Center angle more 

than 90. 

5. The velocity ratios are the same as those of cylindrical 

gears having the same form of pitch line. See Art. 162. 

6. While the pitch surfaces of bevel gears are generally 

made circular cones, they may be elliptical, internal, 
or irregular. 

200. Form of Teeth. The correct tooth profile in bevel gear- 
ing is the spherical involute EF, Fig. 266. Consider the base 




FIG. 266. 



cone OHI to be enclosed in a thin flexible covering such that the 
edges meet along the line OE. Taking hold of one edge, and keep- 
ing the covering taut, unwrap the cover. The involute surface 
OEF is thus generated by the element OF, Every point of the 



BEVEL GEARS 



293 



involute EF is equidistant from the center and therefore the 
curve EF must lie on the surface of a sphere HAL Hence the 
curve EF is called a spherical involute. The difficulty of laying 
out tooth curves on a spherical surface is apparent since the sur- 
face of the sphere cannot be developed. An approximate method 
first published by Tredgold, is commonly used. In this method a 
conical surface CBD is substituted for the spherical surface CAD. 
The cone CBD, which is called the back cone, is tangent to the 
sphere at the circle CD. As shown in Fig. 266, no appreciable 
error is introduced by this substitution. 

201. Layout of Bevel Gears. Fig. 267 shows the layout of 
a pair of right bevel gears, having 18 teeth on the driver and 21 




FIG. 267. 

teeth on the follower. The velocity ratio is then 7 to 6. The 
pitch cones and pitch lines are laid out according to the method 
described in Art. 162. Locate point / the vertex of the driver's 
back cone. 

It is evident that the surface which contains the tooth pro- 
file has a radius of curvature equal to IH. The profile is then 
laid off on a circle of that radius in precisely the same manner as 
that used for spur gearing. While the tooth is shown in its true 
size and shape it is correct for the large end only. It is necessary 



294 TOOTHED WHEELS 

to have the profile at the other end of the tooth in order to 
determine the form for the entire length. Let DK equal the face. 
Develop the back cone LOK and proceed as before. The 
addendum and dedendum in this case are the distances KM and 
KN respectively. Draw PiM 1} P 2 M 2 , QiNi, Q 2 N 2 , DiKi, D 2 K 2 
converging to the center at /. The intersections MI, KI, NI, 
M 2} K 2 and N 2 will be points on the required profile. 

To draw the front view of the gear project points P, D, Q, M, 
K and N upon the vertical axis through point 0. Through the 
points thus located, draw circular arcs which will be the addendum, 
root and pitch circles for the profiles. Make P^P^ = P\P 2 ' 1 
DsD4 = DiD 2 , etc., and complete the profiles. The remaining 
teeth should be laid out in a like manner. 

202. Transmission between Non-intersecting Shafts. The 
transmission of rotary motion between non-parallel, non-inter- 
secting shafts has been accomplished in four ways: 

1. Spiral or Helical Gears 1 n 

r TTT- /-i r Screw Gearing. 

2. Worm Gears J 

3. Skew Bevel Gears. 

4. If the shafts are located far enough apart two pairs 

of ordinary bevel gears may be installed. 

203. Screw Gearing. Screw gearing is a term applied to all 
classes of gears in which the teeth are of screw form. The pitch 
surfaces are cylindrical as in spur gearing, but the teeth are not 
parallel to the axis. Each tooth winds helically like a screw 
thread. Two classes arise: 

1. Helical Gears. When the number of threads or teeth 

on the cylinder is increased to such an extent that 
any one thread does not make a complete turn, 
the resulting gear is called a helical or spiral gear. 

2. Worm Gears. In Fig. 269 cylinder a may be provided 

with one or more threads. If from one to three 
are used the cylinder and threads together are 
called a worm. 

204. Helical Gears, for Parallel Shafts. A helical gear is 
simply a spur gear with the teeth twisted. If a gear should be 
built up of a number of disks and ,if after the teeth had been cut, 



STEPPED GEARS HELICAL GEARS 



295 



across all of the disks at one time, they were then slightly twisted 
relatively to one another, the gear would be made up of a series 
of steps, or in other words it would be a stepped gear. Again if 
the disks were made thin enough lines joining corresponding 
points of the stepped tooth, profiles would be helices and the gear 
would be called a helical gear. 

Single helical gears produce an end-thrust in each shaft. This 
objection is overcome by the use of a double-helical or herring- 
bone gear. 

The profiles of the teeth on the reference plane follow the 
involute form, stub teeth being most generally used. 





FIG. 268. 

Where smoothness of action at high speeds or when a great 
change in velocity is required, helical gears are usually installed. 

The teeth may be either right- or left-handed. In the case 
of the herringbone or of the Weust gear, the teeth of the Weust 
gear are staggered both right- and left-hand teeth are used on 
the same wheel. 

In Fig. 268 let AB and CD be corresponding helices on the 
pitch surfaces of two consecutive teeth. Then EF, measured 
around the surface at right angles to the axis, is the circular pitch. 
The distance EG, measured at right angles to the teeth, is called 
the normal pitch. The distance ED, measured parallel to the 
axis, is called the axial pitch. The angle a denotes the helix 



296 TOOTHED WHEELS 

angle, that is, the angle between the helix and a line parallel to 
the axis. In mating gears the normal pitch and circular pitch 
must be the same. The velocity ratio follows exactly the same 
laws as for ordinary spur gears. 

205. Advantages of Double Helical Gears. When compared 
with spur gears helical gears have the following advantages: 

(a) The tooth comes into contact gradually, the action 

beginning at one end and working over to the other 
end. In spur gearing the contact takes place across 
the entire face of the tooth at once. 

(b) The face of the gear is made long enough so that more 

than one tooth is always in action. This gives a 
continuity of action that is in no wise dependent 
upon the number of teeth in the pinion as in spur 
gearing. 

(c) Due to the continuity of action and the nature of the 

contact the load is transferred from one tooth to 
another without shock. This also gives smooth- 
ness of motion and equalizes the wear, hence the 
tooth profile is not altered. 

(d) High gear ratios may be used. 

(e) All of the advantages of single helical gears are retained. 

In addition to this double helical gears eliminate 
end-thrust. 

(/) Since vibration and noise are eliminated, double helical 
gears may be run at high pitch line speeds. 

206. Helical Gears for Non-parallel, Non-intersecting Shafts. 
As noted in Art. 202, helical gears are used to transmit motion 
between non-parallel, non-intersecting shafts as illustrated in 
Fig. 269. When thus used the gears give point contact only. 

In mating gears the normal pitches, but not necessarily the 
circular pitches, must be the same. The radii of the pitch cylin- 
ders have no direct influence on the velocity ratio, the r.p.m. of 
the shafts being inversely as the number of teeth. 
Let: 

N a and N* = Speed of shafts A A and BB in r.p.m. 
n a and n b = Number of teeth in wheels A A and BB respect- 
ively. 



HELICAL GEARS EXAMPLE 



297 



r a and r b = Radii of pitch cylinders A A and BB respectively. 

a = Helix angle of wheel A A. 

/? = Helix angle of wheel BB. 

6= Angle between shafts. 

d = Shortest distance between shafts. 
p c = Circular pitch. 
p n = Normal pitch. 
Then referring to Fig. 268: 
p n = p c cosa. 

_ 2irr _ 2irr cos a 

/t' "~~ 

PC Pn 

Referring to the two wheels shown in Fig. 263 we have: 
2wr a cos a 



n b 



Pn 

2-irri, cos |3 



Pn 



From which: 



cos a 
cos |8 



N 



Also: 



207. Example. The following data apply to a pair of helical 
gears having pitch cylinders similar to those shown in Fig. 269. 





FIG. 269. 

Angle between shafts . . . = 70. 

Velocity ratio =3 to 1. 

Distance between shafts = 18 inches. 



298 TOOTHED WHEELS 

Two methods are open for attacking the problem : 

(a) We can assume the angle of helix for each cylinder 

arid solve for the pitch radii. 

(6) We can assume the pitch radii and solve for the respect- 
ive angles of helix. 

The first method is the one usually followed. 

It is at once apparent that any number of solutions may be 
obtained since the shaft angle of 70 can be split into an infinite 
number of combinations. 

First solution: 

Assume a =0 = 35. 

From the statement of the problem Tr = 3. 

Then: 

N a r b cos 35 n 

Tr = 3 = , = or r b = 3r a . 

N a r a cos 35 r a 

But: 

r a n = 18 inches. 

Therefore 

r a = 4| inches ; r b = 13^ inches. 

Second solution: Let 

= 30; 0=40. 
Then: 

N a Q r&COS40 A QQ/lA r6 Tb 3 QQO 

YP = 3 = ^=o = 0.8846 or = _ O0 ._ = 3.39. 
Ni, r a cos35 r a r a 0.8846 

As before 

= 18 inches. 



Therefore 

r a = 4 . 1 inches ; r& = 13 .9 inches. 

208. Worm Gearing. When it is desired to obtain high-speed 
reductions between non-intersecting shafts making an angle of 
90 with each other, worm gearing is commonly employed. Figs. 
270 and 271 show the two classes of worm gearing in common use. 

The worm is simply a special form of helical gear in which the 
angle of helix is so large that the tooth becomes a thread winding 
entirely around the pitch cylinder. 

Each tooth curve of the worm wheel is a short portion of a 
helix, the angle of which is the complement of that of the worm. 



WORM GEARING 



299 



In a worm and wheel combination the worm is usually the 
driving element. When it is desired to have the wheel drive 
the worm it is necessary that the angle of the helix of the wheel 
be made greater than the angle of repose of the materials in con- 
tact. This property of the mechanism is used as a locking device 
in some elevators and other classes of machinery in that it pre- 
vents the machine from running backward. 

209. Circular Pitch, Axial Pitch, Pitch. Since worms were 
formerly universally cut in lathes the circular pitch system is 
used in worm and wheel calculations. 




FIG. 270. 

The pitch of worm threads, when measured parallel to the 
axis of the worm, is called Axial Pitch. This, of course, must 
equal the circular pitch of the worm wheel. When the worm is 
straight or of a cylindrical shape, Fig. 270, the axial pitch is con- 
stant for all points of the threads. In the case of the Hindley 
worm gear, Fig. 271, the worm is made smaller at its center than 
at the ends, and the axial pitch varies for every point, since the 
angle of helix changes constantly throughout the length of the 
worm. 



300 



TOOTHED WHEELS 



The term pitch refers to the distance between corresponding 
points on adjacent threads, measured in an axial direction. 

If measured at right angles to the threads it is called the nor- 
mal pitch. 

The terms two-pitch and three-pitch when used in connection 
with worm gearing refer to the number of threads per inch meas- 
ured axially on the worm. 




FIG. 271. 

Lead refers to the axial distance traversed by a point in one 
complete revolution of the worm. 

210. Velocity Ratio. In worm gearing the velocity ratio does 
not depend upon the diameter of the gears, but is the ratio between 
the number of threads on the worm and the number of teeth on 
the worm wheel. Thus, for a single thread worm the velocity 
ration of the worm to the wheel is the reciprocal of the number 
of teeth on the worm wheel. For example, if a wheel has 36 
teeth it would require 36 turns of the worm to turn the wheel 
once and the velocity ratio would be ^. 



WORM GEARS SKEW BEVEL GEARS 301 

For two-thread and three-thread worms the velocity ratio 
would be A = T and A = T2 respectively. 

Thirty-one teeth, with few exceptions, is the least number of 
teeth that it is advisable to use on a worm wheel, in order to avoid 
undercutting. Hence, if the velocity ratio is less than thirty the 
worm must be provided with more than one thread. For example, 
let the worm shaft make 18 revolutions while the wheel shaft turns 
once. Two solutions are possible: 

Teeth on worm wheel 36 : 54 

Teeth on worm 2 : 3 

211. Comparison of Hindley and Cylindrical Worm Gears. 
The efficiency and load carrying capacity of the Hindley worm 
gear is slightly better than that of the cylindrical worm gear. 
However, the use of the Hindley gear presents the following 
restrictions : 

(1) The center distance of the worm and gear must be 

exact. 

(2) The worm axis must be in proper alignment with the 

gear. 

(3) The plane of the gorge circle of the worm must be 

located directly over the center of the wheel. 

Cylindrical worm gears are often made longer than is required. 
This construction allows the worm to be moved along the shaft, 
thus presenting a new surface in the event of wear. 

It should be noted, Fig. 270, that the teeth of the worm on the 
centra] section are those of an involute rack, while those of the 
wheel are of the involute form. The pressure angle used in lay- 
ing out the teeth is generally 14 J. 

212. Skew Bevel Gears. When the distance between the 
shafts is small skew bevel gears alone will answer the purpose. 

Nearly all of these gears are cast from patterns, the cut gears 
of this type not being even good approximations of the correct 
forms. 

No form of tooth has yet been found that can be cut and still 
possess the qualities of strength, reversibility and low pressure 
angle. 



302 TOOTHED WHEELS 

The three main systems so far attempted are: 

(a) Willis Epicycloidal System. The tooth profiles are here 
generated by a hyperboloid rolling inside of one and 
outside of the other pitch hyperboloid. Prof. 
MacCord has shown that teeth so generated are not 
tangent in all positions. 

(6) Olivier Involute System. The tooth profiles are here 
helical con volutes. It is difficult to find a satis- 
factory position for the teeth since they vanish at 
the gorge and since the obliquity of action increases 
rapidly as we leave the gorge. 

(c) Beale Skew Gear. This is a modification of the Olivier 
system in which the teeth do not vanish at the 
gorge. The teeth are undercut and are hard to 
make. 

213. Comparison of Systems. 

(a) Spiral or Helical Gears. Gears of this type work satis- 

factorily if the center distance is slightly altered or 
if either of the gears is shifted slightly along the 
shaft. However, they wear rapidly because of point 
contact. 

(b) Worm Gears. A worm mating with a straight-faced 

helical gear has the same advantages and disadvan- 
tages as spiral gears. A straight worm mating with 
a hobbed wheel gives line contact. In this case, 
however, the shafts and gears must be definitely 
fixed in position. 

(c) Skew Bevel Gears. These gears have straight teeth 

and give line contact. They are little used, however, 
since it is very difficult to produce correctly shaped 
teeth. 

(d) If the shafts are located far enough apart two pairs 

of ordinary bevel gears may be installed. 



CHAPTER XI 
CAMS 

214. General Principles. A cam is a machine element, so 
shaped that, by its oscillating or rotating motion, it gives a pre- 
determined motion to another link called the follower. 

It should be noted that: 

(1) The cam is usually the driving link. 

(2) The velocity of the cam is generally uniform. 

(3) The motion of the follower is either reciprocating or 

angular about a fixed axis. In either case the 
follower usually moves in a non-uniform, irregular 
manner. 

(4) Equal movements of the cam are not usually accom- 

panied by equal movements of the follower. 

(5) Line contact is almost always present between the cam 

and follower. 

(6) Cams are used to produce motions not easily gotten 

by other means. 

(7) To decrease wear, the irregularity is usually confined 

to the cam and the follower is provided with a roller 
whenever possible. 

(8) Cams in which the follower is compelled to follow a 

fixed predetermined path due to the nature of the 
cam groove are called positive motion cams. 

When springs are used to return the follower to 
its original position they must be strong enough 
to perform their work properly. In this case the 
effort necessary to compress or extend the spring 
must be added to the effort necessary to do the 
work. 

In some cases springs are necessary but positive 
motion cams are to be preferred. 
303 



304 CAMS 

(9) Rotary cams are divided into two general classes: 

(a) Cams in which the only requirement is that the 
movement of the follower shall begin at a 
certain position of the cam shaft and that 
the movement of the follower shall end at 
a second position of the cam shaft, the 
follower having moved a certain distance. 
(6) Cams in which the follower shall at all times 

occupy predetermined positions. 
Cams of class (a) are far more numerous and 
they allow greater liberty in the design of the 
cam groove. 

Cam grooves of class (6) can be altered some- 
what by increasing the diameter of the cam. 

215. Layout of Cams. The first step in designing a cam is 
to determine the required movement of the follower at the point 
where the work is to be done in order that the given operation 
may be performed. 

The movement thus determined should be plotted full size 
where possible as shown in Fig. 272. The height of the chart 




360 



FIG. 272. 

should be the same as the total motion of the follower. Since 
the cam chart is the development of the cam equal divisions on 
the pitch line represent equal angular displacements of the cam. 
Through the points of division thus determined vertical lines are 
laid down on which are laid off from the base line the total move- 
ment for the chosen position. The divisions thus located are 
numbered and the same measurement and number is placed 
on the left-hand vertical for use in the cam layout, as shown by 
the line AB in Fig. 272. 

The irregular line joining the points, thus located, is known 
as the base curve. It will be seen that the rising and falling of 
the base curve relative to the base line represents the movement 



BASE CURVES 305 

of the cam roller or follower. Hence, in the cam the base curve 
plays a part analogous to that of the pitch surface in gears. The 
angle which the base curve makes with the base line is known 
as the pressure angle. It may be noted that the velocity of the 
follower is proportional to the tangent of this angle. Wherever 
possible, the maximum pressure angle of the set on the working 
stroke should not exceed 30. The base curve thus determines 
the motion of the follower and also the pressure angle. 

In the preceding discussion but one cam was considered. In 
many machines several cams, whose functions are dependent on 
each other, are used. When this is the case the base curves of 
all cams are usually plotted on the same chart, which is called a 
timing diagram. These diagrams are of value in determining 
the proper sequence of events in the mechanism and also in provid- 
ing against possible interference. All cams of a set usually have 
the same diameter, since this facilitates the layout. 

216. Base Curves. Where the only requirement is that the 
movement of the follower shall begin at a certain position of the 
cam shaft and that the movement of the follower shall end at a 
second position of the cam shaft, the follower having moved a 
certain distance, the choice of a suitable base curve is left to the 
designer. 

The following base curves are in common use: 

(a) Straight Line. This base curve gives an abrupt starting 

and stopping velocity, which produces a consider- 
able shock in the follower. After starting, the ve- 
locity of the follower is constant, hence the acceler- 
ation, while theoretically infinite at the start and 
end of the stroke, is zero during the stroke. 

(b) Straight-line Curve Combination. This base curve is 

similar to the straight line with the exception that 
the ends are rounded off. The addition of the 
curve does away with the actual shock of the straight 
line, but it gives a very sudden action. The straight 
line and straight-line curve combinations are objec- 
tionable, and their use should be avoided. 

(c) Crank or Harmonic Motion Curve. The construction 

of this curve is shown in Fig. 273. AB is the total 
motion of the follower and A \A^ is the developed 



306 



CAMS 

circumference of the pitch circle of the cam between 
two chosen positions of the cam shaft. Draw a 
semicircle on the line CD and divide the semi- 
circumference into any number of equal parts. 
Divide the line A\A<i into the same number of equal 
divisions and at the points thus located erect per- 
pendiculars. These parts will represent equal inter- 
vals of angular displacement of the cam, and if the 
cam is moving with uniform angular velocity these 
parts will also represent equal intervals of time. 
The construction is shown in the figure. This 
curve gives harmonic motion to the follower, that 
is, the velocity follows the law V = ceo sin eo where 
c is one-half the travel of the follower. The acceler- 




FIG. 273. 



FIG. 274. 



ation is zero at the middle of the stroke. On either 
side of the center the acceleration increases accord- 
ing to the law A = ceo 2 cos eo. 

(d) Parabola. Fig. 274 shows the case where the follower 
is to rise half the distance AB with uniform positive 
acceleration, and to complete its travel to B with 
uniform retardation, that is, with uniform negative 
acceleration. The curve is made up of parts of 
two parabolas, A%E and EB\. The usual and most 
convenient construction for drawing the parabola 
is shown in the figure. In this construction, the 
line EC is divided into any convenient number of 
equal parts and each of the division points is con- 
nected by a straight line to A^. The line A^G is 
divided into the same number of equal parts and 
at each division point a perpendicular is erected. 
The intersections 1, 2, 3, etc., are points on the 
required curve. 



PRESSURE ANGLE FACTORS 307 

(e) Elliptical Cwrw. The method of constructing this 
curve is similar in every respect to that used for 
the harmonic motion curve, the only difference 
being the substitution of an ellipse for the semi- 
circle shown in Fig. 273. The single advantage 
of this curve is that it gives slower starting and 
stopping velocities to the follower. The velocity 
at the center of the stroke is higher and the acceler- 
ation is variable, reaching its maximum at the ends 
of the stroke. 

Mr. W. B. Yates recommends a ratio of the ver- 
tical axis of the ellipse to the horizontal axis of 1 
to If. If this ratio be increased the follower will 
start and stop with less acceleration while the ve- 
locity at the center will be considerably higher. 

217. Maximum Pressure Angle Factors. Fig. 275 shows a 
layout of the base curves in common use, plotted for a maximum 
slope of 30. 




FIG. 275. 

AB represents the total motion of the follower, and this 
length is taken as unity for comparison. It will be noted that 
the length of the base curves vary widely. For example, take 
the case of the crank curve for a maximum pressure angle of 30. 
Here the length required is equal to 2.72 times the throw or travel 
of the follower. In the case of the elliptical curve for the same 
maximum pressure angle the length required is 3.95. 

Table 1 1 and the accompanying chart 1 gives these values 
for various pressure angles and for various types of curves. 

1 From "Cams" published by John Wiley & Sons, Inc. Printed by 
permission of Franklin De. R. Furman. 



308 



CAMS 



TABLE I 
TABLE OF PRESSURE ANGLE FACTORS 



No. 


Name of Base Curve 


Factor for Maximum Pressure Angle 


20 


30 


40 


50 


60 


1 


All logarithmic 


No ge 


neral f 


actors 






2 


Logarithmic combination 


No ge 


neral f 


actors 






3 


Straight line. 


2.75 


1.73 


1.19 


0.84 


0.58 


4 


Straight line-curve combination 
(Radius equal to \ followers mo- 
tion) 


2.92 


2.00 


1.56 


1.31 


1.16 


5 


Elliptical curve. (Ratios of semi- 
axes 2 to 4) 


3.32 


2.17 


1.45 


1.00 


0.68 


6 


Straight lire-curve combination. 
(Radius equal to followers mo- 
tion) 


3.10 


2.27 





1.77 


1.73 


7 


Crank curve 


4.43 


2.72 


1.87 


1.32 


0.91 


8 


Parabola 


5.50 


3.46 


2.38 


1.68 


1.15 


9 


Tangential curve Case 1. (Length 
of straight surface not specified) 


No ge 


neral f 


actors 






10 


Circular curve Case 1. (Symmetric- 
al circular arcs) 


5.67 


3.73 


2.75 


2.14 


1.73 


11 


Elliptical curve. (Ratio of semi- 
axes 7 : 4) 


6.25 


3.95 


2.75 


1.95 


1.35 


12 


Cube curve Case 1. (Unsymmetric- 
al cube curve) 


6.68 


4.20 


2.90 


2.04 


1.40 


4.13 + 
2.55 


2.60+ 
1.60 


1.79 + 
1.11 


1.26 + 
0.78 


0.87+ 
0.53 


13 


Cube curve Case 3. (Symmetrical 
cube curves) 


8.22 


5.20 


3.56 


2.52 


1.73 


14 


Circular curve Case 2. (Unsym- 
metrical circular arcs) 


5.67 


3.73 


2.75 


2.14 


1.73 


4.26+ 
1.41 


2.80+ 
0.93 


2.06+ 
0.69 


1.60+ 
0.54 


1.30+ 
0.43 


15 


Cube curve Case 2. (Cube curve 
and circular arc) 


7.60 


4.83 


3.37 


2.42 


1.73 


6.18 + 
1.42 


3.90 + 
0.93 


2.68 + 
0.69 


1.89 + 
0.53 


1.30+ 
0.43 


16 


Tangential curve Case 2. (Length 
of straight surface specified) 


No ge 


neral f 


actors 







SIZE OF CAMS 



309 



218. Size of Cams. If a table similar to Table I is not avail- 
able, a tedious cut and try method must be employed to determine 
the length of the base line. By the use of Table I the length of 
the base line can be determined as follows: 

f=Cam factor as given in Table I; 
t = Total motion or throw of follower; 
n = Degrees of rotation of the cam; 
r Radius of pitch circle of cam. 




Cam Factor* 



As an example, consider the crank curve, Fig. 273, and assume 
a maximum allowable pressure angle equal to 30. Since the 
throw A B is always known, multiply this value by the cam factor 
/, in this case 2.72, to obtain the length BiB 2 which, in turn, isknown 

as a fraction of the entire circumference, equal to 



Hence: 



n 



QAf) 

Circumference = fXtX . 
n 



or 



n 
360/f_ /< 

' O/.O . 

n 



310 CAMS 

219. Rotary Cams. Radial Follower. Many cams of this 
type are laid out with entire disregard for technical design. While 
satisfactory results may often be obtained by experienced designers, 
a great objection remains, that is, there is no control of the velocity 
and acceleration of the follower rod. 

Fig. 277 shows the theoretical lay-out of a one-step rotary 
cam having a radial follower and conforming to the following 
specifications : 

Cam to be based on the harmonic motion base curve. 

Maximum pressure angle = 30. 
The follower shall move: 

Up 4 units in 90. 
Rest through 45. 
Down 4 units in 90. 
Rest through 135. 

Solution : 

From Table I, /=2.72 per unit rise. 
Rise =4 units. 



Therefore since the rise takes place in 90. 
umfer< 
chart. 



onrv 

Circumference = 4 X 2.72 X - = 43.52 units = length of 



Radius of pitch circle = ^ =6.93 units. 

ZTT 

Fig. 276 shows the lay-out of the cam chart. 
Lay-out of cam. 

With 0, Fig. 277, as a center and a radius of 6.93 units 
draw the pitch circle PCDE and divide the area into 
divisions as follows: PC = 90; CD =45; Z)# = 90; and 
#P=135. Divide angle POC into 6 equal angles by 
drawing radial lines. Lay off APB equal to APB, Fig. 
276, and through the division points 1, 2, 3, etc., projected 
on A'P'B' draw circular arcs having as a center. The 
intersections of these arcs with the corresponding radial 
lines give points on the required cam surface. 

The maximum angle occurs at the point where the pitch circle 
cuts the cam surface in practically all cases. This pressure angle 



ROTARY CAM LAY-OUT 



311 



is the angle between a normal and a radial line drawn at the point 
under consideration. In this case the maximum pressure angle 
is at 3' and is equal to 30. 




-10- 
f* A' 




FIG. 277. 




FIG. 278. 

i 

220. Working Surface of Cams. If the follower is fitted with 
a roller instead of the wedge, as shown in Fig. 278, then point A, 
Fig. 277, becomes the center of the roller shaft. Hence the work- 



312 CAMS 

ing surface of a cam is always smaller than the pitch surface, 
except in the case of the wedge follower,- when the working sur- 
face and pitch surface coincide. The working surface is gotten 
by drawing circular arcs with consecutive points of the pitch 
surface as centers, the radius of the arcs being equal to the radius 
of the roller. The working surface is then the envelope to the 
circular arcs. Fig. 278, derived from Fig. 277, shows the con- 
struction. 

221. Two-step Cam. In the case of a two-step cam, or of a 
cam having unequal pressure angles on the two strokes, the pitch 
circle used should always be the larger of the two calculated. 
For this reason the maximum pressure angle will equal that speci- 
fied on one step and will be smaller than specified on the other. 

As an example, let it be required to determine the diameter of 
the pitch circle and the pressure angles for a one-step radial cam 
whose specifications follow : 

Cam to be based on the parabolic base curve. 

The follower shall move: 

Up 3 units in 120. Max. pressure angle not to exceed 20. 
Down 3 units in 90. Max. pressure angle not to exceed 40. 
Rest through 150. 

Required radius of pitch circle for working stroke 

360/E = 360X5.5X3 

2X3.14X120" 



Required radius of pitch circle for return stroke 

360X2.38X3 
= 2X3. 14X90 = 

Hence the pitch circle will have a radius of 7.88 units, and the 
maximum pressure angle on the working stroke will be 20. 

To find the maximum pressure angle on the return stroke sub- 

360 ft 
stitute 7.88 for r in the formula, r== , and solve for/ which 



in this case will equal 4.126. Referring to Table I we find this 
value to correspond to a maximum pressure angle of approxi- 
mately 26. 

In case of a two-step cam the calculations for each step should 
be carried out. The calculations should then be compared, using 



ROTARY CAM. OFFSET FOLLOWER 



313 



the longest chart dimension as a standard. In a large number of 
cases the base chart length will have to be changed in order that 
no pressure angle shall exceed the maximum specified. 

222. Size of Follower Roller. The radius of the follower 
roller should always be less than the shortest radius of curvature 
of the pitch surface. If this rule is followed a smooth curved 
working surface will result, and the follower will always have the 
motion for which it was designed. 

223. Rotary Cam. Offset Follower. The method of con- 
struction for cams of this type is entirely different from that for 




FIG. 279. 

cams having a radial follower. The pitch circle here serves no 
purpose other than that of obtaining a cam of reasonable size 
and will be omitted in this discussion. 

Fig. 279 shows the theoretical lay-out of a one-step rotary 
cam having an offset follower which moves in a straight line. 
The specifications follow: 



314 CAMS 

Cam to be based on the harmonic motion base curve. 
The follower shall move: 

Up 3 units in 90. 
Down 3 units in 90. 
Rest through 180. 

The lowest point of the followers motion to be two units to 
the right and four units above the cam shaft center. 

Solution. Draw OX parallel and OY perpendicular to the line 
of the follower's motion. With as a center and with a radius 
of 2 units draw the circle C-I-J. The center line of the follower's 
motion produced will be tangent to this circle at all times. Divide 
the arc C-I into 6 equal divisions at points C, D, E, etc., and at 
these points draw tangents to the arc. By the usual construction 
as shown, locate the divisions on the line AB which is equal to 
the stroke. The construction for the pitch surface of the cam is 
completed by making D1' = C1; E2' = C2; F3' = C3, etc., and 
by drawing a smooth curve through the points A', 1', 2 f , etc., 
thus located. 

The working surface is next located by choosing a suitable 
radius for the roller and using the pitch surface as a line of centers 
in laying down circular arcs to which the working surface is the 
envelope. The addition of a suitable hub and keyway com- 
pletes the cam. 

The angle a, Fig. 279, is the pressure angle at the lowest point 
of the stroke. 

224. Rotary Cam. Swinging Follower. Cams of this type 
are usually constructed according to one of two systems : 

(a) Those in which the follower arc if continued would 

pass through the center of the cam ; 
(6) Those in which the extremities of the follower arc lie 

on a radial line. 

In the first class the pressure angle will be greater on one stroke 
than on the other, hence cams of this type are used where heavy 
work is to be done on one stroke only. In cams of the second 
class the pressure angles on both strokes are nearly equal, hence 
cams of this type will be chosen where the same amount of work 
is to be done on both strokes. 



ROTARY CAM. SWINGING FOLLOWER 



315 



Fig. 280 shows the lay-out of a rotary cam having a swinging 
follower equipped with a roller and conforming to the following 
specifications : 

Follower to be 4J units long and to move: 

Up in 120 following the parabolic base curve. 

Rest through 30. 

Down in 90 following the harmonic motion base curve. 

Rest through 120. 




FIG. 280. 

The extremities of the follower arc are to lie on a radial line. 

The follower is to swing through 35 and the lowest point of 
its travel is to be 3f units from the center of the cam shaft. 

Solution. The relative motions of the cam and follower will 
not be changed if we hold the cam stationary and move the fixed 
center OA, of the follower arm in a circle about the center of the 



316 CAMS 

cam shaft C. This procedure will be followed for cams having 
swinging follower arms. 

With a radius of 4J units strike an arc of 35 and measure 
the chord. From C lay off CA equal 3| units and from A lay 
off AB equal to the length of the chord as determined above. 
With points A and B as centers and a radius of 4J units draw 
arcs intersecting at OA thus locating the fixed center of the follower 
arm. With C as a center and a radius CO A draw a circle which 
will be the line of centers considering the cam to be fixed and the 
follower to revolve. Lay down the parabolic curve as shown 
in the figure. Make the angle O^CF equal to 120 and divide 
the arc OAF into 6 equal divisions at OA, Oi, 02, etc. With these 
points as centers and using a radius equal to OAA draw the dashed 
arcs shown. With point C as a center and radii CA, Cl, C8, etc., 
draw arcs intersecting the dashed arcs, just laid down in points 1', 
2', 3', etc., which will be points on the required pitch surface. 
A similar construction is used for the down stroke, the complete 
construction being shown in the figure. The working surface 
is gotten as usual. 

While the method just described is not quite correct, it is suf- 
ficiently accurate for all practical purposes. The correct method 
would be to draw the follower in its exact positions and then 
design the cam to meet the roller at these points. 

225. Positive Motion Cams. Any of the preceding cams 
could be made into positive motion cams by the addition of a 
roller where not already in place and a second working surface 
on the opposite side of the roller. The radius of the roller cannot 
be made larger than the smallest radius of curvature of the cam 
grooves. 

Cams so generated are called Face Cams. The disks are 
generally made circular and the groove is cut on one side of the 
disk only. 

226. Rotary Cams. Tangential Follower. Rotary cams hav- 
ing tangential followers are divided into two general classes: 

(a) Sliding tangential follower. 

(6) Pivoted' or swinging tangential follower. 

The method of procedure in both cases is the same, (a) being 
a special case of (6). 



ROTARY CAMS TANGENTIAL FOLLOWER 



317 



Let it be required to lay out a cam that is to operate the slider 
Y, Fig. 281, by means of the swinging tangential follower X, 
which is pivoted at Oo, the center of the cam shaft to be located 
atC. 

r 




FIG. 281. 
The slider Y is to: 

Rest while the cam turns through 90. 

Up 3 units with harmonic motion while the cam turns 

through 120. 
Down 3 units with harmonic motion while the cam turns 

through 90. 
Rest while the cam turns through 60. 

Solution. The foot of the slider is a hemisphere with its 
center at A, which is the lowest position of the slider. Lay off 



318 CAMS 

AB equal to 3 units and divide it into 6 harmonic spaces at A, 
1, 2, 3, 4, 5 and B. The points thus located will be the successive 
centers of the hemisphere. Lay out the necessary portions of 
the arcs as shown, and draw the upper edge of the follower tangent, 
thus locating lines, a, 6, c, ... g. The lines h, i, j . . . n, which 
are the successive positions of the bottom of the follower are 
next drawn parallel to the upper edges and at a distance below 
them equal to the thickness of the follower arm. A vertical 
line through the cam shaft center C cuts these lines in the points 
D, 6, 7, 8, 9, 10 and E. Draw the circle of centers having C for 
a center and a radius equal to COo. 
Lay off the angles: 



Then, consider the cam to remain stationary while the remainder 
of the mechanism revolves counter-clockwise about it. After 
the rest period of 90 the pivot point of the swinging arm will 
lie at OD and the point D will lie at D f . Hence GoD' will be a 
new position of the tangential line of the follower. 

To locate other positions, divide the angles D CO E and D'CE' 
into six equal parts, thus locating the successive centers OQ, Oj, 
8 , . . . E and the radial lines C6', C7', C8' . . . CE'. Lay off 
C6' equal to C6 and join 6 f to Oe, etc. A smooth curve tangent 
to these lines and joining an arc of radius CK will be the working 
surface of the cam. 

The construction for the remainder of the stroke is arrived 
at in the same manner, the construction being shown in the 
figure. 

The wearing surface between the top of the pivoted follower 
and the slider is easily determined. With Oo as a center and a 
radius to the upper tangential point M, draw an arc cutting the 
upper edge of the follower at G. FG is then the surface exposed 
to wear. 

To determine the wearing surface between the follower and the 
cam it is necessary to find the locus of the point of contact, shown 
by the dashed curve, Fig. 281. Points on the curve are located 
as follows: N' is the point of tangency between the follower edge 
Oil' and the cam. Therefore, lay off OoN equal to OiN' and 
one point is located. 



ROTARY CAMS PLANE SLIDING CAMS 



319 



When the locus is completed draw limiting arcs with OQ as a 
center thus locating points H and /. HI is then the limit of wear. 
This part of the follower is often provided with a replaceable 
shoe as indicated in the figure. 

227. Rotary Cams. Sliding Tangential Follower. As stated 
in Art. 226, cams of this type are laid out in precisely the same 
manner as cams having a swinging tangential follower. The only 
difference lies in the fact that the centers OQ, OD, OQ, etc., lie at 
infinity and the lines joining these centers to D, D' , 6, etc., must, 
therefore, in all cases be perpendicular to the radial division lines 
CD, CD', C6 f , etc. The cam is drawn tangent to these lines 
as before. 

The layout of a figure to meet these conditions is left to the 
student. 

In all cams having a tangential follower the working surface 
of the cam must be made tangent to each tangential line in turn, 
in order that irregular motion may be avoided. This is not always 
possible where the data are chosen at will. 

It is evident that cams of this type must always be convex 
externally. 

228. Plane Sliding Cams. Sliding Follower. The flat plate 
AC, Fig. 282, has a reciprocating horizontal motion, in its own 
plane, between fixed guides. It carries a curved face which works 




FIG. 282. 

in contact with the lower end of the follower AB. The follower 
is guided in a vertical direction, rising and falling as the cam 
reciprocates. 

Let it be required to lay out a plane horizontal sliding cam 
having a maximum pressure angle of 40 and working with a slid- 
ing follower that is guided in a vertical direction. 



320 



CAMS 



The follower to: 

Rise 2 units with uniform acceleration, and retardation. 
Rest while the cam moves 1 unit. 

Down 2 units with uniform acceleration and retardation. 
Rest while the cam moves 2 units. 

Solution . The pitch surface of the cam is obtained by assum- 
ing that the cam is fixed and that the follower moves toward 
the right. 

The maximum pressure angle factor for a 40 parabolic curve 
is 2.38. Therefore, the length required for a rise of 2 units is 
2 times 2.38 = 4.76 units. 

To any scale, Fig. 282, lay out AB = 2 units, AD =4.76 units; 
D#=lunit; EF = 4.76 units; FC = 2 units. Complete the rect- 
angle and lay out the pitch and working surfaces as shown in the 
figure. 

229. Plane Sliding Cam. Pivoted Follower. Fig. 283 shows 
the layout of a plane sliding cam having a .pivoted follower 




FIG. 283. 

4 units long, the condition of the problem being the same as given 
in Art. 228. The layout of this cam is in all respects similar 
to the layout of the sliding cam fitted with a sliding follower, the 
lone difference being that the center of motion of the follower, 
in the case of the sliding follower it is at infinity, while in the 
case of the pivoted follower it is at the pivot point. 

Solution. Starting at a point A lay out the distance AB, AD, 
DE, EF, and FC. With A as a center and a radius of 4 units 
locate the point OA, which will be the required pivot center. With 
OA as a center and a radius of 4 units strike the arc A B which will 
be the path of the center of the roller pin. Divide the arc A B 
into any number of spaces conforming to the parabolic curve 
(6 being used in this case). Divide the lines AD and EF into the 



CYLINDRICAL CAMS 



321 



same number of equal spaces. Through the points of division 
thus located, draw arcs parallel to the arc AB. These arcs will 
be the consecutive positions of the path of the center of the roller 
pin. Through the points of division of the arc AB draw hori- 
zontal lines, thus locating points A, 1' , 2', etc., of the pitch sur- 
face of the cam. The working surface is located in the usual 
manner. 

230. Cylindrical Cams. A cylindrical cam may be used to 
give the same motion as a sliding cam. A cam of this form may 
be looked upon as a plane sliding cam bent to a cylindrical shape. 
As the cylinder is rotated on its axis, the follower is given precisely 
the same motion as when the plane sliding cam is given a motion 
of translation. 

231. Cylindrical Cam. Sliding Follower. Let it be required 
to lay out a cylindrical cam, whose follower is to be fitted with 
a cylindrical roller. The cam is to be designed to meet the follow- 
ing conditions: 

Follower to move: 

Parallel to the axis of the cam. 

Out 2 units in 135 based on Elliptical Curve 

In 2 units in 135 based on Elliptical Curve. 

Rest through 90. 

Maximum pressure angle = 40. 

Considering the plane sliding cam to be the development of 
a cylindrical cam, the length of the pitch cam chart, and the 
diameter of the pitch cylinder of the cam are readily determined. 







FIG. 284. 

It should be noted that the pressure angle increases as the 
inner end of the follower roller is approached and that the pitch 
cylinder must pass through the point P, Fig. 284, where the pres- 
sure angle is a maximum. To arrive at the outside radius of the 
cam cylinder, add to the radius of the pitch cam the length of the 



322 



CAMS 



contact line, PQ, of the cam roller, Fig. 285. The length of the 
cam chart, Fig. 284 is equal to the outside circumference of the 
cam cylinder. 

When the cam chart, Fig. 284, is wound around the cylinder 
the line AB will be parallel to the axis as shown in Fig. 285. After 
locating the proper divisions on the end view of the cylinder pro- 
ject lines down and across from corresponding points as at 2, 2' 




FIG. 285. 

and the intersections will be points on the required curve, i.e., the 
curve that must be followed by the center of the cutting tool. 
The following points should be noted : 

1. If the follower arm is provided with a roller, clearance 

should be allowed in the groove, since each side of 
the groove would tend to make the roller revolve 
in a different direction. 

2. This clearance is taken up at the end of the stroke. A 

knock results which may be injurious to the machine 
if running at a high speed. 

3. Clearance must be provided between the bottom of the 

roller and the bottom of the cam groove as shown at 
PR, Fig. 285. 

4. If a cylindrical roller is used pure rolling contact does 

not exist between the working surface of the cam 
and the follower roller. This condition results in 
wear. 



CYLINDRICAL CAMS 



323 



Pure rolling contact may be secured by using a 
conical roller as indicated in Fig. 286. This, how- 
ever, gives a vertical pressure component which tends 
to force the roller out of the groove and produces 
friction in the follower guides. 




FIG. 286. 

5. If a cylindrical roller is used the working surface is a 
right helicoid while with a conical roller it is an 
oblique helicoid. 



CHAPTER XII 



WRAPPING CONNECTORS 



232. Introductory. Belts, Ropes, and Chains are grouped 
together as wrapping connectors. Such machine elements are used 
either (a) to transmit rotary motion from one link to another or 
(6) to transmit force by direct tension. All three types are used 
to transmit motion. Belts are not used for the direct transmis- 
sion of force. 

233. Transmission of Motion. Equivalent Link. Let links 
2 and 8, Fig. 287, represent two pulleys, and let pulley 2 drive 




FIG. 287. 

pulley 3 by means of a rope, belt, or chain. Then if V is the 
velocity of the belt 

V = 7"2C02 = ^"361)3 

or 



where 9*2 and 7*3 are the radii to the neutral plane of the belt, 
provided that there is no slip or stretch of the belt. The neutral 
plane is usually assumed at half the thickness of the belt, although 

324 



BELTS, ROPES, CHAINS 



325 



it may vary somewhat from this position. Exactly the same 
motion is obtained by means of the mechanism shown in Fig. 




FIG. 288. 

288. Here the belt has been replaced by means of a rack, link 
4, meshing with two gear wheels, links 2 and 5, whose pitch radii 
are r^ and r^. Therefore, for pur- 
poses of analysis a belt, rope or chain 
can be replaced by a rack, provided 
that there is no slip. 

Belts and ropes depend upon fric- 
tion to furnish the necessary driving 
force, and there is nearly always some 
slip. Such drives are therefore unsuit- 
able for use where the velocity ratio 
transmitted must be accurately main- 
tained. The motion in short is un- 
constrained. Chains, on the other 
hand, cannot slip and are therefore 
much used where exact velocity ratios 
are required. 

234. Friction between Belt and 
Pulley. The maximum force to be 
transmitted by a chain depends only on 
the strength of the chain. With belts 
and ropes the maximum force depends FKJ. 289. 

not only on the strength of the mem- 
ber, but must not exceed the value which will cause slipping. Con- 
sider an element of a belt running on a pulley of radius r, Fig. 289. 




326 WRAPPING CONNECTORS 

Let dL = length of the element; 

0=arc of contact between belt and pulley in radians; 
r= radius of pulley; 
TI = tension on tight side of belt; 
T2 = tension on loose side of belt ; 
T= tension on element considered; 
w = mass of belt per unit length; 
ju= coefficient of friction. ^ 
Then the mass of the element considered is 

wdl = wrdB ......... (1) 

The forces acting on the element are: 

Tension T acting along the tangent to the left; 
Tension T-\-dt acting along the tangent to the right. 
Centrifugal force 



(2) 



Normal pressure = dP between pulley and belt. 
Friction = fj.dP acting along the tangent to the left. 



Resolving the tensions into components along the radius and along 
the tangent, 

Radial component = 2T sin -^ = TdQ. 

& 

Tangential component = T cos = T to the left, 
and 

(T+dt) cos ^= T+dt to the right. 



Then for equilibrium 

Tdd = dP+wV 2 dO ...... (3) 

and 

(4) 



Eliminating dP between Equations (3) and (4) 



TdO=-dT+wV 2 d0 ...... (6) 



Or 



POWER TRANSMITTED BY BELT 327 

Integrating between limits and 0, T 2 and T\> 



Or 

-SHS ..... 

The net driving force acting on the pulley is 

T l -T 2 = (T 2 -wV 2 )(e^-l) ..... (9) 

Equation (9) gives the maximum value of the force which can be 
transmitted without slip. In addition the tension T\ must not 
exceed the safe working strength of the belt. In general the 
value of Ti T 2 found by Equation (9) will be different for the 
two pulleys. In practically all cases the limiting value is to be 
found at the smaller pulley. 

235. Power Transmitted by Belt. The net pull acting on 
the circumference of the pulley is T\ T 2 . The work performed 
per second is 

W=(T l -T 2 )V ....... (1) 

If the belt is on the point of slipping 

Ti-T 2 = (T 2 -wV 2 )(e'-l) = (T l -wV 2 )(l-e-o) . (2) 
Then 

W=(l-e- e )(TiV-wV*) ........ (3) 

Evidently if 



There is therefore a maximum speed above which power cannot 
be transmitted. Also if 7 = 0, TF = 0. It follows that there 

Ifi 
must be some speed 7 between 7 = and 7 = */ at which the 

power transmitted becomes a maximum. 
Considering T\ as constant 

dW_, 
dV~~ 

At the speed of maximum power -jy = and therefore 

Ti=3wV 2 
or 

V = J^. (5) 



328 



WRAPPING CONNECTORS 



236. Length of Belts. Cone Pulleys. In determining the 
length of a belt two cases arise (a) where the belt is crossed, 
(6) where it is uncrossed. In Fig. 290 is shown a crossed belt 
connecting two pulleys of different diameters. The length of 
the belt is easily seen to be 

l = 2d cos 0+n(7r+20)+r 2 (7T+20) 



7-1+7*2 



sin = 
therefore 



- (n+r 2 ) 2 +(n+r 2 ) 






(1) 




FIG. 290. 

It will be noted that the length of the belt depends only on the 
distance between centers and the sum of the radii of the pulleys. 

Fig. 291 shows the case of the open belt. Here the length Z 
is given by the equation 

l = 2d cos 6+n (7r+20)+r 2 (7T-20), 
where 



sin 
therefore 



d 



Z=2Vd2_( ri _ r2 )2 + ( ri+r2 ) 7r+2 (r 1 -r 2 ) sin' 
In this equation 7*1 r 2 appears as well as 7*1 +r 2 . 



. (2) 



LENGTH OF BELTS 



329 



Equations (1) and (2) are important in the design of cone 
pulleys. Such pulleys are designed so that the length of the belt 
should remain constant. Therefore in the case of crossed belts 
the only condition to be satisfied is that the sum of the diameters 
of the corresponding pulleys on the two cones should be constant. 
For open belts the conditions are more complex. For example, 
suppose that the largest pulley on one cone has a radius of 18 
inches and the smallest on the other has a radius of 6 inches, the 
distance between shaft centers being 8 feet 4 inches. It is required 
to determine the diameters of two other pulleys whose ratio 
should be 2 : 1. 




FIG. 291. 



Putting ri = 2r2 we have 



This is a transcendental equation and can be solved only by trial. 
Since, however, 6 is a small angle we may write 



100 100' 



and also 



without introducing any appreciable error. 
Then 



276.8 = 2(100- 
Solvingr 2 =8.1". 



330 



WRAPPING CONNECTORS 



237. Arrangement of Twisted Belts. Reversibility. Belts 
may be used to transmit motion either between parallel shafts 
or between non-parallel non-intersecting shafts. The condition 
for proper running is that the center line of the belt approaching 
each pulley must lie in the central plane of the pulley. The angle 





FIG. 292. 



FIG. 293. 



at which the belt leaves the pulley is immaterial. For example, 
the arrangement shown in Fig. 292, will operate satisfactorily 
as long as the rotation is in the direction shown. If, however, 
the rotation is reversed, the belt will immediately run off the 
pulleys. 

It is impossible to arrange a reversible belt drive between non- 
parallel shafts using only two pulleys. By the use of guide 



BELT DRIVES. IDLERS 



331 



pulleys or idlers reversible drives become feasible. Reversible 
drives are shown in Figs. 293, 294, and 295. In all these arrange- 
ments it will be noted that the center line of the belt entering 
each pulley lies in the central plane of the pulley. The condition 
for reversibility is simply that the belt must enter and leave each 
pulley so that the center line of the belt lies in the central plane 
of the pulley. 




FIG. 294. 



238. Idlers. Idlers are used to guide the belt where the planes 
of the pulleys do not coincide, to support long belts, and to regulate 
the tension of the belt. The use of idlers as guide pulleys was 
illustrated in the preceding article. In the case of long belts 
idlers are sometimes used to support the belt at intervals and 
thus prevent an undue amount of sag. In the arrangements 
shown in Figs. 296, 297, 298, and 299 the idlers serve to regulate 



332 



WRAPPING CONNECTORS 




FIG. 295. 




IDLER 



FIG. 296. 




SWINGING IDLER 
FIG. 297. 




ADJUSTABLE IDLER 
FIG. 298. 




SWINGING IDLER 
FIG. 299. 



BELT DRIVES. ROPE DRIVES 



333 



the tension, and incidentally to increase the arc of contact between 
belt and pulley. 

239. Crowned Pulleys. Most belt pulleys are crowned, that 
is, the rims are made convex as shown in Fig. 300. A belt always 
tends to climb to the highest portion of the pulley surface. There- 
fore with a crowned pulley the belt tends to center itself instead 
of running off the pulley. Suppose the belt, Fig. 301, to be run- 
ning on the left side of the crowned pulley. Owing to the lateral 
stiffness of the belt the portion which is approaching the pulley 
is thrown to the right as indicated by the dotted lines. This 
portion then runs on the pulley in a position nearer the central 
plane. In other words the belt tends to center itself on the pulley. 




FIG. 300. 



FIG. 301. 



240. Materials of Belts. Most belts are made of leather, 
but sometimes canvas or other textiles are used. Textile belts 
are usually treated with some water-proofing material. Rubber 
belts are sometimes used, the rubber composition being forced 
into a strong textile material. The rubber is used chiefly to give 
a bettter grip on the pulley. 

Steel belts have been used in Germany with good success. To 
date such belts have not passed the experimental stage in this 
country. 

241. Rope Drives. While belts give satisfactory service 
where the shafts are relatively close together and where the belt 
is not exposed to the weather, ropes are greatly to be preferred 
for transmitting power over long distances. 

The general principles of rope transmission are similar to those 
of belt drives, except that since the rope has little lateral stiff- 
ness grooved pulleys must be used to keep the rope from run- 
ning off. 



334 



WRAPPING CONNECTORS 



If a number of pulleys are to be driven by ropes there are two 
systems in general use. 

(a) The English or multiple system. 

(6) The American or continuous system. 

In the multiple arrangement a separate rope is run from a 
common drum to each driven pulley as indicated in Fig. 302. 
In the continuous system a single rope is led consecutively over 
all the pulleys to be driven, as shown in Fig. 303. If only one 




FIG. 302. 



FIG. 303. 



driven shaft is used the rope is given a series of loops around the 
driving and driven drums. In order to carry the rope from one 
end of the driving drum to the opposite end of the driven drum, 
a pair of guide pulleys is necessary. The guide pulleys are usually 
arranged to maintain a constant tension in the slack rope. This 
device is shown in Fig. 304. 1 

242. Friction between Rope and Pulley. In order to increase 
the force which can be transmitted a V-shaped groove is often 
used, as shown in Fig. 305. The analysis of the forces acting 
on the rope is similar to that developed for belts. Let dL, Fig. 
306 represent the length of an element of the rope. Then the mass 
of this element is 

dm = wdL = wrdQ. 

The centrifugal force is 



Leutwiler, Machine Design. 



ROPE DRIVES 



335 



If 20 is the angle of the groove, the vertical forces acting on the 
element are 



2T sin 2dP sin <t>-wV 2 d8 = Q, 




FIG. 306. 



and the horizontal forces are 

Eliminating dP 

TdO-- sin 
M 



336 
Therefore 

Integrating 
Or 



WRAPPING CONNECTORS 



dT 



sm<f>~T-wV 2 ' 



sin 9 j 2 

T^wV 2 



It will be noted that this equation is exactly the same as the 



corresponding equation for belts except that 
for /i. 



. 



is substituted 



The angle 2$ is usually from 45 to 60, and therefore 



1 



sin 



ranges from 2.61 to 2.0? 

243. Chain Drives. Chain drives are used where the velocity 
ratio of driving and driven wheels must be accurately maintained, 
or where the forces involved are too large to be satisfactorily 
transmitted by belts or ropes. 

Chains and chain wheels are made in a large variety of forms 
according to the service to be performed. It is not the intention 
of the authors to enter into detailed description of the various 
makes of chain drives. For such material the reader is referred 
to the catalogues of the different chain manufacturers. 

There is no definite relation between the tensions on the tight 

and slack sides of a chain drive. The 
tension on the tight side is limited 
simply by the strength of the chain, 
and that on the slack side by the 
amount of sag which is permissible. 

244. Hoisting Tackle. Various 
combinations of ropes with fixed and 
movable pulleys are used in hoisting 
devices. Some of these arrangements' 
are shown in Figs. 307-311. 

In Fig. 307 a weight W is hung 
from a movable pulley M. The end 
i vv j of the rope is fastened at a fixed 

FIG. 307. point A, and the rope passes around 




HOISTING TACKLE 



337 



the movable pulley M and over the fixed pulley F. A force P 
is applied at the free end of the rope to overcome the load W. 
If the weight W is lifted 1 foot there will evidently be 1 foot of 
slack in each of the ropes 1 and '//////////s 

2. Therefore 2 feet of rope 
must be hauled in at P to take 
up this slack, and the velocity 
ratio of P to W is 2 to 1. 

In Fig. 308 the movable 
block M consists of three pulleys. 
A rope makes three loops around 
the three pulleys in M and the 
three pulleys of the fixed block 
F. If the weight W is lifted 1 
foot, each of the six supporting 
strands is given 1 foot of slack. 
Therefore, 6 feet of rope must 
be hauled in at P to take up 
this slack, and the velocity ratio 
of P to W is therefore 6 to 1. 

In general if the moving 
pulley is supported by n ropes 
the velocity ratio is n to 1. 

In Fig. 309 the weight W 
is hung from a movable pulley 
MI. A rope passes around MI, 
one end being fastened at a 
fixed point 02, and the other end 
to a second moving pulley M%. 
The ends of the rope which sup- 
ports MI are similarly fastened 
at a fixed point 62 and to a third 
moving pulley M$. One end of 
the rope which supports M% is 
fastened at Os and the other end p IG> 303. 

passes over a fixed pulley F. A 

pull P on the free end of the rope is used to raise the weight W. 
Evidently to raise the weight 1 foot M 2 must raise 2 feet, M 3 
must raise 4 feet, and 8 feet of rope must be hauled in at P. 




338 



WRAPPING CONNECTORS 



In Fig. 310 each of the moving pulleys MI and MI is supported 
by three ropes. To raise the weight W 1 foot M^ must rise 3 

feet, and 9 feet of rope must be 
hauled in at P. 

The various arrangements 
shown in Figs. 307-311 can bs 
combined in different ways to 
give any required velocity ratio. 

245. Differential Pulley. In 
Fig. 311 the weight W is carried 
by a movable pulley M. A 




FIG. 309. 





FIG. 310. 



FIG. 311. 



chain passes around the movable pulley and the doule fixed 
pulley FiF 2 as shown. If a force P be applied on the free 



DIFFERENTIAL PULLEY 339 

loop of the chain the double fixed pulley will turn in the direc- 
tion shown. 

Let n = radius of Fi, 

and 

r2 = radius of %. 

Then if the block F\Fz makes one revolution a length of chain 
2-jrri will be wound on the fixed pulley FI and a length 2irT2 will 
be wound off the pulley 2- The loop carrying the movable 
pulley M will thus be shortened by an amount 2ir(r\ r 2 ), and 
M will be raised a distance Tr(r\ r 2 ). The force P acts through 
a distance 2irri. The velocity ratio of P to W is therefore 



7r(ri-r 2 ) ri-r 2 ' 

This device is known as a differential hoist, since the ratio of 
haul to lift depends on the difference r\ r 2 . 



340 



NOTE A 



NOTE A. IRREGULAR GEARS 

In Fig. 312 let the curve A represent the pitch line of an 
arbitrarily chosen irregular gear rotating about a center P\. It 

is required to lay out a 
second gear B such that 
B makes one revolution 
while A makes n revolu- 
tions. 

The first problem 
which arises is the de- 
termination of the 
center of rotation of 
gear B. Let P 2 repre- 
sent this center, and 
let the distance 




Let R=f (6) be the 

equation of the pitch line of gear A. Then p = CR = F(<f>) 
will be the equation of the pitch line of gear B. 

If gear A rotates through an angle dO gear B rotates through 
an angle d<t> such that 



Since gear B rotates through an angle while gear A makes one 

2ir 



revolution, 



. C 2 " R ,. 27T 
<= I ^ p"0 = . 

Jo LH n 



(2) 



If the form of the function R=f(6) is known Equation (2) can be 
integrated and theoretically the distance C can be determined. 
Practically the calculation of C is very difficult even when R is a 
simple function of B. As an example take R = a sin 2 6. Then 
Equation (2) becomes 



4- +- + 

C 4 C 2 6*4 C 37r " h 8*6'4 



9 - 

r 2 0^4.2 C 



a 3 7.5.3.1 



_ 

C 4 * 



' w 



IRREGULAR GEARS 



341 



It can be shown that the solution of Equation (3) is 

^ I _j 

C* 11 n2 ,yi3 ,y}4 ,yj5 /vj6* 



If, for example, n=4 



a = 2_jJ_ 4_ _ J_4_ JL 7 

C 4 16~ h 64 256^1024 4096 ' ' 



.... (4) 



= 0.3611. 



After the distance C is determined the gears can be laid out 
according to the method explained in Art. 35, or the equation 
of the pitch line of gear B can be found by means of Equation (1). 

From Equation (1) 

dd = d<pj (6) 



or 



C- 



dO. 



(7) 



If 9 can be expressed in terms of p Equation (7) can be inte- 
grated giving the equation of the pitch line of gear B. 

Fig. 313 shows 
the layout of a pair 
of irregular gears 

according to the V 

following data: ^* 



where 



l". 



Gear A makes 
four revolutions to 
one revolution of B. 

It is evident from 
the forms of the 
gears that it would 
be impracticable to 
supply them with 
teeth. Such pitch 
lines, however, could 
be made to roll to- 
gether without diffi- 
culty. 




FIG. 313. 



342 NOTE B 

NOTE B. PROPOSITIONS ON VELOCITY POLYGONS 

Proposition 1. Given the velocity image of a link AB, Fig. 
314, to find the instantaneous center of rotation of that link. 
Let ah be the revolved image of A B and let be the pole of veloc- 
ities. From A and B draw lines parallel to aO and bO respectively. 
The intersection P of these lines is the instantaneous center. 
The triangle ABP is similar to Oab. is the image of a point 
on the link which has no velocity that is the instantaneous 
center. 

If the image of AB reduces to a point (ab) the instantaneous 
center is at infinity in the direction 0(ab). 

Proposition 2. Given the images of two links, AB and CD, 
Fig. 315, to find the instantaneous center of relative motion. 






FIG. 314. 

Let ab and cd be the revolved images of AB and CD. Choose 
at random any point p and draw pa, pb, pc, and pd. From A 
and B draw lines parallel to pa and pb respectively. Let these 
lines intersect at Pi. From C and D, draw lines parallel to pc 
and pd intersecting at P%. Join P\Pz. Choose a second random 
point q and repeat the process. The intersection R of P\P^ and 
QiQ2 is the instantaneous center of relative motion. 

Proof. p and q may be regarded as the images of P\P^ QiQz. 
The relative motion is unaffected by any motion given to both 
links simultaneously. Let be the pole of velocities. Suppose 
the whole system to be given a velocity pO, thus bringing PI and 
P2 to rest. PI and PI thus become the instantaneous centers 
for the two links, and the center of relative velocities must lie 



PROPOSITIONS ON VELOCITY POLYGONS 



343 



on the line P\Pi. Similarly the center of relative velocity must 
lie on the line Q\Qz. 

Proposition 3. Given the velocity polygon for a chain to 
construct the velocity polygon when a different link is held 
stationary. The method of doing this is best shown by an example. 
Given the six-link chain shown in Fig. 316. Let the velocity of 
A be known. The velocity polygon is readily constructed as 
shown. The problem is to construct directly a second polygon 
for the same chain with link 3 held stationary. First, locate 
P the instantaneous center 13. Link 3 can be brought to rest 




* * 




FIG. 316. 



by giving the whole mechanism a rotation = co 31 about P. 
Then each point will be given an additional velocity which is 
equal to co 31 times the distance of the point from the center P. 
The direction of this velocity (revolved) is parallel to the line 
from P to the point in question. Thus, the additional velocity 
of CisPCco 31 . 

Oa 



C0 31 = 



PA' 



The result of adding this rotation is to bring the images of A and 
B to 0, the image of C to c', of D to d', E to e', P 2 to p 2 ', P to p 4 '. 
The polygon with link 3 held stationary is then as shown in 
dotted lines. 



344 



NOTE C 



NOTE C. LOCUS OF THE CENTER OF ACCELERATION 

Consider the epicyclic gear train shown in Fig. 317. Let the 
arm OA, link 2, revolve about with angular velocity ft and 
angular acceleration ft'. This will cause the gear 3 to revolve 

with angular velocity o> = ft and angular acceleration co' = ft'. 

Consider any point P on gear 3, at a distance p from the center. 
Let the angle OAP=d 



- \ ,J 




FIG. 317. 
Then the acceleration of P is given by the equation: 



For the center of acceleration of the gear 8: 



Resolve the four components of A p in horizontal and vertical 
directions. Then for the horizontal components : 

Lft'-pco 2 sin0-po/cos0 = ..... (1) 

For the vertical components : 

Lft 2 -pco 2 cos0+pw'sm0 = ..... (2) 

Let p sin 8 = x and p cos 0=y. 



LOCUS OF ACCELERATION CENTER 345 

Then since LQ' = r' and 



' = Q, ....... (3) 

o' = ....... (4) 

Let ^ = k. 

CO 

Then from Equation (3) 

r y xk = Q. 
Therefore 



From Equation (4) 



Hence 



*./=*. ........ (5) 



j iv yiv | w U. V.O/ 



or 

j;-y-ry+y*+*? = (7) 

Equation (7) is the equation of a circle, which is the locus of 
P. The center of this circle is on the vertical (y) axis. The 
equation can be put in the form: 



where a is the radius of the circle; 

Therefore 

r 2 
-26=--r, 



r/r+L\ 72 r 2 

and b= 






4r] 

L\ 



4rL 



-r\2 



~ 



2 L 



346 



NOTE D 



The position of P on the circle depends on the value of k. 
If fc = 0, then, from (5) 

ry = or y = r. 

That is, the circle must be tangent to the fixed gear at B. 
It has been shown in Art. 32 that the motion of any rigid link 
in a machine may be reproduced with absolute accuracy by rolling 

the moving centrode on the fixed 
centrode. Let FF and MM, Fig. 318, 
be the fixed and moving centrodes for 
the given link. Let be the center 
of curvature of the fixed centrode and 
A that of the moving centrode. Then 
for a very small displacement a rigid 
link might be put in connecting 
and A without in the slightest degree 
altering the motion. If this is done 
it is evident that the argument de- 
veloped for the epicyclic gear train 
applies equally well to the rolling cen- 
trodes. The radii r and R now become 

the radii of curvature of the centrodes. The locus of the center 
of acceleration for the moving link is then a circle BCDE whose 

radius is: s r-^ and which is tangent to the two centrodes at B. 




NOTE D. ON THE SOLUTION OF LINEAR DIFFERENTIAL 
EQUATIONS 

Definitions. An ordinary differential equation of the nth 
order is an equation expressing a relation between an independent 
variable x, a dependent variable y, and the first n derivatives 
of y with respect to x. The general form of such an equation is 

dv d 2 y d 

n i ** & 



A linear differential equation is a differential equation in which 
only the first powers of the dependent variable and its various 
derivatives occur, and in which no product of two derivatives 
or of the dependent variable with any of the derivatives appears. 



LINEAR DIFFERENTIAL EQUATIONS 347 

The general form of a linear differential equation of the nth order 
is 

d n ~ 2 y d 2 y dy 



where P n _i, P n -2 - Po,R are functions of x only. 

A linear differential equation with constant coefficients is 
one in which the functions P n _i, P n _ 2 . . . PO are replaced by 
constants. Equations of this type appear repeatedly in engineer- 
ing problems such as those treated in the text. The object of 
this note is to enable students to handle such equations intelli- 
gently. 

Solution of Differential Equations. A differential equation 
is said to be solved when the dependent variable is expressed as 
a function of the independent variable. Thus 

/(*,) =0, (1) 

is a solution of the differential equation 

dy d?y d 



provided that when the various derivatives are formed and sub- 
stituted in Equation (2) the latter equation is satisfied. 

Solutions are classified as 

(1) General solutions. 

(2) Particular solutions. 

(3) Complete solutions. 

(4) Singular solutions. 

Complete solutions and singular solutions are not required 
in any of the problems arising in the text, and therefore will not 
be considered. The general solution of a differential equation 
always includes a number of arbitrary constants equal to the order 
of the highest derivative occurring in the equation. A particular solu- 
tion is derived from the general solution by giving particular 
values to the arbitrary constants. 

For example: 



348 NOTE D. DIFFERENTIAL EQUATIONS 

is a differential equation of the fourth order. By successive 
integrations we obtain 






.... (4) 

where A, B, C and D are arbitrary constants. Equation (4) is 
the general solution of Equation (3). If the values A = 12, 
B = 6, (7 = and D = 9, are substituted in Equation (4) we obtain 
a particular solution, 



Linear Differential Equation, Second Member Zero. Let 



be chosen as a differential equation which is to be solved. Assume 
that a solution can be written in the form 



* ......... (2) 

Forming the various derivatives 



Substituting these values in Equation (1), 

Q. . (3) 



CHARACTER OF THE ROOTS 349 

Equation (3) is satisfied provided that Wi is a root of the equa- 
tion 

m n +Am n - l +Bm n - 2 + . . . Lm 2 +Mm+N = Q. . . (4) 

Equation (4) is called the auxiliary equation. Equation (2) is 
a solution of Equation (1). It is not, however, a general solution, 
since it contains only one arbitrary constant C\. But Equation 
(4) has n roots m\, m 2 . . . m n . Any of these values satisfies 
Equation (4) and therefore also satisfies Equation (3). The 
general solution of Equation (1) is then 

y = Cie m i x +C 2 e m * x +Cze m * x . . . C n e m ^. . . (5) 

Character of the Roots. The roots mi, m 2 . . . m n may be 
either real or imaginary. If all the roots are real and different, 
Equation (5) is the simplest form of the general solution. If a 
pair of the roots, say mi and m 2 , are conjugate imaginaries, the 
general solution can be thrown into a more convenient form. 
Let 



and 

m 2 = a bi t 

where i V 1 , then 



Now 
and 



or 

tsinfcc 

Therefore 

cosbx+i sin bx = e~ b , 
and 

cos bxi sin bx=e bix . 
Hence 



cos bx+(C 2 -Ci) i sin bx] 

= e ax [C'i cos bx +C' 2 sin bx], 

where C / i = Ci-fC 2 

and 



350 NOTE D. DIFFERENTIAL EQUATIONS 

Since C\ and 2 are arbitrary constants and may be real or imagin- 
ary, C'i and C f 2 are also arbitrary constants. Therefore if two 
of the roots m\ and ni2 form a pair of conjugate imaginaries, 



ni2 = a bi, 
the corresponding terms of the general solution become 

C\e ax cos bx, 
and 

sin bx. 



The general solution now becomes 

cos bx+C 2 sin bx)+Cze x . . . C n eV. 



If two or more of the roots mi, ni2 . . . m n are equal, the form 
of the general solution must be somewhat altered. For example, 
if the roots m\, ni2 and m^ are equal the solution 



x -{- . . . C n e m n x 
= (Ci+C 2 +C 3 ) 



is no longer a general solution, since the sum Ci-\-C2+Cs is sim- 
ply one arbitrary constant, and the number of arbitrary constants 
is thus reduced to n 2. In this case the general solution takes 
the form 



As this form does not occur in any of the problems considered 
in the text the proof will not be developed here. 

Second Member not Zero. Complementary Function. Par- 
ticular Integral. The most general form of linear differential 
equation with constant coefficients is 



where A, B . . . N are constants and R is a function of x. To 
solve equation (1) put 

y = yo+P ........ (2) 



PARTICULAR INTEGRAL 351 




^+M^-+NP=R. (5) 

dx 2 dx 

Equation (4) is called the complementary equation and yo is called 
the complementary function. As shown previously 



*+ . . . C n e m n x , .... (6) 
where m\,m<2. . . . m n are the roots of the auxiliary equation 
m n +Am n - 1 +Bm n - 2 + . . . Lm 2 +Mn+N = Q. 

It will be noted that yo contains n arbitrary constants. The 
general solution is therefore 



where P is any particular solution of Equation (1) or Equation 
(5), that is a solution containing no arbitrary constants. 

The form of the particular solution P depends on the form 
of the function R. Methods for finding P are developed in 
standard works on differential equations. In the text the only 
cases which arise are those where R is a constant or a polynomial 
in powers of x. For such cases the method of finding the par- 
ticular integral is best illustrated by an example. 

Let 

(1) 



be the equation to be solved. 

The complementary equation is 



5^0=0, ........ (2) 



352 NOTE D. DIFFERENTIAL EQUATIONS 

and the auxiliary equation is 

ra 4 -A 4 = ......... (3) 

The four roots of Equation (3) are 

mi=A, 

m 2 =-A, 



The complementary function is therefore 

yo = C l e AX +C 2 e- AX +C 3 cosAx+C4smAx. . . (4) 
To find the particular integral P write 

P = a+bx+cx 2 +dx*+ex 4 +fx 5 +gx Q ..... (5) 
Then 

rfiP 

^ = 24e+120/z+360<7Z 2 ........ (6) 

and 



-A 4 a-A 4 bx-A*cx 2 . . . =Bx 2 +Cx+D . (7) 
Equating coefficients 



0, etc. 

These equations admit of an infinite number of solutions. 
The simplest solution, however, is 



and 

d=e=f=g . . . = 0. 



CONCLUSION 353 



Therefore 



-. 
A 4 ' 



and 
Finally 



cos Az+C 4 sin Ax 

(9) 



Conclusion. The solution of an ordinary linear differential 
equation with constant coefficients consists of the following steps: 

(1) Write the complementary equation by making the second 

member equal to zero. 

(2) Write the auxiliary equation and find the n roots mi, 

W2, mz . . . m n . 

(3) Write the complementary function 

x + . . . C n e m x . 

(4) Find a particular integral P. 

(5) Add P and yo. This sum gives the general solution 

of the equation. 

For further discussion of the properties and solution of dif- 
ferential equations, the reader is referred to the standard works 
on the subject. 



354 FORCES IN GASOLINE ENGINE 

NOTE E.* INVESTIGATION OF FORCES IN GASOLINE ENGINE 

Figs. 319 to 332 and Tables I to IX give the results of an 
investigation of the forces in a standard six-cylinder motor, as 
determined by the students of the junior class at the University 
of Illinois. 

Data. The data for the engine as determined by actual 
measurements are as follows: 

Diameter of cylinder, 3| inches. 

Stroke, 4J inches. 

Clearance, 26.1 per cent. 

Weight of piston, 1.94 pounds. 

Weight of rod, 1.75 pounds. 

Length of r,od, 8| inches. 

Distance from wrist pin to center of gravity of rod, 63^ 
inches. 

Radius of gyration of rod 2.95 inches. 

The following data were taken from information furnished 
by the manufacturers: 
R.P.M.-1500. 

Intake opens 15 after beginning of suction troke. 
Intake closes 42 after beginning of compression stroke. 
Exhaust opens 45 before end of working stroke. 
Exhaust closes 10 after beginning of suction stroke. 
Ignition takes place when crank is on dead center. 
Pressure when exhaust closes atmospheric. 
Explosion pressure 325 pounds per square inch. 

The following data were assumed: 

Equation of compression curve PV 1 ' 28 = constant. 
Equation of expansion curve PF 1 ' 23 = constant. 
Theoretical explosion pressure 392 pounds per square inch. 
Timing Diagram. Indicator Card. Total Pressure Card. 
With the given data the timing diagram, Fig. 319, was first drawn, 
showing the succession of events in the cylinder. Next the posi- 
tions of the piston corresponding to the different events were 

* The original drawings from which the following illustrations were made 
were laid out to a scale of 500 pounds = 1 inch. If the reader desires to deter- 
mine the scale of any figure, a comparison of any numbered ordinate of that 
figure with its accompanying table will be necessary. 



TIMING DIAGRAM. INERTIA FORCES 



355 



determined and laid down in Fig. 320. On the stroke of the 
piston as a base, the indicator diagram, Fig. 320, is drawn. The 
calculations for the pressure on the compression and working 
strokes are given in Tables I and II. The pressures are next 
multiplied by the area of the piston, and the resulting forces are 
plotted on a scale 500 pounds = 1 inch, as the total pressure diagram 
Fig. 321. The last two columns of Tables I and II give the total 
pressure, and the ordinate of the total pressure diagram, Fig. 321. 




FIG. 319. Timing Diagram. 

Inertia Forces. Net Piston Forces. The acceleration of the 
piston is determined by Klein's construction for each 15 of revolu- 
tion of the crank, and the results are checked by calculation from 
the equation 



A = 



0+y cos 20 



The calculations are given in Table III. Multiplying the acceler- 
ations by the mass of the piston as shown in Table IV, the inertia 
force of the piston is now determined, and is plotted to scale 



356 



FORCES IN GASOLINE ENGINE 



1 inch = 500 pounds, in Fig. 322. In the same figure the inertia 
force of the rod is determined by the method of Art. 93 and is 
plotted in its proper position to a scale 3 inches = 500 pounds. 



m 



m 



Exhaust 
Closes 



Intake 
Opens 





FIG. 320. 

A smooth curve drawn through the ends of the vectors represent- 
ing the rod inertia force shows the variation of this force. Table 
V gives the calculations for the rod inertia force. The paths of 
the center of gravity G and of the auxiliary point mz of the kine- 
tically equivalent system are also shown in Fig. 322. 



PRESSURE CARD 



357 



In Fig. 323 the total pressure diagram, Fig. 321, is developed 
on a base line representing the total motion of the piston (dis- 
regarding reversals). In this figure forces tending to help the 
motion of the piston are plotted above the base line, and forces 
resisting the motion are plotted below the base line. 



2500 



Maximum Explosion 



Pressure 2490 Ib. 



2400_ 



uction Pre 



ion 
msl 



sure 



105.8 Ib. 
138.0 Ib. 



2300 



Exhaust Pressure 



2200 



2100 



2000 



1900 



1800 



1700 



1600 



1500 



1400 



1300 



1200 



\ 



1100 



1000 



\ 



900 



700 



Exhaust 



500 



400 



300 




FIG. 321. Total Pressure Card 

The force diagram, Fig. 323, is next transferred to Fig. 324, 
and there combined with the curve representing the inertia force 
of the piston. The resultant net piston force is shown by the 
heavy lines. 

Wrist Pin Pressure. Side Thrust. The total pressure on 
the wrist pin is the combination of the thrust in the connecting 
rod with that portion of the inertia force of the rod which acts 
at the wrist pin. The resultant pressure is found by the method 



358 



FORCES IN GASOLINE ENGINE 




INTERTIA FORCES DEVELOPED CARD 



359 





I 

I 



360 



, FORCES IN GASOLINE ENGINE 




'< O 



COMBINED FORCES TURNING EFFORT 



361 







362 



FORCES IN GASOLINE ENGINE 



of Art. 93 tabulated in Tables VI to IX, and plotted to a scale 
1 inch = 250 pounds in Fig. 325. In this figure the wrist pin is 
drawn full size, and the pressures are plotted as vectors measured 
outward from the circumference. Fig. 325 shows plainly the 
portions of the pin where wear is to be expected. 

The side thrust on the cylinder wall is determined as in Art. 
93, entered in the last column of Tables V to IX, and plotted in 
Fig. 326. 

Crank-pin Pressure. Bearing Pressure. Turning Effort. 
The crank-pin pressure is determined as in Art. 93, entered in 
Tables VI to IX, and plotted in Fig. 327. In this figure the crank 




FIG. 327. Crank Pin Pressures. 

pin is drawn full size and the pin pressures are drawn as vectors 
measured outward from the circumference. As the crank-pin 
is rotating the vectors are drawn, not in their true direction, but 
in their direction relative to a line turning with the pin. In this 
way it is clearly indicated that the pressure is always on the same 
side of the pin, thus demonstrating the cause of uneven wear of 
the pin. 

The crank-pin pressure is resolved into radial and tangential 
components. The radial component gives the pressure on the 
main bearing. The values of the bearing pressure are entered 
in Tables VI to IX, and plotted in Fig. 328. In this figure the 
journal is drawn full size, and vectors are drawn outward from the 



BEARING PRESSURE 



363 



circumference representing the pressure in magnitude and direc- 
tion. 

The tangential component of the crank-pin pressure is the 
turning effort. This is entered in Tables VI to IX, and is plotted 
in Fig. 329 on the developed crank circle as a base. A turning 
effort which assists the motion is plotted above the base line, 
while one which tends to hinder the motion is plotted below the 
base line. 




FIG. 328. Main Bearing Pressure. 

Combined Turning Effort. All the forces so far considered 
nave been those developed in a single cylinder, rod, and crank. 
The total turning effort of the engine is the resultant of the efforts 
of the separate cylinders. In Fig. 330 is shown the order in which 
events occur in the different cylinders. With the aid of this 
diagram it is possible to lay out a common base line for the turn- 
ing efforts of all the cylinders and to plot the combined effort 
as shown in Fig. 331. In Fig. 331, the ordinate is the turning 
effort and the abscissa the movement of the crank. The area 
under the curve is therefore the work performed. When the 
curve rises above the mean ordinate excess work is performed 



364 



FORCES IN GASOLINE ENGINE 



and the engine speeds up. When the curve falls below the mean 
ordinate the engine slows down. 

Shaking Forces. The centrifugal force of the crank arm and 
pin, the inertia force of the rod, and the inertia force of the piston 
all tend to cause vibrations of the engine. In Fig. 332 the shaking 
forces for a single cylinder are shown. From a pole the vector 
OA is laid off representing the centrifugal force of the crank 
arm and pin. From A is drawn AB representing the inertia 
force of the rod, and from B is drawn BC equal to the inertia 
force of the piston. The resultant OC is the total shaking force. 
The vectors are laid off for every 15 of crank travel and a smooth 




FIG. 330. Diagram Showing Sequence of Functions. 

curve R is drawn through the ends of the vectors. The radius 
of the large circle represents the mean value of the shaking force. 
If a counterweight is placed on the crank of size sufficient to 
balance the centrifugal force of the crank arm and pin, the remain- 
ing shaking forces are only the inertia forces of the rod and piston. 
Curve T, Fig. 332, shows the reduction of shaking force due to 
counterbalancing the crank. If a still larger counterweight is 
used such that the centrifugal force of this weight equals the mean 
shaking force, the result is shown in curve S. It is evident, there- 
fore that by suitable counterweights the shaking forces can be 
greatly diminished. 

It should be noted that curves R, S and T give the shaking 
forces for a single cylinder. The shaking forces for all six cylin- 
ders form a balanced system, see Art. 115, 



SHAKING FORCES 



365 




366 



FORCES IN GASOLINE ENGINE 








COMPRESSION CURVE 



367 



bfi 

rS 



I 
3 



hC 

3 






8 










co 



* 



00 










^ 



X 



s 



$ 



& 






Drawn H. A. Gulley 
Traced H. A. Gulley 
Checked Edmonds 
Approved A. C. H. 


TABLE I 
FALLS MOTOR MODEL "S" 


Date May 28, 1920 


Scale 


Coll. of Eng. Univ. of 111. 
Mech. Eng. Dept. 


Order No. 


Sheet No. 



368 



FORCES IN GASOLINE ENGINE 





? 

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to 




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00 


05 


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Drawn Edmonds 


TABLE II 


Date May 24, 1920 


Traced Edmonds 


BALANCING PROBLEM 


Scale 


Checked Edmonds 


College of Engineering U. of Illinois 


Order No. 


Approved A. C. H. 


Department of Mechanical Engineering 


Sheet No. 



EXPANSION CURVE ACCELERATION OF PISTON 369 



+ 



9. 



00 



00 



I I fe 

3 8 

s I 

5 s 

H 9 



O 



+ 



*2 






2 



?; 









8 



i 



^ 



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^4 



s 



? 






8 



CO 



00 



Drawn Staley 
Traced Staley 
Checked Gulley 
Approved A. C. H. 


TABLE III 
BALANCING PROBLEM 


Date May 24, 1920 


Scale 


Coll. of Eng. Univ. of 111. 
Mech. Eng. Dept. 


Order No. 


Sheet No. 



370 



FORCES IN GASOLINE ENGINE 



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Checked Edmonds 
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TABLE V 
BALANCING PROBLEM 


Date May 24, 1920 


Scale 


Coll. of Eng. Univ. of 111. 
Mech. Eng. Dept. 


Order No. 


Sheet No. 



INERTIA FORCES OF CONNECTING ROD 



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TABLE IV 
BALANCING PROBLEM 



Coll. of Eng. Univ. of 111. 

Mech. Eng. Dept. 



Date May 25, 1920 



Scale 



Order No. 



Sheet No. 



372 



FORCES IN GASOLINE ENGINE 



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TABLE VI 
BALANCING PROBLEM 


Date May 24, 1920 


Scale 500 Ib. = 1 in. 


College of Engineering U. of Illinois 
Department of Mechanical Engineering 


Order No. 


Sheet No. 



FORCES ACTING DURING EXHAUST STROKE 373 





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TABLE VII 


Date May 24, 1920 


Traced Staley 


BALANCING PROBLEM 


Scale 


Checked Edmonds 


Coll. of Eng. Univ. of 111. 


Order No. 


Approved A. C. H. 


Mech. Eng. Dept. 


Sheet No. 



374 



FORCES IN GASOLINE ENGINE 



TABLE VIII 
FORCES ACTING DURING SUCTION STROKE 
All values given are vector lengths in inches. Scale 1 inch = 500 Ib. 


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TABLE VIII 
BALANCING PROBLEM 


Date May 26, 1920 


Scale 


Coll. of Eng. Univ. of 111. 
Mech. Eng. Dept. 


Order No. 


Sheet No. 



FORCES ACTING DURING COMPRESSION STROKE 375 



TABLE IX 

FORCES ACTING DURING COMPRESSION STROKE 
All values given are vector lengths in inches. Scale 1 inch = 500 Ib. 


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TABLE IX 
BALANCING PROBLEM 


Date May 28, 1920 


Order No. 


Coll. of Eng. Univ. of 111. 
Mech. Eng. Dept. 


Scale 1 inch =500 Ib. 


Sheet No. 



INDEX 



Acceleration, Cams, 126 

, Components, Projection of, 102 

, Compound supplementary, 93 

, Definition, 86 

, Epicyclic gear, 96 

, Gear train, 95 

, Gyroscope, 211 

, Harmonic motion, 154 

, Normal or radial, 88 

, Relative, 97 

, Shaft governor, 97, 202 

, Steam engine, 157, 355, 369 

, Tangential, 88 

Acceleration center, 104 

, Locus of, 344 
image, 103 
Acceleration polygon, 105 

, Blake stone crusher, 115 

, Epicyclic gear, 107 

, Four-line construction, 121 

, Four-link chain, 109 

, Gear train, 107 

, Joy valve gear, 118 

, Pilgrim step motion, 118 

, Shaper mechanism, 114 

, Steam engine mechanism, 111 

, Stephenson link, 119 

, Velocity scale, 106 

, Walschaert valve gear, 123 

Accelerating force, 129 
Addendum, 259 
circle, 260 

, Limit of 285 

Analysis of mechanism, 24 
Angle of obliquity, 261, 279 



Angular momentum, 218 

, Gyroscopic couple by, 221 

velocity, Ratio, 58 
, Relative, 61 

Annular gear, 273 
, Layout of, 283 

Approach, arc of, 261 

Arc, rectification of, 262 

of action, 259 

, Relation to pitch, 269 
approach, 261 
- recess, 261 

, Relation to addendum, 268 

, - - pitch, 268 

, Relation to addendum, 268 

, pitch, 268 

Atkinson gas engine, 141 
Auxiliary equation, 349 
Axial pitch, Helical gear, 295 

, Worm gear, 299 

Axode, Fixed, 49 
, Moving, 49 

of relative motion, 49 

Back cone, 293 

Backlash, 259 

Balance, Conditions of, 159 

, Eight-cylinder engine, 167 

, Four crank engine, 166 

, Harmonic motion, 156 

, Offset engine, 175 

, Opposed engine, 170 

, Partial, 159 

, Radial engine, 167 

, Rotary engine, 171 



377 



378 



INDEX 



Balance, Running, 153 

, Single-crank engine, 160 

, Six-crank engine, 166 

, Static, 153 

, Three-crank engine, 162 

, Twelve-cylinder engine, 167 

Balancing Reciprocating masses, 156, 

159 

rotating masses, 147 
Base circle, 260, 278 
- curve, 304, 308 

, Elliptical, 307 

, Harmonic, 305 

, Parabolic, 306 

, Straight-line, 305 

, Straight line-curve, 305 

Bearing pressure, 362 

Belt, 324 

, Equivalent link, 325 

, Friction, 325 

, Length, 328 

, Materials, 333 

, Neutral plane, 324 

, Power transmitted, 327 

, Reversibility, 330 

, Bevel gear, Classification, 291 

, Friction, 257 

, Layout, 293 

, Teeth, 292 

Bevel gear tram, 55 

, Direction of rotation, 55 

, Epicyclic, 55 

, Velocity ratio, 55, 56 

Blake stone crusher, 115 
Bicycle sprocket, 34 

Cams, acceleration in, 126 
, Base curve, 304 
, Cylindrical, 321 
, Offset follower, 313 
, Pitch Surface of, 310 
, Plane sliding, 319 
, Positive motion, 303 
, Pressure angle, 305 
, Radial follower, 309 
, Roller follower, 310 
, Size of, 309 



Cams, Swinging follower, 314 

, Tangential follower, 316 

, Timing diagram, 305 

, Two-step, 311 

, Working surface, 310 

Cam factor, 307 

Cast teeth, 290 

Center of acceleration, 104 

, Locus of, 344 

Centers, Law of three, 36 
Centrifugal couple, 145 

force, 145 

, Moment of, 188 

, Unbalanced disk, 228 

Centrode, Fixed, 43 

, Gear as, 46 

, Moving, 43 

of relative motion, 44 
, Rolling of, 46 
Chain drive, 336 

, Velocity, ratio of, 325 
, Constrained, 10 
, Formation of, 13 
, Kinematic, 10 
, Locked, 10 
, Unconstrained, 10 
Chordal pitch, 259 
Circular pitch, 259, 262 

, Helical gear, 295 

, Limit of, 269 

, Worm gear, 299 
Clearance, 260, 267, 279 
Combined motions, Acceleration in, 

90 
Complementary equation, 351 

function, 351 

Compound supplementary accelera- 
tion, 93 

Compression curve, 354, 367 
Conditions for balance, harmonic 

motion, 156 
, Pistons, 159 

, Rotating masses, 153 

Cone pulley, 328 

Cones, rolling, 251 

Connecting rod, Inertia force, 136 

, Influence on acceleration, 156 



INDEX 



379 



Constant acceleration cam, 306 

Constrained motion, 2 

Contact, Cams, 303 

, Gears, 290 

, Helical gears, 302 

, Involutes, 278 

, Line, 5 

, Multiple, 9 

, Surface, 5 

, Worm gears, 302 

Coriolis' law, 90, 113 

Counterbalancing, 160 

Couple, centrifugal, 145 

, Gyroscopic, 210, 221 

polygon, 147, 150 
Crank pin pressure, 362 
Criterion of constraint, 17 

, Application, 22 

, Exceptions, 22, 23, 24 

, with higher pairs, 21 

, with lower pairs, 19 
Critical speed, Disappearance of, 233 
, Inclined disk, 242 
, Shaft in fixed bearings, 228, 238 

, Shaft loaded at middle, 227 

, Uniformly loaded shaft, 230, 

238 

Crown wheels, 257 
Crowned pulley, 333 
Cut teeth, 291 
Cycloid, 263 
Cycloidal curves, 263 
- teeth, 267 
Cylindrical cam, 321 

Davis engine, 223 
Dedendum, 260 

circle, 260 
Describing circle, 260 

, Size of, 270 
Diametral pitch, 259, 262 
Differential equations, 346 
, Auxiliary equation, 349 

, Classes of solutions, 347 
, Complementary equation, 351 

, function, 351 
, General solution, 349 



Differential equations, Linear equa- 
tion, 347 

, Particular solution, 351 
-pulley 338 
Displacement equivalent to rotation, 

30 
Double he'ical gear, 295 

, Advantages, 296 

Duplex engine, 161 

Eight-cylinder engine, 167 
Epicyclic gear, Acceleration in, 107 

, Velocity polygon, 71 

, Velocity ratio, 53 
Epicycloid, 263 
, Normal to, 266 
Equivalence of mechanisms, 46 
Expansion curve, 354, 368 

of link, 32 

Face of teeth, 260 
Flank of teeth, 260 
Fly-ball governor, 177 

, Elementary theory, 183 

, Horizontal, 187 

, Limiting speed, 184, 186, 190 

, Oscillations, 194 

, Stability of, 187, 191, 193 

, Weighted, 186 
Follower, 303 
, Offset, 313 
, Radial, 309 
, Roller, 310 
, Swinging, 314 
, Tangential, 316 
, Wedge, 309 
Force reduction, 179 
Four-crank engine, 166 
Four-line construction, for accelera- 
tions, 121 

, for velocities, 82, 121 

Four-link chain, Accelerations, 109 

, Instantaneous centers, 37 

Friction, Belt and pulley, 325 

Governors, 194, 207 
, Rope and pulley, 335 
wheels, Bevel, 257 



380 



INDEX 



Friction wheels, Crown, 257 
, Grooved, 257 
, Spur, 257 
, Velocity ratio, 250 

Gear teeth, Annular, 273, 283 

, Approximate cycloidal, 274 

, involute, 286 

, Cast, 290 
, Cut, 291 

, Cycloidal, 267 

, Helical, 295 

, Involute, 279 

, Layout cycloidal, 271 

, involute, 279 

, Rack, 272, 284 

, Stub, 288 
Train, Acceleration, 95 

, Bevel, 55, 56 

, Epicyclic, 53, 61 

, Hypocyclic, 54 

, Reverted, 53 

, Velocity ratio, 51 

Gorge circle, 253, 301 
Governors, Classification, 177 

See also Fly-ball governors, Shaft 

governors 
Grant's Odontograph, Cycloidal, 274 

, Involute, 286 

Grooved friction wheels, 257 
Guide pulley, 330 
Gyroscopic couple, 210, 221 

Harmonic motion, 154 

cams, 305 

Helical gear, 294 
, Advantages, 296 

, Double, 295 

for non-parallel shafts, 296 

parallel shafts, 295 

, Velocity ratio, 296 
Herringbone gear, 295, 296 
Higher pair, 3 

, Constraint by, 4, 22 

-, Instantaneous center, 40, 59 

, Properties, 5 

Hindley worm, 299, 301 



Humpage gear, 56 
Hyperboloid of revolution, 253 
Hypocyclic gear train, 54 
Hypocyclic straight line motion, 28 
Hypocycloid, 263 
, Normal to, 266 



Idle gear, 52 

Idler pulley, 331 

Image, Acceleration, 103 

, Velocity, 67 

Indicator card, 354 

Inertia forces, Connecting rod, 136, 

371 
, Harmonic motion, 154 

, Piston, 135, 355, 370 

, Resultant, 129 

, Shaft governor, 203 
governor, 201 
, Forces in, 203 

, Oscillation, 207 
Instantaneous center, 31 
- Gears, 41 

, Higher pairs, 40 
, Relative motion, 35 

, Rolling cylinders, 257 
, Rotation about, 32 

, Table of, 38 

Interchangeable wheels, 270, 290 
Interference, 285 
Internal gear, 273 
Inversion of mechanism, 26 

pair, 6, 27 

Involutes, 277 
, Normal to, 278 
, Sliding contact, 278 
, Spherical, 293 
Involute teeth, 279 

, Annular, 283 

, Bevel, 292 

, Comparison with cycloidal, 290 

, Proportions of cast, 290 

, cut, 291 

, Rack, 284 

, Stub, 288 

Irregular gears, 339 



INDEX 



381 



Joint, 15 

, Binary, ternary, 15 

Joy, valve gear, 74, 118 

Kinetic load on shaft, 145 
Kinetically equivalent system, 130, 

136, 141 
Klein's construction, 112 

Law of three centers, 36 
, Application, 37 

, Special cases, 42 
Layout of gear teeth, 271, 279 

, Approximate, 282, 274 

, Bevel, 293 
Lead of worm, 300 
Line of action of resultant, 131 

, Calculation of, 132, 

141 

Linear velocity ratio, 62 
Link, Kinematic, 10 
, Binary, ternary, 10 
, Expansion of, 32 
, Skeleton, 13, 25 
Lower pair, 3 

, Properties of, 5 

Machine defined, 2 
Mass reduction, 151, 180 
Mechanism defined, 13 
Miter gears, 292 
Movable pulley, 336 

Non-circular wheels, 339 
Non-intersecting shafts, 253, 256, 294 
Normal acceleration, 88 

, Construction for, 99 

Normal pitch, 295 

Obliquity, Angle of, 261, 279 
Odontograph, Cycloidal, 274 
, Involute, 286 
Offset engine, 175 
Offset follower, 313 
Opposed engine, 170 
Oscillations, Flyball governor, 194 
, Gyroscope, 213 



Oscillations, Shaft governor, 207 

Pair, 2 

, Closed, 4 

, Higher, 3 

, Incomplete, 4 

, Inversion of, 6, 27 

, Lower, 3 

, Multiple Contact, 9 

, Properties, 2 

Pairing element, 6 

, Expansion, 7 

, Representation, 13 
Partial balance, 159 
Particular integral, 351 
Period of vibration, 229, 237, 242, 248 
Pilgrim step motion, 75, 118 
Pin gearing, 267 

Piston, Calculation of acceleration, 
157 

velocity, 57 

Pitch, Axial, 295, 299 
, Chordal, 259 
, Circular, 259, 262, 295, 299 
, Diemetral, 262, 259 
, Normal, 295 
- of worm, 300 

circle, 258 

point, 258 

surface cams, 304 

gears, 258 

irregular gears, 339 
Plane motion, 29 

Point of contact, Path of, 278 
Polygon of accelerations, 105 

- couples, 147, 150 
Positive motion cam, 303, 316 
Power transmitted, by belt, 327 

, by rope, 335 
Precession, 213, 223 
Pressure angle, 305, 307 
Primary force, 158 
Product of inertia, 182 
Pulley, Cone, 328 
, Crowned, 333 
, Differential, 338 
, Guide, 330 



382 



INDEX 



Pulley, Idler, 331 
, Movable, 336 
, Rope, 333 
Pulley block, 337 

Rack, Cycloidal, 272 
, Involute, 284 
Radial engine, 167 
Recess, Arc of, 261 
Reciprocating masses, 154 
Rectification of arc, 262 
Rectilinear motion, 3 
Reduction of force, 179 

mass, 180 

Reference plane, 147, 155 
Relative acceleration, 97 

angular velocity, 59 

motion, Center of, 35 
, Centrode of, 44 

velocity, 59 
Resultant couple, 147 
Reversibility of belt drive, 330 
Reverted gear train, Velocity ratio, 53 
Roller follower, 310 

, Size of, 312 

Rolling contact, Centrodes, 43 

, Cones, 251 

, Cylinders, 250 

, Hyperboloids, 253 

Rope drive, American system, 334 

, English System, 334 

, Power transmitted, 335 

hoists, 336, 337 

pulley, 333, 334 
Rotary engine, 171 

Rotation about fixed center, 30, 88 

about instantaneous center, 31 
Running balance, 153 

Scale of velocities, 106 
Screw gearing, 294 
Secondary force, 158 
Shaft governor, Centrifugal, 200 

, Forces in, 203 

, Inertia type, 201 

, Osculations, 207 

Shaking forces, 365 



Shaper mechanism, Accelerations, 114 

, Instantaneous centers, 38, 64 

, Inversion, 28 

, Velocity polygon, 76 

Shear in shaft, 230, 234 
Ship, gyroscopic stabilization of, 223 
Side thrust, 357, 361 
Six-crank engine, 166 
Skeleton link, 13, 25 
Skew bevel gear, 301 
Sliding cam, 319, 320 
Spherical involute, 293 
Spring-controlled governor, 187, 191 
Stability of governor, 187, 191, 193 
Static balance, 153 
Steam engine mechanism, Accelera- 
tion of piston, 157 

. Acceleration polygon, 111 

, Centrodes, 46 

, Finite rod, effect of, 157 

, Inertia forces, 134 

, Instantaneous centers, 37 

Stephenson chain, 14 

- link motion, 25, 79, 119 
Stepped gear, 295 

Stub tooth gear, 288 

, Comparison with standard, 

289 
Surging, 213 

Tangential acceleration, 88 

- followers, 316 
Thickness of tooth, 259 
Three centers, law of, 36 
Three-crank engine, 162 
Three-line construction, for acceler- 
ations, 119 

, for velocities, 77 

Time efficiency, 85 
Timing diagram, 305, 354 
Tooth space, 259 
Total pressure card, 354 
Turning effort, 361, 363, 366 
Twelve-cylinder engine, 167 
Two-step cam, 311 

Velocity curve, 85 



INDEX 



383 



Velocity image, 67 
, Revolved, 69 
, polygon, 69 

, Cams, 70 

, Epicyclic gear train, 71 

, Four-line construction, 82 

, Four-link mechanism, 73 

, Gear train, 71, 72 

, Joy valve gear, 74 

, Pilgrim step motion, 75 

, Shaper mechanism, 76 
, Stephenson link, 79 

, Three-line construction, 77 

, Walschaert valve gear, 84 

, Wanzer needle bar, 80, 81 

, Watt engine, 80 

ratio, Angular, 58 
, Belt drive, 324 
, Differential pulley, 339 

, Epicyclic gear, 53 

, Friction wheels, 250 

, Gear train, 51, 62 

, Helical gears, 296 

, by instant centers, 58 



Velocity ratio, Linear, 62 

, Reverted gear train, 53 

, Rope hoists, 336, 337, 338 

, Worm gear, 300 

Vibrations, elastic, Disk, 246 
, Period of, 229, 237, 242, 248 
, Shaft in fixed bearing, 229, 239 
, loaded at middle, 228 
, uniformly loaded, 234, 239 

Walschaert valve gear, 84, 123 
Wanzer needle bar, 80, 81, 143 
Watt chain, 14 

engine, 80 
Wearing surface, 318 
Wedge, 24 

Weighted governor, 185 
Working surface of cam, 310 
Worm, 294 

, Hindley, 299, 301 

, Double and single thread, 300 

gear, 294, 298 

, Velocity ratio, 300 
Wrist pin pressure, 357 




Wiley Special Subject Catalogues 

For convenience a list of the Wiley Special Subject 
Catalogues, envelope size, has been printed. These 
are arranged in groups each catalogue having a key 
symbol. (See special Subject List Below). To 
obtain any of these catalogues, send a postal using 
the key symbols of the Catalogues desired. 



1 Agriculture. Animal Husbandry. Dairying. Industrial 

Canning and Preserving. 

.*; ' .'^oloioololfi .TfnortoittsA c 

2 Architecture. Building. Concrete and Masonry. 

3 Business Administration and Management. Law. 

Industrial Processes: Canning and Preserving; Oil and Gas 
Production; Paint; Printing; Sugar Manufacture; Textile. 

CHEMISTRY 
4a General; Analytical, Qualitative and Quantitative; Inorganic; 

Organic. 
4b Electro- and Physical; Food and Water; Industrial; Medical 

and Pharmaceutical; Su^ar. 
meo ;m8tnfid:M bns naia^Q 3niri3*J$ D^l 

CIVIL ENGINEERING 

5a Unclassified and Structural Engineering. 

5b Materials and Mechanics of Construction, including; Cement 
and Concrete; Excavation and Earthwork; Foundations; 
Masonry. 

5c Railroads; Surveying. 

5d Dams; Hydraulic Engineering; Pumping and Hydraulics; Irri- 
gation Engineering; River and Harbor Engineering; Water 
Supply. .noi>sli 



CIVIL ENGINEERING Continued 

5e Highways; Municipal Engineering; Sanitary Engineering; 
Water Supply. Forestry. Horticulture, Botany and 
Landscape Gardening. 



6 Design. Decoration. Drawing: General; Descriptive 
Geometry; Kinematics; Mechanical. 

ELECTRICAL ENGINEERING PHYSICS 
7 General and Unclassified; Batteries; Central Station Practice; 
Distribution and Transmission; Dynamo-Electro Machinery; 
Electro-Chemistry and Metallurgy; Measuring Instruments 
and Miscellaneous Apparatus. 



8 Astronomy. Meteorology. Explosives. Marine and 

Naval Engineering. Military. Miscellaneous Books. 

MATHEMATICS 

9 General; Algebra; Analytic and Plane Geometry; Calculus; 

Trigonometry; Vector Analysis. 

MECHANICAL ENGINEERING 

lOa General and Unclassified; Foundry Practice; Shop Practice. 
lOb Gas Power and Internal Combustion Engines; Heating and 

Ventilation; Refrigeration. 

bns 



lOc Machine Design and Mechanism; Power Transmission; Steam 
Power and Power Plants; Thermodynamics and Heat Power. 
1 1 Mechanics. . 

12 Medicine. Pharmacy. Medical and Pharmaceutical Chem- 

istry. Sanitary Science and Engineering. Bacteriology and 
Biology. 

MINING ENGINEERING 

13 General; Assaying; Excavation, Earthwork, Tunneling, Etc.; 

Explosives; Geology; Metallurgy; Mineralogy; Prospecting; 
Ventilation. 

14 -Food and Water. Sanitation. Landscape Gardening. 
Design and Decoration. Housing, House Painting. 



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