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6 & 8 BOUVERIE ST., E. C. 





The present treatise dealing with the Principles of Mechanism 
and Mechanics of Machinery is the result of a number of years' 
experience in teaching the subjects and in practising engineering, 
and endeavors to deal with problems of fairly common occur- 
rence. It is intended to cover the needs of the beginner in the 
study of the science of machinery, and also to take up a number 
of the advanced problems in mechanics. 

As the engineer uses the drafting board very freely in the 
solution of his problems, the author has devised graphical solu- 
tions throughout, and only in a very few instances has he used 
formulae involving anything more than elementary trigonometry 
and algebra. The two or three cases involving the calculus may 
be omitted without detracting much from the usefulness of the 

The reader must remember that the book does not deal with 
machine design, -and as the drawings have been made for the 
special purpose of illustrating the principles under discussion, 
the mechanical details have frequently been omitted, and in cer- 
tain cases the proportions somewhat modified so as to make the 
constructions employed clearer. 

The phorograph of Professor Rosebrugh has been introduced 
in Chapter IV, and appeared in the first edition for the first time 
in print. It has been very freely used throughout, so that most 
of the solutions are new, and experience has shown that results 
are more easily obtained in this way than by the usual methods. 

As the second part of the book is much more difficult than the 
first, it is recommended that in teaching the subject most of the 
first part be given to students in the sophomore year, all of the 
second part and possibly some of the first part being assigned in 
the junior year. 

The thanks of the author are due to Mr. J. H. Parkin for his 
careful work on governor problems, some of which are incorpor- 
ated, and for assistance in proofreading; also to the various firms 
and others who furnished cuts and information, most of which 
is acknowledged in the body of the book. 




The present edition has been entirely rewritten and enlarged 
and all of the previous examples carefully checked and corrected 
where necessary. The cuts have been re-drawn and many new 
ones added; further, the Chapter on Balancing is new. Ques- 
tions at the end of each chapter have been added. 

R. W. A. 

February, 1917. 








General discussion Parts and purpose of the machine Definitions 
Divisions of the subject Constrained motion Turning and 
sliding motion Mechanisms Inversion of the chain Examples 
Sections 1 to 26. 



Plane motion Data necessary to locate a body and describe its 
motion Absolute and relative motion The virtual center Fixed 
and permanent centers Location of the virtual center Sections 
27 to 41. 



Application of virtual center Linear and angular velocities Appli- 
cation to various links and machines Graphical representation 
Piston velocity diagrams Pump discharge Sections 42 to 56. 



Phorograph method of determining velocities Fundamental prin- 
ciples Images of points Application Image of link gives angular 
velocity Sense of rotation Phorograph a vector diagram Steam 
engine Whitworth quick-return motion Valve gears, etc. Sec- 
tions 57 to 80. 



Forms of drives used Spur gearing Proper outlines and condi- 
tions to be fulfilled Cycloidal teeth Involute teeth Parts and 




proportions of teeth Definitions Racks Internal gears Inter- 
ference Stub teeth Module Helical teeth Sections 81 to 103. 



Various types of such gearing Bevel gearing Teeth of bevel 
gears Spiral tooth bevel gears Skew bevel gearing Pitch sur- 
faces General solution of the problem Applications Screw 
gearing Worm and worm-wheel teeth General remarks Sec- 
tions 104 to 125. 



Kinds of gearing trains Ordinary trains Ratio Idlers Ex- 
amples Automobile gear box Screw-cutting lathe Special 
threads Epicyclic or planetary gearing Ratio Weston triplex 
block Drill, etc. Ford transmission Sections 126 to 139. 


CAMS 136 

Purpose of cams Stamp mill cam Uniform velocity cam Design 
of cam for shear General problem of design Application to gas 
engine Sections 140 to 144. 



Classification of forces acting in machines Static equilibrium 
Solution by virtual centers Examples Solution by phorograph 
Shear Rock crusher Riveters, etc. Sections 145 to 152. 



Variations in available energy Crank effort Torque Steam en- 
gine Crank effort from indicator diagrams Types of engines 
Internal combustion engines Effect of various arrangements 
Sections 153 to 161. 



Input and output Meaning of efficiency Methods of expressing 
efficiency Friction Friction factor Sliding pairs Turning 
Pairs Complete machines Sections 162 to 174. 





Methods of governing Purpose of the governor Fly-ball 
governors Powerfulness Sensitiveness Isochronism A c t u a 1 
design Characteristic curves Spring governor Inertia gover- 
nor Distribution of weight Sections 175 to 198. 



Cause of speed fluctuations Illustration in case of steam engine 
Kinetic energy of machines and bodies Reduced inertia of bodies 
and machines Speed fluctuations Graphical determination for 
given machine Practical application in a given case Sections 
199 to 213. 



Purpose of flywheels Discussion of Methods Dimensions of 
wheels Coefficient of speed fluctuation The E-J diagram 
Numerical examples on several machines Sections 214 to 223. 



General effects of acceleration Normal and tangential accelera- 
tion Graphical construction Machines Disturbing forces 
Stresses due to inertia Examples Numerical problem on an en- 
gine Sections 224 to 243. 



Discussion on balancing Balancing of rotating masses Single 
mass Several masses Reciprocating and swinging masses 
Primary balancing Secondary balancing Short connecting rod 
Four crank engine Locomotive Motor cycle engine Sections 
244 to 254. 



INDEX . . . 333 


The following are some of the symbols used in this book, with 
the meanings usually attached to them. 
w = weight in pounds. 

g = acceleration of gravity = 32.2 ft. per second per 

m = mass = - 

v = velocity in feet per second. 
n = revolutions per minute. 

co = radians per second = ~~^~ 

TT = 3.1416. 

a = angular acceleration in radians per second per second. 
= crank angle from inner dead center. 
/ = moment of inertia about the center of gravity. 
k = radius of gyration in feet = \/I/m 
J = reduced inertia referred to primary link. 
T = torque in foot-pounds. 

P, P f , P" represent the point P and its images on the velocity 
and acceleration diagrams respectively. 




1. General. In discussing a subject it is important to know 
its distinguishing characteristics, and the features which it has 
in common with other, and in many cases, more fundamental 
matters. This is particularly necessary in the case of the machine, 
for the problems connected with the mechanics of machinery 
do not differ in many ways from similar problems in the mechan- 
ics of free bodies, both being governed by the same general laws, 
and yet there are certain special conditions existing in machinery 
which modify to some extent the forces acting, and these condi- 
tions must be studied and classified so that their effect may be 

Again, machinery has recently come into very frequent use, 
and is of such a great variety and number of forms, that it de- 
serves special study and consideration, and with this in mind it 
will be well to deal with the subject specifically, applying the 
known laws to the solution of such problems as may arise. 

2. Nature of the Machine. In order that the special nature 
of the machine may be best understood, it will be most con- 
venient to examine in detail one or two well-known machines 
and in this way to see what particular properties they possess. 
One of the most common and best known machines is the recip- 
rocating engine, (whether driven by steam or gas is unimportant) 
which consists of the following essential, independent parts: 
(a) The part which is rigidly fixed to a foundation or the frame- 
work of a ship, and which carries the cylinder, the crosshead 
guides, if these are used, and at least one bearing for the crank- 
shaft, these all forming parts of the one rigid piece, which is for 
brevity called the frame, and which is always fixed in position. 
(6) The piston, piston rod and crosshead, which are also parts 
of one rigid piece, either made up of several parts screwed to- 
gether as in large steam and gas engines, or of a single casting 



as in automobile engines, where the piston rod is entirely omitted 
and the crosshead is combined with the piston. It will be con- 
venient to refer to this part as the piston, and it is to be noticed 
that the piston always moves relatively to the frame with a 
motion of translation, 1 and further always contains the wristpin, 
a round pin to facilitate connection with other parts. The pis- 
ton then moves relatively to the frame and is so constructed as 
to pair with other parts of the machine such as the frame and 
connecting rod now to be described, (c) The connecting rod 
is the third part, and its motion is peculiar in that one end of it 
describes a circle while the other end, which is paired with the 
wristpin, moves in a straight line, which latter motion is 
governed by the piston. All points on the rod move in parallel 
planes, however, and it is said to have plane motion, as has also 
the piston. The purpose of the rod is to transmit the motion of 
the piston, in a modified form, to the remaining part of the 
machine, and for this purpose one end of it is bored out to fit 
the wristpin while the other end is bored out to fit a pin on the 
crank, which two pins are thus kept a fixed distance apart and 
their axes are always kept parallel to one another, (d) The 
fourth and last essential part is the crank and crankshaft, or, 
as it may be briefly called, the crank. This part also pairs with 
two of the other parts already named, the frame and the connect- 
ing rod, the crankshaft fitting into the bearing arranged for it 
on the frame and the crankpin, which travels in a circle about the 
crankshaft, fitting into the bored hole in the connecting rod 
available for it. The stroke of the piston depends upon the 
radius of the crank or the diameter of the crankpin circle, and is 
equal to the latter diameter in all cases where the direction of 
motion of the piston passes through the center of the crankshaft. 
The flywheel forms part of the crank and crankshaft. 

In many engines there are additional parts to those mentioned, 
steam engines having a valve and valve gear, as also do many 
internal-combustion engines, and yet a number of engines have 
no more than the four parts mentioned, so that these appear to 
be the only essential ones. 

3. Lathe. Another well-known machine may be mentioned, 
namely, the lathe. All lathes contain a fixed part or frame or 

1 By a motion of translation is meant that all points on the part considered 
move in parallel straight lines in the same direction and sense and through 
the same distance. 


bed which holds the fixed or tail center, and which also contains 
bored bearings for the live center and gearshafts. Then there is 
the live center which rotates in the bearings in the frame and which 
drives the work, being itself generally operated by means of a 
belt from a countershaft. In addition to these parts there is the 
carriage which holds the tool post and has a sliding motion along 
the frame, the gears, the lead screw, belts and other parts, all 
of which have their known functions to perform, the details of 
which need not be dwelt upon. 

4. Parts of the Machine. These two machines are typical of 
a very large number and from them the definition of the machine 
may be developed. Each of these machines contains more than 
one part, and in thinking of any other machine it will be seen 
that it contains at least two parts : thus a crowbar is not a machine, 
neither is a shaft nor a pulley; if they were, it would be difficult 
to conceive of anything which was not a machine. The so-called 
" simple machines," the lever, the wheel and axle, and the wedge 
cause confusion along this line because the complete machine 
is not inferred from the name: thus the bar of iron cannot be 
called a lever, it serves such a purpose only when along with it 
is a fulcrum; the wheel and axle acts as a machine only when it 
is mounted in a frame with proper bearings; and so with the 
wedge. Thus a machine consists of a combination of parts. 

5. Again, these parts must offer some resistance to change of 
shape to be of any value in this connection. Usually the parts 
of a machine are rigid, but very frequently belts and ropes are 
used, and it is well known that these serve their proper purpose 
only when they are in tension, because only when they are used 
in this way do they produce motion since they offer resistance 
to change of shape. No one ever puts a belt in a machine in a 
place where it is in compression. Springs are often used as in 
valve gears and governors, but they offer resistance wherever 
used. Thus the parts of a machine must be resistant. 

6. Relative Motion. Now under the preceding limitations a 
ship or building or any other structure could readily be included, 
and yet they are not called machines, in fact nothing is a machine 
in which the parts are incapable of motion with regard to one 
another. In the engine, if the frame is stationary, all the other 
parts are capable of moving, and when the machine is serving 
its true purpose they do move; 'in a bicycle, the wheels, chain, 
pedals, etc., all move relatively to one another, and in all machines 


the parts must have relative motion. It is to be borne in mind 
that all the parts do not necessarily move, and as a matter of 
fact there are very few machines in which one part, which is 
referred to briefly as the frame, is not stationary, but all parts 
must move relatively to one another. If one stood on the frame 
of an engine the motion of the connecting rod would be quite 
evident if slow enough; and if, on the other hand, one clung to 
the connecting rod of a very slow-moving engine the frame would 
appear to move, that is, the frame has a motion relative to the 
connecting rod, and vice versa. 

7. In a bicycle all parts move when it is going along a road, 
but still the different parts have relative motion, some parts 
moving faster than others, and in this and in many other similar 
cases, the frame is the part on which the rider is and which has 
no motion relative to him. In case of a car skidding down a 
hill, all parts have exactly the same motion, none of the parts 
having relative motion, the whole acting as a solid body. 

8. Constrained Motion. Now considering the nature of the 
motion, this also distinguishes the machine. When a body moves 
in space its direction, sense and velocity depend entirely upon the 
forces acting on it for the time being, the path of a rifle ball 
depends upon the force of the wind, the attraction of gravity, 
etc., and it is impossible to make two of them travel over exactly 
the same path, because the forces acting continually vary; a 
thrown ball may go in an approximately straight line until struck 
by the batter when its course suddenly changes, so also with a ship, 
that is, in general, the path of a free body varies with the 
external forces acting upon it. In the case of the machine, how- 
ever, the matter is entirely different, for the path of each part is 
predetermined by the designer, and he arranges the whole machine 
so that each part shall act in conjunction with the others to 
produce in each a perfectly defined path. 

Thus, in a steam engine the piston moves in a straight line 
back and forth without turning at all, the crankpin describes a 
true circle, each point on it remaining in a fixed plane, normal 
to the axis of the crankshaft during the rotation, while also the 
motion of the connecting rod, although not so simple is perfectly 
definite. In judging the quality of the workmanship in an 
engine one watches to see how exact each of these motions is 
and how nearly it approaches to what was intended ; for example, 
if a point on the crank does not describe a true circle in a fixed 


plane, or the crosshead does not move in a perfectly straight line 
the engine is not regarded as a good one. 

The same general principle applies to a lathe; the carriage 
must slide along the frame in an exact straight line and the spindle 
must have a true rotary motion, etc., and the lathe in which these 
conditions are most exactly fulfilled brings the highest price. 

These motions are fixed by the designer and the parts are 
arranged so as to constrain them absolutely, irrespective of the 
external forces acting; if one presses on the side of the crosshead 
its motion is unchanged, and if sufficient pressure is produced 
to change the motion the machine breaks and is useless. The 
carriage of the lathe can move only along the frame whether 
the tool which it carries is idle or subjected to considerable 
force due to the cutting of metal; should the carriage be pushed 
aside so that it would not slide on the frame, the lathe would be 
stopped and no work done with it till it was again properly 
adjusted. These illustrations might be multiplied indefinitely, 
but the reader will think out many others for himself. 

This is, then, a distinct feature of the machine, that the relative 
motions of all parts are completely fixed and do not depend in any 
wajr upon the action of external forces. Or perhaps it is better 
to say that whatever external forces are applied, the relative 
paths of the parts are unaltered. 

9. Purpose of the Machine. There remains one other matter 
relative to the machine, and that is its purpose. Machines 
are always designed for the special purpose of doing work. In 
a steam engine energy is supplied to the cylinder by the steam 
from the boiler, the object of the engine is to convert this energy 
into some useful form of work, such as driving a dynamo or 
pumping water. Power is delivered to the spindle of a lathe 
through a belt, and the lathe in turn uses this energy in doing 
work on a bar by cutting a thread. Energy is supplied to the 
crank on a windlass, and this energy, in turn, is taken up by the 
work done in lifting a block of stone. Every machine is thus 
designed for the express purpose of doing work. 

10. Definition of the Machine. All these points may now be 
summed up in the form of a definition: A machine consists of 
resistant parts, which have a definitely known motion relative 
to each other, and are so arranged that a given form of available 
energy may be made to do a desired form of work. 


11. Imperfect Machines. Many machines approach a great 
state of perfection, as for example the cases quoted of the steam 
engine and the lathe, where all parts are carefully made and the 
motions are all as close to those desired as one could make them. 
But there are many others, which although commonly and 
correctly classed as machines, do not come strictly under the 
definition. Take the case of the block and tackle which will 
be assumed as attached to the ceiling and lifting a weight. 
In the ideal case the pulling chain would always remain in a given 
position and the weight should travel straight up in a vertical 
line, and in so far as this takes place the machine may be con- 
sidered as serving its purpose, but if the weight swings, then 
motion is lost and the machine departs from the ideal conditions. 
Such imperfections are not uncommon in machines; the endlong 
motion of a rotor of an electrical machine, the "flapping" of a 
loose belt or chain, etc., are familiar to all persons who have seen 
machinery running; and even the unskilled observer knows that 
conditions of this kind are not good and are to be avoided where 
possible, and the more these incorrect motions are avoided, the 
more perfect is the machine and the more nearly does it comply 
with the conditions for which it was designed. 


12. Divisions of the Subject. It is convenient to divide the 
study of the machine into four parts: 

1. A study of the motions occurring in the machine without 
regard to the forces acting externally; this study deals with the 
kinematics of machinery. 

2. A study of the external forces and their effects on the 
parts of the machine assuming them all to be moving at uniform 
velocity or to be in equilibrium; the balancing forces may then 
be found by the ordinary methods of statics and the problems 
are those of static equilibrium. 

3. The study of mechanics of machinery takes into account the 
mass and acceleration of each of the parts as well as the external 

4. The determination of the proper sizes and shapes to be 
given the various parts so that they may be enabled to carry 
the loads and transmit the forces imposed upon them from 
without, as well as from their own mass. This is machine design, 


a subject of such importance and breadth as to demand an en- 
tirely separate treatment, and so only the first three divisions 
are dealt with in the present treatise. 


13. Plane Motion. It will be best to begin on the first division 
of the subject, and to discuss the methods adopted for obtaining 
definite forms of motion in machines. In a study of the steam 
engine, which has already been discussed at some length, it is 
observed that in each moving part the path of any point always 
lies in one plane, for example, the path of a point on the crankpin 
lies on a plane normal to the crankshaft, as does also the path 
of any point on the connecting rod, and also the path of any 
point on the crosshead. Since this is the case, the parts of an 
engine mentioned are said to have plane motion, by which state- 
ment is simply meant that the path of any point on these parts 
always lies in one and the same plane. In a completed steam 
engine with slide valve, all parts have plane motion but the 
governor balls, in a lathe all parts usually have plane motion, 
the same is true of an electric motor and, in fact, the vast majority 
of the motions with which one has to deal in machines are plane 

14. Spheric Motion. There are, however, cases where different 
motions occur, for example, there are parts of machines where a 
point always remains at a fixed distance from another fixed point, 
or where the motion is such that any point will always lie on the 
surface of a sphere of which the fixed j*fnt /is the center, as in 
the universal and ball and socket joints. Such motion is called 
spheric motion and is not nearly so common as the plane motion. 

15. Screw Motion. A third class of motions occurs where a 
body has a motion of rotation about an axis and also a motion 
of translation along the axis at the same time, the motion of 
translation bearing a fixed ratio to the motion of rotation. This 
motion is called helical or screw motion and occurs quite 

In the ordinary monkey wrench the movable jaw has a, plane 
motion relative to the part held in the hand, the plane motion 
being one of translation or sliding, all points on the screw have 
plane motion relative to the part held, the motion being one of 
rotation about the axis of the screw, and the screw has a helical 
motion relative to the movable jaw, and vice versa. 




It has been noticed already that plane motion is frequently 
constrained by causing a body to rotate about a given axis or by 
causing the body to move along a straight line in a motion of 
translation, the first form of motion may be called turning motion, 
the latter form sliding motion. 

16. Turning Motion. This may be constrained in many ways 
and Fig. 1 shows several methods, where a shaft runs in a fixed 
bearing, this shaft carrying a pulley as shown in the upper left 


Truck A 



FIG. 1. Forms of turning pairs. 

figure, while the lower left figure shows a thrust bearing for the 
propeller shaft of a boat. In the figure (a), there is a pulley P 
keyed to a straight shaft S which passes through a bearing B, 
and if the construction were left in this form it would permit 
plane turning motion in the pulley and shaft, but would not 
constrain it, as the shaft might move axially through B. If, 
however, two collars C are secured to the shaft by screws as 
shown, then these collars effectually prevent the axial motion and 
make only pure turning possible. On the propeller shaft at (6) 
the collars C are forged on the shaft, a considerable number being 


used on account of the great force tending to push the shaft 
axially. Thus in both cases the relative turning motion is neces- 
sitated by the two bodies, the shaft with its collars forming one 
and the bearing the other, and these together are called a turning 
pair for obvious reasons, the pair consisting of two elements. 

It is evident that the turning pair may be arranged by other 
constructions such as those shown on the right in Fig. 1, the form 
used depending upon circumstances. The diagram (c) shows 
in outline the method used in railroad cars, the bearing coming 
in contact with the shaft only for a small part of the cir- 
cumference of the latter, the two being held in contact purely 
because of the connection to the car which rests on top of B, 
and the collars C are here of slightly different form. At (d) 
is a vertical bearing which, in a somewhat better form is often 
used in turbines, but here again it is only possible to insure turn- 
ing motion provided the weight is on the vertical shaft and 
presses it into B. In this case there is only one part correspond- 
ing to the collar C, which is the part of B below the shaft. At 
(e) is a ball bearing used to support a car on top of a truck, the 
weight of the car holding the balls in action. 

17. Chain and Force Closure. In the cases (a) and (6), turn- 
ing motion will take place by construction, and is said to be 
secured by chain closure, which will be referred to later, while 
in the cases (c), (d) and .(e) the motion is only constrained so long 
as the external forces act in such a way as to press the two 
elements of the pair together, plane motion being secured by 
force closure. In cases, such as those described, where force 
closure is permissible, it forms the cheaper construction, as a 
general rule. 

18. Sliding Motion. The sliding pair also consists of two 
elements, and if a section of these elements is taken normal to 
the direction of sliding the elements must be non-circular. As 
in the previous case the sliding pair in practice has very many 
forms, a few of which are shown in Fig. 2, (a), (6), (c) and (d) 
being forms in common use for the crossheads of steam engines, 
(6) and (c) being rather cheaper in general than the others. 
At (e), (/) and (</) are shown forms which are used in automobile 
change gears and other similar places where there is little sliding; 
(e) consists of a gear with a long keyway cut in it while the other 
element has a parallel key, or "feather," fastened to it, so that 
the outer element may slide along the shaft but cannot rotate 



upon it. The construction of the forms (e) and (/) is evident. 
The reader will see very many forms of this pair in machines and 
should study them carefully. 

FIG. 2. Forms of sliding pairs. 

In the automobile engine and in all the smaller gas and gasoline 
engines, the sliding pair is circular, because the crosshead is 
omitted and the connecting rod is directly attached to the piston, 
the latter being circular and not constraining sliding motion. 



In this case the sliding motion is constrained through the con- 
necting rod, which on account of the pairing at its two ends 
will not permit the piston to rotate. The real sliding pair, of 
course, consists of the cylinder and piston, both of which are 
circular, and constrainment is by force closure. 

In the case of sliding pairs also it is possible to have chain 
closure where constraint is due to the construction, as in the cases 
illustrated in Fig. 2; in these cases the motion being one of 
sliding irrespective of the directions of the acting forces. Fre- 

FIG. 3. Sliding pairs. 

quently, however, force closure is used as in the case (6) shown at 
Fig. 3 which represents a planer table, the weight of which 
alone keeps it in place. Occasionally through an accident the 
planer table may be pushed out of place by a pressure on the 
side, but, of course, the planer is not again used until the table is 
replaced, for the reason that the design is such that the table 
is only to have plane motion, a condition only possible if the 
table rests in the grooves in the frame. In Fig. 3 (a) the same 



table is constrained by chain closure and the tail sliding piece of 
the piston rod in Fig. 3 (c) by force closure as is evident. 

19. Lower and Higher Pairs. The two principal forms of 
plane constrained motion are thus turning and sliding, these 
motions being controlled by turning and sliding pairs respect- 
ively, and each pair consisting of two elements. Where contact 
between the two elements of a pair is over a surface the pair is 
called a lower pair, and where the contact is only along a line 
or at a point, the pair is called a higher pair. To illustrate this 
the ordinary bearing may be taken as a very common example 
of lower pairing, whereas a roller bearing has line contact and 
a ball bearing point contact and are examples of higher pairing, 
these illustrations are so familiar as to require no drawings. The 


contact between spur gear teeth is along a line and therefore an 
example of higher pairing. 

In general, the lower pairs last longer than the higher, because 
of the greater surface exposed for wear, but the conditions of the 
problem settle the type of pairing. Thus, lower pairing is used on 
the main shafts of large engines and turbines, but for automobiles 
and bicycles the roller and ball bearings are common. 


20. Formation of Machines. Returning now to the steam 
engine, Fig. 4, its formation may be further studied. The 
valve gear and governor will be omitted at present and the 
remaining parts discussed, these consist of the crank, crankshaft 
and flywheel, the connecting rod, the piston, piston rod and 



crosshead, and finally the frame and cylinder. Taking the 
connecting rod b it is seen to contain two turning elements, one 
at either end, and the real function of the metal in the rod is 
to keep these two elements parallel and at a fixed distance apart. 
The crank and crankshaft a contains two turning elements, one 
of which is paired with one of the elements on the connecting rod 
6, and forms the crankpin, and the other is paired with a corre- 
sponding element on the frame d, forming the main bearing. 
It is true that the main bearing may be made in two parts, both 
of which are made on the frame, as in center-crank engines, or 
one of which may be placed as an outboard bearing, but it will 
readily be understood that this division of the bearing is only a 

FIG. 5. Two-cycle gasoline 

FIG. 6. 

matter of practical convenience, for it is quite conceivable that 
the bearing might be made in one piece, and if this piece were 
long enough it would serve the purpose perfectly. Thus the 
crank consists essentially of two turning elements properly 

Again, the frame d contains the outer element of a turning 
pair, of which the inner element is the crankshaft, and it also 
contains a sliding element which is usually again divided into two 
parts for the purpose of convenience in construction, the parts 
being the crosshead guides and the cylinder. But the two parts 
are not absolutely essential, for in the single-acting gasoline 


engine, the guides are omitted and the sliding element is entirely 
in the cylinder. Of course, the shape of the element depends 
upon the purpose to which it is put; thus in the case last referred 
to it is round. 

Then, there is the crosshead c, with the turning element 
pairing with the connecting rod and the sliding element pairing 
with the sliding element on the frame. The sliding element 
is usually in two parts to suit those of the frame, but it may be 
only in one if so desired and conditions permit of it (see Fig. 2) . 

Thus, the steam engine consists of four parts, each part con- 
taining two elements of a pair, in some cases the elements being 
for sliding, and in others for turning. 

Again, on examining the small gasoline engine illustrated in 
Fig. 5, it will be seen that the same method is adopted here as in 
the steam engine, but the crosshead, piston and piston rod are all 
combined in the single piston c. Further, in the Scotch yoke, 
Fig. 6, a scheme in use for pumps of small sizes as well as on fire 
engines of some makes and for other purposes, there is the 
crank a with two turning elements, the piston and crosshead c 
with two sliding elements, and the block 6, and the frame d, 
each with one turning element and one sliding element. 

21. Links and Chains. The same will be found true in all 
machines having plane motion; each part contains at least two 
elements, each of which is paired with corresponding elements 
on the adjacent parts. For convenience each of these parts 
of the machine is called a Unk, and the series of links so con- 
nected as to give a complete machine is called a kinematic chain, 
or simply a chain. It must be very carefully borne in mind that 
if a kinematic chain is to form part of a machine or a whole 
machine, then all the links must be so connected as to have definite 
relative motions, this being an essential condition of the machine. 

In Fig. 7 three cases are shown in which each link has two 
turning elements. Case (a) could not form part of a machine be- 
cause the three links could have no relative motion whatever, as 
is evident by inspection, while at (6) it would be quite impossible 
to move any link without the others having corresponding 
changes of position, and for a given change in the relative posi- 
tions of two of the links a definite change is produced in the 
others. Looking next at case (c) , it is observed at once that -both 
DC and OD could be secured to the ground and yet AB, BC, and 
OA moved, that is a definite change in AB produces no necessary 



change in OD or in CD, or one link may move without all the 
others undergoing motion or relative change of position. Such 
an arrangement could not form part of a machine because the 
relative motions of the parts are not fixed but variable according 
to conditions. At (d) is a chain which can be used, because if 
any one link move relatively to any other, all the links move re- 
latively, or if one link, say OD, is fastened to the ground and 
OA moved, then must all the other links move. 

22. Mechanisms. When a chain is used as a machine, usually 
one of the links acts as the frame and is fixed to a foundation or 
other stationary body. In studying the motions of various 
links it is not necessary to know the exact shape of the links at all, 


for the motion is completely known if the location and form of 
the pairs of elements is known. Thus, the actual link may be 
replaced by a straight bar which connects the elements of the 
link together, and it will always be assumed that this bar never 
changes its shape or length during motion. Thus, the chain 
will be represented by straight lines and a chain so represented 
having the relative motions of all links completely constrained 
and having one link fixed will be called a mechanism. 

23. Simple and Compound Chains. If the links of a chain 
have only two elements each, the chain is said to be simple, 
but if any link has three or more elements, as AB or BD, in Fig. 
7 (d), the chain is compound. 

24. Inversion of the Chain. Since in forming a mechanism 
one link of the chain is fixed, it would appear that since any of 
the links may be fixed in a given chain, it may be possible to 



change the nature of the resulting mechanism by fixing various 
links successively. Take as an example the mechanism shown 
at (1) Fig. 8, d being the fixed link; here a would describe a circle, 
c would swing about C and b would have a pendulum motion, 
but with a moving pivot B. If b is fixed instead of d, a still 
rotates, c swings about B and d now has the motion b originally 
had, or the mechanism is unchanged. 

If a is fixed then the whole mechanism may rotate, 6 and d 
rotating about A and respectively as shown, and c also rotating, 
the form of the mechanism being thus changed to one in which all 
the links rotate. If, on the other hand, c is fixed, then none of 
the links can rotate, but b and d simply oscillate about B and C 
respectively. The reader will do well to make a cardboard model 
to illustrate this point. 

FIG. 8. Inversion of the chain. 

The process by which the nature of the mechanism is altered by 
changing the fixed link is called inversion of the chain, and in 
general, there are as many mechanisms as there are links in the 
chain of which it is composed, although in the above illustra- 
tion there are only three for the four links. 

25. Slider-crank Chain. This inversion of the chain is very 
well illustrated in case of the chain used in the steam engine, 
which will be referred to in future as the slider-crank chain. 
The mechanism is shown in Fig. 9 with the crank a, connecting 
rod b and piston c, the latter having one sliding and one turning 
element and representing the reciprocating masses, i.e., piston, 
piston rod and crosshead. The frame d is represented by a 
straight line and although it is common, yet the line of motion of c 
does not always pass through 0; however, as shown at (1), it 
represents the usual construction for the ordinary engine. If 
now, instead of fixing d, b is fastened to the foundation, b being 



the longer of the two links containing the two turning elements, 
then a still rotates, c merely swings about Q and d has a swinging 
and sliding motion, and -if c is a cylinder and a piston is attached 

FIG. 9. Inversion of slider-crank chain. 

to d the result is the oscillating engine as shown at (2) Fig. 9, 
and drawn in some detail in Fig. 10. 

If instead of fixing the long rod b with the two turning elements, 
the shorter rod a is fixed as shown at (3) , then b and d revolve 

FIG. 10. Oscillating engine. 

about P and respectively, and c also revolves sliding up and 
down on d. If 6 is driven by means of a belt and pulley at 
constant speed, then the angular velocity of d is variable and the 
device may be used as a quick-return motion; in fact, it is em- 
ployed in the Whitworth quick-return motion. The practical 



form is also shown, Fig. 11, and the relation between the mechan- 
ism and the actual machine will be readily discovered with the 
help of the same letters. 

FIG. 11. Whitworth quick-return motion. 

In the Whitworth quick-return motion, Fig. 11, the pinion 
is driven by belt and meshes with the gear 6. The gear rotates 
on a large bearing E attached to the frame a of the machine, and 
through the bearing E is a pin F, to one side of the center of E, 

FIG. 12. Gnome aeroplane motor. 

carrying the piece d, the latter being driven from 6 by a pin c 
working in a slot in d. The arm A is attached to a tool holder 
at B. 



The Gnome motor used on aeroplanes is also an example of 
this same inversion. It is shown in Fig. 12 and the cylinder 
shown at the top with its rod and piston form the same mechanism 
as the Whitworth quick-return motion, a being the link between 
the shaft and lower connecting-rod centers. Study the mechanism 
used with the other cylinders. 

The fourth inversion found by fixing c is rarely used though it is 
found occasionally. It is shown at (4) Fig. 9. 

There are thus four inversions of this chain and it might be 
further changed slightly by placing Q to one side of the link d 

FIG. 13. Shaper mechanism. 

so that the/line of motion of Q, Fig. 9 (1), passses above 0, giving 
the scheme used in operating the sleeves in some forms of gasoline 
engines, etc. 

A somewhat different modification of the slider-crank chain is 
shown at Fig. 13 a device also used -as a quick-return motion in 
shapers and other machines. On comparing it with the Whit- 
worth motion shown at Fig. 11, and the engine shown at Fig. 10, 
it is seen that the mechanism of Fig. 9 may be somewhat altered 
by varying the proportions of the links. The mechanism illus- 
trated at Fig. 13 should be clear without further explanation. 
D is the driving pinion working in with the large gear b, the tool 



is attached to B which is driven from c by the link A. It is 
readily seen that B moves faster in one direction than the other. 
Further, an arrangement is made for varying the stroke of B 
at pleasure by moving the center of c closer to, or further from, 
that of 6. 

26. Double Slider-crank Chain. A further illustration of a 
chain which goes through many inversions in practice is given 
in Fig. 14 and contains two links, b and d, with one sliding and 
one turning element each, also one link a 
with two turning elements and one c with 
two sliding elements. When the link d 
is fixed, c has a reciprocating motion and 
such a setting is frequently used for small 
pumps driven by belt through the crank 
a (Fig. 14), c being the plunger. A detail 
of this has already been given in Fig. 6. 

With a fixed the device becomes Oldham's coupling which is 
used to connect two parallel shafts nearly in line, Fig. 15. In 
the figure b and d are two shafts which are parallel and rotate 
about fixed axes. Keyed to each shaft is a half coupling with a 
slot running across the center of its face and between these half 
couplings is a peice c with two keys at right angles to each other, 
one on each side, fitting in grooves in b and d. As b and d 

FIG. 14. 

FIG. 15. Oldham's coupling. 

revolve, c works sideways and vertically, both shafts always turn- 
ing at the same speed. Points on c describe ellipses and a modi- 
fication of the device has been used on elliptical chucks and on 
instruments for drawing ellipses. 


1. Define the term machine and show that a gas engine, a stone crusher 
and a planer are machines. Is a plough or a hay rake or hay fork a ma- 
chine? Why? 

2. What are the methods of constrainment employed in the following: 


Line shafting, loose pulley, sprocket chain, engine crankshaft, lathe spindle, 
eccentric sheave, automobile clutch, change gear, belt. Which are by force 
and which by chain closure? 

3. Make a classification of the following with regard to constrainment 
and the form of closure: Gas-engine piston, lathe carriage, milling-machine 
head, ordinary D-slide valve, locomotive crosshead, valve rod, locomotive 
link. Give a sketch to illustrate each. Why would force closure not do 
for a connecting rod? 

4. What form of pairing is used in the cases given in the above two ques- 
tions? Is lower or higher pairing used in the following, and what is the 
type of contact: Roller bearing, ball bearing, vertical-step bearing, cam and 
roller in sewing machine, gear teeth, piston? 

6. Define plane, helical and spherical motion. What form is used in the 
parts above mentioned, and in a pair of bevel gears? 

6. In helical motion if the pitch of the helix is zero, what form of motion 
results; also what form for infinite pitch? 

7. What is the resulting form of motion if the radius for a spherical 
motion becomes infinitely great? 

8. Show that all the motions in an ordinary engine but that of the gover- 
nor balls are plane. What form of motion do the latter have? 

9. Define and illustrate the following terms : Element, lower pair, higher 
pair, link, chain, mechanism and compound chain. 

10. List the links and their elements and give the form of motion and 
method of constrainment in the parts of a locomotive side rod, beam engine, 
stone crusher (Fig. 95) and shear (Fig. 94). 

11. Explain and illustrate the inversion of the chain. Show that the 
epicyclic gear train is an inversion of the ordinary train. 


27. Plane Motion. It is now desirable to study briefly certain 
of the characteristics of plane motion, a term which may be 
defined by stating that a body has plane motion when it moves in 
such a way that any given point in it always remains in one and 
the same plane, and further, that the planes of motion of all 
points in the body are parallel. Thus, if any body has plane 
motion relative to the paper, then any point in the body must 
remain in a plane parallel to the plane of the paper during the 
motion of the body. 

A little consideration will show that in the case of plane motion 
the location of a body is known when the location of any line in 
the body is known, provided this line lies in a plane parallel to 
the plane of motion or else in the plane of motion itself. The 
explanation is, that since all points in the body have plane motion, 
then the projection of the body on the plane is always the same 
for all positions and hence the line in it simply locates the body. 
For example, if a chair were pushed about upon the floor and had 
points marked R and L upon the bottoms of two of the legs, 
then the location of the chair is always known if the positions 
of R and L, that is, of the (imaginary) line RL is known. If, 
however, the chair were free to go up and down from the floor 
it would be necessary to know the position of the projection of 
RL on the floor and also the height of the line above the floor 
at any instant. Further, if it were possible for the chair to be 
tilted backward about the (imaginary) line RL, the position 
of the latter would tell very little about the position of the chair, 
as the tips of its legs might be kept stationary while tilting the 
chair back and forth, the position of RL being the same for various 
angular positions of the chair. 

If the case where a body has not plane motion is considered, 
then the line will tell very little about the position of the body. 
In the case of an airship, for example, the ship may stand at 
various angles about a given line, say the axis of a pair of the 



wheels, the ship dipping downward or rising at the will of the 

28. Motion Determined by that of a Line. Since the location 
of a body having plane motion is known when the location of 
any line in the body is known, then the motion of the body will 
be completely known, if the motion of any line in the body is 
known. Thus let C, Fig. 16, represent the projection on the 
plane of the paper of any body having plane motion, AB being 
any line in this body, and let AB be assumed to be in the plane of 
the paper, which is used as the plane of reference. Suppose now 
it is known that while C moves to C", the points A and B move over 
the paths A A' and BB', then the motion of C during the change 
is completely known. Thus at some intermediate position the 
line is at AiBi and the figure of C can at once be drawn about 
this line, and this locates the posi- 
tion of the body corresponding to 

the location A i#i of the line AB. 

It will therefore follow that the 
\motion of a body is completely 
V known provided only that the 
/ motion of any line in the body is FlG 16 

v known. This proposition is of 

much importance and should be carefully studied and understood. 

29. Relative Motion. It will be necessary at this point to 
grasp some idea of the meaning of relative motion. We have 
practically no idea of any other kind of motion than that referred 
to some other body which moves in space, the moon is said to 
move simply because it changes its position as seen from the earth, 
or a train is said to move as it passes people standing on a rail- 
road crossing. Again, one sees passengers in a railroad car as 
the train moves out and says they are moving, while each pas- 
senger in turn looks at other passengers sitting in the same car 
and says the latter are still. Again, a brakeman may walk 
backward on a flat car at exactly the same rate as the car goes 
forward, and a person on the ground who could just see his head 
would say he was stationary, while the engine driver would say 
he was moving at several miles per hour. If one stood on shore 
and saw a ship go out one would say that the funnel was moving, 
and yet a person on the ship would say that it was stationary. 

These conflicting statements, which are, however, very com- 
mon, would lead to endless confusion unless the essential differ- 


ences in the various cases were grasped, and it will be seen that 
the real difference of view results from the fact that different 
persons have entirely different standards of comparison. Stand- 
ing on the ground the standard of rest is the earth, and anything 
that moves relative to it is said to be moving. The man on the 
flat car would be described as stationary because he does not 
move with regard to the chosen standard the earth, but the 
engine driver would be thinking of the train, and he would say 
the man moved because he moved relative to his standard the 
train. It is easy to multiply these illustrations indefinitely, 
but they would always lead to the same result, that whether a 
body moves or remains at rest depends altogether upon the 
standard of comparison, and it is usual to say that a body is at 
rest when it has the same motion as the body on which the 
observer stands, and that it is in motion when its motion is 
different to that of the body on which the observer stands. 
On a railroad train one speaks of the poles flying past, whereas a 
man on the ground says they are fixed. 

30. Absolute and Relative Motion. When the standard which 
is used is the earth it is usual to speak of the motions of other 
bodies as absolute (although this is incorrect, for the earth 
itself moves) and when any standard which moves on the earth 
is used, the motions of the other bodies are said to be relative. 
Thus the absolute motion of a body is its motion with regard 
to the earth, and the relative motion is the motion as compared 
with another body which is itself moving on the earth. Unless 
these ideas are fully appreciated the reader will undoubtedly meet 
with much difficulty with what follows, for the notion of relative 
motion is troublesome. 

In this connection it should be pointed put that a body secured 
to the earth may have motion relative to another body which is 
not so secured. Thus when a ship is coming into port the dock 
appears to move toward the passengers, but to the person on 
shore the ship appears to come toward the shore, thus the motion 
of the ship relative to the dock is equal and opposite to the motion 
of the dock relative to the ship. 

31. Propositions Regarding Relative Motion. Certain proposi- 
tions will now be self-evident, the first being that if two bodies 
have no relative motion they have the same motion relative to 
every other body. Thus, two passengers sitting in a train have 
no relative motion, or do not change their positions relative to 


one another, and hence they have the same motion or cnange of 
position relative to the earth, or to another train or to any other 
body: the converse of this proposition is also true, or two bodies 
which have the same change of position relative to other bodies 
have no relative motion. 

32. Another very important proposition may be stated as 
follows: The relative motions of two bodies are not affected by 
any motion which they have in common. Thus the motion of 
the connecting rod of an engine relative to the frame is the 
same whether the engine is a stationary one, or is on a steamboat 
or locomotive, simply because in the latter cases the motion of 
the locomotive or ship is common to the connecting rod and 
frame and does not affect their relative motions. 

The latter proposition leads to the statement that if it be 
desired to study the relative motions in any machine it will not 
produce any change upon them to add the same motion to all 
parts. For example, if a bicycle were moving along a road it 
would be found almost impossible to study the relative motions 
of the various parts, but it is known that if to all parts a motion 
be added sufficient to bring the frame to rest it will not in any 
way affect the relative motions of the parts of the bicycle. 
Or if it be desired to study the motions in a locomotive engine, 
then to all parts a common motion is added which will bring 
one part, usually the frame, to rest relatively to the observer, or to 
the observer and to all parts of the machine such a motion is 
added as to bring him to rest relative to them, in fact, he stands 
upon the engine, having added to himself the motion which all 
parts of the engine have in common. So that, whenever it is 
found necessary to study the motions of machines all parts of 
which are moving, it will always be found convenient to add to the 
observer the common motion of all the links, which will bring 
one of them to rest, relative to him. 

To give a further illustration, let two gear wheels a and 6 run 
together and turn in opposite sense about fixed axes. Let a run 
at + 50 revolutions per minute, and b at 80 revolutions per 
minute; it is required to study the motion of b relative to a. 
To do this add to each such a motion as to bring a to rest, that 
is, 50 revolutions per minute, the result being that a turns 
+ 50 50 = 0, while b turns 80 50 = 130 revolutions 
per minute or b turns relative to a at a speed of 130 revolutions 
per minute and in opposite sense to a. Here there has simply 


been added to each wheel the same motion, which does not affect 
their relative motions but has the effect of bringing one of the 
wheels to rest. To find the motion of a relative to 6, bring b 
to rest by adding + 80 revolutions per minute, so that a goes 
+ 50 + 80 = 130 revolutions per minute, or the motion of b 
relative to a is equal and opposite to that of a relative to b. 

33. The Instantaneous or Virtual Center. It has already been 
pointed out in Sec. 27 that when a body has plane motion, the 
motion of the body is completely known provided the motion 
of any line in the body in the plane of motion is known, that is, 
provided the motions or paths of any two points in the body are 
known. Now let c, Fig. 17, represent any body moving in the 
plane of the paper at any instant, the line AB being also in the 
plane of the paper, and let FA / and BE repre- 
sent short lengths of the paths of A and B 
respectively at this instant. The direction of 
motion of A is tangent to the path FA at A, 
and that of B is tangent to the path BE at 
B, the paths of A and B giving at once the 
direction of the motions of these points at the 
instant. Through A draw a normal AO to 
the direction of motion of A, then, if a pin is 
stuck through any point on the line AO into 
the plane of reference and c is turned very 
slightly about the pin it will give to A the 
direction of motion it actually has at the instant. The same 
argument applies to BO a normal to the path at B, and hence 
to the point where AO and BO intersect, that is, if a pin is put 
through the point in the body c and into the plane of reference, 
where the body is in the position shown, the actual motion of 
the body is the same as if it rotated for an instant about this 
pin. is called the instantaneous or virtual center, because it 
is the point in the body c about which the latter is virtually 
turning, with regard to the paper, at the instant. 

In going over this discussion it will appear that may be found 
provided only the directions of motion of A and B are known at 
the instant. The only purpose for which the paths of these 
points have been used was to get the directions in which A and B 
were moving at the instant, and the actual path is unimportant 
in so far as the finding of is concerned. It will further appear 
that the point will, in general, change for each new position 


of the body, because the directions of motion of A and B will 
be such as to change the location of 0. Should it happen, 
however, that A and B moved in parallel straight lines, would 
be at infinity or the body c would have a motion of translation; 
on the other hand, if the points A and B moved around concen- 
tric circles, would be fixed in position, being the common center 
of the two circles, and c would simply rotate about the fixed 
point 0. 

34. Directions of Motion of Various Points. The virtual 
center so found gives much information about the motion of the 
body at the given instant. In the first place it shows that the 
direction of motion of G, with respect to the paper, which has 
been selected as the reference plane, is perpendicular at OG and 
that of H is perpendicular to OH, since the direction of motion 
of any point in a rotating body is perpendi&ilar to the radius 
to the point; thus, when the virtual center is known, the direc- 
tion of motion of every point in the* body is known. It is not 
possible to put down at random to direction of motion of G as 
well as those of A and B because that of G is fixed when those of A 
and B are given; the virtual center does not, however, give the 
path of G but only its direction. 

35. Linear Velocities. In the next place the virtual center 
gives the relative linear velocities of all points in c at the instant. 
Let the body c be turning at the rate of n revolutions per minute, 
corresponding to co radians per second, the relation being co = 

-gQ- At the instant the velocity v of a point situated r ft. from 

will evidently be v = ^ = rco ft. per second, and, since 

co is the same for the whole body, the linear velocity of any point 
is proportional to its distance from the center 0. 

Thus if V A , VB, V Q be used to denote the velocities of the points 
A, B and G respectively then it follows that V A = OA-co, V B = 
OB-u and V G = OG-u, and it will also be clear that even though 
oj is unknown the relations between the three velocities* are 
known and also the sense of the motion. 

36. Information Given by Virtual Centers. The virtual center 
for a body may, therefore, be found, provided only that the 
directions (not necessarily the paths) of motion of two points 
in it are known, and having found this center the directions of" 
motion of all points in the body are known, and their relative 


velocities; and also the actual velocities in magnitude, sense and 
direction will be known if the angular velocity is known. (This 
should be compared with the phorograph discussed in a later 
chapter.) It is to be further noted that the virtual center 
is a double point ; it is a point in the paper and also in c, and the 
motion of any point in c with regard to the paper being perpen- 
dicular to the radius from to that point so also the motion of 
any point in the paper with regard to c is perpendicular to the 
line joining this point to 0. 

Another point is to be noticed, that if the various virtual centers 
are known, then at once the relative motion of c to the paper 
is known. Thus the virtual center of one body with regard to 
another gives always the motion of the one body with regard to 
the other. 

37. The Permanent Center. It has already been pointed out 
that the instantaneous or virtual center is the center for rota- 
tion of any one body with regard to another at a given instant, 
and that the location of this center is changing from one instant 
to the next. There are, however, very many cases where one 
body is joined to another by means of a regular bearing, as in 
the case of the crankshaft of an engine and the frame, or a wagon 
wheel and the body of the axle, or the connecting rod and crank- 
pin of an engine. A little reflection will show that in each of these 
cases the one body is always turning with regard to the other, and 
that the center or axis of revolution has a fixed position with 
regard to each of the bodies concerned, thus in these cases the 
virtual center remains relatively fixed and may be termed the 
permanent center. 

38. The Fixed Center. The permanent center must not be 
confused with the fixed center, which term would be applied to a 
center fixed in place on the earth, but is intended to include only 
the case where the virtual center for the rotation of one body with 
regard to another is a point which remains at the same place in 
each body and does not change from one instant to another. 
The center between the connecting rod and crank and between 
the crankshaft and frame are both permanent, the latter being 
also fixed usually. 

39. Theorem of the Three Centers. Before applying the 
virtual centers in the solution of problems of various kinds, a 
very important property connected with them will be proved. 
Let a, b and c, Fig. 18, represent three bodies all of which have 


plane motion of any nature whatever, and which motion is for 
the time being unknown. Now, generally a has motion relative 
to b, and b has motion relative to c, and similarly c with regard 
to a, in brief all three bodies move in different ways, hence from 
what has been said in Sec. 33, there is a virtual center of a ~ b 1 
which may be called ab, and this is of course also the center 
of b ~ a. Further, there is a virtual center of 6 ~ c, that is be, 
and also a center of c <~ a, which is ca, thus for the three bodies 
there are three virtual centers. Now it will be assumed that 
enough information has been given about the motions of a, b 
and c to determine ab and ac only, and it is required to find be. 

Since be is a point common to both bodies b and c, let it be 
supposed to lie at P' f then P' is a point in both the bodies b and c. 
As a point in b its motion 
with regard to a will be nor- 
mal to _jPl^jO&4_ that is, in 
the direction P'A, because 
the motion of a point in one 
body with regard to another 
body is normal to the line 
joining this point to the vir- 
tual center for the two bodies FIG. 18. 
(Sec. 34). As a point in c, 

the motion of P f ~ a is normal to P f - ac or in the direction P'B, 
j3o that P' has two different motions with regard to a at the 
same time, which is impossible or P' cannot be the virtual 
center of 6 ^ c. Since, however, this is not the point, it shows 
at once that the point be is located somewhere along the line 
ab - ca, or say at P, because it is only such a point as P which 
has the same motion with regard to a whether considered as a 
point in b or in c; thus the center be must lie on the same straight 
line as the centers ab and ac. It is not possible to find the ex- 
act position of be, however, without further information, all 
that is known is the line on which it lies. 

This proposition may be thus stated: If in any mechanism 
there are any three links a, /, g, all having plane motion, then for 
the three links there are three virtual centers of, fg and ag, and 
these three centers must all He on one straight line. 

Two of the centers may be permanent but not the third; in 

1 The sign ~ means "with regard to." 


the steam engine taking the crank a, the connecting rod b and the 
frame d, the centers ab and ad are permanent, but bd is not. 

40. The Locating of the Virtual Centers. The chapter will 
be concluded by finding the virtual centers in a few mechanisms 
simply to illustrate the method, the application being given in 
the next chapter. As an example, consider the chain with four 
turning pairs, which is first taken on account of its simplicity. 
It is shown in Fig. 19, and consists of four links, a, b, c and d, 
of different lengths, d being fixed, and by inspection the four 
permanent centers ab, be, cd and ad, at the four corners of the 
chain, are at once located. It is also seen that there are six 
possible centers in the mechanism, viz., ab, be, cd, da, bd and ac, 

FIG. 19. 

these being all the possible combinations of the links in the 
chain when taken in pairs, and of these six, the four permanent 
ones are found already, and only two others, ac and bd, remain. 
There are two methods of finding them, the first of which is the 
most instructive, and will be given first for that reason. 

By the principle of the virtual center bd may be found if the direc- 
tions of motion of any two points in b ~ d are known. On exam- 
ining ab remember that it is a point in a and also in 6; as a point 
in a it moves with regard to d about the center ad and thus in a 
direction normal to ad - ab or to a itself. And as a point in 6 
it must have the same motion with regard to ~d as it has when 
considered as a point in a; that is, the motion of ab in b with re- 
gard to d is in the direction perpendicular to a. Hence, from 
Sec. 33, the virtual center will lie on the line through ab in the 
direction of a, that is, in a produced. Again be is a point in b 
and c, and as a point in c it moves with regard to d in a direction 



perpendicular to cd - be, or in the direction be - F, and this must 
also be the direction of be as a point in b ~ d, so that the virtual 
center of b ~ d must also lie in the line through be normal to 
be - F, or in c produced. Hence, bd is at the intersection of a 
and c produced. 

This could have been solved by the theorem of the three 
centers, for there will be three centers, ad, ab and bd, for the three 
bodies a, b and d, and these must lie in one straight line, and as 
both ad and ab are known, this gives the line on which bd lies. 
Similarly, by considering the three bodies, b, c and d } and know- 
ing the centers be and cd, another line on which bd lies is iso- 
lated, and hence bd is readily found. To find the center ac it 
is possible to proceed in either of the ways already explained, 
and thus find ac at the intersection of the lines b and d produced. 

41. Sliding Pairs. One other example may be solved, and in 
order to include a sliding pair consider the case shown in Fig. 20, 
in which a is the crank, b the connect- 
ing rod, c the crosshead, piston, etc., cdo 
and d the fixed frame. As before 
there are six centers ad, ab, be, cd, ac, 
bd, of which ad, ab, and be are perma- 
nent and found by inspection. 

To find the center cd it is noticed 
that c slides in a horizontal direction ad 
with regard to d, that is, c has a motion FIG. 20. 

of translation in a horizontal straight 

line, or, what is the same thing, it moves in a circle of infinite 
radius, and the center of this circle must, as before, lie in a line 
normal to the direction of motion of c ~ d. Hence cd lies in 
a vertical line through be or through any other point in the 
mechanism such as ad, and at an infinite distance away. The 
figure shows cd above the mechanism, but it might be below 
just as well. . 

Having found (cd,j the other centers ac and bd may be found by 
the theorem of the three centers. Thus bd lies on be - cd and on 
ad - ab and 'is therefore at their intersection, and similarly -ac 
lies at the intersection of be - ab and ad - cd. 


1. Define motion. What data define the position and motion of a point, 
line, plane figure and body, the latter three having plane motion? 


2. Two observers are looking at the same object; one sees it move while 
to the other it may appear stationary ; explain how this is possible. 

3. When a person in an automobile, which is gaining on a street car, 
looks at the latter without looking at the ground, the car appears to be 
coming toward him. Why? 

4. Explain the difference between relative and absolute motion and state 
the propositions referring to these motions. 

6. The speeds of two pulleys are 100 revolutions per minute in the clock- 
wise sense and 125 revolutions per minute in the opposite sense, respectively; 
what is the relative speed of the former to the latter? 

6. In a geared pump the pinion makes 90 revolutions per minute and the 
pump crankshaft 30 revolutions in opposite sense; what is the motion of the 
pinion relative to the shaft? 

7. Distinguish between the instantaneous and complete motion of a body. 
What information gives the former completely? What is the virtual center? 

8. What is the virtual center of a wagon wheel (a) with regard to the 
earth, (6) with regard to the wagon? 

9. A vehicle with 36-in. wheels is moving at 10 miles an hour; what are 
the velocities in space and the directions of motion of a point at the top of 
the tire and also of points at the ends of a horizontal diameter? Is the 
motion the same for the latter two points? If not, find their relative motion. 

10. A wheel turns at 150 revolutions per minute; what is its angular 
velocity in radians per second? Also, if it is 20 in. diameter, what is the 
linear velocity of the rim? 

11. Give the information necessary to locate the virtual center between 
two bodies. 

12. What is the difference between the virtual, permanent and fixed 
center? State and prove the theorem of the three centers. 

13. Find the virtual centers for the stone crusher or any other somewhat 
complicated machine. 




42. Applications of Virtual Center. Some of the main appli- 
cations of the virtual center discussed in the last chapter are to 
the determination of the velocities of the various points and links 
in mechanisms, and also of the forces acting throughout the 
mechanism due to external forces. The latter question will be 
discussed in a subsequent chapter 'and the present chapter will 
be confined to the determination of velocities and to the repre- 
sentation of these velocities. 

43. Linear and Angular Velocities. There are two kinds of 
velocities which are required in machines, the linear velocities of 
the various points and the angular velocities of the various links, 
and it will be best to begin with the determination of linear 

44. Linear Velocities of Points in Mechanisms. The linear 
velocities of the various points may be required in one of two 
forms, either the absolute velocities may be required or else it 
may be only desired to compare the velocities of two points, that 
is, to determine their relative velocities. The latter problem may 
be always solved without knowing the velocity of any point in 
the machine, the only thing necessary being the shape of the 
mechanism and which link is fixed, while for the determination 
of the absolute velocity of a point in a mechanism that of some 
point or link must be known. 

Again, the two points to be compared may be in one link, or in 
different links, and the solution will be made for each case and an 
effort will be made to obtain solutions which are quite general. 

45. Points in the Same Link. The first case will be that of the 
four-link mechanism, frequently referred to, containing four 
turning pairs and shown in Fig. 21, and the letter d will be used 
to indicate the fixed link. As a first problem, let the velocity 
of any point A i in a be given and that of another point A 2 in the 
same link be required. The six virtual centers have been found 
and marked on the drawing, and the link a has been selected for 
the first example because it has a permanent center which is ad. 



Now the velocity of AI which is assumed given, is the absolute 
velocity, that is the velocity with regard to the earth. From A i 
lay off A iE in any direction to represent the known velocity of A i 
and join ad-E and produce this line outward to meet the line 
A zF, parallel to AiE, in the point F. Then A 2 F will represent 
the linear velocity of A 2 on the same scale that AiE represents 
the velocity of A\. (It is assumed in this construction that ad, 
A i and A 2 are in the same straight line.) The reasoning is 
simple, for a turns with regard to the earth about the center ad 
and hence, since AI and A 2 are on the same link, their linear 
velocities are directly proportional to their distances from ad 
(Sec. 35) so that, 

Linear velocity of AI ad AI A\E 
Linear velocity of A 2 ad A z 


If the linear velocity of A i is given, so that A\E can be drawn 
to scale,, then the construction gives the numerical value of the 
velocity of A 2 , but if the velocity of A i is not given then the above 
method simply gives the relative velocities of AI and A 2. 

Next, let it be required to find the velocity of a point B 2 in b, 
Fig. 22, the velocity of ^i in the same link being given and let d 
be the fixed link as before. Now since it is the absolute velocity 
of BI that is given, the first point is to find the center of bd about 
which b is turning with regard to the earth. The velocity of B\ 
then bears the same relation to that of B% as the respective dis- 
tances of these points from bd, or 

Linear velocity of B\ _ bd - B\ 
Linear velocity of Bz bd Bz 

It is then only necessary to get a simple graphical method of 
obtaining this ratio and the figure indicates one way. First, 


with center bd draw arcs of circles through B\ and Bz cutting 
bd- ab in B'i and B' 2 . Then if B\G be drawn in any direction 
to represent the given velocity of BI, it may be readily shown that 
B'zH parallel to B'iG, will represent the linear velocity of B 2 , or 
the ratio of B\G to B'^H is the ratio of the velocities of BI 
and B 2 . 

The only difference between this and the last case is that in the 
former case the center ad used was permanent, whereas in this 
case the center bd used is a virtual center. 

46. Points in Different Links. If it were required to compare 
the linear velocity of the point AI in a with that of BI in 6 the 
method would be as indicated in Fig. 22. The two links con- 
cerned are a and b and d is the 
fixed link and these links have 
the three centers ad, ab, bd, 
all on one line, also ab is a 
point common to a and b, 
being a point on each link. 
Treating it as a point in a, 
proceed as in the first ex- 
ample to find its velocity. 
Thus set off A iE in any direc- p IG> 22. 

tion to represent the linear 

velocity of AI, then ab-F parallel to A\E will represent the ve- 
locity of ab to the same scale. Now treat ab as a point in b and 
its velocity is given as ab-F, so that the matter now resolves 
itself into finding the velocity of a point BI in 6, the velocity 
of the point ab in the same link being given. 

It must again be remembered that ab F represents the 
absolute velocity of the point ab, that is, the velocity of this 
point, using the fixed frame of the machine as the standard. With 
regard to the frame the link b is turning about the center bd, thus 
for the instant b turns relative to d about bd, and the velocities 
of all points in it at this instant are simply proportional to their 
distances from bd. The velocity of B i is to the velocity of ab in 
the ratio bd-'Bi to bd-ab, and in order to get this ratio con- 
veniently, draw the arc BiB'i with center bd, then join bd-F 
and draw B\G parallel to ab-F to meet bd-F in G, then B'iG 
represents the velocity of B i in the link b on the same scale that 
AiE represents the velocity of AI. 

Notice that in dealing with the various links in finding relative 



velocities it is .necessary to use the centers of the links under con- 
sideration with regard to the fixed link; thus the centers ad and 
bd and the common center ab are used. The reason ad and bd 
are employed, is because the velocities under consideration are 
all absolute. 

To compare the velocity of any point A! in a with that of C\ 
in c, Fig. 23, it would be necessary to use the centers ad, ac and 
cd. Proceeding as in the former case the velocity of ac is found 
by drawing the arc AiL with center ad and drawing LN in any 
direction to represent the velocity of A i on a chosen scale, than 
the line ac- M parallel to LN meeting ad N produced in M 
will represent the velocity of ac. Join cd - M, and draw the arc 
CiC'i with center cd, then C'iK parallel to ac- M, will represent 
the linear velocity of 

FIG. 23. 

A general proposition may be stated as follows: The velocity 
of any point A in link a being given to find the velocity of F in 
/, the fixed link being d. Find the centers ad, fd and af, and 
using ad and the velocity of A, find the velocity of af, and then 
treating af as a point in / and using the center fd, find the velocity 

47. Relative Angular Velocities. Similar methods to the 
preceding may be employed for finding angular velocities in 

Let any body having plane motion turn through an angle 8 

about any axis, either on or off the body, in time t, then the 


angular velocity of the body is defined by the' relation = . As 

all links in a mechanism move except the fixed link, there are in 
general as many different angular velocities as there are moving 
links. The angular velocities of the various links a, b } c } etc., 


will be designated by co a , cot', co c , etc., respectively, the unit being 
the radian per second. 

As in the case of linear velocities, angular velocities may be 
expressed either as a ratio, in which case the result is a pure 
number, or as a number of radians per second, the method de- 
pending on the kind of information sought and also upon the 
data given. Unless the data includes the absolute angular 
velocity of one link it is quite impossible to obtain the absolute 
velocity of any other link and it is only the ratio between these 
velocities which may be found. 

48. Methods of Expressing Velocities. In rinding the rela- 
tive angular velocities between two bodies it is most usual to 

express the result as a ratio, thus , which result is, of course, a 

t pure number, such a method is very commonly employed in 
connection with gears, pulleys and other devices. If a belt con- 
nects two pulleys of 30 in. and 20 in. diameter their velocity ratio 
will be 2 %o = %) that is, when standing on the ground and count- 
ing the revolutions with a speed counter, one of the wheels will 
be found to make two-thirds the number of revolutions the other 
one does, and this ratio is alWiys the same irrespective of the 
absolute speed of either pulley. 

It happens, however, that it may be necessary to know the 
relative angular velocities in a different form, that is, it may be 
desired to know how fast one of the wheels goes considering the 
other as a standard ; the result would then be expressed in radians 
per second. Suppose a gear a turns at 20 revolutions per minute, 
w a = 2.09 radians per second, and meshes with a gear b running 
at 30 revolutions per minute, co 6 = 3.14 radians per second, the 
two wheels turning in opposite sense, then the velocity of a 
with regard to b is co u b = 2.09 ( 3.14) = +5.23 radians 
per second, that is, if one stood on gear 6 and looked at gear a, 
the latter would appear to turn in the opposite sense to b and at a 
rate of 5.23 radians per second. If, on the other hand, one stood 
on a, then dd, since co& co a = 3.14 2.09 = 5.23, b would 
appear to turn backward at a rate of 5.23 radians per second, the 
relative motion of a to b being equal and opposite to that of b 
with regard to a. 

The first method of reckoning these velocities will alone be 
employed in this discussion and the construction will now be 



49. Relative Velocities of Links. Given the angular velocity 
of a link a to find that of any other link b. Find the three centers 
ad, bd and ab, then as a point in a, ab has the linear velocity 
(ad ab) co a and as a point in b, ab has the velocity (bd ab) co&. 
But as ab must have the same velocity whether considered as a 

point in a or in b, then (ad ab) co a = (bd ab) u b , or -- _ 

C0 a 

. The construction is shown in Fig. 24 and will require 
bd ab 

very little explanation. Draw a circle with center ab and radius 
ab ad, which cuts ab bd in G, lay off bd F in any direc- 

FIG. 24. 

tion to represent co on chosen scale, then draw GE parallel to 
bd F to meet ab F in E, and GE will represent the angular 
velocity of b or u b on the same scale. 

Similar processes may be employed for the other links b and c, 
and no further discussion of the point will be given here. The 
general constructions are very similar to those for finding linear 

As in the case of the linear velocities the following general 
method may be conveniently stated : The angular velocity of any 
link a being given to find the angular velocity of a link /, d being 
the fixed link. Find the centers ad, fd and af, then the angular 
velocity co a is to / in the same ratio that/d af is to ad af. 

50. Discussion on the Method. Although the determination 
of the linear and angular velocities by means of the virtual center 


is simple enough in the cases just considered, yet when it is em- 
ployed in practice there is frequently much difficulty in getting 
convenient constructions. Many of the lines locating virtual 
centers are nearly parallel and do not intersect within the limits 
of the drafting board, and hence special and often troublesome 
methods must be employed to bring the constructions within 
ordinary bounds. Further, although it is common to have 
given the motion of one link such as a, and often only the 
motion of one other point or link say /, elsewhere in the mech- 
anism is desired, requiring the finding of only three virtual 
centers, ad, af and df, yet frequently in practice these cannot be 
obtained without locating almost all the other virtual centers 
in the mechanism first. This may involve an immense amount of 
labor and patience, and in some cases makes the method 

51. Application to a Mechanism. A practical example of a 
more complicated mechanism in common use will be worked out 
here to illustrate the method, only two more centers af and bf 
being found than those necessary for the solution of the problem. 
Fig. 25 shows the Joy valve gear as frequently used on locomotives 
and other reversing engines, more especially in England: a rep- 
resents the engine crank, b the connecting rod, and c the piston, 
etc., as in the ordinary engine, the frame being d. One end of a 
link e is connected to the rod b and the other end to a link /, 
the latter link being also connected to the engine frame, while to 
the link e a rod g is jointed, which rod is also jointed to a sliding 
block h, and at its extreme upper end to the slide valve stem V. 
The part ra on which h slides is controlled in direction by the 
engineer who moves it into the position shown or else into the 
the dotted position, according to the sense of rotation desired in 
the crankshaft, but once this piece m is set, it is left stationary 
and virtually becomes fixed for the time. 

A very useful problem in such a case is to find the velocity of 
the valve and stem V for a given position and speed of the crank- 
shaft. The problem concerns three links, a, d and g, the upper 
end of the latter link giving the valve stem its motion, so that 
the three centers ad, ag, and dg are required. First write on all 
the centers which it is possible to find by inspection, such as ad, 
ab, be, be, cd, ac, ef, etc., and then proceed to find the required 
centers by the theorem of three centers given in Sec. 39. The 
centers ag and dg cannot be found at once and it will simplify 



the work to set down roughly in a circle (not necessarily accur- 
ately) anywhere on the sheet points which are approximately 
equidistant, there being one point for each link, in this case eight. 
Now letter these points a, b, c, d, e, f, g, h, to correspond with the 
links. As a center such as ab is found join the points a and b in 
the lower diagram and it is possible to join a fairly large number 






FIG. 25. Joy valve gear. 

of the points at once, then any two points not joined will repre- 
sent a center still to be found. The figure shows by the plain 
lines the stage of the problem after the centers ab, ac, ad, ae, af, 
ag, be, bd, df, be, cd, de, df, dg, dh, ef, eg, gh, have been found, 
which represents the work necessary to find the above three 
centers ag, ad and dg. 

When all points on the lower figure are joined, all the centers 


have been found, and the figure shows by inspection what centers 
can be found at any time, for it is possible to find any center 
provided there are two paths between the two points correspond- 
ing to the center. It may happen that there will appear to be 
two paths between a given pair of points, but on examination it 
may be found that the paths are really coincident lines, in which 
case they will not fix the center and another path is necessary. 
The lower diagram in Fig. 25 shows that the centers ab, ac, ad, de, 
df, are known, while the centers ah, bg, ch, are not known, 
and that the center fg can probably be found as there are the two 
paths fa ag and fd dg between them as well as the path fe 
eg. The center gc could not, however, be found before gd } as 
there would then be only one path ga ac between the points. 

Having now found the centers ad, dg, and ag, proceed as in the 
previous cases to find the velocity of V or the valve from the 
known velocity of a. If the velocity of the crankpin ab is given, 
revolve ad ab into the line ad dg and lay off a'b' B 
to represent the velocity of ab on any scale. Join ad B, then 
ag A parallel to a'b' B gives the velocity of ag. Next join 
dg A and with center dg draw the arc VV, then V'C parallel 
to ag A will represent the velocity of the valve V. The whole 
process is evidently very cumbersome and laborious and is fre- 
quently too lengthy to be adopted. The reader's attention is 
called to the solution of the same problem by a different method 
in Chapter IV. 

52. It must not be assumed that the methods here described 
are not used, because, in spite of the labor involved they are 
frequently more simple than any other method and a number of 
applications of the virtual center are given farther on in the 
present treatise. It is always necessary to do whatever work is 
required in solving problems, the importance of which frequently 
justifies large expenditure of time. In many cases the method 
described in the next chapter simplifies the work, and the reader's 
judgment will tell him in each case which method is likely to 
suit best. 

53. Graphical Representation of Velocities. It is frequently 
desirable to have a diagram to represent the velocities of the 
various points in a machine for one of its complete cycles, as the 
study of such diagrams gives very much information about the 
nature of the machine and of the forces acting on it. Two 



methods are in fairly common use (1) by means of a polar dia- 
gram, (2) by means of a diagram on a straight base. 

To illustrate these a very simple case, the slider-crank 
mechanism, Fig. 26, will be selected, and the linear velocities of 
the piston will be determined, a problem which may be very 
conveniently solved by the method of virtual centers. Let 
the speed of the engine be known, and calculate the linear velocity 
of the crankpin ab; for example, let a be 5 in. long, and let the 

FIG. 26. Piston velocities. 

speed be 300 revolutions per minute, then the velocity of the 

300 5 
crankpin = 2-jr X ^TT X T^ = 13.1 ft. per second. Now be is a 

point on both the piston c and on the rod b and clearly the velocity 
of be is the same as that of c, the latter link having only a motion 
of translation, and further, the velocity of the crankpin ab is 
known, which is also the same as that of the forward end of the 
connecting rod. The problem then is: given the velocity of a 
point ab in b to find the velocity of be in the same link, and from 
what has already been said (Sec. 35), the relation may be written: 

velocity of piston velocity of be bd be 
velocity of crankpin ~~ velocity of ab ~ bd ab 


But by similar triangles 

bd be _ ac ad 
bd ab ~ ad ab 
so that 

velocity of piston _ ad ac 
velocity of crankpin ad ab 

and as ad ab is constant for all positions of the machine, it is 
evident that ad ac represents the velocity of the piston on the 
same scale as the length of a represents the linear velocity of the 
crankpin. Or, in the case chosen, if the mechanism is drawn full 
size then ad ab = 5 in., and the scale will be 5 in. = 13.1 ft. 
per second or 1 in. = 2.62 ft. per second. 

54. Polar Diagram. Now it is convenient to plot this velocity 
of the piston either along a as ad E if the diagram is to show 
the result for the different crank positions, or vertically above the 
piston as be F, if it is desired to represent the velocity for 
different positions of the piston. If this determination for the 
complete revolution is made, there are obtained the two diagrams 
shown, the one OEGOHJO is called a polar diagram, being the 
pole. The diagram consists of two closed curves passing through 
and both curves are similar; in fact the lower one can be' ob- 
tained from the upper by making a tracing of the latter and turn- 
ing it over the horizontal line ad be. The longer the connect- 
ing rod the more nearly are the curves symmetrical about the 
vertical through 0, and for an infinitely long rod the curves are 
circles, tangent to the horizontal line at 0. 

55. Diagram on a Straight Base. The diagram found by 
laying off the velocities above and below the piston positions is 
KFLMK, and, as the figure shows, is egg-shaped with the small 
end of the egg toward 0, and the whole curve symmetrical about 
the line of travel of the piston. Increasing the length of the rod 
makes the curve more nearly elliptical, and with the infinitely 
long rod it is a true ellipse. 

If the direction of motion of the piston does not pass through 
ad, then the curve FKML is not symmetrical about the line of 
motion of the piston, but takes the form shown at Fig. 27, where 
the piston's direction passes above ad, a device in which it is 
clear from the velocity diagram that the mean velocity of the 
piston on its return stroke is greater than on the out stroke, and 
which may, therefore, be used as a quick-return motion in a shaper 


or other similar machine. Engines are sometimes made in this 
way, but with the cylinders only slightly offset, and not as much 
as shown in the figure. 

56. Pump Discharge. One very useful application of such 
diagrams as those just described may be found in the case of 
pumping engines. Let A be the area of the pump cylinder in 
square feet, and let the velocity of the plunger or piston in a 
given position be v ft. per second, as found by the preceding 
method, let Q cu. ft. per second be the rate at which the pump 
is discharging water at any instant, then evidently Q = Av and 
as A is the same for all piston positions, Q is simply proportional 

FIG. 27. Off-set cylinder. 

to v, or the height of the piston velocity diagram represents 
the rate of delivery of the pump for the corresponding piston 

If a pipe were connected directly to the cylinder, the water in 
it would vary in velocity in the way shown in the velocity dia- 
gram (a), Fig. 28, the heights on this diagram representing piston 
velocities and hence velocities in the pipe, while horizontal dis- 
tances show the distances traversed by the piston. The effect 
of both ends of a double-acting pump is shown; this variation in 
velocity would produce so much shock on the pipe as to injure 
it and hence a large air chamber would be put on to equalize the 

Curve (6) shows two pumps delivering into the same pipe, 
their cranks being 90 apart, the heavy line showing that the 
variation of velocity in the pipe line is less than before and re- 
quires a much smaller air chamber. At (c) is shown a diagram 
corresponding to three cranks at 120 or a three-throw pump, 



in which case the variation in velocity in the pipe line would be 
much smaller still, this velocity being represented by the height 
up to the heavy line. All the curves are drawn for the case of a 
very long connecting rod, or of a pump like Fig. 6. 

Thus the velocity diagram enables the study of such a problem 
to be made very accurately, and there are many other useful 
purposes to which it may be put, and which will appear in the 
course of the engineer's experience. Angular velocities may,, of 
course, be plotted the same way as linear velocities. 

&th Stcoke ^ ^ 

(a) Single Cylinder Pump 

Piston Positions 

Resultant Velocity 

(6) Two Cylinders. Cranks at 90 

Resultant Velocity 



(c) Three Cylinders, Cranks at 120 
FIG. 28. Rate of discharge from pumps. 

Another method of finding both linear and angular velocities 
is described in the next chapter, and a few suggestions are made 
as to further uses of these velocities in practice. 


1. In the mechanism of Fig. 21, a, b, c and d are respectively 3, 15, 10 and 
18 in. long, d is fixed and a turns at 160 revolutions per minute. Find the 
velocity of the center of each link with a at 45 to d. 

2. Find also the angular velocities of the links in the same case as above. 

3. JAn 8-in by 10-in. engine has a connecting rod 20 in. long and a speed / 
of 250 revolutions per minute. Find the velocity of the center of the rod and*, 
the angular velocity of the latter in radians per second (a) for the dead 
points, (6) when the crank has moved 45 from the inner dead point, (c) for 
90 and 135 crank angles. 


4. Find the linear velocity of the left end of the jaw in Fig. 84, knowing 
the angular velocity of the camshaft. 

6. Plot a diagram on a straight base for every 15 of crank angle for 
the piston and center of the rod in question 3; also plot a polar diagram of 
the angular velocity of the rod. 

6. In the mechanism of Fig. 27 with a = 6, b = 18 in., and the line of 
travel of c, 9 in. above ad, plot the velocity diagram for c. 

7. What maximum speed will be obtained with a Whitworth quick- 
return motion, Fig. 37, with d = % in., a = 2 in., 6 = 2% in. and e = 13 
in., the line of motion of the table passing through ad ? 

8. Find the velocity of the tool in one of the riveters given in Chapter IX, 
assuming the velocity of the piston at the instant to be known. 

9. Draw the polar diagram for the angular velocity of the valve steam in 
the mechanism of Fig. 40. 


57. Uses of Velocity Diagrams. In the previous chapter 
something has been said about the methods of finding the veloci- 
ties of various points and links in mechanisms, and a few 
applications of the methods were given. As a matter of fact 
there are a very great number of cases in which such velocity 
diagrams are of great value in studying the conditions existing 
in machines. Such problems, for example, as the value of a quick- 
return motion, or of a given type of valve gear or link motion; 
or again, problems involving the action of forces in machines, 
such as the turning moment produced on the crankshaft by a 
given piston pressure, or the belt pull necessary to crush a cer- 
tain kind of stone in a stone crusher, and many other similar 

All of the above problems may be solved by the determination 
of the velocities of various parts and hence the matter of finding 
these velocities deserves some further consideration, more espe- 
cially in view of the fact that a somewhat simpler method than 
that described in Chapter III, and which enables the rapid solu- 
tion of all such problems, is available and may now be discussed. 
The graphical method of solution is usually the best, because it 
is simple and because the designing engineer always has drafting 
instruments available for such a method, and further because 
motions in machines are frequently so complex as to render 
mathematical solutions altogether too cumbersome. 

58. Method to be Used. In all machines there is one part 
which has a definitely known motion, and frequently this motion 
is one of rotation about a fixed center at uniform speed, as in the 
case of the flywheel of an engine, or the belt wheel of a stone 
crusher on punch or machine tool, this part is called the link of / 
reference. Provided the motion of this link is known, it is 
possible to definitely determine the motions of all other parts, 
but if its motion is not known, then all that is possible is the 
determination of the relative motions of the various parts; the 
method described here may be used in either case. 

4 49 



The construction about to be explained has been called by its 
discoverer 1 the phorograph method, and, as the name suggests, 
is a method for graphically representing the motions. It is a 
vector method of a kind similar to that used in determining the 
stresses in bridges and roofs with the important differences that 
the vector used in representing stresses are always parallel to the 
member affected, while the vector representing velocity is in 
many cases normal to the direction of the link concerned and 

further that the diagram is 
drawn on an arbitrarily selected 
Jink of reference which is itself 

59. The Phorograph. The 
phorograph is a construction 
by which the motions of all 
points on a machine may be 
represented in a convenient 
graphical manner. As dis- 
cussed here the only applica- 
tion made is to plane motion 
although the construction may 
readily be modified so as to 

make it apply to non-plane motion, but in most cases of the 
latter kind any graphical construction becomes complicated. 
The method is based pn very few important principles and 
these will first be explained. 

60. Relative Motion of Points in Bodies. First Principle. The 
first principle is that any one point in a rigid body can move 
relatively to any other point in the same body only in a direc- 
tion at right angles to the straight line joining them; that is to 
say, if the whole body moves from one position to another, then 
the only motion which the one point has that the other has not is 
in a direction normal to the line joining them. To illustrate 
this take the connecting rod of an engine, a part of which is shown 
at a in Fig. 29 and let the two points B and C, and hence the 
line BC, lie in the plane of the paper. Let the rod move from 
a to a', the line taking up the corresponding new position B'C'. 
During the motion above described C has moved to C' and B 

1 PROFESSOR T. R. ROSEBRUQH of the University of Toronto discovered 
the method and first gave it to his classes about 25 years ago, but the 
principle has not appeared in print before. 

FIG. 29. 


to B'. Now draw BB l} parallel to CC' and C'B l} parallel to BC, 
then an inspection of the figure shows at once that if the rod had 
only moved to B\C' the points B and C would have had exactly 
the same net motion, that is, one of translation through CC' = 
BBi in the same direction and sense, and hence B and C would 
have had no relative motion. But when the rod has moved to 
a, B has had a further motion which C has not had, namely 


Thus during the motion of a, B has had only one motion not 
shared by C, or B has moved relatively to C through the arc 
BiB', and at each stage of the motion the direction of this arc 
was evidently at right angles to the radius from C', or at right 
angles to the line joining B and C. 

Thus when a body has plane motion any point in the body 
can move relatively to any other point in the body only at right 
angles to the line joining the two points. It follows from this 
that if the line joining the two points should be normal to the 
plane of motion, then the two points could have no relative 

61. Second Principle. Let Fig. 30 represent a mechanism 
having four links, a, b, c and d, joined together by four turning 

FIG. 30. 

pairs 0, P, Q and R as indicated. This mechanism is selected 
because of its common application and the reader will find it 
used in many complicated mechanisms. For example, it forms 
half the mechanism used in the beam engine, when the links are 
somewhat differently proportioned, a being the crank, 6 the con- 
necting rod and c one-half of the walking beam. The same chain 
is also used in the stone crusher shown at Fig. 95 and in many 
other places. 

The second principle upon which the phorograph depends may 
now be explained by illustrating with the above mechanism. 


In this mechanism the fixed link is d which will be briefly 
referred to as the frame. Thus and R are fixed bearings or 
permanent centers, while P and Q move in arcs of circles about 
O and R respectively. Choose one of the links as the link of 
reference, usually a or c will be most suitable as they both have 
a permanent center while b has not; the one actually selected is 
a. Imagine that to a there is attached an immense sheet of 
cardboard extending indefinitely in all directions from O, and 
for brevity the whole sheet will be referred to as a. 

A consideration of the matter will show that on the cardboard 
on the link a there are points having all conceivable motions and 
velocities in magnitude, direction and sense. Thus, if a circle 
be drawn on a with center at 0, all points on the circle will have 
velocities of the same magnitude, but the direction and sense will 
be different; or if a vertical line be drawn through O } all points 
on this line will move in the same direction, that is, horizontal, 
those above moving in opposite sense to those below and all 
points having different velocities. If any point on a be selected, 
its velocity will depend on its distance from 0, the direction of its 
motion will be normal to the radial line joining it to 0, and its 
sense will depend upon the relative positions of the point and 
on the radial line.. The above statements are true whether a 
has constant angular velocity or not, and are also true even 
though moves. 

From the foregoing it follows that it will be possible to 
find a point on a having the same motion as that of any point Q 
in any link or part of the machine, which motion it is desired to 
study; and thus to collect on a a set of points each representing 
the motion of a given point on the machine at the given instant. 
Since the points above described are all on the link a, their 
relative motions will be easily determined, and this therefore 
affords a very direct method of comparing the velocities of the 
various points and links at a given instant. If the motion of a 
is known, as is frequently the case, then the motions of all points 
on a are known, and hence the motion of any point in the mechan- 
ism to which the determined point on a corresponds; whereas, 
if the motion of a is unknown, only the relative motions of the 
different points at the instant are known. 

A collection of points on a certain link, arbitrarily chosen as 
the link of reference, which points have the same motions as 
points on the mechanism to which they correspond, and about 


which information is desired, is called the photograph of the 
mechanism, because it represents graphically (vectorially) the 
relative motions of the different points in the mechanism. 

62. Third Principle. The third point upon which this graph- 
ical method depends is that the very construction of the mechan- 
ism supplies the information necessary for finding in a simple 
way the representative point on the reference link corresponding 
to a given point on the mechanism; this representative point may 
be conveniently called the image of the actual point. Looking at 
the mechanism of Fig. 30, and remembering the first principle 
as enunciated in Sec. 60, it is clear that if it is desired to study the 
motion and velocity of such a point as Q, then the mechanism 
gives the following information at once: 

1. The motion of Q relative to P is normal to QP since P 
and Q are both in the link b, and as P is also a point in link a, 

the motion of Q ~ P in a is normal to QP, or the motion of Q 
in b with regard to a point P in a is known. 

2. Since Q is also a point in c the motion of Q ~ R is normal to 
QR. But R is a point in the fixed link d and hence R is stationary 
as is, so that the motion of Q ~ R is the same as the motion 
of Q ~ 0. Hence the motion of Q with regard to a second point 
in a is known. 

As will now be shown these facts are sufficient to determine on 
a point Q', a having the same motion as Q and the method of 
doing this will now be demonstrated in a general way. 

63. Images of Points. Let there be a body K, Fig. 31, con- 
taining two points E and F, and let K have plane motion of any 


nature whatsoever, the exact nature of its motion being at pres- 
ent unknown. On some other body there is a point G also 
moving in the same plane as K\ the location of G is unknown 
and the only information given about it is that its instantaneous 
motion relative to E is in the direction G 1 and its motion 
relative to F is in the direction G 2. It is required to find a 
point G' on K which has the same motion as (?; the point G f is 
called the image of G. 

Referring to the first principle it is seen that the motion of any 
point in K ~ E is perpendicular to the lir*e joining this point to 
E, for example the motion of F ~ E is perpendicular to FE. Now 
a point G' is to be found in K having the same motion as G, and as 
the direction of motion of G ~ E is given, this gives at once the 
position of the line joining E to the required point; it must be per- 
pendicular to G 1 and pass through E. The point could not 
lie at H for instance, because then 'the direction of motion would 
be perpendicular to HE, which is different to the specified direc- 
tion G 1. Thus G f lies on a line EG' perpendicular to G 1 
through E. 

Similarly it may be shown that G' must lie on a line through 
F perpendicular to G 2, and hence it must lie at the intersec- 
tion of the lines through E and F or at G' as shown in Fig. 31. 
Then G' is a point on K having the same motion as G in some 
external body. 

64. Possible Data. A little consideration will show that it is 
not possible to assume at random the sense or magnitude of the 
motions, but only the two directions. The point G' could, how- 
ever, be found by assuming the data in another form; for example, 
if the angular velocity of K were known and also the magnitude 
direction and sense of motion of G ~ E, G' could be located, and 
then the motion of G~F could be determined, the reader will 
readily. see how this is done. In general the data is given in 
the form stated first, as will appear later. 

Now as to the information given, the discussion farther on will 
show this more clearly but to introduce the subject in a simple 
way let the virtual center of the body at the given instant, with 
regard to the earth, be H and let the body be rotating at this 
instant in a clockwise sense at the rate of n revolutions per min- 
ute or co = -^^radians per second. Then the motion of G in 
space is in the direction normal to G'H, and it is moving to the 


right with a velocity G'H.u ft. per second, where G'H is in feet. 
Further, the motion of G ~ E is in the sense (71 and the veloc- 
ity of G ~ E is EG'.u ft. per second. 

65. Application to Mechanisms. The application of the above 
. principles to the solution of problems in machinery will illustrate 
the method very well, and in doing this' the principles upon which 
the construction depends should be carefully studied, and atten- 
tion paid to the fact that if too much is assumed the different 
items may not be consistent. 

The simple mechanism with four links and four turning pairs 
will be again selected as the first example, and is shown in Fig. 
32, the letters a, 6, c, d, 0, P, Q and R having the same significance 
as in former figures and a is chosen as the link of reference, or 
more conveniently, the primary link, a rough outline being shown 
to indicate its wide extent. In future this outline will be omitted. 
It is required to find the linear velocities of the point S the center 
of 6, of T in c and of Q, also the angular velocities of b and c com- 
pared to a while the mechanism is passing through the position 
shown in the figure. 

Points will first be found on a having the same motions as Q 
and R } these points being the images of Q and R, and are indi- 
cated by accents; thus Q f is a point on a having the same motion 
in every respect as Q actually has. 

Inspection at once shows that since P is a point in a therefore 
P 1 the image of P will coincide with the latter, and if w be the 
angular velocity of a (where co may be constant or variable), 
then the linear velocity of P at the instant is OP.u = aco ft. per 
second, where a is the length in feet of the link a. The direc- 
tion of motion of P is perpendicular to OP and its sense must 
correspond with co. Such being the case, the length OP or a 
represents aco ft. per second, so that the velocity scale iso>:l. 
Again since R is stationary it is essential that R' be located at 
O the only stationary point in the link a. 

The point Q' may be found thus: The direction of motion 
of Q ~ P is perpendicular to QP or b, and hence, from the prop- 
osition given in Sec. 60 to 63, Q f must lie in a line through P f 
(which coincides with P) perpendicular to the motion of Q ~ P, 
that is in a line through P' in the direction of b, or on b produced. 
Again, the direction of motion of Q ~ R is perpendicular to 
QR or c, and since R' at has the same motion as R, both being 
fixed, this is also the direction of motion of Q ~ R' } so that Q' lies 


on a line through R r perpendicular to the motion of Q ~ R, that is, 
on the line through R' in the direction of c. Now as Q f has been 
shown to lie on b or on b produced, and also on the line through 
parallel to c, therefore it lies at the point shown on the diagram 
at the intersection of these two lines. 

66. Images are Points on the Primary Link. It may be well 
again to remind the reader that the point Q r is a point on a but 
that its motion is identical with that of Q at the junction of the 
links 6 and c. If the angular velocity of a is co radians per second, 
then the linear velocity of Q f on a is Q'O.u ft. per second and its 
direction in space is perpendicular to Q'O, and from the sense of 
rotation shown on Fig. 32 it moves to the left. Since the motion 
of Q' is the same as that of Q then Q also moves to the left in 
the direction normal to Q'O and with the velocity Q' ft. per 

67. Images of Links. Since P' and Q f are the images of P and 
Q on b, P'Q' may be regarded as the image of b, and will in future 
be denoted by &'; similarly R'Q'(OQ') will be denoted by d '. By 
a similar process of reasoning it may be shown that since S bisects 
PQ, so will S' bisect P'Q' and also T r may be found from the 
relation R'T' : T'Q' = RT : TQ. 

Since the latter point is of importance and of frequent occur- 
rence, it may be well to prove the method of locating S'. The 
direction of motion of S ~ P is clearly the same as that of Q ~ P, 
that is, perpendicular to PQ or b, but the linear velocity of Q ~ 
P is twice that of S ~ P, both being on the same link and $ 
bisecting PQ. But the motion of P' is the same as P and of Q' is 
the same as Q; hence the motion of Q' ~ P' is exactly the same as 
that of Q~P, so that the velocity of S' ~ P' is one-half that 


of Q' ~ P', or S' will lie on P f Q f and in the center of the latter 
V) line. 

68. Angular Velocities. The diagram may be put to further 
use in determining the angular velocities of b and c when that of 
a is known or the relation between them when that of a is not 
known. Let co& and o> c , respectively, denote the angular velocities 
of 6 and c in space, the angular velocity of the primary link a 
being co radians per second. Now Q and P are on one link b and 
the motion of Q ~ P is perpendicular to QP, and hence the 
velocity of Q ~ P is QP.u b = 6 ft. per second where o>& is the 
angular velocity of 6, and co& is as yet unknown. Again Q r and 
P' are points on the same link a, which turns with the known 
angular velocity co, and hence the velocity of Q' ~ P' is Q'P'.co = 
b' co ft. per second. But from the nature of the case, since Q' has 
the same motion as Q, and P' the same motion as P, the velocity 
of Q ~ P is equal to that of Q' ~ P' } that is, 6w& = 6'w or 


69. Image of Link Represents Its Angular Velocity. The 

above discussion shows that if the angular velocity of a is con- 
stant then the lengths of the images b' and c' represent the angular 

velocities of the links b and c to the scale JT and - respectively, 

(j C 

since b and c are the same for all positions of the mechanism. On 

the other hand, even though o> is variable, at any instant = v' 

co o 

etc., so that there is a direct method of getting the relation 
between the angular velocities in such cases. 

70. Sense of Rotation of Links. The diagram further shows 
the sense in which the various links are turning, and by the 
formulas for the angular velocities these are readily inferred. 

Thus u b = -TV, and starting at the point P, P'Q' = b' is drawn to 

the left and PQ = b to the right, hence the ratio -j- is negative, 

or the link b is at the instant turning in opposite sense to a or 
in a clockwise sense. In the case of the link c the lines R'Q' and 



RQ are drawn upward from R and R f , that is is positive and 


hence a and c are turning in the same sense. 

71. Phorograph a Vector Diagram. The figure OP'Q'R' is 
evidently a vector diagram for the mechanism, the distance of 
any point on this diagram from the pole being a measure of the 
velocity of the corresponding point in the mechanism. The 
direction of the motion in space is normal to the line joining the 
image of the point to 0, and the sense of the motion is known 
from the sense of rotation of the primary link. Further, the 
lengths of the sides of this vector diagram, b'(P'Q'), c'(R'Q') 
and d'(OR') are measures of the angular velocities of these links 
the sense of motion being determined as explained. As d is at 
rest, OR' has no length. 

In Fig. 33 other positions and proportions of a similar mechan- 
ism are shown, in which the solution is given and the results will 
be. as follows: 

FIG. 33. 

At (1) the ratio -r is positive as is also or all links are turning 
in the same sense; at (2) the link a is parallel to c and hence Q 1 
andP 1 lie at P, so that 6' = or w& = -rco = 0, that is, at this in- 
stant the link b has no angular velocity and is either at rest or 
has a motion of translation. Evidently it is not at rest since the 
velocity of all points on it are not zero but are = ao> ft. 
per second. As shown at (3) the links a and b are in one straight 
line and in the phorograph Q' and R f both lie at 0, so that Q'R' = 
c' = 0, and hence u e = 0, in which case the link c is for the 
instant at rest, since both Q' and R' are at 0, the only point at 

b' c' 

rest in the figure. At (4) both the ratios -r and are negative 

o c 

so that 6 and c both turn in opposite sense to a and therefore 
in the same sense as one another. At (5) the parallelogram 
used commonly on the side rods of locomotives, is shown and the 


phorograph shows that Q' and P' coincide with P so that 6' = 
or the side rod b has a motion of translation, as is well known. 
There is the further well-known conclusion that since c 1 = c the 
links a and c turn with the same angular velocity. 

It is to be noted that if the image of any link reduces to a 
single point two explanations are possible: (a) if this point falls 
at the link is stationary for the instant as for d in the former 
figure, and also as for c in (3) of Fig. 33 ; but (6) if the point is 
not at 0, the inference is that all points in the link move in the 
same way or the link has a motion of translation at the instant, 
as for the link b in (2). 

The method will now be employed in a few typical cases. 

72. Further Example. The mechanism shown in Fig. 34 is a 
little more complicated than the .previous ones. Here P', Q f 

FIG. 34. 

and R f are found as before, and since the motions oiS^Q and S > >P 
are perpendicular respectively to SQ and SP, therefore S'P' 
and S'Q' are drawn parallel respectively to SP and SQ, thus 
locating S'. Also T' is located from the relation R'T' : T'Q' = 
RT : TQ (Sec. 67). Next since the motions of U~S&ud U ~ Tare 
given, draw S'U' parallel to SU and U'T' parallel to UT, their 
intersection locating U f . Assuming a to turn at angular veloc- 
ity co in the sense shown, then the angular velocity of SU is 

Cf' JJt TT' rfl' 

w in the same sense as a and that of UT is ~rjm w in opposite 

sense to a (Sec. 70). The linear velocity of U is Ot/'.co ft. per 

73. Image is Exact Copy of Link. There is an important 
point which should be emphasized here and it is illustrated in 
finding the image of the link b. The method shows that the rela- 
tion, of S' to P'Q' is the same as that of S to PQ, or the image 



of the link is an exact copy of the link itself, and although it may 
be inverted and is usually of different size, all lines on the image 
are parallel with the corresponding lines on the mechanism. 
Whenever the image of the link is inverted it simply means that 
the link, of which this is the image, is turning at the given in- 
stant in the opposite sense to the link of reference; if the image is 
the same size as the original link, then the link has the same 
angular velocity as the link reference. 

74. Valve Gear. The mechanism shown in Fig. 35 is very 
commonly used by some engine builders for operating the slide 
valve, OP being the eccentric, RS the rocker arm pivoted to 
the frame at R, ST the valve rod and T the end of the valve 
stem which has a motion of sliding. OP has been selected as 
the link of reference and P', Q', R' and S' are found as before. 

FIG. 35. Valve gear. 

The construction forces T to move horizontally in space, and 
therefore T' must lie on a line through perpendicular to the 
motion of T, that is T' is on a vertical line through 0, and further 
T' lies on a line through S' parallel to ST, which fixes T'. 

Following the instructions given regarding former cases, it is 
evident that the velocity of the valve is OT'.u ft. per second, 
co being the angular velocity of OP. While the other velocities 
are not of much importance yet the figure gives the angular 

velocity of ST as ~om~' w in the opposite sense to a and the 

linear velocity of S is greater than that of T in the ratio OS':OT'. 

75. Steam Engine. The steam engine mechanism is shown in 

Fig. 36, (a) where the piston direction passes through and (6) 

where it passes above with the cylinder offset, but the same 



letters and description will apply to both. Evidently Q f lies 
on P'Q' through P', parallel to PQ, that is, on QP produced", and 
also since the motion of Q in space is horizontal, Q f will lie on the 
vertical through 0. 

The velocity of the piston is OQ'.co in both cases and the angular 

velocity of the connecting rod is ^ -co in the opposite sense to that 

of the crank, since P'Q' lies to the left of P' while PQ lies to the 
right of P, and it is interesting to note that in both cases when 
the crank is to the left of the vertical line through 0, the crank 
and rod turn in the same sense; further that the rod is not turn- 

FIG. 36. 

ing when the crank is vertical because Q' and P' coincide and 
hence b' = 0. Again, the piston velocity will be zero when Q' 
lies, at 0, which will occur when a and b are in a straight line; 
the maximum piston velocity will be when OQ' is greatest and this 
will not occur for the same crank angle in the two cases (a) 
and (b). 

If a comparison is made between the two figures of Fig. 36 
it will be clear that the length OQ' in the upper figure is greater 
than the corresponding length in the lower one, or the upper 
piston is moving at this instant at a higher rate than the lower 
one, but if the whole revolution be examined the reverse will be 
true of other crank positions; in fact, the lower construction is 
frequently used as a quick-return motion (see Fig. 27). 



76. Whitworth Quick-return Motion. The Whitworth quick- 
return motion, already described and shown at Fig. 11, is illus- 
trated at Fig. 37. There are four links a, b, d and e and two 
sliding blocks, c and /, d being the fixed and a the driving 
link which rotates at speed co, and is selected as the primary 
link. P' and Q' are found by inspection. Further, S' lies on a 
vertical line through 0, and R' on a line through Q' parallel to 
QR, but the exact positions of S' and R' are unknown. 


FIG. 37. Whitworth quick-return motion. 

Now P is a point on both a and c. Choose T on b exactly 
below P on a, T thus having a different position on b for each 
position of a. As all links have plane motion, the only motion 
which T can have relative to P is one of sliding in the direction 
of &, or the motion of T ~ P is in the direction of b; hence T' lies 
on a line through P' perpendicular to b. But T is a point on the 
link b and hence T f must be on a line through Q' parallel with b; 
thus T' is determined, and having found T' it is very easy to find 
R l by dividing Q'T' externally at R' so that 

Q'R' = QR. 
Q'T' ~~~~ QT 

The dotted lines show a simple geometrical method for finding 
this ratio, and it is always well to look for some such method as it 
enormously reduces the time involved in the problem. 

Now since S moves horizontally, S' will be located on the 
vertical line through and also on the line R'S' through R' 
parallel to RS. This return motion is frequently used on shapers 


and other machines, the tool holder being attached to the block 
/, so that the tool holder is moving with linear velocity OS'.u and 

the angular velocity of the link b is ~7yn~ '<* in the same sense as a. 

It must be noticed that although P and T coincide, their images 
do not, for T has a sliding motion with regard to P at the rate 
P'T'.c>) ft. per second, and hence both cannot have the same 
velocity. If P' and T' coincided 
then P and T would have the 
same velocity. The velocity dia- 
gram for S for the complete 
revolution of a is shown in Fig. 38. 

77. Stephenson Link Motion. 
The Stephenson link motion 
shown in Fig. 39 involves a 
slightly different method of at- 
tack. The proportions have been 
considerably distorted to avoid F ">' %$&** '" 
confusion of lines. The primary 

link is the crankshaft containing the crank C and the eccen- 
trics E and F, and the scale here will be altered so that 
OC' = 2 X OC, OE' and OF' being similarly treated. The 
scale will then be OC' = ft. per second or J^co : 1. 

The points C", E', F', H', D r and /' are readily located. Further 
choose M on the curved link AGB directly below K on the rocker 
arm LDK and draw lines E'A' } F'B', H'G' and D'K', of unknown 
lengths but parallel respectively to EA, FB, HG and DK. It 
has already been seen (Sec. 73) that the image of each link is 
similar to and similarly divided to the link itself (it is, in fact, a 
photographic image of it) ; hence the link AGB must have an 
image similar to it, that is, the (imaginary) straight lines A'G' and 
G'B' must be parallel to AG and GB and the triangle A'G'B' 
must be similar to AGB. But the lines on which A', G f and B' lie 
are known; hence the problem is simply the geometrical one of 
drawing a triangle A'G'B' similar and parallel to AGB with its 
vertices on three known lines. The reader may easily invent a 
geometrical method of doing this with very little effort, the pro- 
cess being as simple as the one shown in Fig. 37, but the con- 
struction is not shown because the figure is already complicated. 

Having now located the points A f , G' and B' the curved link 
A'G'B' may be made by copying AGB on an enlarged scale and on 



it the point M' may be located similarly to M in the actual link. 
For the purpose of illustrating the problem the image of the 
curved link has been drawn in on the figure although this is not 
at all necessary in locating M'. Since K slides with regard to M , 
then K'M' is drawn normal to the curved link at M' which 
locates K r and then L' is found from the relation LD : DK = 
L'D' : D'K'. 

The construction shows that the curved link is turning in the 

A'B f 

same sense as the crank with angular velocity M"~T17 ' w since 


the scale is such that OE' = 20E. Again, the valve is moving to 


Roach Rod 

FIG. 39. Stephenson link motion. 

the right at the velocity represented by OZ/, and further the 
velocity of sliding of the block in the curved link is represented 
by K'M', both of these on the same scale that OC r = 2.00 
represents the linear velocity of the crankpin. The method 
gives a very direct means of studying the whole link motion. 

78. Reeves Valve Gear. The Corliss valve gear used on the 
Reeves engine is shown diagrammatically in Fig 40, there being 
no great attempt at proportions. Very little explanation is 
necessary; O, R and S are fixed in space, S being the end of the 
rocking valve stem; hence R r and S' are at 0, and the link OP is 
driven direct from the crankshaft through the eccentric connec- 



FIG, 40, Reeves valve gear. 

FIG. 41, 


tion. The point Q on the sliding block e is directly over a mova- 
ble point T on the lever / which lever is keyed to the valve stem. 
The image of the link / is found by drawing S'T' parallel to ST and 
T f is located by drawing Q'T' perpendicular to S'T' or to ST. In 

S'T f 
this position the angular velocity of the valve is ^ times that 

of the link OP, and from this data the linear velocity of the valve 
face is readily found 

79. Joy Valve Gear. This chapter will be concluded by one 
other example here, although the method has been used a good 
deal throughout the book and other examples appear later on. 
The example chosen is the Joy valve gear, shown in Fig. 41, this 
gear having been largely used for locomotives and reversing 
engines. Referring to the figure, a is the crank, b the connecting 
rod, c the crosshead, e or RST, / and g or SWV are links connected 
as shown. The link g, to which the valve stem is connected at 
V, is pivoted to a block h, which ordinarily slides in a slotted link 
fixed in position, but in order to reverse the engine the slotted 
block is thrown over to the dotted position. 

There should be no difficulty in solving this problem, the only 
point causing any hesitation being in drawing OW, which should 
be normal to the direction of motion of the block h. The velocity 
of the valve is OF 1 co in opposite sense to P and of such a 
point as S is OS 1 w in a direction at right angles to OS 1 . 

80. Important Principles. It may be well to call attention to 
certain fundamental points connected with the construction dis- 
cussed. In the first place, it will be seen that the method is a 
purely vector one for representing velocities, and is thus quite 
analogous to the methods of graphic statics.. As in the latter 
case the idea seems rather hard to grasp but the application will 
be found quite simple, and even in complicated mechanisms 
there is little difficulty. Graphical methods for dividing up 
lines and determining given ratios are worth the time spent in 
devising them. 

It should be remembered that in the phorograph the image of 
a link is a true image of the actual link, that is, it is exactly 
similar to it, and similarly divided, and is always parallel to the 
link but may be inverted relative to it. The image may be the 
same size or larger or smaller than the link depending on how 
fast it is revolving; it is in fact exactly what might be seen by 
looking through a lens at the link. If this statement is kept in 


mind, it will greatly aid in the solution of problems and the 
understanding of the method. 


1. Prove that any point in a body can only move relatively to any other 
point, in a direction perpendicular to the line joining them. 

2. Define the phorograph and state the principles involved. 

3. A body a rotates about a fixed center 0; show that all points in it have 
different velocities, either in magnitude, sense or direction. 

4. Show that the phorograph is a vector diagram. What quantities 
may be determined directly by vectors from it? 

6. If the image of a link is equal in length to the link and in the same 
sense, what is the conclusion? What would it be if the image was a point? 
*/ 6. In the mechanism (3) of Fig. 9, let d turn at constant speed, as in the 
Gnome motor; find the phorograph and the angular velocity of the rod 6. 

7. In an engine of 30 in. stroke the connecting rod is 90 in. long, and the 
engine runs at 90 revolutions per minute. Find the magnitude and sense of 
the angular velocity of the rod for crank angles 45, 135, 225 and 315. 

8. Make a diagram of a Walschaert valve gear and find the velocity and 
direction of motion of the valve for a given crank position. 

9. Plot the angular velocity of the jaw and the linear velocity of the 
center G of the crusher given in Chapter XV, Fig. 168. See also Fig. 95. 


81. Forms of Drives. In machinery it is frequently necessary 
to transmit power from one shaft to another, the ratio of the 
angular velocities of the shafts being known, and in very many 
cases this ratio is constant; thus it may be desired to transmit 
power from a shaft running at 120 revolutions per minute to 
another running at 200 revolutions per minute. Various methods 
are possible, for example, pulleys of proper size may be attached 
to the shafts and connected by a belt, or sprocket wheels may be 
used and connected by a chain, as in a bicycle, or pulleys may be 
placed on the shafts and the faces of the pulleys pressed together, 
so that the friction between them may be sufficient to transmit 
the power, a drive used sometimes in trucks, or, again, toothed 
wheels called gear wheels may be used on the two shafts, as in 
street cars and in most automobiles. 

Any of these methods is possible in a few cases, but usually 
the location of the shafts, their speeds, etc., make some one of 
the methods the more preferable. If the shafts are far apart, a 
belt and pulleys may be used, but as the drive is not positive the 
belt may slip, and thus the relative speeds may change, the speed 
of the driven wheel often being 5 per cent, lower than the diam- 
eters of the pulleys would indicate. Where the shafts are fairly 
close together a belt does not work with satisfaction, and then a- 
chain and sprockets are sometimes used which cannot slip, and 
hence the speed ratio required may be maintained. For shafts 
which are still closer together either friction gears or toothed 
gears are generally used. Thus the nature of the drive will 
depend upon various circumstances, one of the most important 
being the distance apart of the shafts concerned in it, another 
being the question as to whether the velocity is to be accurately 
or only approximately maintained, and another being the power 
to be transmitted. 

. 82. Spur Gearing. The discussion here deals only with drives 
of the class which use toothed gears, these being generally used 
between shafts which must turn with an exact velocity ratio which 



must be known at any instant, and they are generally used when 
the shafts are fairly close together. It will be convenient to 
deal first with parallel shafts, which turn in opposite senses, the 
gear wheels connecting which are called spur wheels, the larger 
one being commonly known as the gear, and the smaller one as 
the pinion. Kinematically, spur gears are the exact equivalent 
of a pair of smooth round wheels of the same mean diameter, and 
which are pressed together so as to drive one another by friction. 
Thus if two shafts 15 in. apart are to rotate at 100 revolutions per 
minute and 200 revolutions per minute, respectively, they may 
be connected by two smooth wheels 10 in and 20 in. in diameter, 
one on each shaft, which are pressed together so that they will not 
slip, or by a pair of spur wheels of the same mean diameter, both 
methods producing the desired results. But if the power to be 
transmitted is great the friction wheels are inadmissible on ac- 
count of the great pressure between them necessary to prevent 
slipping. If slipping occurs the velocity ratio is variable, and 
such an arrangement would be of no value in such a drive as is 
used on a street car, for instance, on account of the jerky motion 
it would produce in the latter. 

83. Sizes of Gears. In order to begin the problem in the 
simplest possible way consider first the very common case of a 
pair of spur gears connecting two shafts which are to have a con- 
stant velocity ratio. This is, the ratio between the speeds HI 
and n 2 is to be constant at every instant that the shafts are re- 
volving. Let I be the distance from center to center of the shafts. 
Then, if friction wheels were used, the velocities at their rims 
will be irdiUi and Trd 2 n 2 in. per minute, where di and dz are the 
diameters of the wheels in inches, and it will be clear that the 
velocity of the rim of each will be the same since there is to be no 


since .... 

and i riUi = r 2 n 2 where ri and r 2 are the radii. 

But n + r 2 = r^^~. 

Hence - r 2 + r 2 = I 

or r 2 = ^p I in. 

and ri = - I in. 



Now, whatever actual shape is given of these wheels, the motion 
of the shafts must be the same as if two smooth wheels, of sizes 
as determined above, rolled together without slipping. In 
other words, whatever shape the wheels actually have, the re- 
sulting motion must be equivalent to that obtained by the roll- 
ing together of two cylinders centered on the shafts. In gear 
wheels these cylinders are called pitch cylinders, and their pro- 
jections on a plane normal to their axes, pitch circles, and the 
circles evidently, touch on a line joining their centers, which 
point is called the pitch point. 

84. Proper Outlines of Bodies in Contact. Let a small part 
of the actual outline of each wheel be as shown in the hatched 
lines of Fig. 42, the projections on the wheels being required to 

FIG. 42. 

prevent slipping of the pitch lines. It is required to find the 
necessary shape which these projections must have. 

Let the actual outlines of the two wheels touch at P and let 
P be joined to the pitch point C; it has been already explained 
that there must be no slipping of the circles at C. Now P is a 
point in both wheels, and as a point in the gear 6 it moves with 
regard to C on the pinion a at right angles to PC, while as a 
point on the pinion a it moves with regard to the gear 6 also at 
right angles to PC. Whether, therefore, P is considered as a 
point on a or b its motion must be normal to the line joining it to 
C. A little consideration will show that, in order that this con- 
dition may be fulfilled, the shape of both wheels at P must be 
normal to PC. 


In order to see this more clearly, examine the case shown in 
the lower part of the figure, where the projections are not normal 
to PiC at the point PI where they touch. From the very nature 
of the case sliding must occur at PI, and where two bodies slide 
on one another the direction of sliding must always be along the 
common tangent to their surfaces at the point of contact, that 
is, the direction of sliding must in this case be PiP r . But PI 
is the point of contact and is therefore a point in each wheel, 
and the motion of the two wheels must be the same as if the two 
pitch circles rolled together,' having contact at C. Such being 
the case, if the two projections shown are placed on the wheels, 
the direction of motion at their point of contact should be 
perpendicular to PiC, whereas here it is perpendicular to PiC". 
This would cause slipping at C, and would give the proper 
shape for pitch circles of radii AC' and BC', which would corre- 
spond to a different velocity ratio. Thus C f should lie at C and 
PiP' should be normal to PiC. 

Another method of dealing with this matter is by means of 
the virtual center. Calling the frame which supports the bear- 
ings of a and b, the link d, then A is the center ad and B is bd 
while the pitgh point C is ab. It is shown at Sec. 33 that the 
motion of b with regard to a at the given instant is one of rota- 
tion about the center ab and hence the motion of P in b is normal 
to PC. Where the two wheels are in contact at P there is rela- 
tive sliding perpendicular to PC, that is, at P the surfaces must 
have a common tangent perpendicular to PC. The shape 
shown at PI is incorrect because from Sec. 33 the center ab must 
lie in a normal through PI and also on ad bd from the 
theorem of the three centers, Sec. 39; so that it would lie at 

C'. But if ab were at C\ then = -T-~r, which does not give 

the ratio required. 

85. Conditions to be Fulfilled. From the foregoing the follow- 
ing important statements follow : The shapes of the projections 
or teeth on the wheels must be such that at any point of contact 
they will have a common normal passing through the fixed pitch 
point, and while the pitch circles roll on one another the pro- 
jections or teeth will have a sliding motion. These projections 
on gear wheejs are called teeth, and for convenience in manu- 
facturing, all the teeth on each gear have the same shape, al- 
though this is not at all necessary to the motion. The teeth 



on the pinion are not the same shape as those on the gear with 
which it meshes. 

There are a great many shapes of teeth, which will satisfy 
the necessary condition set forth in the previous paragraph, but 
by far the most common of these are the cycloidal and the in- 
volute teeth, so called because the curves forming them are 
cycloids and involutes respectively. 

86. Cycloidal Teeth. Select two circles PC and P'C, Fig. 43, 
and suppose these to be mounted on fixed shafts, so that the 

FIG. 43. Cycloidal teeth. 

centers A and B of the pitch circles, and the centers of the de- 
scribing circles PC and P'C, as well as the pitch point C, all lie in 
the same straight line, which means that the four circles are 
tangent at C. Now place a pencil at P on the circle PC and let 
all four circles run in contact without slipping, that is, the cir- 
cumferential velocity of all circles at any instant is the same. As 
the motion continues P approaches the pitch circles eC and fC, 
and if the right-hand wheel is extended beyond the circle fCh, the 
pencil at P will describe two curves, a shorter one Pe on the 
wheel eCg and a longer one Pf on the wheel fCh, the points e and 
/ being reached when P reaches the point C, and from the condi- 
tions of motion arc PC = arc eC = fC. 

Now P is a common point on the curves Pe and Pf and also 
a point on the circle PC, which has the common point C with 
the remaining three circles. Hence the motion of P with regard 


to eCg is perpendicular to PC, and of P with regard to fCh is 
perpendicular to PC, that is, the tangents to Pe and Pf at P 
are normal to _PC^ or the two curves have a common tangent, 
and hence a common normal PC at their point of contact, and 
this normal will pass through the pitch point C. Thus Pe and 
Pf fulfil the necessary conditions for the shapes of gear teeth. 
Evidently the points of contact along these two curves lie along 
PC, since both curves are described simultaneously by a point 
which always remains on the circle PC. Now these curves are 
first in contact at P and the point of contact travels down the 
arc PC relative to the paper till it finally reaches C where the 
points, P, C, e and / coincide, so that since Pe is shorter than Pf, 
the curve Pe slips on the curve Pf through the distance Pf Pe 
during the motion from P to C. 

Below C the pencil at P would simply describe the same curves 
over again only reversed, and to further extend these curves, a 
second pencil must be placed at P' on the right-hand circle P'C, 
which pencil will, in moving downward from C, draw the curves 
P'g and P'h, also fulfilling the necessary conditions, the points 
of contact lying along the arc CP r and the amount of slipping 
being P'g - P'h. 

Having thus described the four curves join the two formed on 
wheel eCg, that is gP r and Pe, forming the curve Peg'P'i and the 
two curves on fCh as shown at PfhiP'i, and in this way long curves 
are obtained which will remain in contact from P to P' , the point 
of contact moving relative to the paper, down the arcs PCP f , 
and the common normal at the point of contact always passing 
through C. The total relative amount of slipping is Pfh\P f \ 
Peg'P'i. If now tw r o pieces of wood are cut out, one having its 
side shaped like the curve PeP'i and pivoted at A, while the 
other is shaped like PfP'i and pivoted at B', then from what has 
been said, the former may be used to drive the latter, and the 
motion will be the same as that produced by the rolling of the 
two pitch circles together; hence these shapes will be the proper 
ones for the profiles of gear teeth. 

87. Cycloidal Curves. The curves Pe, Pf, P'g and P'h, which 
are produced by the rolling of one circle inside or outside of 
another, are called cycloidal curves, the two Pe and P'h being 
known as hypocycloids, since they are formed by the describing 
circle rolling inside the pitch circle, while the two curves Pf and 
P'g are known as epicycloidal curves, as they lie outside the pitch 



circles. Gears having these curves as the profiles of the teeth 
are said to have cycloidal teeth (sometimes erroneously called 
epicycloidal teeth), a form which is in very cpmmon use. So 
far only one side of the tooth has been drawn, but it will be 
evident that the other side is simply obtained by making a 
tracing of the curve PeP\ on a piece of tracing cloth, with center 
A also marked, then by turning the tracing over and bringing 
the point A on the tracing to the original center A on the draw- 

FIG. 44. Cycloidal teeth. 

ing, the other side of the tooth on the wheel eCg may be pricked 
through with a needle. The same method may be employed 
for the teeth on wheel fCh. 

The method of drawing these curves on the drafting board is not 
difficult, and may be described. Let C ... 5 in Fig. 44 repre- 
sent one of the pitch circles and the smaller circles the describing 
circle. Choose the arc C5 of any convenient length and divide 
it into an equal number of parts the arcs C-l, 1-2, etc., each 
being so short as to equal in length the corresponding chords. 
Draw radial lines from A as shown, and locate points G 


distance from the pitch circle equal to the radius of the describ- 
ing circle, and from these points draw in a number of circles 
equal in size to the describing circle. Now lay off the arc IM = 
arc 1C, and arc 2N equal the arc C2 or twice the arc Cl, and the 
arc 3R equal three times arc Cl, etc., in this way finding the 
points C, M, N, R, S, which are points on the desired epicycloidal 
curve. Similarly the hypocycloidal curve below the pitch circle 
may be drawn. 

88. Size of Describing Circle. Nothing has so far been said 
of the sizes of the describing circles, and, indeed, it is evident 
that any size of describing circle, so long as it is somewhat smaller 
than the pitch circle, may be used, and will produce a curve ful- 

FIG. 45. 

filling the desired conditions, but it may be shown that when the 
describing circle is one-half the diameter of the corresponding 
pitch circle the hypocycloid becomes a radial line in the pitch 
circle, and for reasons to be explained later this is undesirable. 
The maximum size, of the describing circle is for this reason 
limited to one-half that of the corresponding pitch circle and 
whenlTset of gears are to run together, the describing circle for 
the set is usually half the size of a gear having from 12 to 15 teeth. 
This will enable any two wheels of the set to work properly 

The proof that the hypocycloid is a radial line if the describing 
circle is half the pitch circle, may be given as follows : Let ABC, 
Fig. 45, be the pitch circle and DPC the describing circle, P being 
the pencil, and BP the line described by P as P and B approach 
C. The arc BC is equal to the arc PC by construction, and hence 



the angle PEC at the center E of DPC is twice the angle BDC, 
because the radius in the latter case is twice that in the former. 
But the angles BDC and PEC are both in the smaller circle, 
the one at the circumference and the other at the center, and 
since the latter is double the former they must stand on the same 
arc PC. In other words BP is a radial line in the larger circle 
since DP and DB must coincide. 

89. Teeth of Wheels. In the actual gear the tooth profiles 
are not very long, but are limited between two circles concentric 

FIG. 46. 

with the pitch circle in each gear, and called the addendum and 
root circles respectively, for the tops and bottoms of the 
teeth, the distances between these circles and the pitch circle 
being quite arbitrarily chosen by the manufacturers, although cer- 
tain proportions, as given later, have been generally adopted. 
These circles are shown on Fig. 46 and they limit the path of 
contact to the reversed curve PCPi and the amount of slipping 
of each pair of teeth to PR - PD + P^E - P^F = PR + PiE - 
(PD + PiF), the distances being measured along the profiles of 
the teeth in all cases. Further, since the common normal to the 
teeth always passes through C, then the direction of pressure 
between a given pair of teeth is always along the line joining their 
point of contact to C, friction being neglected, the limiting direc- 
tions of this line of pressure thus being PC and PiC. 


The arc PC is called the arc of approach, being the locus of 
the points of contact down to the pitch point C, while the arc 
CPi is the arc of recess, PI being the last point of contact. Sim- 
ilarly, the angles DAC and CAE are called the angle of approach 
and angle of recess, respectively, for the left-hand gear. The 
reversed curve PCPi is the arc of contact and its length depends 
to some extent on the size of the describing circles among other 
things, being longer as the relative size of the . describing circle 
increases. If this arc of contact is shorter than the distance be- 
tween the centers of two adjacent teeth on the one gear, then only 
one pair of teeth can be in contact at once and the running is 
uneven and unsatisfactory, while if this arc is just equal to the 
distance between the centers of a given pair of teeth on one gear, 
or the circular pitch, as it is called (see Fig. 52), one pair of teeth 
will just be going out of contact as the second pair is coming in, 
which will also cause jarring. It is usual to make PC PI at least 
1.5 times the pitch of the teeth. ~This will, of course, increase the 
amount of slipping of the teeth. 

With the usual proportions it is found that when the number 
of teeth in a wheel is less than 12 the teeth are not well shaped 
for strength of wear, and hence, although they will fulfil the 
kinematic conditions, they are not to be recommended in practice. 

90. Involute Teeth. The second and perhaps the most com- 
mon method of forming the curves for gear teeth is by means of 
involute curves. Let A and B, Fi$ A |7, represent the axes of the 
gears, the pitch circles of which touch at C, and through C draw 
a secant DCE at any angle 6 to the normal to AB, and with 
centers A and B draw circles to touch the secant in D and E. 

Now (Sec. 83) = ^n = T^> so ^ na ^ ^ ne new circles have the 

71 2 

same speed ratios as the original pitch circles. If then a string 
is run from one dotted circle to the other and used as a belt 
between these dotted or base circles as they are called; the proper 
speed ratio will be maintained and the two pitch circles will 
still roll upon one another without slipping, having contact at C. 
Now, choose any point P on the belt DE and attach at this 
point a pencil, and as the wheels revolve it wilt evidently mark 
on the original wheels having centers at A and B, two curves 
Pa and Pb respectively, a being reached when the pencil gets 
down to E, and b being the starting point just as the pencil leaves 



D, and since the point P traces the curves simultaneously they 
will always be in contact relation to the paper at some point 
along DE, the point of contact traveling downward with the 
pencil at P. Since P can only have a motion with regard to the 

FIG. 47. Involute teeth. 

wheel aE normal to the string PE, and its motion with regard 
to the wheel Db is at right angles to PD, it will be at once evident 

FIG. 48. Involute teeth. 

that these two curves have a common normal at the point where 
they are in contact, and this normal evidently passes through C. 
Hence the curves may be used as the profiles of gear teeth 
(Sec, 85). 


4^-u 7 -" 

The method of describing these curves on the drafting board 
is as follows: Draw the base circle 6-5 with center B, Fig. 48, 
and lay off the short arcs 6-1, 1-2, 2-3, 3-4, etc., all of equal 
length and so short that the arc may be regarded as equal in 
length to the chord. Draw the radial lines Bb, B - 1, B -2, etc., 
and the tangents bD, 1-E, 2-F, 3-G, 4-H any length, and 
lay off 4 H = arc 5 4, 3 G = arc 3 5 which equals twice arc 
5-4, 2-F equal three times arc 5-4, 1-E equal four times 
arc 5-4, etc.; then D, E, F } G, H and 5 are all points on the 
desired curve and the latter may now be drawn in and extended, 
if desired, by choosing more points below b. 

91. Involute Curves. The curves Pa and Pb, Fig. 47, are called 
involute curves, and when they are used as the profiles of gear 
teeth the latter are involute teeth. The angle 6 is the angle of 
obliquity, and evidently gives the direction of pressure between 
the teeth, so that the smaller this angle becomes the less will be 
the pressure between the teeth for a given amount of power trans- 
mitted. If, on the other hand, this angle is unduly small, the 
base circles approach so nearly to the pitch circles in size that the 
curves Pa and Pb have very short lengths below the pitch circles. 
Many firms adopt for the angle 14J^, in which case the diameter 
of the base circle is 0.968 (about 3 ^2) that of the pitch circle. 
If the teeth are to be extended inside the base circles, as is usual, 
the inner part is made radial. With teeth of this form the dis- 
tance between the centers A and B may be somewhat increased 
without affecting in any way the regularity of the motion. 
Recently some makers of gears for automobile work have in- 
creased the angle of obliquity to 20, in this way making the 
teeth much broader and stronger. Stub teeth to be discussed 
later, are frequently made in this way, largely for use on auto- 
mobiles. A discussion of the forms of teeth appears in a later 

92. Sets of Wheels with Involute Teeth. Gears with involute 
teeth are now in very common use, and if a set of these is to be 
made, any two of which are capable of working together, then 
all must have the same angle of obliquity. The arc of contact 
is usually about twice the circular pitch and the number of teeth 
in a pinion should not be less than 12 as the teeth are liable to 
be weak at the root unless the angle of obliquity is increased. 

A more complete drawing of a pair of gears having involute 
teeth is shown in Fig. 49. Taking the upper gear as the driver, 


the line of 'Contact will be along DPCE, but the addendum circles 
usually limit the length of this contact to some extent, contact 
taking place only on the part of the obliquity line DE inside the 
addendum line. The larger the addendum circles the longer the 
lines of contact will be and the proportions are such in Fig 49 
that contact occurs along the entire line DE. No contacts can 
possibly occur inside the base circles. 

nterference Line 
Upper Gear 

Interference Line 
Lower Gear 

FIG. 49. Involute teeth. 

93. Racks. When the radius of one of the gears becomes 
infinitely large the pitch line of it becomes a straight line tangent 
to the pitch line of the other gear and it is then called a rack. 
The teeth of the rack in the cycloidal system are made in exactly 
the same manner as those of an ordinary gear, but both the de- 
scribing circles roll along a straight pitch line, generating cycloidal 
curves, having the same properties as those on the ordinary 

For the involute system the teeth on the rack simply have 
straight sides normal to the angle of obliquity, each side of such 
teeth forming the angle 6 with the radius line AC drawn from 
the center of the pinion to the pitch point. 

94. Annular or Internal Gears. In all cases already discusesd 
the pair of gears working together have been assumed to turn in 
opposite sense, resulting in the use of spur gears, but it not in- 
frequently happens that it is desired to have the two turn in the 
same sense, in which case the larger one of the gears must have 
teeth inside the rim and is called an annular or internal gear. 
An annular gear meshes with a spur pinion, and it will be self- 
evident that the annular gear must always be somewhat larger 
than the pinion. 

A small part of annular gears both on the cycloidal and in- 
volute systems is shown at Fig. 50 and the odd appearance of the 



involute internal gear teeth is evident; such gears are frequently 
avoided by the use of an extra spur gear. 

Cycloidal Teeth 

Involute Teeth 

FIG. 50. Internal gears. 

95. Interference. The previous discussion deals with the cor- 
rect theoretical form of teeth required to give a uniform velocity 

FIG. 51. Interference. 

ratio, but with the usual proportions adopted in practice for 
the addendum, pitch and root circles, it is found that in certain 


cases parts of teeth on one of the gears would cut into the teeth 
on the other gear, causing interference. This is most common 
with the involute system and occurs most where the difference 
in size of the gears in contact is greatest; thus interference is 
worst where a small pinion and a rack work together, but it 
may occur, to some extent with all sizes of gears. 

An example will make this more clear. The drawing in Fig. 51 
represents one of the smaller pinions geared with a rack in the 
involute system and it is readily seen that the point of the rack 
tooth cuts into the root of the tooth on the gear at H and that in 
order that the pair may work together it will be necessary either 
to cut away the bottom of the pinion tooth or the top of the 
rack tooth. This conflict between the two sets of teeth is called 

Looking at the figure, and remembering the 'former discussion 
(Sec. 90) on involute teeth, it is seen that contact will be along 
the line of obliquity from C to E and that points on this line CE 
produced have no meaning in this regard, so that if BC denote 
the pitch line of the rack, the teeth of the rack can only be use- 
fully extended up to the line ED, whereas the actual addendum 
line is FG. Thus, the part of the rack teeth between ED and FG, 
as shown hatched on one tooth on the right, cannot be made the 
same shape as the involute would require but must be modified 
in order to clear the teeth of the pinion. The usual practice 
is to modify the teeth on the rack, leaving the lower parts of the 
teeth on the pinion unchanged, and the figure shows dotted how 
the teeth of the rack are trimmed off at the top to make proper 
allowance for this. 

Interference will occur where the point E, Fig. 51 (a), falls below 
the addendum line FG, the one tooth cutting into the other at H 
on the line of obliquity. Where a pinion meshes with a gear which 
is not too large, then the curvature of the addendum line of the 
gear may be sufficient to prevent contact at the point H, in which 
case interference will not occur. As has already been explained, 
interference occurs most when a pinion, meshes with a gear 
which is very much larger, or with a rack. Where a large 
gear meshes with a rack as in the diagram at Fig. 51 (6), the 
interference line DE is above the addendum line FG and hence 
no modification is necessary. 

In Fig. 49 the interference line for the lower gear is inside the 
addendum line and hence the points of these teeth must be cut 



away, but the points of the teeth on the upper gear would be 
correct as the interference line for it coincides with its adden- 
dum line. 

96. Methods of Making Gears. Gear wheels are made in 
various ways, such as casting from a solid pattern, or from a 
pattern on a moulding machine containing only a few teeth, 
neither of which processes give the most accurate form of tooth. 
The only method which has been devised of making the teeth 
with great accuracy is by cutting them from the solid casting, 
and the present discussion deals only with cut teeth. In order 
to produce these, a casting or forging is first accurately turned 
to the outside diameter^^lheKeeth, that is to the diameter of 
the addendum line, and the metal forming the spaces between the 
teeth is then carefully oBKmt by machine, leaving accurately 
formed teeth if the work is well done. Space does not permit 
the discussion of the machinery for doing this class of work, for 
various principles are used in them and a number of makes of 
the machines will produce theoretically correct tooth outlines. 
The reader will be able to secure information from the builders 
of these machines himself. 

FIG. 52. 


97. Parts of Teeth. The various terms applied to gear teeth, 
either of the involute or cycloidal form, will appear from Fig. 
52. The addendum line is the circle whose diameter is that of 
the outside of the gear, the dedendum line is a circle indicating 
the depth to which the tooth on the other gear extends; usually 
the addendum and dedendum lines are equidistant from the 
pitch line. The teeth usually are cut away to the root circle 
which is slighly inside the dedendum circle to allow for some 
clearance, so that the total depth of the teeth somewhat exceeds 


the working depth or distance between the addendum and de- 
dendum circles. The dimension or length of the tooth parallel 
to the shaft is the width of face of the gear, or often only the 
face of the gear, while the face of the tooth is the surface of the 
latter above the pitch line and the flank of the tooth is the surface 
of the tooth below the pitch line. The solid part of the tooth 
outside the pitch line is the point and the solid part below the 
pitch line is the root. 

Two systems of designating cut teeth are now in use, the one 
most commonly used being by Brown and Sharpe and it will 
first be described. 

Let d be the pitch diameter of a gear having t teeth, h\ the depth 
of the tooth between pitch and addendum circles, and /i 2 the depth 
below the pitch circle, so that the whole depth of the tooth is 
h = hi + h 2 , while the working depth is 2hi. The distance 
measured along the circumference of the pitch circle from center 
to center of teeth is called the circumferential or circular pitch 
which is denoted by p and it is evident that pt = ird. In the case 
of cut teeth the width W of the tooth and also of the space along 
the pitch circle are equal, that is, the width of the tooth measured 
around the circumference of the pitch circle is equal to one-half 
the circular pitch. The statements in the present paragraph are 
true for all systems. 

98. System of Teeth Used by Brown and Sharpe. Brown and 
Sharpe have used very largely the term diametral pitch which is 
defined as the number of teeth divided by the diameter in inches 
of the pitch circle, and the diametral pitches have been largely 
confined to whole numbers though some fractional numbers have 
been introduced. Thus a gear of 5 diametral pitch means one 
in which the number of teeth is five times the pitch diameter in 
inches, that is such a gear having a pitch diameter of 4 in. would 

have 20 teeth. Denoting the diametral pitch by q then q = - 

and from this it follows that pq = TT or the product of the diame- 
tral and circular pitches is 3.1416 always. The circular pitch 
is a number of inches, the diametral pitch is not. 

The standard angle of obliquity used by Brown and Sharpe 

is 14J^ and further hi= in.,/i2 = + ^ in., clearance = ^ 

in., and the width W of the tooth is , so that there is no side 


clearance or back lash between the sides of the teeth. 


99. Stub Tooth System. Recently the very great use of 
gears for automobiles and the severe service to which these 
gears have been put has caused manufacturers to introduce what 
is often called the " Stub Tooth " system in which the teeth are 
not proportioned as adopted by Brown and Sharpe. Stub teeth 
are made on the involute system with an obliquity of 20 usually, 
and are not cut as deep as the teeth already described. The di- 
mensions of the teeth are designated by a fraction, the numerator 
of which indicates the diametral pitch used, while the denomina- 
tor shows the depth of tooth above the pitch line. A % gear 
is one of 5 diametral pitch and having a tooth of depth hi = 
J<7 in. above the pitch line (in the Brown and Sharpe system 
Tii would be J^ in. for the same gear). 

The usual pitches with stub tooth gears are , j^f, %, %, %Q, 
%l> 1 ?f 2 an d 1 M4- Some little difference of opinion appears 
to exist with regard to the clearance between the tops and roots 
of the teeth, the Fellows Gear Shaper Co. making the clear- 
ance equal to one-quarter of the depth hi. Thus, a % gear 
would have the same hi as is used in the Brown and Sharpe 
system for a 7 diametral pitch gear, that is hi = 0.1429 in., and 
a clearance equal to 0.25 X J<f = 0.0357 in., which is much 
greater than the 0.0224 in. which would be used in the Brown 
and Sharpe system on a 7 pitch gear. 

100. The Module. In addition to the methods already ex- 
plained of indicating the size of gear teeth, by means of the cir- 
cular and diametral pitch, the module has also to some extent 
been adopted, more especially where the metric system of meas- 
urement is in use. The module is the number of inches of 
diameter per tooth, and thus corresponds with the circular pitch, 
or number of inches of circumferences per tooth, and is clearly 
the reciprocal of the diametral pitch. Using the symbol m for 
the module the three numbers indicating the pitch are related 
as follows : 

1 1 TT 

m = -; also a = - 

q' m p 

. The module is rarely expressed in other units than millimeters. 

101. Examples. A few illustrations will make the use of the 
formulas clear, and before working these it is necessary to re- 
member that any pair of gears working together must have the 
same pitch and a set of wheels constructed so that any two may 


work together must have the same pitch and be designed on the 
same system. 

Let di and d z be the pitch diameters of two gears of radii 
ri and r 2 respectively and let these be placed on shafts I in. apart 
and turning at HI and n 2 revolutions per minute. Then from 
Sec. 83, where spur gears only are used, 


= __ 

HI -h n z 

Suppose I = 9 in. between centers of shafts which turn at 
100 revolutions and 200 revolutions per minute; then, substitut- 
ing in the above formula n = inn 7-57^ X 9 = 6 in. and r 2 = 

J.UU "T~ ^UU 


100 -H200 * 9 = 3 m ' J or tlie gears w ^ ke 1- m> anc * ^J n ' ^" 
ameter respectively. If cut to 4 diametral pitch the numbers of 
teeth will be ti = 4 X 12 = 48 and t 2 = 4 X 6 = 24. The cir- 
cular pitch is 4 = 0.7854 in. Further, hi = Y in. and the 
outside diameters of the gears are 12 J^ in. and 6% in., the tooth 
clearances = ~~^ = 0.0393 in. The nodule would be J in. 

6 in. 
~ 24 teeth* 

If the gears have stub teeth of four-fifths size, then the numbers 
of teeth will be 48 and 24 as before, but hi will be % = 0.2 in., so 
that the outside diameters will be 12.4 in. and 6.4 in. respectively, 
the clearance will be J4 X J4 = 0.0625 in. and the total depth 
of the teeth 0.4625 in. as compared with 0.5393 in. for the teeth 
on the Brown and Sharpe system. 

Inasmuch as it is rather more usual to use the outside diameter 
of a gear than the pitch diameter in shops where they are made, 
it is very desirable that the reader become so familiar with the 
proportions as to be able to know instantly the relations between 
the different dimensions of the gears in terms of the outside di- 
ameter aud the pitch. 


102. Discussion of the Gear Systems. The involute form of 
tooth is now more generally used than the cycloidal form. In the 
first place the profile is a single curve instead of the double one 
required with the cycloidal shape. Again, because of its construc- 
tion, it is possible to separate the centers of involute gears without 
causing any unevenness of running, that is, if the gears are de- 
signed for shafts at certain distance apart this distance may be 
slightly increased without in the least altering the velocity ratio 
or disturbing the evenness of the running, this is an advantage 
not possessed by cycloidal teeth. 

In cycloidal teeth the direction of pressure between a given 
pair of teeth is variable, being always along the line joining the 
pitch point to the point of contact, and when the point of con- 
tact is the pitch point, the direction of pressure is normal to the 
line joining the shaft centers. In the involute teeth the pressure 
is always in the same direction being along the line of obliquity, 
and thus the pressure between the teeth and the force tending 
to separate the shafts is somewhat greater in the involute form, 
although there is no very great advantage in cycloidal teeth 
from this point of view. The statements in this paragraph as- 
sume that there is no friction between the teeth. 

Interference is somewhat greater in involute teeth. 

As regards the Brown and Sharpe proportions and the stub 
teeth, of course the large angle of obliquity of the latter teeth 
increases the pressure for the power transmitted. The stub tooth 
gears are, however, stronger and there is very little interference 
owing to the shorter tooth. The teeth would possibly be a 
little cheaper to cut, and this as well as their greater strength 
would give them a considerable advantage in such machines as 

103. Helical Teeth. A study of such drawings as are shown at 
Fig. 46, etc., shows that the smaller the depth of the teeth the 
less will be the amount of slipping and therefore the less the 
frictional loss. But this is also accompanied by a decrease in the 
arc of contact and hence the number of teeth in contact at any 
one time will, for a given pitch, be decreased, which may cause 
unevenness in the motion. If, however, the whole width of the 
gear be assumed made up of a lot of thin discs, and if, after the 
teeth had been cut across all the discs at once, they were then 
slightly twisted relatively to one another, then the whole width 
of the gear would be made up of a series of steps and if these 


steps were made small enough the teeth would run across the 
face of the gear as helices, and the gear so made would be called 
a helical gear. The advantage of such gears will appear very 
easily, for instead of contact taking place across the entire width 
of the face of a tooth at one instant, the tooth will only gradually 
come into contact, the action beginning at one end and working 
gradually over to the other, and in this way very great evenness 
of motion results, even with short teeth of considerable pitch. 

The profile of the teeth of such gears is made the same, on a 
plane normal to the shaft, as it would be if they were ordinary 
spur gears. 

FIG. 53. Double helical gears. 

Helical gears are a necessity in any case where high speed and 
velocity ratio are desired, and the modern reduction gear now 
being much used between steam turbines and turbine pumps and 
dynamos would be a failure, on account of the noise and vibra- 
tion, if ordinary spur gears were used. Such reduction gears are 
always helical and frequently two are used with the teeth run- 
ning across in opposite directions so as to avoid end thrust. 
Some turbines have been made in which such gears, running at 
speeds of over 400 revolutions per second, have worked without 
great noise. 

A photograph of a De Laval double helical gear is shown in 
Fig. 53, this gear being used to transmit over 1,000 hp. without 
serious noise. In the figure the teeth run across the face of the 
gear at about 45, and are arranged to run in oil so as to prevent 
undue friction. 



A shaft running at 320 revolutions per minute is to drive a second one 
20 in. away at 80 revolutions per minute, by means of spur gears; find their 
pitch diameters. 

X f2. If both shafts were to turn in the same sense at the speeds given and 
were 4 in. apart, what would be the sizes of the gears? 

3. What is the purpose of gear teeth and what properties must they pos- 

4. Define cycloid, hypocycloid and epicycloid. 

fa. Draw the gear teeth, cycloidal system, for a gear of 25 teeth, 2^ pitch 
with 3-in. describing circle, (a) for a spur gear, (6) for an annular gear. 

16. A pair of gears are to connect two shafts 9 in. apart, ratio 4 to 5; the 
diametral pitch is to be 2 and the describing circles 1^ in. diameter. Draw 
the teeth. 

7. Define the various terms used in connection with gears and gear teeth. 

8. Define involute curve, angle of obliquity, base circle. Shoy that all 
involute curves from the same base circle are identical. / 

V9. Lay out the gears in problem 6, for involute teeth, obliquity 14^. 

10. What is meant by interference in gear teeth and what is the cause of 
it? Why does it occur only under some circumstances and not always? 

11. How may interference be prevented? What modification is usually 
made in rack teeth? 

\12. What effect has variation of the angle of obliquity on involute teeth? 
13. What are the relative merits of cycloidal and involute teeth? 
\14. Obtain all the dimensions of the following gears: (a) Two spur 
wheels, velocity ratio 2, pitch 2, shafts 9 in. apart. (6) Outside diameter 
of gear 4 in., diametral pitch 8. (c) Gear of 50 teeth, 4 pitch, (d) Wheel 
of 103^ in. outside diameter, 40 teeth, (e) Pair of gears, ratio 3 to 4, 
smaller 6 in. pitch diameter with 30 teeth. 

16. What is a stub tooth, and what are the usual, -proportions? What 
advantages has it? 

16. Describe the various methods of giving the sizes of gear teeth and 
find the relation between them. 

v 17. Two gears for an automobile are to hav^e.a velocity ratio 4 to 5, shaft 
centers 4^ in., 6 pitch; draw the corree| teeth jp the 20 stub system. 
18. Explain the construction of the helical ejfear and state its advantages. 


104. Gears for Shafts not Parallel. Frequently in practice the 
shafts on which gears are placed are not parallel, in which case 
the spur gears already described in the former chapter cannot 
be used and some other form is required. The type of gearing 
used depends, in the first place, on whether the axes of shafts 
intersect or not, the most common case being that of intersecting 
axes, such as occurs in the transmission of automobiles, the con- 
nection between the shaft of a vertical water turbine and the 
main horizontal shaft, in governors, and in very many other well- 
known cases. 

On the other hand it not infrequently happens that the shafts 
do not intersect, as is true of the crank- and camshafts of many 
gas engines, and of the worm-gear transmissions in some motor 
trucks. In many of these cases the shafts are at right angles, as 
in the examples quoted, but the cases where they are not are 
by no means infrequent and the treatment of the present chapter 
has been made general. 

105. Types of Gearing. Where the axes of the shafts inter- 
sect the gears connecting them are called bevel gears. Where 
the axes of the shafts do not intersect the class of gearing depends 
upon the conditions to be fulfilled by it. If the work to be done 
by the gearing is of such a nature that point contact between the 
teeth is sufficient, then screw or spiral gearing is used, a form of 
transmission very largely used where the shafts are at rightangles, 
although it may also be used for shafts at other angles. One 
peculiarity of this class is that the diameters of the gears are not 
determined by the velocity ratio required, and in fact it would 
be quite possible to keep the velocity ratio between a given pair 
of shafts constant and yet to vary within wide limits the relative 
diameters of the two gears used. 

Where it is desired to maintain line contact between the teeth 
of the gears on the two shafts, then the sizes of the gears are 
exactly determined, as for spur gears, by the velocity ratio and 




also the angle and distance between the shafts. Such gears are 
called hyperboloidal or skew bevel gears and are not nearly so 
common as the spiral gears, but are quite often used. 

The different forms of this gearing will now be discussed, and 
although a general method of dealing with the question might 
be given at once, it would seem better for various reasons to 
defer the general case for a while and to deal in a special way with 
the simpler and more common case afterward giving the general 


106. Bevel Gearing. The first case is where the axes of the 
shafts intersect, involving the use of bevel gearing. The inter- 
secting angle may have any value from nearly zero to nearly 
180, and it is usual to measure this angle on the side of the 
point of intersection on 
which the bevel gears are 
placed. A very common 
angle of intersection is 90 
and if in such a case both 
shafts turn at the same speed 
the two wheels would be 
identical and are then called 
mitre gears. The type of 
bevel gearing corresponding 
to annular spur gearing is 
very unusual on account of 
the difficulty of construction, 
and because such gears are 
usually easily avoided, how- 
ever they are occasionally 

Let A and B, Fig. 54, rep- 
resent the axes of two shafts intersecting at the point C at 
angle 6, the speeds of the shafts being, respectively, ni and n 2 
revolutions per minute; it is required to find the sizes of the 
gears necessary to drive between them. Let E be a point cf 
contact of the pitch lines of the desired gears and let its dis- 
tances from A and B be TI and r 2 , these being the respective 
radii. Join EC. 

Now from Sec. 83 it will be seen that ntti = r 2 n 2 since the 



pitch circles must have the same velocity, there being no .slipping 

between them, and hence = - - is constant, that is at any point 

?"2 n\ 

where the pitch surfaces of the gears touch, must be constant, 

a condition which will be fulfilled by any point on the line EC. 
In the case of bevel gearing, therefore, the pitch cylinders re- 
ferred to in Sec. 83 will be replaced by pitch cones with apex at 
Cj which cones are generated by the revolution of the line CE 
about the axes A and B. Two pairs of frustra are shaded in 
on Fig. 54 and both of these would fulfil the desired conditions 
for the pitch surfaces of the gear wheels, so that in the case of 
bevel gearing the diameters of the gears are not fixed as with 
spur wheels, but the ratio between these diameters alone is fixed 
by the velocity ratio desired. The angles at the apexes of the 
two pitch cones are 20i and 20 2 as indicated. If 6 = 90 and 
HI = n 2 , then 61 = 6 2 = 45 and this gives the case of the mitre 

In going over the discussion it will be observed that when 
6 and ni/n z are known, the angles 0i and 2 and hence the pitch 

cones are fully determined 
but the designer has still 
the option of selecting one 
of the radii, for example 
7*1, at his pleasure^ after 
which the other one, r 2 , is 

107. Proper Shape of 
Teeth on Bevel Gears. 
It is much beyond the pur- 
pose of the present treatise 
to go into a discussion of 
the exact method of ob- 
taining the form of teeth 

for bevel wheels, 'because such a method is indeed complicated, 
and the practical approximation produces very accurate teeth 
and will be described. Let Fig. 55 represent one of the wheels 
with angle 26, at the vertex of the pitch cone and radii r\ and r/ 
selected in accordance with the previous discussion, the power 
to be transmitted, and other details fixed by the place in which 
the gear is to be used. The diameter 2r*i is the pitch diameter 

FIG. 55. Bevel gears. 



of the gear, while the distance CF is called the cone distance 
and FH the back cone distance, 61 is the pitch angle and other 
terms are the same as are used for spur gears and already 
explained. The lines DG and FH are normal to CF and in- 
tersect the shaft at G and H respectively. 

The practical method is to make the teeth at F the same shape 
and proportions as they would be on a spur gear of radius FH, 
while at D they correspond to the teeth on a spur gear of radius 
DG, and a similar method is used for any intermediate point. 
The teeth should taper from F to D and a straight edge passing 
through C would touch the tooth at any point for its entire length. 
Either the involute or cycloidal system may be used. 

FIG. 56. Spiral tooth bevel gears. 

108. Spiral Tooth Bevel Gears. Within the past few years the 
Gleason Works, and possibly others, have devised a method for 
cutting bevel gears with a form of " spiral 7 ' tooth of the same 
general nature as the helical teeth used with spur gears and de- 
scribed at Sec. 103. A cut of a pair of these from a photograph 
kindly supplied by the Gleason Works is given at Fig. 56, and 
shows the general appearance of the gears. Nothing appears to 
be gained in the way of reducing friction, but they run very 
smoothly and noiselessly and the greater steadiness of motion 
makes them of value in automobiles and other similar machines, 
in which they are mainly used at present. 




109. Following the bevel gearing the next class logically is 
the hyperboloidal gearing and the treatment of this includes the 
general case of all gearing having line contact between the teeth. 
Let AO and BP, Fig. 57, represent the axes of two shafts which are 
to be geared together, the line OP being the shortest line between 
the axes and is therefore their common perpendicular. Let the 
axes of the shafts be projected in the ordinary way on two planes, 
one normal to OP and the other passing through OP and one of 
the axes AO, the projections on the former plane being AO, OP 
and PB while those on the second plane are A'O', O'P' and P'B'. 





n l 





FIG. 57. 

On the latter plane the shaft axes will appear as parallel straight 
lines with O'P' as their common perpendicular, while on the 
former plane OP appears as a single point where AO and BP 
intersect. The angle A OB = 6 is the angle between the shafts 
and the distance O'P' is the distance between them and when 
and O'P' = h are known the exact positions of the shafts are 
given. The speeds of the shafts n\ and n z must also be known, 
as well as the sense in which they are to turn. 

In stating the angle between the shafts it is always intended 
to mean the angle in which the line of contact must lie, thus in 
Fig. 57 the sense of rotation would indicate that the line of con- 
tact CQ must lie somewhere in the angle AOB and not in AiOB 
so that the angle B = A OB is used instead of A\OB. Should 
the shaft AO turn in the opposite to that shown, then the line 


of contact would be in the angle, AiOB, such as CiQ (since an- 
nular gears are not used for this type) and then the angle A iOB 
would be called 9. 

110. Data Assumed. It is assumed in the problem that the 
angle 6, the distance h, and the speeds n\ and n 2 or the ratio 
tti/7i 2 , are all given and it is required to design a pair of gears for 
the shafts, such that the contact between the teeth shall be along 
a straight line, the gears complying with the above data. 

111. Determination of Pitch Surfaces. Let the line of contact 
of the pitch surfaces be CQ and let it be assumed that this line 
passes through and is normal to OP, so that on the right-hand 
projections A'O', C'Q r and B'P' are all parallel. The problem 
then is to locate CQ and the pitch surfaces to which it corre- 
sponds, the first part of the problem being therefore to deter- 
mine hi, hzj 61 and d 2 , Fig. 57, and this will now be done. 

Select any point R on CQ, Fig. 57, R being thus one point of 
contact between the required gears, and from R drop perpendicu- 
lars RT and RV on OA and BP respectively. These perpendicu- 
lars, which are radii of the desired gears, have the resolved parts 
ST and UV, respectively, parallel to OP, and the resolved parts 
RS and R U perpendicular both to OP and to the respective 
shafts. These resolved parts are clearly shown 'in the figure, 
and a most elementary knowledge of descriptive geometry will 
enable the reader to understand their locations. Further, it is 
clear the RT 2 = RS* + ST' 2 and RV 2 = RU* + U'V*. 

At the point of contact R, the correct velocity ratio must be 
maintained between the shafts, and as R is a point of contact it is 
a point common to both gears From the discussion in Sec. 84 
it will be clear that, as a point on the gear located on OA the motion 
of R in a plane normal to the line of contact CQ must be identical 
with the motion of the same point R considered as a point on 
the wheel on BP, that is, in the plane normal to the line of contact 
CQ, the two wheels must have the same motion at the point of 
contact R. Sliding along CQ is not objectionable, however, 
except from the point of view of the wear on the teeth and causes 
no uneyenness of motion any more than the axial motion of 
spur gears would do, it being evident that the endlong motion of 
spur gears will in no way affect the velocity ratio or the steadiness 
of the motion. In designing this class of gearing, therefore, no 
effort is made to prevent slipping along the line of contact CQ. 

Imagine now that the motion of R in each wheel in the plane 


normal to CQ is divided into two parts, namely, those normal to 
each plane of reference in the drawing. Taking first the motion 
of R parallel to OP (that is, normal to the first plane) and in the 
plane normal to CQ, its motion as a point in the wheel on OA 
in the required direction is proportional to RS X n\ and as a 
point in the wheel on BP its motion in the same direction is 
proportional to RU X n 2 . So that the first condition to be 
fulfilled is that 

RS X ni = RU X n 2 

But RS = OR sin 6 1 and RU = OR sin 2 

Hence OR sin 0i X m = OR sin 2 X n 2 

or ni sin 0i = n 2 sin 62 

In the second place, consider the motion of R in the plane 
normal to CQ but in the direction normal to OP. As a point in 
the wheel on OA its motion in the required direction is propor- 
tional to S'T' X ni X cos 0i, while as a point in the wheel on PB 
its motion is proportional to U'V X n 2 X cos 2 . 

The second condition therefore is 

S'T' XniX cos 0i = U'V' Xn 2 X cos 2 
htfii cos 0i = /i 2 n 2 cos 2 . 

112. Equations for Finding the Line of Contact. Two other 
conditions may be written as self-evident, and assembling the 
four sets at one place, for convenience, gives 

0i + 02 = (1) 

h 1 + h 2 = h (2) 

HI sin 0i = n z sin 2 (3) 

and hini cos 0i = h 2 n 2 cos 2 (4) 

These four equations are clearly independent, and since 0, h, n\ 
and n 2 or ni/n 2 are given, the values of 0i, 2 , hi and h 2 are known 
and hence the location of the line of contact CQ. 

113. Graphical Solution for Line of Contact. The most simple 
solution is graphical and the method is indicated in Fig. 58 where 
OA and OB represent the projections of the axes of the shafts 
on a plane normal to their common perpendicular. Lay off 
OM and ON along the directions of OB and OA respectively to 
represent to any scale the speeds n z and HI, if the latter are given 
absolutely, but if not, make the ratio OM/ON = n 2 /ni choosing 



one of the lines, say OM, of any convenient length. It is im- 
portant to lay off OM and ON in the proper sense, and since the 
shafts turn in opposite sense, these are laid off in opposite sense 
from 0. Join MN and draw OK perpendicular to M N. Then 
NK/KM = fa/hi, and to find their numerical values take any 
distance NL to represent h, join LM, and draw KJ parallel 
LM. Then fa = JL, fa = JN, ONM = 6 1 and OMN = 2 . 

FIG. 58. 

The proof of the construction is as follows: Since 77^7- = - 

UN n\ 

and since OK = OM sin OMK = ON sin ONK, it follows that 
nz _ OM _ sin ONK 
ni " ON '" sin OMK' 

Comparing this with equation (3), Sec. 112, it is dear that 

ONK = 0! and OMK = 2 so that OC is parallel to MN. Again, 

NK = ON cos 0i and M K = OM cos 2 from which 

NK _ ON cos 0i _ HI cos 0! _ fa 

MK ~ OM ' cos 2 ~~ nz cos 02 ~~ fa 



by comparing with equation (4). The construction for finding 

the numerical values of hi and h 2 requires no explanation. 

114. General and Special Cases. A few applications will 

show the general nature of the solution found. 

Case 1. Shafts inclined at any angle and at distance h 

apart. This is the general case already solved and 0i, 2 , hi 

and h 2 are found as indicated. 

Case 2. Shafts inclined at angle = 90 and at distance h 

apart. Care must be taken not to confuse the method and type 

of gear here described with the spiral 
gear to be discussed later. Choose the 
axes as shown in Fig. 59, lay off to 
scale ON = HI and OM = n 2 and join 
MN', then draw OK perpendicular to 
MN from which (Sec. 113) NK:KM 
= h 2 :hi. In this' case hiUi cos 0i = 
h 2 n 2 cos 2 gives 

hini cos 02 sin 0i n 2 

= = = tan 0i = , 

cos 0i cos 0i n\ 

and hence 

hi /^2\ 
h 2 \nj 

To take a definite case, suppose n\ 
= 2n 2 then 


= I/ 

FIG. 59. 

and if the distance apart of the shafts ; 
h, is 20 in. then hi = 4 in. and h z = 16 in., and the angle 0i is 
given by 


tan 0i = -- =K = 

0! = 26 34' and 2 = 90 - 6 l = 63 26', 

so that the line of contact is located. 

Case 3. Parallel shafts at distance h apart. This gives the 
ordinary case of the spur gear. Here 0=0 and therefore 
0i = = 2 , hence, sin 0i = = sin 2 and cos 0i = 1 = cos 2 , 


so that there are only two conditions to satisfy: hi + h 2 = h 
and hini = /i 2 n 2 . Solving these gives 

^1 7 

and substituting in hi + h z = /i gives 

n 2 


formulas which will be found to agree exactly with those of Sec. 
83 for spur gears. 

Case 4. Intersecting shafts. Here h = 0, therefore hi = 
and h 2 = 0. Referring to Fig. 60, draw OM = n z and ON = HI 

FIG. 60. 

then M N is in the direction of the line of contact OC. There are 
only two equations here to satisfy: 0i + 6 2 = 6 and n\ sin 6\ = 
n z sin 62 and these are satisfied by MN. Then draw OC parallel 
to MN (compare this with the case of the bevel gear taken up at 
the beginning of the chapter). 

Case 5. Intersecting shafts at right angles. Here 6 = 90. 
Further let n 2 = HI then 2 = 45, thus the wheels would be 
equal and are mitre wheels. 



115. Pitch Surfaces. Returning to the general problem in 
which the location of the line of contact CQ is found by the 
method described for finding hi, h%, &i and 2 . Now, just as in 
the case of the spur and bevel gears, a short part of the line of 
contact is selected to use for the pitch surfaces of the gears, ac- 
cording to the width of face which is decided upon, the width of 
face largely depending upon the power to be transmitted, and 
therefore being beyond the scope of the present discussion. 

It is known from geometry that if the line CQ were secured 
to AO, while the latter revolved the former line would describe a 
surface known as an hyperboloid of revolution and a second hy per- 

boloid would be described by securing the line CQ to BP, the 
curved lines in the drawing, Fig. 61, showing sections of these 
hyperboloids by planes passed through the axes AO and BP. 
As the process of developing the hyperboloid is somewhat diffi- 
cult and long, the reader is referred to books on descriptive 
geometry or to other works for the method. In the solution of 
such problems as the present one, however, it is quite unneces- 
sary to draw the exact forms of the curves, and at any time the 
true radius to the curve maybe computed as explained at Sec. 111. 
Or referring to Fig. 57 the radius of the wheel on OA at the point 
T on its axis is computed from the relation RT 2 = RS 2 -f ST 2 
where RT is the radius sought. In this way any number of 
radii may be computed and the true form of the wheels drawn in. 



Should the distance h be small, then sections of the hyper- 
boloids selected as shown at D and E must be used, the distances 
of these from the common normal depending upon the size of the 
teeth desired, the power to be transmitted, the velocity ratio, 
etc., in which case true curved surfaces will have to be used, more 
especially if the gears are to have a wide face. If the face is not 
wide, it may be possible to substitute frustra of approximately 
similar conical surfaces. 

If the distance h is great enough, and other conditions permit 
of it, it is customary to use the gorges of the hyperboloids as 
shown at F and G and where the width of gear face is not great, 
cylindrical surfaces may often be substituted for the true curved 
surfaces. For the wheels F and G the angles 0i and 2 give the 
inclination of the teeth and the angles of 
the teeth for D and E may be computed 
from 61 and 2 . 

116. Example.---To explain more fully, 
take Case (2), Sec. 114, for which 6 =' 
90, ^ = 2n 2 , and h = 20 in., then the 
formulas give hi = 4 in. and h 2 = 16 in. 
Let it be assumed that the drive is such as 
to allow the use of the gorge wheels corres- 
ponding to F and G, then the wheel on OA 
will have a diameter di = 2hi = 8 in. 
and that on BP will be d z = 32 in. diame- 
ter. Further the angles have been determined to be 0i = 26 34', 
6 2 = 63 26'. As the numbers of teeth will depend on the power 
transmitted, etc., it will here be assumed that gear F has ti = 20 
teeth. Then the circular pitch, measured on the end of the gear, 

7T X 8 

from center to center of teeth along the pitch line is pi = - ^Q 

= 1.256 in., which distance will be different from the correspond- 
ing pitch in the other gear G which has 40 teeth. As the gears 
are to work together the normal distance from center to center of 
teeth on the pitch surface must be the same in each gear. This 
distance is called the normal pitch, and is the shortest distance 
from center to center of teeth measured around the pitch sur- 
face; it is, in fact, the distance from center to center of the 
teeth around the pitch line measured on a plane normal to the 
line of contact (CQ in Fig. 57) and agrees with what has already 
been said, that the motion in this normal plane must the be 



same in each gear. Calling the normal pitch p, then for both 
gears p = pi cos 0i = p 2 cos 2 = 1.256 cos 26 34' = 1.123 in. 
For the gear G the number of teeth t% = 40 since n\ = 2n 2 and 
pi = 2.513 in. while p = 1.123 in. A sketch of the gear F is 
given at Fig. 62. 

117. Form of Teeth. Much discussion has arisen over the 
correct form of the teeth on such gears, and indeed it is almost 
impossible to make a tooth which will be theoretically correct, 
but here again one is to be guided by the fact the correct condi- 
tions must be fulfilled in the plane normal to the line of contact. 
Hence on this normal plane the teeth should have the correct 
involute or cycloidal profile. 

In this type of gearing there is a good deal of slip along the 
line of contact (CQ) resulting in considerable frictional loss and 
wear, but such gearing, if well made will run very smoothly and 
quietly. Although it is difficult to construct there are cases 
where the positions of the shafts make its use imperative. 



118. Screw Gearing. In speaking of gears for shafts which 
were not parallel and did not interest two classes were mentioned : 
(a) hyperboloidal gears, and (6) spiral or screw gears and this 
latter class will now be discussed, the former having just been 
dealt with. In screw gearing there is no necessary relation 
between the diameters of the wheels and the velocity ratio 
HI/ n 2 between the shafts; thus it is frequently found that while 
the camshaft of a gas engine runs at half the speed of the crank- 
shaft, the two screw gears producing the drive are of the same 
diameters, while if skew level gears were used the ratio of diam- 
eters would be 1 to 4 (Sec. 114(cT)) and bevel and spur gears for 
the same work would have a ratio 1 to 2. 

119. Worm Gearing. The most familiar form of this gearing 
is the well-known worm and worm wheel which is sketched in 
Fig. 63, and it is to be noticed that the one wheel takes the form 
of a screw, this wheel being distinguished by the name of the 
worm. The distance which any point on the pitch circle of the 
worm wheel is moved by one revolution of the worm is called 
the axial pitch of the worm, and if this pitch corresponds to the 
distance from thread to thread along the worm parallel to its 



axis, the thread is single pitch. If the distance from one thread 
to the next is one-half of the axial pitch the thread is double 
pitch, and if this ratio is one-third the pitch is triple, etc. The 
latter two cases are illustrated at (a) and (6), Fig. 64. 

Fig. 63. Worm gearing. 



FIG. 64. Double and triple pitch worms. 

120. Ratio of Gearing. Let pi be the axial pitch of the worm 
and D be the pitch diameter of the wheel measured on a plane 
through the axis of the worm and normal to the axis of the wheel. 


Then the circumference of the wheel is ivD, and since, by defini- 
tion of the pitch, one revolution of the worm will move the gear 

forward pi in., hence there will be revolutions of the worm for 

one revolution of the wheel, or this is the ratio of the gears. 
Let t be the number of teeth in the gear, then if the worm is single 

pitch t = or the ratio of the gears is simply the number of 

teeth in the wheel. If the worm is double pitch, then pi the 
distance from center to center to teeth measured as before is 
given by pi = 2p', where p' is the axial distance from the center 

of one thread to the center of the next one, and t = - and as 


the ratio of the gears. is , in the double pitch worm this is 

equal to ~ > and for triple pitch it is -^> etc. 

^ o 

121. Construction of the Worm. A brief study of the matter 
will show that as the velocity ratio of the gearing is fixed by the 
pitch of the worm and the diameter of the wheel, hence no matter 
how large the worm may be made it is possible still to retain the 
same pitch, and hence the same velocity ratio, for the same worm 
wheel. The only change produced by changing the diameter of 
the worm is that the angle of inclination of the spiral thread is 
altered, being decreased as the diameter increases, and vice versa. 
The angle made by the teeth across the face of the wheel must 
be the same as that made by the spiral on the worm, and if the 
pitch of the worm be denoted by pi and the mean diameter of 
the thread on the latter by d, then the inclination of the thread 

is given by tan 6 = -,> and this should also properly be the in- 


clination of the wheel teeth. From the very nature of the case 
there will be a great deal of slipping between the two wheels, for 
while the wheel moves forward only a single tooth there will be 
slipping of amount wd, and hence considerable frictional loss, so 
that the diameter of the worm is usually made as small as possi- 
ble consistent with reasonable values of 6. The worm is often 
immersed in oil, but still the frictional loss is frequently above 25 
per cent. 

When both the worm and wheel are made parts of cylinders, 
Fig. 65, then there will only be a very small wearing surface on 
the wheel, but as this is unsatisfactory for power transmission, 



the worm and wheel are usually made as shown in section in the 
left-hand diagram in Fig. 65 where the construction increases 
the wearing surface. The usual method of construction is to 
turn the worm up in the lathe, cutting the threads as accurately 
as may be desired, then to turn the wheel to the proper outside 
finished dimensions. The cutting of the teeth in the wheel rim 
may then be done in various ways of 
which only one will be described, that 
by the use of a hob. 

122. Worm and Worm-wheel Teeth. 
A hob is constructed of steel and is 
an exact copy of the worm with which 
the wheel is to work, and grooves are 
cut longitudinally across the threads so 
as to make it after the fashion of a 
milling cutter; the hob is then hardened and ground and is 
ready for service. The teeth on the wheel may now be roughly 
milled out by a cutter, after which the hob and gear are brought 
into contact and run at proper relative speeds, the hob milling 
out the teeth and gradually being forced down on the wheel till 
it occupies the same relative position that the worm will even- 

FIG. 65. 

FIG. 66. Proportions of worrrO 

tually take. In this way the best form of worm teeth are cut 
and the worm and wheel will work well together. 

The shape of teeth on the worm wheel is determined by the 
worm, as above explained. If a section of the worm is taken 
by a plane passing through its axis, the section of the threads is 
made the same as that of the rack for an ordinary gear, and more 
usually the involute system is used with an angle 14J^. A sec- 


tion of the worm thread is shown at Fig. 66 in which the propor- 
tions used by Brown and Sharpe are the same as in a rack. 

123. Large Ratio in this Gearing. Although the frictional 
losses in screw gearing are large, even when the worm works 
immersed in oil, yet there are great advantages in being able to 
obtain high velocity ratios without excessively large wheels. 
Thus if a worm wheel has 40 teeth, and is geared with a single- 
threaded worm, the velocity ratio will be jr while with a double- 

2 1 
threaded worm it will be jr = ^ so that it is very convenient 

for large ratios. It also finds favor because ordinarily it cannot 
be reversed, that is, the worm must always be used as the driver 
and cannot be driven by the wheel unless the angle 8 is large. 
In cream separators, the wheel is made to drive the worm. 

124. Screw Gearing. Consider now the case of the worm and 
wheel shown in Fig. 65, in which both are cylinders, and suppose 
that with a worm of given size a change is made from a single 
to double thread, at the same time keeping the threads of the 
same size. The result will be that there will be an increase in 
the angle 6 and hence the threads will run around the worm and 
the teeth will run across the wheel at greater angle than before. 
If the pitch be further increased there is a further increase in 6 
and this may be made as great as 45, or even greater, and if at 
the same time the axial length of the worm be somewhat de- 
creased, the threads will not run around the worm completely, 
but will run off the ends just in the same way as the teeth of 
wheels do. 

By the method just described the diameter of the worm is un- 
altered, and yet the velocity ratio is gradually approaching unity, 
since the pitch is increasing, so that keeping to a given diameter 
of worm and wheel, the velocity ratio may be varied in any way 
whatever, or the velocity ratio is independent of the diameters 
of the worm and wheel. When the pitch of the worm is increased 
and its length made quite short it changes its appearance from 
what it originally had and takes the form of a gear wheel with 
teeth running in helices across the face. A photograph of a pair 
of these wheels used for driving the camshaft of a gas engine is 
shown in Fig. 67, and in this case the wheels give a velocity ratio 
of 2 to 1 between two shafts which do not intersect, but have an 
angle of 90 between planes passing through their axes. This 



form of gear is very extensively used for such purposes as afore- 
said, giving quiet steady running, but, of course, the frictional 
loss is quite high. 

Some of the points mentioned may be made clearer by an illus- 
tration. Let it be required to design a pair of gears of this 
type to drive the camshaft of a gas engine from the crankshaft, 
the velocity ratio in this case being 1 to 2, and let both gears be 
of the same diameter, the distance between centers being 12 in. 
From the data given the pitch diameter of each wheel will be 

FIG. 67. Screw gears. 

12 in. and since for one revolution of the camshaft the crankshaft 
must turn twice, the pitch of the thread on the worm must be 
MX TrX 12 = 18.85 in. For the gear on the crankshaft (cor- 
responding to the worm) the " teeth" will run across its face at 

18 85 
an angle given by tan 6 = - ' 10 = 0.5, or = 26 34', and this 

7T X 1^ 

angle is to be measured between the thread or tooth and the plane 
normal to the axis of rotation of the worm (see Fig. 64). The 
angle of the teeth of the gear on the camshaft (corresponding to 


the worm wheel) will be 90 - 26 34' = 63 26' measured in the 
same way as before (compare this with the gear in Sec. 116). 

It will be found that the number of teeth in one gear is double 
that in the other, also the normal pitch of both gears must be 
the same. The distance between adjacent teeth is made to suit 
the conditions of loading and will not be discussed. 

Spiral gearing may be used for shafts at any angle to one 
another, although they are most common in practice where the 
angle is 90. A more detailed discussion of the matter will not 
be attempted here and the reader is referred to other complete 
works on the subject. 

125. General Remarks on Gearing. In concluding this chap- 
ter it is well to point out the differences in the two types of gear- 
ing here discussed. In appearance in many cases it is rather diffi- 
cult to tell the gears apart, but a close examination will show 
the decided difference that in hyperboloidal gearing contact be- 
tween the gears is along a straight line, while in spiral gearing 
contact is at a point only. A study of gears which have been in 
operation shows this clearly, the ordinary spiral gear as used in 
a gas engine wearing only over a very small surface at the centers 
of the teeth. It is also to be noted that the teeth of hyper- 
boloidal gears are straight and run perfectly straight across the 
face of the gear, while the teeth of spiral gears run across the face 
in helices. 

Again in both classes of gears where the spiral gears have the 
form shown at Fig. 67, the ratio between the numbers of teeth on 
the gear and pinion is the velocity ratio transmitted, but in the 
case of the spiral gears the relative diameters may be selected 
as desired, while in the hyperboloidal gears the diameters are 
fixed when the angle between the shafts and the velocity ratio 
is given. 


1. Two shafts intersect at 80, one running at double the speed of the 
other. Design bevel gears for the purpose, the minimum number of teeth 
being 12 and the diametral pitch 3. 

2. What would be the sizes of a pair of miter gears of 20 teeth and 1>^ iia. 
pitch? If the face is two and one-half times the pitch, find the radii of the 
spur gears at the two ends, from which the teeth are determined. 

3. Two shafts cross at angle 6 = 45 and are 10 in. apart, velocity ratio 2, 
locate the line of contact of the teeth. 

4. Find the diameters of a pair of gorge wheels to suit question 3; also 
the sizes of the gears if the distance OS = 12 in. 


5. If the angle between the shafts is 90 in the above case, find the sizes 
of the gorge wheels. 

6. Explain fully the difference between spiral and skew bevel gears. 

7. A worm gear is to be used for velocity ratio of 100, the worm to be 6 in. 
diameter, 1% in. pitch, and single-thread; find the size of the gear and the 
angle of the teeth. 

8. What would be the dimensions above for a double-threaded worm ? 


126. Trains of Gearing. In ordinary practice gears are usually 
arranged in a series on several separate axles, such a series being 
called a train of gearing, so that a train of gearing consists of 
two or more toothed wheels which all have relative motion at the 
same time, the relative angular velocities of all wheels being 
known when that of any one is given. A train of gearing may 
always be replaced by a single pair of wheels of suitable diame- 
ters, but frequently the sizes of the two gears are such as to 
make the arrangement undesirable or impracticable. 

When the train consists of four or more wheels, and when two 
of these of different sizes are keyed to the same intermediate 
shaft, the arrangement is a compound train. This agrees with 
the definition of a compound chain given in Chapter I, because 
one of the links contains over two elements, this being the pair 
of gears on the intermediate shaft. The compound trains are 
in very common use and are sometimes arranged so that the axes 
of the first and last gears coincide, in which case the train is said 
to be reverted ; a very common illustration of this is the train of 
gears between the minute and hour hands of a clock, the axes 
of both hands coinciding. 

127. Kinds of Gearing Trains. If one of the gears in the train 
is prevented from turning, or is held stationary, and all of the other 
gears revolve relatively to it, usually by being carried bodily 
about the fixed gear as in the Weston triplex pulley block or 
the differential on an automobile when one wheel stops and the 
other spins in the mud, the arrangement is called an epicyclic 
train. Such a train may be used as a simple train of only two 
wheels, but is much more commonly compounded and reverted 
so that the axis of the last wheel coincides with that of the first. 

For the ordinary train of gearing the velocity ratio is the num- 
ber of turns of the last wheel divided by the number of turns 
of the first wheel in the same time, whereas in the epicyclic train 
the velocity ratio is the number of turns of the last wheel in the 



train divided by the number of turns in the same time of the 
frame carrying the moving wheels. 

128. Ordinary Trains 1 of Gearing. It will be well to begin 
this discussion with the most common class of gearing trains, 
that is those in which all the gears in the train revolve, and the 
frame carrying their axles remains fixed in space. The outline 
of such a train is shown in Fig. 68, where the frame carrying the 
axles is shown by a straight line while only the pitch circles of 
the gears are drawn in; there are no annular gears in the train 
shown, although these may be treated similarly to spur gears. 
Let HZ be the number of revolutions per minute made by the 

FIG. 68. 

last gear and n\ the corresponding number of revolutions per 
minute made by the first gear, then, from the definition already 
given, the ratio of the train is 

n 2 the number of revolutions per minute of the last gear 
n\ the number of revolutions per minute of the first gear. 

The figure shows a train consisting of six spur gears marked 1, 
a, b, c, e and 2, and let 1 be considered the first gear and 2 the 
last gear. 

The following notation will be employed : HI, n a = n b , n c = n e 
and n 2 will represent the revolutions per minute, n, r a , n, r c , r e 
and r 2 the radii in inches, and t\ } t a , tb, t c , t e and t% the numbers of 
teeth for the several gears used. The gears a and b and also the 
gears c and e are assumed to be fastened together, so that the 
train is compounded, a statement true of any 'train of over two 
gears. Any pair of gears such as 6 and c, which mesh with one 
another, must have the same type and pitch of teeth, but both 
the type and pitch may be different for any other pair which mesh 
together, such as e and 2; the only requirement is that each gear 

1 In what follows in this chapter reference is made to spur and bevel 
gears only. 


must have teeth corresponding with those in the gear with 
which it meshes. 

129. Ratio of the Train. Now, from the results given in Sec. 
101, evidently 

n a TI ti n c Tb U 

= = , ana = = - 

n\ T a l a Ub T c t c 


= = -. 

n e ~ TZ ~ tz 

Therefore, the ratio of the train 

n\ n\ n a n c 

n a n c nz . , 

= since n a = n b and n c = n. 
HI nb n e 

and therefore 

R = r -l X T - X - = r -~^~- 

T a T c TZ T a X T c X TZ 
__ti tb te tiXtbXt e 

i /N , /N / 

ta tc t 2 t a Xt c X tz 

Calling the first wheel in each pair (i.e., 1, b and e) the driver, 
then the formula for the ratio of the train may be written thus, 
The ratio of any gear train is the product of the radii of the drivers 
divided by the product of the radii of the driven wheels, or the 
ratio of the train is the product of the numbers of teeth in the 
drivers divided by the product of the numbers of teeth in the 
driven wheels. 

The same law may be readily shown to apply although some of 
the gears are annular, and indeed is true when a pair of gears is 
replaced by an open or a crossed belt or a pair of sprockets and 
a chain. 

To take an illustration, let the train shown in Fig. 68 have 
gears of the following sizes: 

7*1 = 6 in., r a '= 3% in., n = 4 in. r c = 2% in., r e = 5 in. and 
7*2 = 3 in. and let the diametral pitches be 4, 6 and 8 for the pairs 
1 and a, b and c, and e and 2 respectively. Then ti = 48, t a = 30, 
t b = 48, t c = 32, t e = "80 and t 2 = 48 teeth. 

The ratio of the train is then 

R = 3% X 2% X 3 = 4 fr m the 



48 X 48 X 80 
= QQ y QO X 48 = fr m the numbers of teeth. 

Further, if wheel 1 turn at a speed of HI = 50 revolutions per 
minute the speeds of the other gears will be n a = n b = 80 revolu- 
tions, n c = n e = 120 revolutions and n 2 = 200 revolutions per 

If the distance between the axes of gears 1 and 2 were fixed 
by some external conditions at the distance apart corresponding 
to the above train, then the whole train could be replaced by a 
pair of gears having radii of 19.53 in. and 4.88 in., and these would 
give the same velocity ratio as the train, but would often be 
objectionable on account of the large size of the larger gear. 

The sense of rotation of the various gears may now be ex- 
amined. Looking again at Fig. 68, it is observed that for two 
spur wheels (which have one contact) the sense is reversed, where 
there are two contacts, as between 1 and c the sense remains 
unchanged, and with three contacts such as between 1 and 2 
the sense is reversed and the rule for determining the relative 
sense of rotation of the first and last wheels may be stated thus: 
In any spur-wheel train, if the number of contacts between the 
first and last gears are even then both turn in the same sense, 
and if the number of contacts is odd, then the first and last 
wheels turn in the opposite sense. Should the train contain 
annular gears, the same rule will apply if it is remembered that 
any contact with an annular gear has the same effect as two 
contacts between spur gear. The same rule also applies in case 
belts are used, an open belt corresponding to an annular gear and 
a crossed one to spur gears. 

The rules both for ratio and sense of rotation are the same for 
bevel gears as for spur gears. 

130. Idlers. It not infrequently happens that in a compound 
train the two gears on an intermediate axle are made of the same 
size and combined into one; thus r a may be made equal to 7-5 or 
t a = tb. This single intermediate wheel, then, has no effect on 
the velocity ratio R, as an inspection of the formula for R will 
show, and is therefore called an idler. The sole purpose in using 
such wheels is either to change the sense of rotation or else to 
increase the distance between the centers of other wheels without 
increasing their diameters. 


131. Examples. The application of the formula may be best 
explained by some examples which will now be given: 

1. A wheel of 144 teeth drives one of 12 teeth on a shaft which 
makes one revolution in 12 sec., while a second one driven by it 
turns once in 5 sec. On the latter shaft is a 40-in. pulley con- 
nected by a crossed belt to a 12-in. pulley; this latter pulley turns 
twice while one geared with it turns three times. Show that the 
ratio is 144 and that the first and last wheels turn in the same 

2. It is required to arrange a train of gearing having a ratio 

of ir 

It is possible to solve this problem by using two gears having 
250 teeth and 13 teeth respectively, but in general the larger 
wheel will be too big and it will be well to make up a train of four 

or six gears. Break the ratio up into factors, thus: R = -y^- = 

50 60 

75- X TT; and referring to the formula for the ratio of a train it 

lO \.a 

is evident that one could be made up of four gears having 50 
teeth, 13 teeth, 60 teeth and 12 teeth, and these would be ar- 
ranged with the first wheel on the first axle, the gears of 60 teeth 
and 13 teeth would be keyed together and turn on the inter- 
mediate axle and the 12-tooth gear would be on the last axle and 
the contacts would be the 50 to the 13 and the 60 to the 12-tooth 

Evidently, the data given allow of a great many solutions for 
this problem, another with six wheels being, 

= 5540601540 

13 "" 1 A 4 A 13 " 12 A 12 A 13 

This would give a train similar to Fig. 68 in which the gears are 
ti = 60, t a = 12, t b = 15, t e = 12, t e = 40 and t 2 = 13 teeth. 

3. To design a train of wheels suitable for connecting the sec- 
ond hand of a watch to the hour hand. Here the ratio is 
R = 720 and the first and last wheels must turn in the same 
sense, and as annular wheels are not used for this purpose, the 
number of contacts must be even. The following two solutions 
would be satisfactory for eight wheels : 

R - 700 - 4X4X5X9 - ? v 48 50 108 
" ' 1 ' 14 X 12 X 10 X 12 


p - yon - 6 X 6X4X5 -Z? v ?0fi280 

1 " 12 A 10 X 13 X 12 

Attention is called to the statement in Sec. 89 that it is un- 
usual to have wheels of less than 12 teeth. 

4. Required the train of gears suitable for connecting the min- 
ute and hour hands of a clock. 

Here R = 12 and the train must be reverted; further, since 
both hands turn in the same sense there must be an even number 
of contacts, and four wheels will be selected. In addition to 
obtaining the correct velocity ratio it is necessary that TI + r a 
tb + 7*2, and if all the wheels have the same pitch ti -f t a = h + fa- 
The following train will evidently produce the correct result: 

19 4X3 48 46 
~~T = 12 X W 

or ti = 48, t a = 12, t b = 45 and t 2 = 15 teeth, the hour hand 
carrying the 48 teeth and the minute hand the 15 teeth. 

132. Automobile Gear Box. Very many applications of 
trains of gearing have been made to automobiles and a drawing 
of a variable-speed transmission is shown in Fig. 69. The draw- 
ing shows an arrangement for three forward speeds (one without 
using the gears) and one reverse. The shaft E is the crankshaft 
and to it is secured a gear A having also a part of a jaw clutch B 
on its right-hand side. Gear A meshes with another G on a 
countershaft M, which carries also the gears H, J and K keyed 
to it, and whenever the engine shaft E operates the gears G, H, 
J and K are running. On the right is shown the power shaft P 
which extends back to the rear or driving axle; this shaft is 
central with E and carries the gears D and F and also the inner 
part C of the jaw clutch for B. 

The gears D and F are forced to rotate along with P by means 
of keys at T, but both gears may be slid along the shaft by means 
of the collars at N and R respectively. In addition to the gears 
already mentioned there is another, L, which always meshes with 
K and runs on a bearing behind the gear K. 

Assuming the driver wishes to operate the car at maximum 
speed he throws F into the position shown and pushes D to the 
left so that the clutch piece C engages with B, in which case P 
runs at the same speed as the -engine shaft E. Second highest 
speed is obtained by slipping D to the right until it comes into 


contact with H, the ratio of gears then being A to G and H to D; 
F remains as shown. For lowest speed D is placed as shown in 
the figure and F slid into contact with J; the shaft P and the car 
are reversed by moving F to the right until it meshes with L, 
the gear ratio being A to G and K to L to F. 

Builders of automobiles so design the operating levers that it is 
possible to have only one set of gears in operation at once. 

FIG. 69. Automobile gear box. 

133. The Screw-cutting Lathe. Most lathes are arranged for 
cutting threads on a piece of work, and as this forms a very 
interesting application of the principles already described, it will 
be used as an illustration. 

The general arrangement of the headstock of a lathe is shown 
in Fig. 70, and in this case in order to make the present discussion 
as simple as possible, it is assumed that the back gear is not in 
use. The cone C is connected by belt to the countershaft which 
supplies the power, the four pulleys permitting the operation of 
the lathe at four different speeds. This cone is secured to the 
spindle S, which carries the chuck K to which the work is at- 



tached and by which it is driven at the same rate as the cone C. 
On the other end of S is a gear e, which drives the gear h through 
one idler g or two idlers / and g. The shaft which carries h 
also has a gear 1 which is keyed to it, and must turn with the shaft 
at the same speed as h. The gear 1 meshes with a pinion a on a 
separate shaft, this pinion being also rigidly connected to and re- 
volving with gear b, which latter gear meshes with a wheel 2 
keyed to the leading screw L. Thus the spindle S is geared to 
the leading screw L through the wheels e, /, g, h, 1, a, 6, 2 of 
which the first four are permanent, while the latter four may 
be changed to suit conditions, and are called change gears. 

The work is attached to the chuck K on S and is supported by 
the center on the tail stock so that it rotates with K. The lead- 


FIG. 70. Lathe head stock. 

ing screw L passes through a nut in the carriage carrying the 
cutting tool, and it will be evident that for given gears on 1, 
a, b, 2 a definite number of turns of S correspond to a definite 
number of turns of L, and hence to a certain horizontal travel 
of the carriage and cutting tool. Suppose that it is desired to 
cut a screw on the work having s threads per inch, the number of 
threads per inch I on the leading screw being given. This requires 
that while the tool travels 1 in. horizontally, corresponding to I 
turns of the leading screw L, the work must revoive s times, or 
if HI represents the revolutions per minute of the work, and n 2 
those of the leading screw, then 

_n2 _l _te ti k 

""* * *"~ ~~~ j /N ^\ j 

n\ s th t a tz 


where t e , fo, etc., represent the numbers of teeth in the gears. 
Evidently / and g are idlers and have no effect on the ratio. 
In many lathes the gears e and h are made the same size so 
that gear 1 turns at the same speed as gear h. 

Then R = ^xr 

t a tz 

This ratio is used in the example here. For many purposes 
also t a = tb. 

Further, if L and S turn in the same sense, and if the leading 
screw has a right-hand thread, as is usual, then the thread cut 
on the work will also be right-hand. The idlers / and g arc 
provided to facilitate this matter, and if a right-hand thread 
is to be cut, the handle m carrying the axes of / and g is moved 
so that g alone connects e and h, while, if a left-hand thread is 
to be cut the handle is depressed so that / meshes with e and g 
with h. The figure shows the setting for a right-hand thread. 

An illustration will show the method of setting the gears to do 
a given piece of work. Suppose that a lathe has a leading screw 
cut with 4 threads per inch, and the change gears have respect- 
ively 20, 40, 45, 50, 55, 60, 65, 70, 75, 80 and 115 teeth. Assume 

t e == th- 

1. It is required to cut a right-hand screw with 20 threads 
per inch. Then -=7X7 where I = 4 and s is to be 20. 

S ta tz 

ti^h 4 1 
Thus T, X t* = 20 = 5 

This ratio may be satisfied by using the following gears ti = 20. 
t a = 50, t b = 40 and tz = 80. Only the one idler g would be 
used to give the right-hand thread. 

2. To cut a standard thread on a 2-in. gas pipe in the lathe. 
The proper number of threads here would be llj^ per inch and 

hence I = 4, s = llj^ and r X .- = TTT? = oo' This could 

la *2 11-72 46 

be done by making ti = 40 and tz = 115, and tb = t a both acting 
as one idler. 

3. If it were required to cut 100 threads per inch then I = 4, 

s = 100 and r X ^ = TT^ = ^> which may be divided into two 

t a 12 

parts, thus 05 = z X gr/ so that making ti = 20, t a = 80, 



t z = 75, would require an extra gear of 12 teeth to take the 
place of 6, as U = 12. 

The axle holding the gears a and 6 may be changed in position 
so as to make these gears fit in all cases between 1 and 2. The 
details of the method of doing this are omitted in the drawing. 

In order to show how much gearing has been used in the mod- 
ern lathe, the details of the gearing for the headstock of the 
Hendey-Norton lathe are given in Figs. 71, 72 and 73, from 
figures made up from drawings kindly supplied by the Hendey 
Machine Co., Torrington, Conn. 

FIG. 71. Hendey-Norton lathe. 

A general view of the Hendey-Norton lathe is shown in 
Fig. 71, and a detailed drawing in Fig. 72 in the latter of which 
is shown a belt cone with four pulleys, P, running freely on the 
live spindle S. Keyed to the same spindle is the gear shown at 
Q, and secured to the cone P is the pinion T, and Q and T mesh, 
when required, with corresponding gears on the back gearshaft 
R. When the cone is driving the spindle directly the pin W, 
shown in the gear Q, is left in the position shown in the drawing, 
thus forcing P to drive the spindle through Q, but when the back 
gear is to be used, the pin W is drawn back out of contact with 
the cone pulley, the shaft R is revolved by means of the handle 
A so as to throw the gears on it into mesh with T and Q and 





then the spindle is driven from the cone, through T and the 
back gears, and back through gear Q which is keyed to the 
spindle. The use of the back gears enables the spindle to be 
run at a much slower speed than the cone pulley. 

For screw cutting the back gear is not used, but a train of 
gears, /, F, Z, X, C, E, F, G, H, K, L, M, N, and B, and an idler 
(or tumble gear as it is called) delivers motion from the live 
spindle to the set of gears A on the lead screw. These gears 
are partly shown on Fig. 72 and partly on Fig. 73 which latter 
is diagrammatic. Of the gears mentioned J is keyed to the live 

Stud D is Geared to Spindle 1 to 1 

Stud D-+ 

48 T 
Gear C 

Position of Gears 
with Handle U 
*n*"l Hole 

Position of Gears 
with Handle U 
in^2 Hole 

Position of Gears 
with Handle U 
in^3 Hole 

FIG. 73. Gears on Hendey-Norton lathe. 

spindle, the idler Y is slipped over into gear when a screw is to 
be cut and causes the gear Z to turn at the same speed as the 
spindle. The gear Z gives motion to the stud D by means of 
the bevel gears and jaw clutch shown below gear T, and the sense 
of motion of D will depend on whether the jaw clutch is put into 
contact with the right- or left-hand bevel gear, these bevel gears 
being to adapt the lathe to the cutting of a right- or left-hand 
thread. Gear X, keyed to stud D, revolves at the same speed 
as Z and therefore as the spindle S and in the same or opposite 
sense to it according to the position of the clutch; with the clutch 
to the left both would turn in the same sense. The remainder 
of the train of gearing is indicated clearly on Fig. 73, and it is 
to be noted that where several of these gears are on the same 
shaft they are fastened together, such as C and E, G and F, 


L, K, H, etc. The numbers of teeth shown in the various gears 
correspond to those used in the 16-in. and 18-in. lathes. 

The handles shown control the gear ratios; thus U controls the 
positions of the gears L, K, and H and the figure shows the three 
possible positions provided by the maker and corresponding to 
the three holes 1, 2 and 3 in Fig. 72. The gear B is provided 
with a feather running in a long key seat cut in the shaft shown, 
and the handle V is arranged so as to control the horizontal 
position of the gear B and its tumbler gear; that is the handle 
V enables the operator to bring B into gear with any of the 12 
gears on the lead screw. The lead screw has 6 threads per inch. 

With the handle U in No. 3 hole and the handle V in the fourth 
hole as shown in the right-hand diagram of Fig. 73 the ratio is 

n 2 _ I _ h t x fc /, t K fe 

" WJL ~ 8 ~ t z X t C X /, X T H X tu 70 

1 48 68 34 34 70 

" S = 12 X 6 = 12 X 6 = 

or the lathe would be set to cut 3J^ threads per inch. 

134. Cutting Special Threads, Etc. When odd numbers of 
threads are to be cut, various artifices are resorted to to get the 
required gearing, sometimes approximations only being employed. 
Thus if it were required to cut threads on a 2-in. gas pipe, which 
has properly llj^ threads per inch, and the lathe had not gears 
for the purpose, it might be possible to cut 11}^ threads per inch 
or 11% threads per inch, either of which would serve such a 
purpose quite well. There are cases, however, where exact 
threads of odd pitches must be cut and an example will show one 
method of getting at the proper gears. 

Let it be required to cut a screw with an exact pitch of 1 mm. 
(0.0393708 in.) with a lathe having 8 threads per inch on the 
leading screw, and assume t e = th. 

A convenient means of working out this problem is the method 
of continued fractions. 

The exact value of the ratio ^ is 

_! Y% 0.125 

R ~ 0.0393708 ~ 0.0393708* 


The first approximation is 

1 1 68,876 

g = 3, the real value being ^ = 3 + 

The second is 

3 + =, the real value being 3 + 

5' 49,328 

5+ 6^876 

and in this way the third, fourth, fifth, etc., approximations are 
readily found. The sixth is 


7_ 127 
^40 " 40 

Thus with a gear of 40 teeth at 1 or ti = 40 and t z = 127 on 
the leading screw, and an idler in place of a and b the thread could 
be cut. 

125 127 

(Note that n HOODOO = 3.17494 while -^ = 3.175, so that the 

U.Uoyo/Uo 4U 

arrangement of gears would give the result with great accuracy.) 

Problems of this nature frequently lend themselves to this 
method of solution, but other methods are sometimes more con- 
venient and the ingenuity of the designer will lead him to devise 
other means. 

135. Hunting Tooth Gears. These are not much used now 
but were formerly employed a good deal by millwrights who 
thought that greater evenness of wear on the teeth would result 
when a given pair of teeth in two gears came into contact the 
least number of times. To illustrate this, suppose a pair of gears 
had 80 teeth each, the velocity ratio between them thus being 
unity; then a given tooth of one gear would come into contact 
with a given tooth of the other gear at each revolution of each 
gear, but, if the number of teeth in one gear were increased to 81 
then the velocity ratio is nearly the same as before and yet a 
given pair of teeth would come into contact only after 80 revolu- 
tions of one of the gears and 81 revolutions of the other. The 
odd tooth is called a hunting tooth. Compare the case where 
the gears have 40 teeth and 41 teeth with the case cited. 




136. Epicyclic Gearing. An epicyclic train has been defined 
at the beginning of the chapter as one in which one of the gears 
in the train is held stationary or is prevented from turning, 
while all the other gears revolve relative to it. The frame 
carrying the revolving gear or gears must also revolve. The train 
is called epicyclic because a point on the revolving gear describes 
epicyclic curves on the fixed one, and the term planetary gearing 
appears to be due to the use of such a train by Watt in his "sun 
and planet" motion between the crankshaft and connecting rod 
of his early engines. 

An epicyclic train of gears is made up in exactly the same 
way as an ordinary train already examined, the only difference 

Epicyclic trains. 

between the two is in the part of the combination that is fixed; 
in the ordinary train the axles on which the gears revolve are 
fixed in space, that is, the frame is fixed and all the gears revolve, 
whereas in the epicyclic train one of the gears is prevented from 
turning and all of the other gears and the frame revolve. This 
is another example of the inversion of the chain explained in 
Chapter I. 

The general purpose of the train is to obtain a very low 
velocity ratio without the use of a large number of gears; thus a 
ratio of ma y ve - r y smi ply be obtained with four gears, the 
largest of which contains 101 teeth. It also has other applications. 
Any number of wheels may be used, although it is unusual to 
employ over four. 

In discussing the train, the term "first wheel" will correspond 
with wheel 1 and " last wheel " with wheel 2 in the train shown in 
Fig. 68, and it will always be the first wheel which is prevented 


from turning. The ratio of the train is the number of turns of 
the last wheel for each revolution of the frame. 

In Fig. 74 two forms of the train, each containing two wheels, 
are shown. In the left-hand figure, wheel 1 is fixed in space and 
the frame F and wheel 2 revolve, whereas in the right-hand figure 
the wheel 1 is fixed only in direction, being connected to links in 
such a way that the arrow shown on it always remains vertical 
(a construction easily effected in practice), that is wheel 1 does 
not revolve on its axis, and the frame F and wheel 2 both revolve 
about the center B. The following discussion applies to either 

137. Ratio of Epicyclic Gearing. Let the gears 1 and 2 con- 
tain ti and t% teeth respectively; then as a simple train the ratio is 

R = - and is negative, since the first and last wheels turn in 

the opposite sense. The method of obtaining the velocity ratio 
of the corresponding epicyclic train may now be explained. 
Assume first that the frame and both wheels are fastened together 
as one body and the whole given one revolution in space; then 
frame F turns one revolution, and also the gears 1 and 2 each 
turn one revolution on their axes (not axles) . But in the epicyclic 
train the gear 1 must not turn at all, hence it must be turned 
back one revolution to bring it back to its original state, and this 
will cause the wheel 2 to make R revolutions in the same sense as 
before, since the ratio R is negative. During the whole operation 
gear 1 has not moved, the frame F has made 1 revolution and the 
last wheel 1 + R revolutions in the same sense, hence the ratio 
of the train is 

^ _ 1 + R _ Revolutions made by the last wheel. 
1 Revolutions made by the frame. 

A study of the problem will show that if R were positive then 
E = 1 R and in fact the correct algebraic "formula is 

E = 1 - R 

and in substituting in this formula care must be taken to attach 
to R the correct sign which belongs to it in connection with an 
ordinary train. 

Owing to the difficulty presented by this matter the following 
method of arriving at the result may be helpful, and in this case 
a train will be considered where R is positive, i.e., there are an 
even number of contacts. 


1. Assume the frame fixed and first wheel revolved once; then: 
Frame makes revolutions. 

First wheel makes + 1 revolutions. 

Last wheel makes + R revolutions. 

But the epicyclic train is one in which the first wheel does not 
revolve, and therefore to bring it back to rest let all the parts be 
turned one revolution in opposite sense to the former motion. 

2. After all parts have been turned backward one revolution 
the total net result of both operations is: 

Frame has made 1 revolutions. 

First wheel has made +1 1 revolutions. 

Last wheel has made -\-R-l revolutions, 

which has brought the wheel 1 to rest; hence 

r> _ -I 

E = -Q-^ j - = 1 R as before. 

138. Examples. The following examples will illustrate the 
meaning of the formula and the application of the train in 

1. Let wheel 1 have 60 teeth and wheel 2 have 59 teeth; 

then ti = 60, t% 59 and therefore R = -^ 

Hence, E = l-R = l-- = l+ = or the 

last wheel turns in the same sense as the frame and at about 
double its speed. 

2. Suppose now that an idler is inserted between 1 and 2, 

keeping R = still, but making it positive. 

Thus, the wheel 2 turns in opposite sense to the frame and at 
3^9 its speed. 

3. If in example (2) wheels 1 and 2 are interchanged, then 


R = and is positive, so that 

p 1 59 1 

"60 = ^60 

in which case the last wheel will turn in the same sense as the 
frame and at %o of its velocity. 

4. To design a train having a positive ratio of in QQQ> that is, 



one in which the last wheel turns in the same sense as the frame 
and at TTT^T^ the speed. 



E = 1 - R = 


or R = 1 - 



\ 1007 \ 

J_\ -??- v *. 

1007 " 100 A 100 

The train thus consists of four gears 1, a, b and 2 and the 
numbers of teeth are ti = 99, t a = 100, t b = 101, t 2 = 100. 

In practice such a train could easily be reverted, although the 
numbers of teeth are not exactly suited to it, and would work 
quite smoothly. The train is frequently made up in the form 


FIG. 75. 

shown in Fig. 75 where the frame takes the form of a loose 
pulley A, carrying axles D on which the intermediate gears run 
and the 99-toothed gear is keyed to B, which is also keyed to the 
frame C. The pulley will turn 10,000 times for each revolution 
that the shaft makes. Should the gears be changed around, so 
that the 100-tooth wheel is fixed, while the 99-tooth wheel is on 
the shaft and gears with the 100-tooth wheel on D, then 



or the shaft will turn slowly in opposite sense to the wheel A 

P _ 1 v _ 

" 101 X 99 " 


The arrangement sketched in Fig. 75, in a slightly modified 
form, is used in screw-cutting machines, but with a much larger 
value of E. In this case there are two pulleys, one as shown at A 
and one keyed to the shaft, while the gears on D are usually re- 
placed by a broad idler. When the die is running up on the stock 
the operation is slow and the belt is on the pulley A, but for other 
operations the speed is much increased by pushing the belt over 
to the pulley keyed to the shaft, the gears then running idly. 

5. Watt's Sun and Planet Motion. In this case gear 2 was 
the same size as 1 and was keyed to the crankshaft, while the gear 
1 was secured to the end of the connecting rod and a link kept the 
two gears at the proper distance apart, as in the right-hand 
diagram of Fig. 74. There was, of course, no crank. 

Here R = - 1 and E = 1 - R = 2. 

Therefore, the crankshaft made two revolutions for each two 
strokes of the piston. 

139. Machines Using Epicyclic Gearing. There are a great 
many illustrations of this interesting arrangement and space 
permits the introduction of only a very few of these. 

(a) The Weston Triplex Pulley Block. A form of this block, 
which contains an epicyclic train of gearing, is shown in Fig. 76. 
The frame D contains bearings which carry the hoisting sprocket 
F, and on the casting carrying the hoisting sprocket are axles 
each carrying a pair of compound gears BC, the smaller one C 
gearing with an annular gear made in the frame D, while the other 
and larger gear B of the pair meshes with a pinion A on the 
end of the shaft S to which the hand sprocket wheel H is attached. 
When a workman pulls on the hand sprocket chain he revolves H 
and with it the pinion A on the other end of the shaft, which in 
turn sets the compound gears BC in motion. As one of these 
gears meshes with the fixed annular gear on the frame D the only 
motion possible is for the axles carrying the compound gears to 
revolve and thus carry with them the hoisting sprocket F. 

In the one ton Weston block the annular wheel has 49 teeth, 
the gear B has 31 teeth, C has 12 teeth and the pinion A has 13 
teeth. For the train, then, evidently R is negative since one 
wheel is annular and 

Rm ^ v? 1 - -973 
" 12 X 13 


E = 1 - R = 1 - (- 9.73) = 10.73. 



So that there must be 10.73 turns of the hand wheel to cause 
one turn of the hoisting wheel F. As these wheels are respec- 
tively 9% in. and 3> in. diameter, the hand chain must be 

moved ~y/ X 10.73 = 33.2 ft. to cause the hoisting chain to 

FIG. 76. Weston triplex block. 

move 1 ft., so that the mechanical advantage is 33.2 to 1, neglect- 
ing friction. 

(6) Motor-driven Portable Drill. A form of air-driven drill is 
shown at Fig. 77 in which epicyclic gearing is used. This drill 
is made by the Cleveland Drill Co., Cleveland, and the figure 
shows the general construction of the drill while a detail is also 
given of the train of gearing employed. 



! Valve j 
! D \ 

FIG. 77. Cleveland air-driven portable drill. 


The outer casing A of the drill is held from revolving by means 
of the handles H and H f , air to drive the motors passing in 
through H. Fastened to the bottom of the crankshaft of the air 
motors is a pinion B which drives a gear C and through it a 
second pinion D, which latter revolves at the same speed as B 
and operates the valve for the motors. The gear C is keyed 
to a shaft which has another gear E also secured to it, the latter 
meshing with pinions F which in turn mesh with the internal 
gear G secured to the frame A. The gears F run freely on shafts 
J which are in turn secured in a flange on the socket S, which 
carries the drill. 

As the motors operate on the crankshaft causing it to revolve, 
the pinion B turns with it and also the gear C and with it the 
gear E. As E revolves it sets the gears F in motion and as these 
mesh with the fixed gear G the only thing possible is for the 
spindles J carrying F to revolve in a circle about the center of 
E and as these revolve they carry with them the drill socket S. 
In one of these drills the motor runs at 1,275 revolutions per 
minute and the numbers of teeth in the gears are t B = 14, t c = 70, 
t E = 15, t P = 15 and to = 45. The speed of the spindle S 
will then be 1,275 + {1 + 4 %5 X 7 % 4 } = 80 revolutions per 

(c) Automobile Transmission. Epicyclic gearing is now 
commonly used in automobiles and two examples are given here, 
in concluding the chapter. The mechanism shown in Fig. 78 ] 
is used in Ford cars for variable speed and reversing. The 
engine flywheel A carries three axles X uniformly spaced around 
a circle and each carrying loosely three gears H, G and K, the 
three gears being fastened together and rotating as one solid 
body. Each of these gears meshes with another one which 
is connected by a sleeve to a drum; thus H gears with B and 
through it to the drum C which is keyed to the shaft P 
passing back to the rear axle. Gear G meshes with the gear F. 
which is attached to the drum E, while K meshes with the gear 
J on the drum 7. The disc M is keyed to an extension of the 
crankshaft as shown and carries one part of a disc clutch, the 
other part of which is carried on casting C. The mechanism for 
operating this clutch is not shown completely. 

The driver of the car has pedals and a lever under his control 

1 A drawing was kindly furnished by the Ford Motor Co., Ford, Ontario, 
for the purpose of this cut. 



and it is beyond the present purpose to discuss the action of 
these in detail, but it may be explained that these control band 
brakes, one about the drum /, another about the drum E and 
a third about the drum C, and in addition the pedals and lever 
control the disc clutch between C and M. 

In this mechanism the gears have the following numbers of 
teeth: t a = 27 teeth, to = 33 teeth, t K = 24 teeth, t B = 27 teeth, 
t F = 21 teeth and tj = 30 teeth. 

FIG. 78. Ford transmission. 

Should the driver wish the car to travel at maximum speed he 
throws the disc clutch into action which connects M and C and 
thus causes the power shaft P to run at the same speed as that 
of the engine crankshaft. If he wishes to run at slow speed he 
operates the pedal which applies the band brake to the drum E, 
causing the latter to come to rest. The gears F, G, H and B 
then form an epicyclic train and for this 


R = 

X ^ = 0.636 and E = 1 - R = 0.36. 

So that the power shaft P will turn forward, making 36 revolu- 
tions for each 100 made by the crank. 



If the car is to be reversed, drum I is brought to rest and the 
train consists of gears /, K, H and B. 


OQ 07 5 
R = X < = 

E = I - R = - 

or the power shaft P will turn in opposite sense to the crank and 
at one-fourth its speed. 

The brake about C is for applying the brakes to the car. 

(d) Automobile Differential Gear. The final illustration is 
the differential used on the rear axle of Packard cars. This is 

FIG. 79. Automobile differential gear. 

shown in Fig. 79 which is from a Packard pamphlet. The power 
shaft P attached to the bevel pinion A drives the bevel gear B 
which has its axis at the rear axle but is not directly connected 
thereto. The wheel B carries in its web bevel pinions C, the axles 
of which are mounted radially in B, and the pinions C may rotate 
freely on these axles. 

The rear axle S is divided where it passes B and on one part 
of the axle there is a bevel gear D and on the other one the bevel 
gear E of the same size as D. When the car is running on a 
straight smooth road the two wheels and therefore the two parts 


S of the rear axle run at the same speed and then the power is 
transmitted from P through A and B just as if the gears C, D 
and E formed one solid body. 

In turning a corner, however, the rear wheel on the outer part 
of the curve runs faster than the inner one, that is D and E 
run at different speeds and gear C rotates slowly on its axle. 
When the one wheel spins in the mud, and the other one remains 
stationary, as not infrequently happens when a car becomes 
stalled, the arrangement acts as an epicyclic train purely. 


1. Find the velocity ratio for a train of gears as follows: A gear of 30 
teeth drives one of 24 teeth, which is on the same shaft with one of 48 teeth; 
this last wheel gears with a pinion of 16 teeth. 

2. The handle of a winch carries two pinions, one of 24 teeth, the other 
of 15 teeth. The former may mesh with a 60-tooth gear on the rope drum 
or, if desired, the 15-tooth gear may mesh with one of 56 teeth on the same 
shaft with one of 14 teeth, this latter gear also meshing with the gear of 60 
teeth on the drum. Find the ratio in each case. 

3. Design a reverted train for a ratio 4 to 1, the largest gear to be not 
over 9 in. diameter, 6 pitch. 

4. A gear a of 40 teeth is driven from a pinion c of 15 teeth, through an 
idler b of 90 teeth. Retaining c as before, also the positions of the centers 
of a and c, it is required to drive a 60 per cent, faster, how may it be done? 

6. A car is to be driven at 15 miles per hour by a motor running at 1,200 
revolutions per minute. The car wheels are 12 in. diameter and the motor 
pinion has 20 teeth, driving through a compound train to the axle; design 
the train. 

6. In a simple geared lathe the lead screw has 5 threads per inch, gear 
e = 21 teeth, h = 42 teeth, 1 = 60 teeth and 2 = 72 teeth; find the thread 
cut on the work. 

7. It is desired to cut a worm of 0.194 in. pitch with a lathe as shown at 
Fig. 70, using these change gears; find the gears necessary. 

8. Make out a table of the threads that can be cut with the lathe in Fig. 
70 with different gears. 

9. Make a similar table to the above for the Hendey-Norton lathe 

10. Design an auto change-gear box of the selective type, with three 
speeds and reverse, ratios 1.8 and 3.2 with % pitch stub gears, shaft centers 
not over 10 in. 

11. A motor car is to have a speed of 45 miles per hour maximum with an 
engine speed of 1,400 revolutions per minute. What reduction will be 
required at the rear axle bevel gears, 36-in. tires? At the same engine 
speed find the road speed at reductions of 4 and 2 respectively. 

12. Design the gear box for the above car with % stub-tooth gears, shafts 
9 in. centers. 


13. Prove that the velocity ratio of an epicyclic train is E = 1 R, 

14. Design a reverted epicyclic train for a ratio of 1 to 2,500. 

16. In a train of gears a has 24 teeth, and meshes with a 12-tooth pinion 
6 which revolves bodily about a, and 6 also meshes with an internal gear c 
of 48 teeth. Find the ratio with a fixed and also with c fixed. 


140. Purpose of Cams. In many classes of machinery certain 
parts have to move in a non-uniform and more or less irregular 
way. For example, the belt shifter of a planer moves in an 
irregular way, during the greater part of the motion of the planer 
table it remains at rest, the open and crossed belts driving their 
respective pulleys, but at the end of the stroke of the table the 
belts must be shifted and then the shifter must operate quickly, 
moving the belts, after which the shifter comes again to rest and 
remains thus until the planer table has completed its next stroke, 
when the shifter operates again. The valves of a gas engine 
afford another illustration, for these must be quickly opened at 
the proper time, held open and then again quickly closed. The 
operation of the needle bar of a sewing machine is well known 
and the irregular way in which it moves is familiar to everyone. 

In the machines just described, and indeed in almost all 
machines in which this class of motion occurs, the part which 
moves irregularly must derive its motion from some other part 
of the machine which moves regularly and uniformly. Thus, 
in the planer all the motions of the machine are derived from the 
belts which always run at steady velocity; further, the shaft 
operating the valves of a gas engine runs at speed proportional 
to the crankshaft while the needle bar of a sewing machine is 
operated from a shaft turning uniformly. 

The problem which presents itself then is to obtain a non- 
uniform motion in one part of a machine from another part which 
has a uniform motion, and it is evident that at least one of the 
links connecting these two parts must be unsymmetrical in 
shape, and the whole irregularity is usually confined to one part 
which is called a cam. Thus a cam may be defined as a link 
of a machine, which has generally an irregular form and by 
means of which the uniform motion of one part of the machine 
may be made to impart a desired kind of non-uniform motion 
to another part. 




Cams are of many different forms and designs depending 
upon the conditions to be fulfilled. Thus in the sewing machine 
the cam is usually a slot in a flat plate attached to the needle 
bar, in the gas engine the cam is generally a non-circular disc 
secured to a shaft, whereas in screw-cutting machines it often 
takes the form of a slot running across the face of a cylinder, and 
many other cases might be cited, the variations in its form being 
very great. Some forms of cams are shown in Fig. 80. 

Several problems connected with the use of cams will explain 
their application and method of design. 

141 Stamp-mill Cam. The first illustration will be that of 
the stamp mill used in mining districts for crushing ores, and a 
general view of such a mill is shown in Fig. 81. Such a mill con- 
sists essentially of a number of stamps A, which are merely 

FIG. 80. Forms of cams. 

heavy pieces of metal, and during the operation of the mill 
these stamps are lifted by a cam to a desired height, and then 
suddenly allowed to drop so as to crush the ore below them. The 
power to lift the stamps is supplied through a shaft B which is 
driven at constant speed by a belt, and as no work is done by 
the stamps as they are raised, the problem is to design a cam 
which will lift them with the least power at shaft B, and after 
they have been lifted the cam passes out of gear and the weights 
drop by gravity alone. 

Now, it may be readily shown that the force required to move 
the stamp at any time will depend upon its acceleration, being 
least when the acceleration is zero, because then the only force 
necessary is that which must overcome the weight of the stamp 
alone, no force being required to accelerate it. Thus, for the 



minimum expenditure of energy, the stamp must be lifted at a 
uniform velocity, and the problem, therefore, resolves itself into 
that of designing a cam which will lift the stamp A at uniform 

The general disposition of the parts involved, is shown in Fig. 
82, where B represents the end of the shaft B shown in Fig. 81, 

FIG. 81. Stamp mill. 

and YF represents the center line of the stamp A, which does 
not pass through the center of B. Let the vertical shank of the 
stamp have a collar C attached to it, which collar comes into 
direct contact with the cam on B\ then the part C is usually 
called the follower, being the part of the machine directly 
actuated by the cam. 

It will be further assumed that the stamp is to be raised twice 



for each revolution of the shaft B, and as some time will be taken 
by the stamp in falling, the latter must be raised its full distance 
while the shaft B turns through less than 180. Let the total 
lift occur while B turns through 102. 

Further, let the total lift of the cam be h ft., that is, let the 
distance 6, Fig. 82, through which the bottom of the follower 
C rises, be h ft. 

The construction of the cam may now be begun. Draw BF 
perpendicular to YF and lay off the angle FBE equal to 102. 
Next divide the distance 6 = h, and also the angle FBE, into 


FIG. 82. Stamp mill cam. 

any convenient number of equal parts, the same number being 
used in each case; six parts have been used in the drawing. 

Now a little consideration will show that since the stamp A 
and also the shaft B are to move at uniform speed, the distances 
0-1,1-2, 2-3, etc. and also the angles FBG, GBH, HBJ, etc., 
must each be passed through in the same intervals of time and all 
these intervals of time must be equal. With center B and radius 
BF draw a circle FGH ...E tangent to the line 6 and draw GM, 
HN, etc., tangent to this circle at G, H, etc. Now while the 
follower is being lifted from to 1 the shaft B is revolved through 
the angle FBG, and then the line GM will be vertical and must 
be long enough to reach from F to 1 or GM should equal FL 
The construction is completed by making HN = F 2, JP 



= F 3, etc., and in this way locating the points 0, M, N, P, Q, R 
and S and a smooth curve through these points gives the face 
of the cam. As a guide in drawing the curve it is to be remem- 
bered that MG, NH, etc., are normals to it. 

A hub of suitable size is now drawn on the shaft, the dimen- 
sions of the hub being determined from the principles of machine 
design, and curves drawn from S and down to the hub complete 
the design; the curve from S must be so drawn that the follower 
will not strike the cam while falling. 

The curve OMN . . . S is clearly an involute having a base 
circle of radius BF, or the curve of the cam is that which would 
be described by a pencil attached to a cord on a drum of radius 

FIG. 83. Uniform velocity cam. 

BF, the cord being unwound and kept taut. The dotted line 
shows the other half of the cam. 

In this case there is line contact between the cam and its 
follower, that is, it is a case of higher pairing, as is frequently, 
though not always, the case with cams. 

142. Uniform Velocity Cam. As a second illustration, take 
a problem similar to the latter, except that the follower is to have 
a uniform velocity on the up and down stroke and its line of 
motion is to pass through the shaft B. It will be further assumed 
that a complete revolution of the shaft will be necessary for the 
up and down motion of the follower. 

CAMS 141 

Let 8, Fig. 83 (a) represent the travel of the follower, the 
latter being on a vertical shaft, with a roller where it comes 
into contact with the cam. Divide 8 into, say, eight equal 
parts as shown, further, divide the angle OBK ( = 180) into the 
same number of equal parts, giving the angles OBI', l'B2', 
etc. Now since the shaft B turns at uniform speed the center 
of the follower is at 1 when Bl' is vertical and at 2 when B2' is 
vertical, etc., hence it is only necessary to revolve the lengths 
Bl, B2, etc., about B till they coincide with the lines Bl', B2', 
etc., respectively. The points 1', 2', 3', will be obtained on 
the radial lines Bl', B2 r , etc., as the distances from B which the 
center of the follower must have when the corresponding line 
is vertical. With centers 1', 2', 3', etc., draw circles to represent 
the roller and the heavy line shown tangent to these will be 
the proper outline for one-half of the cam, the other half being 
exactly the same as this about the vertical center line. Here 
again there is higher pairing and some external force is supposed 
to keep the follower always in contact with the cam. 

A double cam corresponding to the one above described is 
shown at Fig. 83 (6), this double cam making the follower perform 
two double strokes at uniform speed for each revolution of the 

143. Cam for a Shear. The problem may appear in many 
different forms and the case now under consideration assumes 
somewhat different data from the former two, and the shear 
shown in Fig. 84 may serve as a good illustration. Suppose it is 
required to design a cam for this shear; it would usually be desir- 
able to have the shear remain wide open during about one-half 
the time of rotation of the cam, after which the jaw should begin 
to move uniformly down in cutting the plate or bar, and then 
again drop quickly back to the wide-open position. With the 
shear wide open, let the arm be in the position A\B\ where it is 
to remain during nearly one-half the revolution of the cam ; then 
let it be required to move uniformly from AiBi to A 2^2 while 
the cam turns through 120, after which it must drop back again 
very quickly to AiBi. 

An enlarged drawing of the right-hand end of the machine is 
shown at Fig. 85, the same letters being used as in Fig. 84, the 
lines AiBi and A 2 B 2 representing the extreme positions of the 
arm AB. Draw the vertical line QBiB 2 and lay off the angle 
BiQB'z equal to 120; this then is the angle through which the 



camshaft must turn while the arm is moving over its range from 
AiBi to AzB 2 . Now divide the angle BiOB 2 , Fig. 84, into any 
number of equal parts, say four, by the lines OC, OD, and OE; 
these lines are shown on Fig. 85. Next, divide the angle 

FIG. 84. 

BiOBz into the same number of parts as BiOB 2 , that is four, 
by the lines QC', QD' and QE r . 

Now, when the line QBi is vertical as shown, the cam must be 
tangent to AiBi. Next, when the cam turns so that QCi becomes 

FIG. 85. Cam for shear. 

vertical, the arm must rise to C, and hence in this position the 
line OC must be tangent to the cam and the corresponding out- 
line of the cam may thus be found. Draw the arc CC' with 
center Q, and through C' draw a line making the same angle 



ai with QC f that OC does with QC. The line through C' is a 
tangent to the cam. Similarly, tangents to the cam through 
D', E' and B' 2 may be drawn and a smooth curve drawn in 
tangent to these lines, as shown in Fig. 85. 

The details of design for the part B' 2 G may be worked out if 
proper data are given, and evidently the part GFB is circular and 
corresponds with the wide-open position of the shear. 

144. Gas-engine Cam. It not infrequently happens that the 
follower has not a straight-line motion but is pivoted at some 
point and moves in the arc of a circle. This is the case with 
some gas engines and an outline of the exhaust cam, camshaft 

FIG. 86. Gas-engine cams. 

lever and exhaust valve for such an engine is shown at Fig. 86 (a), 
where A is the camshaft and B is the pin about which the fol- 
lower swings. This presents no difficulties not already discussed 
but in executing such a design care must be taken to allow for the 
deviation of the follower from a radial line, and if this is not done 
the cam will not do the work for which it was intended. 

As this problem occurs commonly in practice, it may be as 
well to work out the proper form of cam. The real difficulty 
is not in making the design of the cam, but in choosing the correct 
data and in determining the conditions which it is desired to 
have the cam fulfil. A very great deal of discussion has taken 


place on this point, and as the matter depends primarily on the 
conditions set in the engine, it is out of place here to enter into 
it at any length. Such a cam should open and close the valve 
at the right instants and should push it open far enough, but in 
addition to these requirements it is necessary that the valve 
should come back to its seat quietly, and that in moving it 
should always remain in contact with the cam-actuated operating 
lever. Further, there should be no undue strain at any part 
of the motion, or the pressure of the valve on the lever should be 
as nearly uniform and as low as possible, during its entire 

The total force required to move the valve at any instant is 
that necessary to overcome the gas pressure on top of it, plus 
that necessary to overcome the spring, plus that necessary to lift 
and accelerate the valve if it has not uniform velocity. The gas 
pressure is great just at the moment the valve is opened (the 
exhaust valve is here spoken of) and immediately falls almost to 
that of the atmosphere, while the spring force is least when the 
valve is closed and most when the valve is wide open. The 
weight of the valve is constant and its acceleration is entirely 
under the control of the designer of the cam. Under the above 
circumstances it would seem that the acceleration should be low 
at the moments the valve is opened and closed, and that it 
might be increased as the valve is raised, although the increasing 
spring pressure would prevent undue increase in acceleration. 

Again, the velocity of the valve at the moment it returns to its 
seat must be low or there will be a good deal of noise, and 
the cam should be so designed that the valve can fall rapidly 
enough to keep the follower in contact with the cam, or the noise 
will be objectionable. The general conditions should then be 
that the follower should start with a small acceleration which 
may be increased as the valve opens more, and that it must 
finish its stroke at comparatively low velocity. 

In lieu of more complete data, let it be assumed that the 
valve is to remain open for 120 of rotation of the cam, and is to 
close at low velocity. The travel of the valve is also given and 
it is to remain nearly wide open during 20 of rotation. It will 
first be assumed that the follower moves bira radial line as at Fig. 
86 (6) and correction made later for the deviation due to the arc. 

From the data assumed the valve- is to move upward for 50 
of rotation of the cam and downward during the same interval, 



and as the camshaft turns at constant speed each degree of rota- 
tion represents the same interval of time. Let the acceleration 
be as shown on the diagram Fig. 87 (a) on a base representing 
degrees of camshaft rotation, which is also a time base; then the 
assumed form of acceleration curve will mean that at first the 
acceleration is zero but that this rapidly increases during the 
first 5 of rotation to its maximum va^e at which it remains for 

120 Degrees 

also Seconds 

also Seconds 


also Seconds 

FIG. 87. 

the next 15. It then drops rapidly to the greatest negative value 
where it also remains constant for a short interval and then 
rapidly returns to zero at which it remains for 20, after which 
the process is repeated. Such a curve means a rapidly increasing 
velocity of the valve to its maximum value, followed by a rapid 
decrease to zero velocity corresponding to the full opening of the 
valve and in which position the valve remains at rest for 20. 
The valve then drops rapidly, reaching its seat at the end of 120 
at zero velocity. 



By integrating the acceleration curve the velocity curve is 
found as shown at (6) Fig. 87, and making a second integration 
gives the space curve shown at (c), the maximum height of the 
space curve representing the assumed lift of the cam. These 
curves show that the valve starts from rest, rises and finally comes 
to rest at maximum opening; it then comes down with rapid 
acceleration near the middle of its stroke and comes back on its 
seat again with zero velocity and acceleration and therefore 
without noise. 

FIG. 88. Gas-engine cam. 

Having now obtained the space curve the design of the cam is 
made as follows: 

In Fig. 88, let A represent the camshaft and the circle G 
represent the end of the hub of the cam, the diameter of which is 
determined by considerations of strength. There is always a 
slight clearance left between the hub and follower so that the 
valve may be sure to seat properly and this clearance circle is 
indicated in light lines by C. Lay off radii (say) 10 apart as 
shown; then AD and AE, 120 apart, represent the angle of action 
of the cam. Draw a circle F with center A and at distance from 
C equal to the radius of the roller; then this circle F is the base 

CAMS 147 

circle from which the displacements shown in (c), Fig. 87, are to be 
laid off, and this is now done, one case being shown to indicate the 
exact method. The result is the curve shown in dotted lines 
which begins and ends on the circle F. A pair of compasses are 
now set with radius equal to the radius of the roller of the follower 
and a series of arcs drawn, as shown, all having centers on the 
dotted curve. The solid curve drawn tangent to these arcs is 
the outline of the cam which would fulfil the desired conditions 
provided the follower moved in and out along the radial line from 
the center A as shown at Fig. 86 (6) . 

Should the follower move in the aro of a circle as is the case 
in Fig. 86 (a), where the follower moves in the arc of a circle 
described about B, then a slight modification must be made in 
laying out the cam although the curves shown at (a), (6) and 
(c), Fig. 87, would not be altered. The method of laying out 
the cam from Fig. 87 (c) may be explained as follows: From 
center A draw a circle H (not shown on the drawings) of radius 
AB equal to the distance from the center of camshaft to the 
center of the fulcrum for the lever. Then set a pair of compasses 
with a radius equal to the distance from B to the center of the 
follower, and with centers on H draw arcs of circles outward 
from the points where the radial lines AD, etc., intersect the 
circle F; one of these is shown in Fig. 88. All distances such as 
a are then laid off radially from F but so that their termini will 
be on the arcs just described; thus the point K will be moved 
over to L, and so with other points. The rest of the procedure 
is as in the former case. For ordinary proportions the two cams 
will be nearly alike. 

Should the follower have a flat end without a roller, as is often 
the case, then the circle F is not used at all and all distances such 
as a are laid off on radial lines from the circle C and on each 
radius a line is drawn at right angles to such radius and of length 
to represent the face of the follower. The outline of the cam 
is then made tangent to these latter lines. 

Lack of space prevents further discussion of this very interest- 
ing machine part, which enters so commonly and in such a great 
variety of forms into modern machinery. No discussion has 
been given of cams having reciprocating motion, nor of those 
used very commonly in screw machines, in which bars of various 
shapes are secured to the face of a drum and form a cam which 
may be easily altered to suit the work to be done by simply 


removing one bar and putting another of different shape in its 

After a careful study of the cases worked out, however, there 
should be no great difficulty in designing a cam to suit almost 
any required set of conditions. The real difficulty, in most cases, 
is in selecting the conditions which the cam should fulfil, but 
once these are selected the solution may be made as explained. 


1. Design a disk cam for a stamp mill, for a flat-faced follower, the line 
of the stamp being 4 in. from the camshaft. The stamp is to be lifted 9 in. 
at a uniform rate. 

2. Design a disk cam with roller follower to give a uniform rate of rise and 
fall of 3 in. per revolution to a spindle the axis of which passes through the 
center line of the shaft. 

3. Taking the proportions of the parts from Fig. 84, design a suitable 
cam for the shear. 

4. A cam is required for a 1 in. shaft to give motion to a roller follower 
% in. diameter, and placed on an arm pivoted 6 in. to the left and 2 in. 
above the camshaft. The roller (center) is to remain 2 in. above the cam- 
shaft center for 200 of camshaft rotation, to rise % in. at uniform rate 
during 65, to remain stationary during the next 30, and then to fall uni- 
formly to its original position during the next 65. Design the cam. 

5. Design a cam similar to Fig. 88 to give a lift of 0.375 in. during 45, a 
full open period of valve of 25 and a closing period of 45. Base radius 
of cam to be 0.625 in. and roller 1 in. diameter. 


146. External Forces. When a machine is performing any 
useful work, or even when it is at rest there are certain forces 
acting on it from without, such as the steam pressure on an 
engine piston, the belt pull on the driving pulley, the force of 
gravity due to the weight of the part, the pressure of the water on 
a pump plunger, the pressure produced by the stone which is 
being crushed in a stone crusher, etc. These forces are called 
external because they are not due to the motion of the machine, 
but to outside influence, and these external forces are trans- 
mitted from link to link, producing pressures at the bearings 
and stresses in the links themselves. In problems in machine 
design it is necessary to know the effect of the external forces in 
producing stresses in the links, and further what the stresses are, 
and what forces or pressures are produced at the bearings, for the 
dimensions of the bearings and sliding blocks depend to a very 
large extent upon the pressures they have to bear, and the shape 
and dimensions of the links are determined by these stresses. 

The matter of determining the sizes of the bearings or links 
does not belong to this treatise, but it is in place here to deter- 
mine the stresses produced and leave to the machine designer 
the work of making the links, etc., of proper strength. 

In most machines one part usually travels with nearly uniform 
motion, such as an engine crankshaft, or the belt wheel of a 
shaper or planer, many of the other parts moving at variable 
rates from moment to moment. If the links move with variable 
speed then they must have acceleration and a force must be 
exerted upon the link to produce this. This is a very important 
matter, as the forces required to accelerate the parts of a machine 
are often very great, but the consideration of this question is left 
to a later chapter, and for the present the acceleration of the 
parts will be neglected and a mechanism consisting of light, 
strong parts, which require no force to accelerate them, will be 
assumed in place of the actual one. 



146. Machine is Assumed to be in Equilibrium. It will be 
further assumed that at any instant under consideration, the 
machine is in equilibrium, that is, no matter what the forces 
acting are, that they are balanced among themselves, or the 
whole machine is not being accelerated. Thus, in case of a 
shaper, certain of the parts are undergoing acceleration at various 
times during the motion, but as the belt wheel makes a constant 
number of revolutions per minute there must be a balance be- 
tween the resistance due to the cutting and friction on the one 
hand and the power brought in by the belt on the other. In 
the case of a train which is just starting up, the speed is 
steadily increasing and the train is being accelerated, 
which simply means that more energy is being supplied through 
the steam than is being used up by the train, the balance of 
the power being free to produce the acceleration, and the forces 
acting are not balanced. When, however, the train is up to 
speed and running at a uniform rate the input and output must 
be equal, or the locomotive is in equilibrium, the forces acting 
upon it being balanced. 

147. Nature of Problems Presented. The most general form 
of problem of this kind which comes up in practice is such as 
this: Given the force required to crush a piece of rock, what 
belt pull in a crusher will be required for the purpose? or: What 
turning moment will be required on the driving pulley of a 
punch to punch a given hole in a given thickness of plate? or: 
Given an indicator diagram for a steam engine, what is the result- 
ing turning moment produced on a crankshaft?, etc. Such prob- 
lems may be solved in two ways: (a) by the use of the virtual 
center; (6) by the use of the phorograph, and as both methods 
are instructive each will be discussed briefly. 

148. Solutions by Use of Virtual Centers. This method de- 
pends upon the fundamental principles of statics and the general 
knowledge of the virtual center discussed in Chapter II. The 
essential principles may be summed up in the following three 
statements : 

If a set of forces act on any link of a machine then there will 
be equilibrium, provided: 

1. That the resultant of the forces is zero. 

2. That if the resultant is a single force it passes through a 
point on the link which is at the instant at rest. Such a point 


may, of course, be permanently fixed or at rest, or only tempora- 
rily so. 

3. That if the resultant is a couple the link has, at the instant, 
a motion of translation. 

The first statement expresses a well-known fact and requires 
no explanation. The second statement is rather less known but 
it simply means that the forces will be in equilibrium if their 
resultant passes through a point which is at rest relative to the 
fixed frame of the machine. No force acting on the frame of 
the machine can disturb its equilibrium, for the reason that the 
frame is assumed fixed and if the frame should move in any 
case where it was supposed to remain fixed, it would simply 
mean that the machine had been damaged. Further, a force 
passing through a point at rest is incapable of producing 

The third statement is a necessary consequence of the second 
and corresponds to it. If the resultant is a couple, or two parallel 
forces, then both forces must pass through a point at rest, which 
is only possible if the point is at an infinite distance, or the link 
has a motion about a point infinitely distantly attached to the 
frame, that is the link has a motion of translation. 

Let a set of forces act on any link b of a mechanism in which 
the fixed link is d; then the only point on b even temporarily at 
rest is the virtual center bd, which may possibly be a permanent 
center. Then the forces acting can be in equilibrium only if 
their resultant passes through bd, and if the resultant is a couple 
both forces must pass through bd, which must therefore be at 
an infinite distance, or b must at the instant, have a motion of 
translation. These ppints may be best explained by some 

149. Examples. 1. Three forces PI, P 2 and P 3 , Fig. 89, act on 
the link 6; under what conditions will there be equilibrium? 
In the first place the three forces must all pass through the same 
point A on the link, and treating P 2 as the force' balancing PI 
and P 3 , then in addition to P 2 passing through A it must also 
pass through a point on the link b which is at rest, that is the 
point bd. This fixes the direction of P 2 , by fixing two points on 
it, and thus the directions of the three forces PI, P 2 and P 3 are 
fixed and their magnitudes may be found from the vector triangle 
to the right of the figure. 

2. To find the force P 2 acting at the crankpin, in the direction 



of the connecting rod, Fig. 90, which will balance the pressure PI 
on the piston. In this case PI and P 2 may both be regarded as 
forces acting on the two ends of the connecting rod and the 
problem is thus similar to the last one. PI and P 2 intersect at be; 
hence their resultant P must pass through be and also through 
the only point on b at rest, that is bd, which fixes the position and 

FIG. 89. 

direction of P and hence the relation between the forces may be 
determined from the vector triangle. This enables P 2 to be 
found as in the upper right-hand figure. 

The moment of P 2 on the crankshaft is P 2 X OD, which 
may readily be shown by geometry to be equal to PI X ac 

that is, the turning effect on the crankshaft 

snce -~- = 

1 2 


FIG. 90. 

due to the piston pressure PI is the same as if PI was transferred 
to the point ac on the crankshaft. 

Let P 3 , acting normal to the crank a through the crankpin, 
be the force which just balances PI; it is required to find PS. 
Now P 3 and PI intersect at H, and their balancing force P' 
must pass through H and through bd which gives the direction 



and position of P' and the vector triangle EFG gives P 3 corre- 
sponding to a known value of PI. The force P 3 is called the 
crank effort and may be defined as the force, passing through the 
crankpin and normal to the crank, which would produce the same 
turning moment on the crank that the piston pressure does. 
More will be said about this in the next chapter. 

3. Forces PI and P 2 act on a pair of gear wheels, the pitch 
circles of which are shown in Fig. 91; it is required to find the 
relation between them, friction of the teeth being neglected. 
Since friction is not considered, the direction of pressure between 

FIG. 91. 

the teeth must be normal to them at their point of contact, and 
is shown at P 3 in the figure, this force always passing through the 
point of contact of the teeth and always through the pitch point 
or point of tangency of the pitch circles. 1 For the involute 
system of teeth P 3 is fixed in direction and coincides with the line 
of obliquity, but with the cycloidal system P 3 becomes more 
and more nearly vertical as the point of contact approaches the 
pitch point. Knowing the direction of P 3 from these considera- 
tions, let it intersect PI and P 2 at A and B respectively. On the 
wheel a there are the forces PI, P 3 and P, the latter acting through 
A and ad, and their values are obtained from the vector triangle; 
and on b the forces P 3 , P% and P', the latter acting through B 
1 For a complete discussion on these points see Chapter V. 



and bd, and representing the bearing pressure, are found in the 
same way, the vector polygon on the left giving the values of the 
several forces concerned and hence Pz if PI is known. 

4. The last example taken here is the beam engine illustrated 
in outline in Fig. 92, and the problem is to find the turning 
moment produced on the crankshaft due to a given pressure PI 
acting on the walking beam from the piston. Two convenient 
methods of solution are available, the first being to take moments 
about cd and in this way to find the force Pz acting through be 
which is the equivalent of the force PI at C, the remainder of the 
problem there being solved as in Example 2. 

FIG. 92. 

It is more general, however, and usually -simpler to determine 
the equivalent force on the crankshaft directly. Select any 
point D on PI and resolve PI into two components, one P passing 
through the only point on the beam c at rest, that is cd, the 
other, P 3 , passing through the common point ac of a and c. The 
positions of P and P 3 and their directions are known, since both 
pass through D and also through cd and ac respectively; hence the 
vector triangle on the right gives the forces PS and P. But 
Pa acts through ac on a, and if ad E = h, be drawn from ad 
normal to Pa, the moment of PS about the crankshaft is Pzh, 
which therefore balances the moment produced on the crankshaft 
by the pressure PI on the walking beam. The magnitude of this 
moment is, of course, independent of the position of the point D. 


150. General Formula. The general formula for the solu- 
tion of all such problems by use of virtual centers is as follows : 
A force PI acts through any point B on a link 6; it is required to 
find the magnitude of a force P 2 , of known direction and posi- 
tion, acting on a link e which will exactly balance PI, d being 
the fixed link. Find the centers bd, be and ed. Join B to be 
and bd and resolve PI into P 3 in the direction B be and P 4 in 
the direction B bd; then the moment of P 3 about de must be the 
same as the moment of P 2 about the same point and thus P 2 
is known. 

151. Solution of Such Problems by the Use of the Phoro- 
graph. In solving such problems as are now under considera- 
tion by the use of the phorograph the matter is approached from 
a somewhat different standpoint, and as there is frequent occasion 
to use the method it will be explained in some detail. 

It has already been pointed out that the present investigation 
deals only with the case where the machine is in equilibrium, 
or where it is not, on the whole, being accelerated. This is 
always the case where the energy put into the machine per second 
by the source of energy is equal to that delivered by the 
machine, for example, where the energy per second delivered by 
a gas engine to a generator is equal to the energy delivered to 
the piston by the explosion of the gaseous mixture, friction 
being neglected. 

Suppose now that on any mechanism there is a set of forces 
PI, P 2 , PS, etc., acting on various links, and that these forces are 
acting through points having the respective velocities v\, v%, v s , 
etc., feet per second in the directions of PI, P 2 , PS. The energy 
which any force will impart to the mechanism per second is 
proportional to the magnitude of the force and the velocity 
with which it moves in its own direction; thus if a force of 20 
pd. acts at a point moving at 4 feet per second in the direction 
of the force, the energy imparted by the latter will be 80 ft.-pd. 
per second, and this will be positive or negative according to 
whether the sense of force and velocity are the same or different. 

The above forces will then impart respectively P\v\, P 2 ^ 2 , 
Ps^s, etc., ft.-pd. per second of energy, some of the terms being 
negative frequently and the direction of action of the various 
forces are usually different. The total energy given to the 
machine per second is Pii + P 2 v 2 + PsV 3 + etc., ft.-pd. and if 
this total sum is zero there will be equilibrium, since the net 


energy delivered to the machine is zero. This leads to the im- 
portant statement that if in the machine any two points in the 
same or different links have identical motions, then, as far as 
the equilibrium of the machine is concerned, a given force may 
be applied at either of the points as desired, or if at the two points 
forces of equal magnitude and in the same direction but opposite 
in sense are applied then the equilibrium of the machine will be 
unaffected by these two forces, for the product Pv will be the same 
in each case, but opposite in sense, and the sum of the products 
Pv will be zero. 

To illustrate these points further let any two points B and E' 
in the same or different links in the mechanism have the same 
motion, and let any force P act through B, then the previous 
paragraph asserts that without affecting the conditions of equilib- 
rium in any way, the force P may be transferred from B to B', 
that is to say that if a force P act through a point B in any link, 
and there is found in any other link in the mechanism a point B' 
with the same motion as B, the force P will produce the same effect 
as far as the equilibrium of the mechanism is concerned, whether 
it acts at B or B'. 

It has been shown in Chapter IV that to each point in a median-* 
ism there may be found a point called its image on a selected 
link, which point has the same motion as the point under dis- 
cussion, and thus it is possible to find on a single link a collec- 
tion of points having the same motions as the various points 
of application of the acting forces. Without affecting the con- 
ditions of equilibrium, any force may be moved from its actual 
point of application to the image of this point, and thus the 
whole problem be reduced to the condition of equilibrium of 
a set of forces acting on a single link. There will be equilib- 
rium provided the sum of the moments of the forces about the 
center of rotation relative to the frame is zero. 

152. Examples Using the Phorograph. As this matter is 
somewhat difficult to understand it may best be explained by a 
few practical examples in which the application is given and 
in the solution of the problems it will be found that the only diffi- 
culty offered is in the finding of the phorograph of the mechanism, 
so that Chapter IV must be carefully mastered and understood. 

1. To find the turning effect produced on the crankshaft of 
an engine due to the weight of the connecting rod. Let Fig. 93 
represent the engine mechanism, with connecting rod AB 



having a weight W Ib. and with its center of gravity at G; the 

weight W then acts vertically downward through G. Find A', 

B f and G f , the images of A, B and G on the crank OA ; then since, 

by the principle of the phorograph, the motion of G' is identical 

with that of G, it follows that G f must have exactly the same 

velocity as G, that is to say energy will be imparted to the 

mechanism at the same rate per second by the force W acting 

at G' as it will by the same 

force acting at G, so that the 

force W may be moved to G' 

without affecting the conditions 

of equilibrium, and this has 

been done in the figures. It F Q3 

must not be supposed that W 

acts both at G and G' at the same time ; it is simply transferred 

from G to G'. 

Since G f is a point on the crankshaft, the moment due to the 
weight of the rod is Wh ft.-pd., where h is the shortest distance, 
in feet, from to the direction of the force W. 

2. A shear shown in Fig. 94 is operated by a cam a attached 
to the main shaft 0, the shaft being driven at constant speed by 
a belt pulley. Knowing the force F necessary to shear the bar 

\F R 


FIG. 94. Shear. 

at 8, the turning moment which must be applied at the camshaft 
is required. Let P be the point on the cam a where it touches 
the arm b at Q, then the motion of P with regard to Q is one of 
sliding along the common tangent at P. Choosing a as the link 
of reference, P' will lie at P, R r at 0, R'Q' will be parallel to 
RQ and Q' will lie in P'Q' the common normal to the surfaces 
at P, this locates Q'. Having now two points on b', viz., R' 
and Q', complete the figure by drawing from Q' the line Q'S' 
parallel to QS, also drawing R'S' parallel to RS and thus locat- 
ing S'. The construction lines have not been drawn on the 



diagram. The figure shows the whole jaw dotted in, although 
it is quite unnecessary. Having now found S f a point on a with 
the same velocity at S on 6, the force F may be transferred to 
S f and the moment F X h of F about is the moment which must 
be produced on the shaft in the opposite sense. By finding the 

FIG. 95. Rock crusher. 

moment in a number of positions it is quite easy to find the 
necessary power to be delivered by the belt for the complete 
shearing operation. 

3. A somewhat more complicated machine is shown in Fig. 95, 
which represents a belt-driven rock crusher built by the Fair- 



banks-Morse Co., the lower figure having been redrawn from 
their catalogue, and the upper figure shows the mechanism on a 
larger scale. 

On a belt wheel shown, the belt exerts a net pull Q which 
causes the shaft 0, having the eccentric OA attached to it, to 
revolve. The shaft H carried by the frame has the arm HB 
attached to it, to the left-hand end of which is a roller resting 
on the eccentric OA. The crusher jaw is pivoted on the frame 
at J and a strong link CD keeps the jaw and the arm HB a fixed 
distance apart. As shaft turns, the eccentric imparts a motion 
to the arm HB which in turn causes the jaw to have a pendulum 
motion about J and to exert a pressure P on a stone to be crushed. 
It is required to find the rela- 
tion between belt pull Q and the 
pressure P. 

Select OA as the link of ref- 
erence and make the phorograph 
of double scale as in Fig. 39, mak- 
ing OA' = 2 OA. As the device 
simply employs two chains 
similar to Fig. 32, viz., OABCH 
and JDCH, the images of all the 
points may readily be found and 
these are shown on the figure. 
Then P is transferred from G to G r 
and Q from E to E r and then P 
and Q both act on the one link 
and hence their moments must 
be equal, or Q X OE' = mo- 
ment of P about 0, from which 
P is readily found for a given value of Q. 

4. The application to a governor 1 is shown in Fig. 96 which 
represents one-half of a Proell governor, and it is required to 
find the speed of the vertical spindle which will hold the parts 
in equilibrium in the position shown. In the sketch the arm 
OA is pivoted to the spindle at and to the arm BA at A, the 
latter arm carrying the ball C on an extension of it and being 
attached to the central weight W at B. The weight of each 

revolving ball at C is - Ib. and of the central weight is W Ib. 

FIG. 96. Proell governor. 

A complete discussion of governors is given in Chapter XII. 


Treating OA as the link of reference #nd G as the center of it, 


find the images of A' at A and also B' and C", then transfer -~- 


(one-half the central weight acts on each side) to B f and ^ to C', 


and if it is desired to allow for the weights w a and Wb of the arms 
OA and A B the centers of gravity G and # of the latter are 
found and also their images G' and H f , then Wb is transferred to 
H', but as (r' is at G, w a is not moved. If the balls revolve 
with linear velocity v ft. per second in a circle of radius r ft., 

w v^ 
then the centrifugal force acting on each ball will be P ~~ X 

pds. in the horizontal direction, and this force P is trans- 
ferred to C'. Let the shortest distances from the vertical line 
through to B f , C', G' and H' be hi, h 2 7& 3 and h* respectively, 
and let the vertical distance from C" to OB' be h$, then for equili- 
brium of the parts (neglecting friction), taking moments about 

~2'hi + -^-h 2 + w a h 3 + Wb h 4 = ^- X - X h b 

which enables the velocity v necessary to hold the governor in 
equilibrium in any given position to be found, and from this 
the speed of the spindle may be computed. 

5. The chapter will be concluded by showing two very interest- 
ing applications to riveters of toggle-joint construction. The 
first one is shown in Fig. 97, the drawing on the left showing the 
construction of the machine, while on the right is shown the 
mechanism involved and the solution for finding the pressure P 
at the piston necessary to exert a desired rivet pressure R. 
The frame d carries the cylinder g, with piston /, which is con- 
nected to the rod e by the pin C. At the other end of e is a pin 
A which connects e with two links a and 6, the former of which is 
pivoted to the frame at 0. The link b is pivoted at B to the slide 
c which produces the pressure on the rivet. 

Select a as the primary link because it is the only one having 
a fixed point; then A' is at A } and since B has vertical motion, 
therefore B' will lie on a horizontal line through and also on a 
line through A' in the direction of b, that is, on b produced so 
that B' is found. C' lies on a line through normal to the direc- 
tion of motion of C, that is to the axis of the cylinder g, and since 



it also lies on a line through A' parallel to e, therefore C" is 

By transferring P to C' and R to B' as shown dotted, the rela- 
tions between the forces P and R are easily found, since their 
moments about must be equal, that is, P X OC f = R X OB'. 

By comparing the first and later positions in this and the 
following figures the rapid increase in the mechanical advantage 
of the mechanism, as the piston advances, will be quite evident. 

6. Another form of riveter is shown at Fig. 98 and the solution 
for finding the rivet pressure R corresponding to a given piston 
pressure P is shown along with the mechanism on the right in 

Later Position 

FIG. 97. Riveter. 

two positions. The proportions in the mechanism have been 
altered to make the illustration more clear. The loose link b 
contains four pivots, C } B, A, F; C being jointed to the frame 
at D by the link e; B having a connection to the link c, which link 
is also connected at E to the sliding block e acting directly on the 
rivet. A is connected to the frame at by means of the link a, 
and F is connected to the piston g at G by means of the link /. 
Either links a or e may be used as the link of reference, as 
each has a fixed center, the link a having been chosen. The 
images are found in the following order: C' is on A'C' and on D'C' 
parallel to DC; B' is next found by proportion, as is also F' and 
thus the image of the whole link b. Next E''is on B'E', parallel 
to BE and on O'E' drawn perpendicular to the motion of the 
slide e, while G' is on a line through perpendicular to the motion 

of the piston g and is also on the line F'G' parallel to FG. 



Transfer the force P from G to G', and the force R from E to 
E', and then the moment about of R through E f must equal the 
moment of P through G f , that is, R X OE r = P X OG' from 
which the relation between R and P is computed and this may be 
done for all the different positions of the piston g. 

D Later Position 

FIG. 98. Riveter. 


1. Why are external forces so named? 
the machine? 

What effects do they produce in 

Solve the following by virtual centers: 

2. Determine the crank effort and torque when the crank angle is 45 
in an 8 in. by 10-in. engine with rod 20 in. long, the steam pressure being 40 
pds. per square inch. 

3. In a pair of gears 15 in. and 12 in. diameter respectively the direction 
of pressure between the teeth is at 75^ to the line of centers, which is hori- 
zontal. On the large gear there is a pressure of 200 pds. sloping upward at 
8 and its line of action is 3 in. from the gear center. On the smaller gear is a 
force P acting downward at 10 and to the left, its line of action being 4 in. 
from the gear center. Find P. 

4. In a mechanism like Fig. 37 a = 15 in., b = 24 in., d 4 in. and e = 60 
in. and the link a is driven by a belt on a 10-in. pulley sloping upward at 60. 
Find the relation between the net belt pull and the pressure on / when a is 
at 45. 

6. The connecting rod of a 10 in. by 12-in. engine is 30 in. long and weighs 
30 lb., its center of gravity being 12 in. from the crankpin. What turning 
effect does the rod's weight produce for a 30 crank angle? 


Solve the following by the phorograph: 

6. In a crusher like Fig. 95, using the same proportions as are there given, 
find the ratio of the belt pull to the jaw pressure and plot this ratio for the 
complete revolution of the belt-wheel. 

7. In the Gnome motor, Fig. 178, with a fixed link 2 in. long, and the 
others in proportion from the figure, find the turning moment on the 
cylinder due to a given cylinder pressure. 


153. Variations in Available Energy . In the case of all engines, 
whether driven by steam, gas or liquid, the working fluid delivers 
its energy to one part of the machine, conveniently called the 
piston, and it is the purpose of the machine to convert the energy 
so received into some useful form and deliver it at the shaft to 
some external machine. Pumps and compressors work in exactly 
the opposite way, the energy being delivered to them through 
the crankshaft, and it is their function to transfer the energy 
so received to the water or gas and to deliver the fluid in some 
desired state. 

In the case of machines having pure rotary motion, such as 
steam and water turbines, turbine pumps, turbo-compressors, 
etc., there is always an exact balance between the energy supplied 
and that delivered, and the input to and output from the 
machine is constant from instant to instant. Where reciprocat- 
ing machines, having pistons, are used the case becomes somewhat 
different, and in general, the energy going to or from the piston 
at one instant differs from that at the next instant and so on. 
This of necessity causes the energy available at the crankshaft 
to vary from time to time and it is essential that this latter energy 
be known for any machine under working conditions. 

These facts are comparatively well known among engineers. 
Steam turbines are never made with flywheels because of the 
steadiness of motion resulting from the manner of transforming 
the energy received from the steam. On the other hand, recipro- 
cating steam engines are always constructed with a flywheel, 
or what corresponds to one, which will produce a steadying 
effect and the size of the wheel depends on the type of engine very 
largely. Thus, a single-cylinder engine would have a heavy 
wheel, a tandem compound engine would also have a heavy 
one, while for a cross-compound engine for the same purpose the 
flywheel could be much smaller and lighter. 

Again, a single-cylinder, four-cycle, single-acting gas engine 
would have a much larger wheel than any form of steam engine, 



and the flywheel size would diminish as the number of cylinders 
increased, or as the engine was made double-acting or made to 
run on the two-cycle principle, simply because the input to the 
pistons becomes more constant from instant to instant, and the 
energy delivered by the fluid becomes more steady. 

In order that the engineer may understand the causes of these 
differences, and may know how the machines can best be designed, 
the matter will here be dealt with in detail and the first case 
examined will be the steam engine. 

154. Torque. An outline of a steam engine is shown in Fig. 
99, and at the instant that the machine is in this position let 

J l 

i , , ,, ,, 1 1 ,,~ , rr?u. rr-i 


FIG. 99. 

the steam produce a pressure P on the piston as indicated (the 
method of arriving at P will be explained later), then it is re- 
quired to find the turning moment produced by this pressure on 
the crankshaft. It is assumed that the force P acts through 
the center of the wristpin B. 

Construct the phorograph of the machine and find the image 
B' of B by the principles laid down in Chapter IV. Now in 
Chapter IX it is shown that, for the purposes of determining the 
equilibrium of a machine, any acting force may be transferred 
from its actual point of application to the image of its point of 
application. Hence, the force P acting through B will produce 
the same effect as if this force were transferred to B' on the 
crankdisc, so that the turning moment produced on the crank- 
disc and shaft is P X OB' ft.-pds.,/ind this turning moment will 
be called the torque T. 

Thus T = P X OB' ft.-pds. where OB' is measured in feet. 

155. Crank Effort. Now let the torque T be divided by the 
length a of the crank in feet, then since a is constant for all crank 
positions, the quantity so obtained is a force which is proportional 
to the torque T produced by the steam on the crankshaft. This 


force is usually termed the crank effort and may be defined as 
the force which if acting through the crankpin at right angles 
to the crank would produce the same turning effect that the 
actual steam pressure does (see Sec. 149 (2)). 
Let E denote the crank effort; then 

E X a = T = PxOB' ft.-pds. 

It is evident that the turning moment produced on the crank- 
shaft by the steam may be represented by either the torque T 
ft.-pds. or by the crank effort E pds., since these two always bear a 
constant relation to one another. For this reason, crank efforts 
and torques are very frequently confused, but it must be re- 
membered that they are different and measured in different 
units, and the one always bears a definite relation to the other. 
The graphical solution for finding the effort E corresponding 
to the pressure P is shown in Fig. 99. It is only necessary to lay 
off OH along a to represent P on any convenient scale, and to 
draw HK parallel to A 'B' , and then the length OK will represent 
E on the same scale that OH represents P. The proof is simple. 
Since the triangles OB' A' and OKH are similar, it is evident that: 

OK OB' OB' E . 

OH = OA = ~a~ = P smce X a = P X OB ' 

156. Crank Effort and Torque Diagrams. Having now shown 
how to obtain the crank effort and torque, it will be well to plot 
a diagram showing the value of these for each position of the 
crank during its revolution. Such a diagram is called a crank- 
effort diagram or a torque diagram. In drawing these diagrams 
the usual method is to use a straight base for crank positions, 
the length of the base being equal to that of the circumference 
of the crankpin circle. 

157. Example. Steam Engine. The method of plotting such 
curve from the indicator diagrams of a steam engine is given in 
detail so that it may be quite clear. 

Let the indicator diagrams be drawn as shown in Fig. 100, an 
outline of the engine being shown in the same figure, and the 
crank efforts and torques will be plotted for 24 equidistant posi- 
tions of the crankpin, that is for each 15 of crank angle. The 


straight line OX in Fig. 101 is to be used as the base of the new 
diagram, and is made equal in length to the crankpin circle, 
being divided into 24 equal parts. The corresponding numbers 
in the two figures refer to the same positions. 

The vertical line OL through will serve as the axis for torques 
and crank efforts, but, of course, the scale for crank efforts must 
be different from that for torques. 

Let A i and A 2 represent respectively the areas in square 
inches of the head and crank ends of the piston, the difference 
between the two being due to the area of the piston rod; the 
stroke of the piston is L ft. 

Suppose the indicator diagrams to be drawn to scale s, by 
which is meant that such a spring was used in the indicator that 

FIG. 100. 

1 in. in height on the diagram represents s pds. per square inch 
pressure on the engine piston; thus if s = 60 then each inch in 
height on the diagram represents a pressure of 60 pds. per square 
inch on the piston. The lengths of the head- and crank-end 
diagrams are assumed as li and Z 2 in. (usually li = Z 2 ) and these 
lengths rarely exceed 4 in. irrespective of the size of the engine. 
Now place the diagrams above the cylinder as in Fig. 100 
with the atmospheric lines parallel to the line of motion of the 
piston. The two diagrams have been separated here for the 
sake of clearness, although often they are superimposed with 
the atmospheric lines coinciding. Further, the indicator dia- 
gram lengths have been adjusted to suit the length representing 
the travel of the piston. While this is not necessary, it will fre- 
quently be found convenient, but all that is really required is to 


draw on the diagrams a series of vertical lines showing the points 
on the diagrams corresponding to each of the 24 crank positions; 
these lines are shown very light on the diagrams. Next, draw 
on the diagrams the lines of zero pressure which are parallel 
to the atmospheric lines and at distances below them equal to the 
atmospheric pressure on scale s. 

Having done this preliminary work, it is next necessary to 
find the image A'B' of b for each of the 24 crank positions, one 
of the images being shown on the figure For the crank posi- 
tion 3 shown, it *will be observed that the engine is taking steam 
on the head end and exhausting on the crank end, since the pis- 
ton is moving to the left, and hence at this instant the indicator 
pencils would be at M and N on the head- and crank-end dia- 

456 7JV8 9 lO^^^lZ R' 14 15 16 17 18 19S20 21 22 

FIG. 101. Crank effort and torque diagram. 

grams respectively. It is to be observed that when the crank 
reaches position 21 the piston will again be in the position shown 
in Fig. 100, but, since at that instant the piston is moving to the 
right, the indicator pencils will be at R and Q; some care must 
be taken regarding this point. 

Now let hi in. represent the height of M above the zero line 
and h 2 in. the height of TV; then the force urging the piston for- 
ward is hi X s X AI, while that opposing it is h z X s X A z and 
hence the piston is moving forward under a positive net force of 

P = hi X s X AI - h 2 X s X A 2 pds. 

While it is clear that P is positive in this position, and as a matter 
of fact is positive for most crank positions, yet there are some in 
which it is negative, the meaning of which is that in these posi- 
tions the mechanism has to force the piston to move against 
an opposing steam pressure; the mechanism is able to do this 
to a limited extent by means of the energy stored up in its parts. 
From the value of P thus found, the crank effort E is deter- 
mined by the method already explained and the process repeated 


for each of the 24 crank positions, obtaining in this way 24 
values of E. These will be found to vary within fairly wide 
limits. Then, using the axis of Fig. 101, having a base OX equal 
the circumference of the crankpin circle, plot the values of E 
thus found at each of the 24 positions marked and in this way 
the crank-effort diagram OMNRSX is found, vertical heights 
on the diagram representing crank efforts for the corresponding 
crankpin positions, and these heights may also be taken to repre- 
sent the torques on a proper scale determined from the crank- 
effort scale. 

158. Relation between Crank-effort and Indicator Diagrams. 
From its construction, horizontal distances on the crank-effort 
diagram represent space in feet travelled by the pin, while verti- 
cal distances represent forces in pounds, in the direction of 
motion of the crankpin, and therefore the area under this curve 
represents the work done on the crankshaft in foot-pounds. 
Since the areas of the indicator diagrams represent foot-pounds 
of work delivered to the piston, and from it to the crank, there- 
fore the work represented by the indicator diagrams must be 
exactly equal to that represented by the crank-effort diagram. 
The stroke of the piston has been taken as L ft. and hence the 
length of the base OX will represent TT X L ft., while the length 
of each indicator diagram will represent L ft. Calling p m the 
mean pressure corresponding to the two diagrams and E M the 
mean crank effort, then 2 L X p m = TT X L X E M ft.-pds. or 

E M -p m pds., that is, the mean height of the crank-effort diagram 


in pounds is - times the mean indicated pressure as shown by 

the indicator diagrams. In this way the mean crank-effort 
line LU may be located, and this location may be checked by 
finding the area under the crank-effort diagram in foot-pounds, 
by planimeter and then dividing this by OX will give E M . 

The crank-effort diagram may also be taken to represent 
torques. Thus, if the diagram is drawn on a vertical scale of E 
pds. equal 1 in., and if the crank radius is a ft. then torques may be 
scaled from the diagram using a scale of E X a ft.-pds. equal to 
1 in. 

The investigation above takes no account of the effect of inertia 
of the parts as this matter is treated extensively in Chapter XV 
under accelerations in machinery. 



159. Various Types of Steam Engines. An examination of 
Fig. 101 shows that the turning moment on the crankshaft, in 
the engine discussed, is very variable indeed and this would 
cause certain variations in the operation of the engine which 
will be discussed later. In the meantime it may be stated that 
designers try to arrange the machinery as far as possible to pro- 
duce uniform effort and torque. 

Resultant Torque 
Mean / /~\ Torque 

FIG 102. Torque diagrams for cross-compound engine. 

Steam engines are frequently designed with more than one 
cylinder, sometimes as compound engines and sometimes as 
twin arrangements, as in the locomotive and in many rolling-mill 
engines. Compound engines may have two or three and some- 
times four expansions, requiring at least two, three or four cylin- 
ders, respectively. Engines having two expansions are arranged 
either with the cylinders tandem and having both pistons 

FIG. 103. Torque diagrams for tandem engine. 

connected to the same crosshead, or as cross-compound engines 
with the cylinders placed side by side and each connected through 
its own crosshead and connecting rod to the one crankshaft, 
the cranks being usually of the same radius and being set at 90 
to one another. In Fig. 102 are shown torque diagrams for twin 
engines as used in the locomotive or for a cross-compound engine 
with cranks at 90, the curve A showing the torque corresponding 


to the high-pressure cylinder with leading crank and B that for 
the low-pressure cylinder, while the curve C in plain lines gives 
the resultant torque on the crankshaft, and the horizontal dotted 
line D shows the corresponding mean torque. The very great 
improvement in the torque diagram resulting from this arrange- 
ment of the engine is evident, for the torque diagram C varies very 
little from the mean line D and is never negative as it was with 
the single-cylinder engine. 

On the other hand, the tandem engine shows no improvement 
in this respect over the single-cylinder machine as is shown by 
the torque diagram corresponding to it shown in Fig. 103, the 
dotted curves corresponding to the separate cylinders and the 
plain curve being the resultant torque on the shaft. 

FIG. 104. Torque diagram for triple-expansion engine. 

Increasing the number of cylinders and cranks usually smooths 
out the torque curve and Fig. 104 gives the results obtained from 
a triple-expansion engine with cranks set at 120, in which it is 
seen that the curve of mean torque differs very little from the 
actual torque produced by the cylinders. 

160. Internal-combustion Engines. It will be well in con- 
nection with this question to examine its bearing on internal- 
combustion engines, now so largely used on self-propelled vehicles 
of all kinds. Internal-combustion engines are of two general 
classes, two-cycle and four-cycle, and almost all machines of this 
class are single-acting, and only such machines are discussed 
here as the treatment of the double-acting engine offers no 
difficulties not encountered in the present case. 

In the case of four-cycle engines the first outward stroke of 
the piston draws in the explosive mixture which is compressed in 
the return stroke. At the end of this stroke the charge is ignited 
and the pressure rises sufficiently to drive the piston forward on 


the third or power stroke, on the completion of which the exhaust 
valve opens and the burnt products of combustion are driven out 
by the next instroke of the piston. Thus, there is only one 
power stroke (the third) for each four strokes of the piston, or 
for each two revolutions. An indicator diagram for this type of 
engine is shown in (a) Fig. 105, and the first and fourth strokes 
are represented by straight lines a little below and a little 
above the atmospheric line respectively. 

The indicator diagram from a two-cycle engine is also shown 
in (6) Fig. 105 and differs very little from the four-cycle card 


FIG. 105. Gas-engine diagrams. 

except that the first and fourth strokes are omitted. The action 
of this type may be readily explained. Imagine the piston at its 
outer end and the cylinder containing an explosive mixture, 
then as the piston moves in the charge is compressed, ignited 
near the inner dead point, and this forces the piston out on the 
next or power stroke. Near the end of this stroke the exhaust is 
opened and the burnt gases are displaced and driven out by a 
fresh charge of combustible gas which is forced in under slight 
pressure; this charge is then compressed on the next instroke. 
In this cycle there is one power stroke to each two strokes of the 
piston or to each revolution, and thus the machine gets the same 
number of power strokes as a single-acting steam engine. 

The torque diagram for a four-cycle engine is shown in Fig. 
106 and its appearance is very striking as compared with those 
for the steam engine, for evidently the torque is negative for 
three out of the four strokes, that is to say, there has to be 
sufficient energy in the machine parts to move the piston during 
these strokes, and all the energy is supplied by the gas through 
the one power or expansion stroke. The torque has evidently 
very large variations and the total resultant mean torque is very 
small indeed. 

For the two-cycle engine the torque diagram will be similar to 
the part of the curve shown in Fig. 106 and included in the com- 
pression and expansion strokes, the suction and exhaust strokes 


being omitted. Evidently also the mean torque line will be 
much higher than for the four-cycle curve. 

Returning now to the four-cycle engine it is seen that the turn- 
ing moment is very irregular and if such an engine were used 
with a small flywheel in driving a motorcycle or dynamo, the 
motion would be very unsteady indeed, and would give so much 
trouble that some special means must be used to control it. 
Various methods are taken of doing this, one of the most common 

Suction Stroke i Compression Stroke . Expansion Stroke I Exhaust Stroke 

FIG. 106. Torque diagram for four-cycle gas engine. 

in automobiles, etc., being to increase the number of cylinders. 
Torque diagrams from two of the more common arrangements 
are shown in Fig. 107. The diagram marked (a) gives the 
results for a two-cylinder engine where these are either opposed 
or are placed side by side and the cranks are at 180. Diagram 
(6) gives the results from a four-cylinder engine and corresponds 

FIG. 107. Torque diagrams for multicylinder gasoline engines. 

to the use of two opposed engines on the same shaft or of four 
cylinders side by side, each crank being 180 from the one next it. 
All of the arrangements shown clearly raise the line of mean 
torque and thus make the irregularities in the turning moment 
very much less, and by sufficiently increasing the number of 
cylinders this moment may be made very regular. Some auto- 
mobiles now have twelve-cylinder engines resulting in very 
uniform turning moment and much steadiness, and the arrange- 


ments made of cylinders in aeroplanes are particularly satis- 
factory. Space does not permit the discussion of the matter in 
any further detail as the subject is one which might profitably 
form a subject for a special treatise. 

161. General Discussion on Torque Diagrams. The unsteadi- 
ness resulting from the variable nature of the torque has been 
referred to already and may now be discussed more in detail, 
although a more complete treatment of the subject will be found 
in Chapter XIII under the heading of " Speed Fluctuations in 

For the purpose of the discussion it is necessary to assume the 
kind of load which the engine is driving, and this affects what is 
to be said. Air compressors and reciprocating pumps produce 
variable resisting torques, the diagram representing the torque 
required to run them being somewhat similar to that of the engine, 
as shown in Fig. 101. It will be assumed here, however, that the 
engine is driving a dynamo or generator, or turbine pump, or 
automobile or some machine of this nature which requires a 
constant torque to keep it moving; then the torque required 
for the load will be that represented by the mean torque line in 
the various figures, and this mean torque is therefore also what 
might be called the load curve for the engine. 

Consider Fig. 101 ; it is clear that at the beginning of the revolu- 
tion the load is greater than the torque available, whereas be- 
tween M and N the torque produced by the engine is in excess 
of the load and the same thing is true from R to S, while NR 
and SU on the other hand represent times when the load is in 
excess. Further, the area between the torque curve and MN 
plus the corresponding area above RS represents the total work 
which the engine is able to do, during these periods, in excess 
of the load, and must be equal to the sum of the areas between 
LM, NR and SU and the torque curve. 

Now during MN the excess energy must be used up in some 
way and evidently the only way is to store up energy in the parts 
of the machine during this period, which energy will be restored 
by the parts during the period NR and so on. The net result 
is that the engine is always varying in speed, reaching a maximum 
at N and S and minimum values at M and R } and the amount of 
these speed variations will depend upon the mass of the moving 
parts. It is always the purpose of the designer to limit these 
variations to the least practical amounts and the torque curves 


show one means of doing this. Thus, the tandem compound 
engine has a very decided disadvantage relative to the cross- 
compound engine in this respect. 

Internal-combustion engines with one cylinder are also very 
deficient because Fig. 106 shows that the mean torque is only a 
small fraction of the maximum and further that enough energy 
must be stored up in the moving parts during the expansion 
stroke to carry the engine over the next three strokes. When 
such engines are used with a single cylinder they are always 
constructed with very heavy flywheels in order that the parts 
may be able to store up a large amount of energy without too 
great variation in speed. In automobiles large heavy wheels 
are not possible and so the makers of these machines always use 
a number of cylinders, and in this way stamp out very largely 
the cause of the difficulty. This is very well shown in the figures 
representing the torques from multicylinder engines, and in such 
engines it is well known that the action is very smooth and even 
and yet all parts of the machine including the flywheel are 
quite light. It has not yet been possible, however, to leave off 
the flywheel from these machines. 


1. Using the data given in the engine of Chapter XIII and the indicator 
diagrams there given, plot the crank effort and torque diagrams. For what 
crank angle are these a maximum? 

2. What would be the torque curve for two engines similar to that in 
question 1, with cranks coupled at 90? At what crank angle would these 
give the maximum torque? 

3. Compare the last results with two cranks at 180 and three cranks at 

4. Using the diagram Fig. 161 and the data connected therewith, plot the 
crank-effort curve. 

5. In an automobile motor 3)^ in. bore and 5 in. stroke, the rod is 12 in. 
long; assuming that the diagram is similar to Fig. 161, but the pressures are 
only two-thirds as great, draw the crank-effort curve and torque diagram. 
Draw the resulting curve for two cylinders, cranks at 180; four cylinders, 
cranks at 90; six cylinders, cranks at 60 and at 120 respectively. Try the 
effect of different sequence of firing. 


162. Input and Output. The accurate determination of the 
efficiency of machines and the loss by friction is extremely com- 
plicated and difficult, and it is doubtful whether it is possible 
to deal with the matter except through fairly close approxima- 
tions. All machines are constructed for the purpose of doing 
some specific form of work, the machine receiving energy in 
one form and delivering this energy, or so much of it as is not 
wasted, in some other form; thus, the water turbine receives 
energy from the water and transforms the energy thus received 
into electrical energy by means of a dynamo ; or a motor receives 
energy from the electric circuit, and changes this energy into 
that necessary to drive an automobile, and so for any machine. 
For convenience, the energy received by the machine will be 
referred to as the input and the energy delivered by the machine 
as the output. 

Now a machine cannot create energy of itself, but is only used 
to change the form of the available energy into some other which 
is desired, so that for a complete cycle of the machine (e.g., one 
revolution of a steam engine, or two revolutions of a four-cycle 
gas engine or the forward and return stroke of a shaper) there 
must be some relation between the input and the output. If 
no energy were lost during the transformation, the input and 
output would be equal and the machine would be perfect, as it 
would change the form of the energy and lose none. However, 
if the input per cycle were twice the output then the machine 
would be imperfect, for there would be a loss of one-half of the 
energy available during the transformation. The output can, 
of course, never exceed the input. It is then the province of 
the designer to make a machine so that the output will be as 
nearly equal to the input as possible and the more nearly these 
are to being equal the more perfect will the machine be. 



163. Efficiency. In dealing with machinery it is customary 
to use the term mechanical efficiency or efficiency to denote the 
ratio of the output per cycle to the input, or the efficiency 77 = 

output per cycle , . . . . 

- i " The maximum value of the efficiency is 

input per cycle 

'unity, which corresponds to the perfect machine, and the mini- 
mum value is zero which means that the machine is of no value 
in transmitting energy; the efficiency of the ordinary machine 
lies between these two limits, electric motors having an efficiency 
of 0.92 or over, turbine pumps usually not over 0.80, large steam 
pumping engines over 0.90, etc., while in the case where the clutch 
is disconnected in an automobile engine the efficiency of the 
latter is zero, all the input being used up in friction. 

The quantity 1 77 represents the proportion of the input 
which is lost in the bearings of the machine and in various other 
ways; thus in the turbine pump above mentioned, 77 = 0.80 and 
1 77 = 0.20, or 20 per cent, of the energy is wasted in this 
case in the bearings and the friction of the water in the pump. 
The amount of energy lost in the machine, and which helps to 
heat up the bearings, etc., will depend on such items as the 
nature of lubricant used, the nature of the metals at the bear- 
ings and other considerations to be discussed later. 

Suppose now that on a given machine there is at any instant 
a force P acting at a certain point on one of the links which 
point is moving at velocity v\ in the direction and sense of P; 
then the energy put into the machine will be at the rate of Pvi 
ft.-pds. per second. At the same instant let there be a resisting 
force Q acting on some part of the machine and let the point of 
application of Q have a velocity with resolved part v% in the direc- 
tion of Q so that the energy output is at the rate of Qv 2 ft.-pds. 
per second. The force P may for example be the pressure acting 
on an engine piston or the difference between the tensions on the 
tight and slack sides of a belt driving a lathe, while Q may repre- 
sent the resistance offered by the main belt on an engine or by 
the metal being cut off in a lathe. Now from what has been 

already stated the efficiency at the instant is 77 = - - = ~-^ 

input i .v\ 

and if no losses occurred this ratio would be unity, but is always 
less than unity in the actual case. Now, as in practice Qv% 
is always less than Pvi, choose a force PQ acting in the direction 
of, and through the point of application of P such that PQV\ = 


Qv 2 , then clearly P is the force which, if applied to a friction- 
less machine of the given type, would just balance the resist- 
ance Q, and 


_ _ 

" P Vl " PV 1 == P 

so that evidently the efficiency will be ~ at the instant, and P 
will always be less than P. 

The efficiency may also be expressed in a different form. Thus, 
let Qo be the force which could be overcome by the force P if 
there were no fricticm in the machine; then Pvi = Q v z and 

= ^r and QQ is always greater than Q. 

164. Friction. Whenever two bodies touch each other there 
is always some resistance to their relative motion, this resistance 
being called friction. Suppose a pulley to be suitably mounted 
in a frame attached to a beam and that a rope is over this pulley, 
each end of the rope holding up a weight w Ib. Now, since each 
of these weights is the same they will be in equilibrium and it 
would be expected that if the slightest amount were added to 
either weight the latter would descend. Such is, however, not 
the case, and it is found by experiment that one weight may be 
considerably increased without disturbing the conditions of rest. 

It will also be found that the amount it is possible to add to 
one weight without producing motion will depend upon such 
quantities as the amount of the original weight w, being greater 
as w increases, the kind and amount of lubricant used in the 
bearing of the pulley, the stiffness of the rope, the materials used 
in the bearing and the nature of the mechanical work done on it, 
and upon very many other considerations which the reader will 
readily think of for himself. 

One more illustration might be given of this point. Suppose 
a block of iron weighing 10 Ib. is placed upon a horizontal table 
and that there is a wire attached to this block of iron so that a 
force may be produced on it parallel to .the table. If now a tension 
is put on the wire and there is no loss the block of iron should 
move even with the slightest tension, because no change is being 
made in the potential energy of the block by moving it from place 
to place on the table, as no alteration is taking place in its height. 


It will be found, however, that the block will not begin to move 
until considerable force is produced in the wire, the force possibly 
running as high as 1.5 pds. The magnitude of the force necessary 
will, as before, depend upon the material of the table, the nature 
of the surface of the table, the area of the face of the block of 
iron touching the table, etc. 

These two examples serve to illustrate a very important matter 
connected with machinery. Taking the case of the pulley, it is 
found that a very small additional weight will not cause motion, 
and since there must always be equilibrium, there must be some 
resisting force coming into play which is exactiy equal to that 
produced by the additional weight. As the additional weight 
increases, the resisting force must increase by the same amount, 
but as the additional weight is increased more and more the resist- 
ing force finally reaches a maximum amount, after which it 
is no longer able to counteract the additional weight and then 
motion of the weights begins. There is a peculiarity about this 
resisting force then, it begins at zero where the weights are equal 
and increases with the inequality of the weights but finally reaches 
a maximum value for a certain difference between them, and if 
the difference is increased beyond this amount the weights move 
with acceleration. 

In the case of the block of iron on the table something of the 
same nature occurs. At first there is no tension in the wire and 
therefore no resisting force is necessary, but as the tension in- 
creases the resisting force must also increase, finally reaching a 
maximum value, after which it is no longer able to resist the 
tension produced in the wire and the block moves, and the 
motion of the block will be accelerated if the tension is still 
further increased. This resisting force must be in the direction 
of the force in the wire but opposite in sense, so that it must act 
parallel to the table, that is, to the relative direction of sliding, 
and increases from zero to a limiting value. 

The resisting force referred to above always acts in a way to 
oppose motion of the parts and also always acts tangent to the 
surfaces in contact, and to this resisting force the name of friction 
has been applied. Much discussion has taken place as to the 
nature of the force, or whether it is a force at all, but for the 
present discussion this idea will be adopted and this method of 
treatment will give a satisfactory solution of all problems con- 
nected with machinery. 


Wherever motion exists friction is always acting in a sense 
opposed to the motion, although in many cases its very presence 
is essential to motion taking place. Thus it would be quite 
impossible to walk were it not for the friction between one's 
feet and the earth, a train could not run were there no friction 
between the wheels and rails, and a belt would be of no use in 
transmitting power if there were no friction between the belt 
and pulley. Friction, therefore, acts as a resistance to motion 
and yet without it many motions would be impossible. 

165. Laws of Friction. A great many experiments have been 
made for the purpose of finding the relation between the friction 
and other forces acting between two surfaces in contact. Morin 
stated that the frictional resistance to the sliding of one body 
upon another depended upon the normal pressure between the 
surfaces and not upon the areas in contact nor upon the velocity 
of slipping, and further that if F is the frictional resistance to 
slipping and N the pressure between the surfaces, then F = 
nN where JJL is the coefficient of friction and depends upon the 
nature of the surfaces in contact as well as the materials composing 
these surfaces. 

A discussion of this subject would be too lengthy to place here 
and the student is referred to the numerous experiments and 
discussions in the current engineering periodicals and in books 
on mechanics, such as Kennedy's " Mechanics of Machinery," 
and Unwin's "Machine Design." It may only be stated that 
Morin's statements are known to be quite untrue in the case of 
machines where the pressures are great, the velocities of sliding 
high and the methods of lubrication very variable, and special 
laws must be formulated in such cases. In machinery the nature 
of the rubbing surfaces, the intensity of the pressures, the 
velocity of slipping, methods of lubrication, etc., vary within 
very wide limits and it has been found quite impossible to devise 
any formula that would include all of the cases occurring, or 
even any great number of them, when conditions are so variable. 
The only practical method seems to be to draw up formulas 
for each particular class of machinery and method of lubrication. 
Thus, before it is possible to tell what friction there will be in 
the main bearing of a steam engine, it is necessary to know by 
experiment what laws exist for the friction in case of a similar 
engine having similar materials in the shaft and bearing and 
oiled in the same way, and if the machine is a horizontal Corliss 


engine the laws would not be the same as with a vertical high- 
speed engine; again the laws will depend upon whether the lubri- 
cation is forced or gravity and on a great many other things. 
For each type of bearing and lubrication there will be a law for 
determining the frictional loss and these laws must in each case 
be determined by careful experiment. 

166. Friction Factor. Following the method of Kennedy and 
other writers, the formula used in all cases will be F = fN for 
determining the frictional force F corresponding to a normal pres- 
sure N between the rubbing surfaces, where / is called the fric- 
tion factor and differs from the coefficient of friction of Morin 
in that it depends upon a greater number of elements, and the 
law for / must be known for each class of surfaces, method of 
lubrication, etc., from a series of experiments performed on 
similarly constructed and operated surfaces. 

In dealing with machines it has been shown that they are made 
up of parts united usually by sliding or turning pairs, so that it 
will be well at first to study the friction in these pairs separately. 


167. Friction in Sliding Pairs. Consider a pair of sliding 
elements as shown in Fig. 108 and let the normal component of 
the pressure between these two elements 

be N, and let R be the resultant external 

force acting upon the upper element 

which is moving, the lower one, for the 

present being considered stationary. Let 

the force R act parallel to the surfaces in 

the sense shown, the tendency for the 

body is then to move to the right. Now, from the previous 

discussion, there is a certain resistance to the motion of a the 

amount of which is fN, where / is the friction factor, and this 

force must in the very nature of the case act tangent to the 

surfaces in contact (Sec. 164); thus, from the way in which R is 

chosen, the friction force F = fN and R are parallel. 

Now if R is small, there is no motion, as is well known, for the 
maximum value of F due to the normal pressure N is greater 
than R; this corresponds to a sleigh stalled on a level road, the 
horses being unable to move it. If, however, R be increased 
steadily it reaches a point where it is equal to the maximum 
value of F and then the body will begin to move, and so long as 


R and F are equal, will continue to move at uniform speed because 
the force R is just balanced by the resistance to motion; this 
corresponds to the case where the sleigh is drawn along a level 
road at uniform speed by a team of horses. Should R be still 
further increased, then since the frictional resistance F will be 
less than R, the body will move with increasing speed, the acceler- 
ation it has depending upon the excess of R over F; this corre- 
sponds to horses drawing a sleigh on a level road at an increasing 
speed, and just here it may be pointed out that the friction 
factor must depend upon the speed in some way because the 
horses soon reach a speed beyond which they cannot go. 
These results may be summarized as follows: 

1. If R is less than F, that is R <fN, there is no relative motion. 

2. If R is equal to F, that is R = fN, the relative motion of the 
bodies will be at uniform velocity. 

3. If R is greater than F, that is R>fN, there will be accelerated 
motion, relatively, between the bodies. R is the resultant external 
force acting on the body and is parallel to the surfaces in contact. 

Consider next the case shown in Fig. 109, 
where the resultant external force R acts 
at an angle $ to the normal to the surfaces 
in contact, and let it be assumed that the 
motion of a relative to d is to the right as 
shown by the arrow. The bodies are taken 
to be in equilibrium, that is, the velocity 
of slipping is uniform and without accelera- 
tion. Resolve R into two components 
AB and BC, parallel and normal respectively to the surfaces of 
contact, then since BC = N is the normal pressure between 
the surfaces, the frictional resistance to slipping will be F = fN, 
from Sec. 166, where / is the friction factor, and since there is 
equilibrium, the velocity being uniform, the value of F must be 
exactly equal and opposite to AB, these two forces being in the 
same direction. Should AB exceed F = fN there would be ac- 
celeration, and should it be less than fN there would be no motion. 
Now from Fig. 109, AB = R sin < and also AB = BC tan < 
= N tan $. Hence, since AB = fN, there results the relation 
fN = N tan or / = tan <; this is to say, in order that two 
bodies may have relative motion at uniform velocity, the re- 
sultant force must act at an angle <j> to the normal to the rubbing 
surfaces, and on such a side of the normal as to have a resolved 



part in the direction of motion. The angle <f> is fixed by the fact 
that its tangent is the friction factor /. 

168. Angle of Friction. The angle may be conveniently called 
the angle of friction and wherever the symbol < occurs in the rest 
of this chapter it stands for the angle of friction and is such that 
its tangent is the friction factor /. The angle < is, of course, the 
limiting inclination of the resultant to the normal and if the re- 
sultant act at any other angle less than cf> to the normal, motion 
will not occur; whereas if it should act at an angle greater than 
there will be accelerated motion, for the simple reason that in the 
latter case, the resolved part of the resultant parallel to the sur- 
faces would exceed the frictional resistance, and there would then 
be an unbalanced force to cause acceleration. 

169. Examples. A few examples should make the principles 
clear, and in those first given all friction is neglected except that 

FIG. 110. Crosshead. 

in the sliding pair. The friction in other parts will be considered 

1. As an illustration, take an engine crosshead moving to the 
right under the steam pressure P acting on the piston, Fig. 110. 
The forces acting on the crosshead are the steam pressure P, the 
thrust Q due to the connecting rod and the resultant R of these 
two which also represents the pressure of the crosshead on the 
guide. Now from the principles of statics, P, Q and R must all 
intersect at one point, in this case the center of the wristpin 0, 
since P and Q pass through 0, and further the resultant R must 
be inclined at an angle < to the normal to the surfaces in contact, 
(Sec. 167); thus R has the direction shown. Note that the side 
of the normal on which R lies must be so chosen that R has a 
component in the direction of motion. Now draw AB = P 
the steam pressure, and draw AC and BC parallel respectively to 
R and Q, then BC = Q the thrust of the rod and AC = R the 
resultant pressure on the crosshead shoe. 



If there were no friction in the sliding pair R would be normal 
to the surface and in the triangle ABD the angle BAD would 
be 90; BD is the force in the connecting rod and AD is the 

pressure on the shoe. 

- - Or J 

The efficiency in this position will thus be 
is just as direct to find PQ the force 
necessary to overcome Q if there were no friction by drawing CE 

T) T> ft* 

normal to AB then PQ = BE and rj = -73- = TTJ* 

Jr t5J\. 
2. A cotter is to be designed to connect two rods, Fig. Ill; 

FIG. 111. Cotter pin. 

it is required to find the limiting taper of the cotter to prevent it 
slipping out when the rod is in tension. It will be assumed that 
both parts of the joint have the same friction factor/, and hence 
the same friction angle 4>, and that the cotter tapers only on 
one side with an angle 6. The sides of the cotter on which the 
pressure comes are marked in heavy lines and on the right-hand 
side the total pressure Ri is divided into two parts by the shape 
of the outer piece of the connection. Both the for-ces R\ and 
Rz act at angle < to the normal to their surfaces and, from what 
has already been said, it will be understood that when the cotter 
just begins to slide out they act on the side of the normal shown, 
so that by drawing the vector triangle on the left of height 
AB = P and having CB and BD respectively parallel to Ri 
and R 2) the force Q necessary to force the cotter out is given by 
the side CD. 


In the figure the angle ABC = and ABD = 0. 

Q = P [tan + tan (0 - 6)] 
The cotter will slip out of itself when Q = 0, that is 

tan + tan (0 6) = 0, 
or = 20 

This angle is evidently independent of P except in so far as 
is affected by the tension P in the rod. 

FIG. 112. Lifting jack. 

If the cotter is being driven in then the sense of the relative 
motion of the parts is reversed and hence the forces Ri and R z 
take the directions Ri and RJ and the vector diagram for 
this case is also shown on the right in the figure. The force 
Q' = C'D' necessary to drive the cotter in is Q' = P[tan + 
tan (0 + 0)] and Q' increases with 6. Small values of 6 make 
the cotter easy to drive in and harder to drive out. 

3. An interesting example of the friction in sliding connections 
is given in Fig. 112, which shows a jack commonly used in lifting 
automobiles, etc.; the outlines of the jack only are shown, and 


no details shown of the arrangement for lowering the load. In the 
figure the force P applied to lifting the load Q on the jack is as- 
sumed to act in the direction of the pawl on the end of the handle, 
and this would represent its direction closely although the direc- 
tion of P will vary with each position of the handle. The load 
Q is assumed applied to the toe of the lifting piece, and when the 
load is being raised the heel of the moving part presses against 
the body of the jack with a force R\ in the direction shown and 
the top pressure between the parts is R 2 , both R i and J? 2 being 
inclined to the normals at angle <. 

At the base of the jack are the forces Q and R i, the resultant 
of which must pass through A, while at the top are the forces 
R 2 and P, the resultant of which must pass through B; and if 
there is equilibrium the resultant FI of Q and Ri must balance 
the resultant FI of R% and P, which can only be the case if F\ 
passes through A and B; thus the direction of F\ is known. 

Now draw the vector triangle ECG with sides parallel to F\ t 
R 2 and P, and for a given value of P, so that F\ = EC and 
Rz = CG. Next through E draw ED parallel to Ri and through 
C draw CD parallel to Q from which Q = CD is found. If there 
were no friction the reactions between the jack and the frame 
would be normal to the surfaces at the points of contact, thus 
A would move up to A and B to 5 and the vector diagram would 
take the form EDoC Q G where EG = P as before and D C = Qo 
so that Qo is found. 

The efficiency of the device in this position is evidently 77 = ^r~ 

lt is evident that with the load on the toe, the efficiency is a 
maximum when the jack is at its lowest position because AB is 
then most nearly vertical, while for the very highest positions 
the efficiency will be low. 

4. One more example of this kind will suffice to illustrate the 
principles. Fig. 113 shows in a very elementary form a quick- 
return motion used on shapers and machine tools, and illustrated 
at Fig. 12. Let Q be the resistance offered to the cutting tool 
which is moving to the right and let P be the net force applied 
by the belt to the circumference of the belt pulley. For the 
present problem only the friction losses in the sliding elements 
will be considered leaving the other parts till later. Here the 
tool holder g presses on the upper guide and the pressure on this 
guide is R it the force in the rod e is denoted by FI. Further the 



pressure of 6 on c is to the right and as the former is moving 
downward for this position of the machine, the direction of 
pressure between the two is R 2 through the center of the pin. 

Now on the driving link a the forces acting are P and R 2 , the 
resultant F 2 of which must pass through and A. In the vector 
diagram draw BC equal and parallel to P, then CD and BD 
parallel respectively to F 2 and R 2 will represent these two forces 

Q H 
PIG. 113. Quick-return motion. 

\H Q 

so that R 2 is determined. Again on c the forces acting are R? 
and FI, and their resultant passes through Oi and also through E, 
the intersection of FI and R 2 , so that drawing BG and DG in the 
vector diagram parallel respectively to F 3 and FI gives the force 
FI = DG in the rod e. Acting on the tool holder g are the forces 
FI, Q and R i and the directions of them are known and also the 
magnitude of FI, hence complete the triangle GHD with sides 
parallel to the forces concerned and then GH = Q and HD = R! 
which gives at once the resistance Q which can be overcome at 
the tool by a given net force P applied by the belt. 

If there were no friction in these sliding pairs then the forces 
RI and R% would act normal to the sliding surfaces instead of 
at angles fa and <f> 2 to the normals so that A moves to AQ and 



E to EQ and the construction is shown by the dotted lines, from 
which the value of Q is obtained. The efficiency for this posi- 

The value of rj should be found 

tiori of the machine is 17 = -~r 


for a number of other positions of the machine, and, if desirable, 
a curve may be plotted so that the effect of friction may be 
properly studied. 

Before passing on to the case of turning pairs the attention of 
the reader is called to the fact that the greater part of the problem 
is the determination of the condition of static equilibrium as 
described in Chapter IX, the method of solution being by means 
of the virtual center, in these cases the permanent center being 
used. The only difficulty here is in the determination of the 

direction of the pressures R 
between the sliding surfaces, 
and the following suggestions 
may be found helpful in this 

Let a crosshead a, Fig. 114, 
slide between the two guides 
di and d 2 , first find out, by 
inspection generally, from the 
forces acting whether the 
pressure is on the guide di 
or d 2 . Thus if the con- 
necting rod and piston rod are in compression the pressure 
is on d 2 , if both are in tension it is on di, etc., suppose for this 
case that both are in compression, the heavy line showing the 
surface bearing the pressure. 

Next find the relative direction of sliding. It does not matter 
whether both surfaces are moving or not, only the relative 
direction is required it is assumed in the sense shown, i.e., the 
sense of motion of a relative to cZ 2 is to the left (and, of course, the 
sense of motion of d 2 relative to a is to the right). Now the re- 
sultant pressure between the surfaces is inclined at angle <f> 
to the normal where $ = tan" 1 /, / being the friction factor, so 
that the resultant must be either in the direction of R i or R\. 

Now RI the pressure of a upon d! 2 acts downward, and in 
order that it may have a resolved part in the direction of motion, 
then RI and not RI is the correct direction. If RI is treated as 
the pressure of d z upon a then RI acts upward, but the sense of 



motion of d 2 relative to a is the opposite of that of a relative to 
d z , and hence from this point of view also Ri is correct. 

It is easy to find the direction of RI by the following simple 
rule: Imagine either of the sliding pieces to be an ordinary 
carpenter's wood plane, the other sliding piece being the wood 
to be dressed, then the force will have the same direction as the 
tongue of the plane when the latter is being pushed in the given 
direction on the cutting stroke, the angle to the normal to the 
surfaces being </>. 

170. Turning Paks. In dealing with turning pairs the same 
principles are adopted as are used with the sliding pairs and should 
not cause any difficulty. Let a, Fig. 115, represent the outer 

FIG. 115. 

element of a turning pair, such as a loose pulley turning in the 
sense shown upon the fixed shaft d of radius r, and let the forces 
P and Q act upon the outer element. It must be explained that 
the arrow shows the sense in which the pulley turns relatively 
to the shaft and this is to be understood as the meaning of the 
arrow in the rest of the present discussion. It may be that both 
elements are turning in a given case, and the two elements may 
also turn in the same or in opposite sense, but the arrow indicates 
the relative sense of motion and the forces P and Q are assumed 
to act upon the link on which they are drawn, that is upon a 
in Fig. 115. 



If there were no friction then the resultant of P and Q would 
pass through the intersection A of these forces and also through 
the center of the bearing, so that under these circumstances it 
would be possible to find Q for a given value of P by drawing the 
vector triangle. 

There is, however, frictional resistance offered to motion at the 
surface of contact, hence if the resultant R of P and Q acted 
through 0, there could be no motion. In order that motion may 
exist it is necessary that the resultant produce a turning moment 
about the center of the bearing equal and opposite to the resist- 
ance offered by the friction between the surfaces. It is known 

already that the frictional 
resistance is of such a na- 
ture as to oppose motion, 
and hence the resultant force 
must act in such a way as 
to produce a turning mo- 
ment in the sense of mo- 
tion equal to the moment 
offered by friction in the 
opposite sense. Thus in 
the case shown in the 
figure the resultant must 
pass through A and lie to 
the left of 0. 

In Fig. 116, which shows an enlarged view of the bearing, let 
p be the perpendicular distance from to R, so that the moment 
of R about is R X p. The point C may be conveniently 
called the center of pressure, being the point of intersection of R 
and the surfaces under pressure. Join CO. Now resolve R into 
two components, the first, F t tangent to the surfaces at C, and 
the second, N, normal to the surfaces at the same point. Fol- 
lowing the method employed with sliding pairs, N is the normal 
pressure between the surfaces and the frictional resistance to 
.motion will be fN, where /is the friction factor (Sec. 166), and 
since the parts are assumed to be in equilibrium, there must be 
no unbalanced force, so that the resolved part F of the resultant 
R must be equal in magnitude to the frictional resistance, or 
F fN. But / = tan <, where <j> is the friction angle, so that 

FIG. 116. 

tan <f> = f = -.: 



from which it follows that the angle between N and R must 
be <, and hence the resultant R must make an angle <f> with 
the radius r at the center of pressure C. 

171. Friction Circle. With center draw a circle tangent to 
.R as shown dotted ; then this circle is the one to which the result- 
ant R must be tangent to maintain uniform relative motion, 
and the circle may be designated as the friction circle. The 
radius p of the friction circle is p = r sin 4>, where r is the radius 
of the journal, and this circle is concentric with the journal and 
much smaller than the latter, since < is always a small angle 
in practice. Thus, in turning pairs the resultant must always 

FIG. 117. 

be at an angle (j> to the normal to the surfaces, and this is most 
easily accomplished by drawing the resultant tangent to the 
friction circle, and on such a side of it that it produces a turning 
moment in the sense of the relative motion of the parts. Since 
/ is always small in actual bearings, < is also small, and hence 
tan 4> = sin < nearly, so that approximately p = r sin <f> = r tan < 

= Tf. 

Four different arrangements of the forces on a turning pair 
are shown at Fig. 117, similar letters being used to Fig. 115. 
At (a) P and Q act on the outer element but their resultant R 
acts in opposite sense to the former case and hence on opposite 
side of the friction circle, since the relative sense of rotation is 
the same. In case (c) P and Q act on the inner element and the 
relative sense of rotation is reversed from (a), hence R passes on 
the right of the friction circle; at (fr) conditions are the same as 
(a) except for the relative sense of motion which also changes the 
position of R' } at (d) the forces act on the outer element and the 
sense of rotation and position of R are both as indicated. 

172. Examples. The construction already shown will be 
applied in a few practical cases. 



1. The first case considered will be an ordinary bell-crank 
lever, Fig. 118, on which the force P is assumed to act horizontally 
and Q vertically on the links a and c respectively, the whole 
lever b turning in the clockwise sense. An examination of the 
figure shows that the sense of motion of a relative to b is counter- 
clockwise as is also the motion of c relative to 6, therefore P 
will be tangent to the lower side of the friction circle at bearing 
1, and Q will be tangent to the left-hand side of the friction circle 

FIG. 118. 

at bearing 3, and the resultant of P and Q must pass through A 
and must be tangent to the upper side of the friction circle on 
the pair 2, so that the direction of R becomes fixed. Now draw 
DE in the direction of P to represent this force and then draw 
EF and DF parallel respectively to Q and R and intersecting at 
F, then EF = Q and DF = R. 

In case there was no friction and assuming the directions of 
P and Q to remain unchanged (this would be unusual in practice), 
then P, Q and their resultant, would act through the centers of 
the joints 1, 3 and 2 respectively. Assuming the magnitude of 
P to be unchanged, then the vector triangle DEF' has its sides 
EF' and DF' parallel respectively to the resistance Q and the 
resultant RQ so that there is at once obtained the force QQ = EF'. 

Then the efficiency of the lever in this position is rj = ^- and 


for any other position may be similarly found. 



The friction circles are not drawn to scale but are made larger 
than they should be in order to make the drawing clear. 

2. Let it be required to find the line of action of the force in 
the connecting rod of a steam engine taking into account friction 
at the crank- and wristpins. To avoid confusion the details of 
the rod are omitted and it is represented by a line, the friction 
circles being to a very much exaggerated scale. Let Fig. 119(a) 
represent the rod in the position under consideration, the direc- 

FIG. 119. Steam-engine mechanism. 

tion of the crank is also shown and the piston rod is assumed to 
be in compression, this being the usual condition for this position 
of the crank. Inspection of the figure shows that the angle a. 
is increasing and the angle /? is decreasing, so that the line of 
action of the force in the connecting rod must be tangent to the 
top of the friction circle at 2 and also to the bottom of the' fric- 
tion circle at 1, hence it takes the position shown in the light line 
and crosses the line of the rod. This position of the line of action 
of the force is seen on examination to be correct, because in 
both cases the force acts on such a side of the center of the bear- 
ing as to produce a turning moment in the direction of relative 




Two other positions of the engine are shown in Fig. 119 at 
(6) and .(c), the direction of revolution being the same as before 
and the line of action of the force in the rod is in light lines. In 
the case (6), the rod is assumed in compression and evidently 
both the angles a and /3 are decreasing so that the line of action 
of the force lies below the axis of the rod; while in the position 
shown in (c), the connecting rod is assumed in tension, a is decreas- 
ing, and ]8 is increasing so that the line of the force intersects the 
rod. In all cases the determining factor is that the force must 
lie on such a side of the center of the pin as- to produce a turning 
moment in the direction of relative motion. 

173. Governor Turning Pairs Only. A complete device in 
which turning pairs alone occur is shown at Fig. 120, which 

FIG. 120. Governor. 

represents one of the governors discussed fully in the following 
chapter, except for the effect of friction. The governor herewith 
is shown also at Fig. 125 and only one-half of it has been drawn 
in, the total weight of the two rotating balls is w Ib. while that 
of the central weight including the pull of the valve gear is taken 
as W Ib. In Chapter XII no account has been taken of friction 
or pin pressures while these are essential to the present purpose. 
There will be no frictional resistance between the central weight 
and the spindle and the friction circles at A, B and C are drawn 
exaggerated in order to make the construction more clear, 

It is assumed that the balls are moving slowly outward and 
that when passing through the position illustrated the spindle 
rotates at n revolutions per minute or at co radians per second; 
it is required to find n and also the speed n' of the spindle as the 


balls pass through this same position when travelling inward. 
The difference between these two speeds indicates to some extent 
the quality of the governor, as it shows what change must be 
made before the balls will reverse their motion. 

On one ball there is a centrifugal force -~ pds., where = rco 2 , 

z z Zg 

r being the radius of rotation of the balls in feet, also -~ acts 

horizontally while the weight of the ball -~ lb. acts vertically, 

and their resultant is a force P inclined as shown in the left-hand 
figure. The arms AB and BC are both in tension evidently, 
and as the, balls are moving outward, a. is increasing and /3 is 
decreasing (see Fig. 120) ; hence the direction of the force in the 
arm BC crosses the axis of the latter as shown, F\ representing 
the force. 

Now the direction of the force P is unknown and it cannot 


be determined without first finding -= which, however, depends 


upon n, the quantity sought. An approximation to the slope 
of P may be found by neglecting friction and with this approxi- 
mate value the first trial may be made. With the assumed 
direction of P the point H, where P intersects FI, is determined 
and then the resultant R of FI and P must pass through H and 
also be tangent to the friction circle at A. (If there were no 
friction, R would pass through the center of A, Sec. 170.) Turn- 


ing now to the vector diagram on the right make DE = -^ and 

EL = -^r] then draw DG horizontally to meet EG, which is paral- 


lei to FI in G. The length EG represents the force FI in the arm 
BC y while DG represents the tension on the weight W which is 
balanced by the other half of the governor. 

Next draw GJ and EJ parallel respectively to R and P, whence 
these forces are found. If the slope of P has been properly 
assumed, the point J will be on the horizontal line through L, 
and if J does not lie on this line a second trial slope of P must be 
made and the process continued until / does fall on the horizon- 
tal through L. 

The length LJ then represents ~ = ^r ra)2 from which w is 

readily computed, and from it the speed n in revolutions per 



The dotted lines show the case where the mechanism passes 
through the same position but with the balls moving inward 
and from the length LJ' the value of co' and of n' may be found. 

If only the relation between n and n r is required, then = A/FT/ 

n \ LJ 

The meaning of this is that if the balls were moving outward due 
to a decreased load on the prime mover to which the governor 
was connected then they would pass through the position shown 
when the spindle turned at n revolutions per minute, but if the 
load were again increased causing the balls to move inward the 
speed of the spindle would have to fall to n' before the balls 
would pass through the position shown. Evidently the best, 
governor is one in which n and n' most nearly agree, and the 
device would be of little value where they differed much. 

FIG. 121. 

In reading this problem reference should also be made to the 
chapter on governors. 

174. Machine with Turning and Sliding Pairs. This chapter 
may be very well concluded by giving an example where both 
turning and sliding pairs are used, although there should be no 
difficulty in combining the principles already laid down in any 
machine. The machine considered is the steam engine, the barest 
outlines of which are shown in Fig. 121. The piston is assumed 
combined with the crosshead and only the latter is shown, and 
in the problem it has been assumed that the engine is lifting a 
weight from a pit by means of a vertical rope on a drum, the 
resistance of the weight being Q Ib. Friction of the rope is not 
considered. The indicator diagram gives the information 
necessary for finding the pressure P acting through the piston 


on the crosshead, and the problem is to find Q and the 

From the principles already laid down, the direction of Ri 
the pressure on the crosshead is known, also the line of action 
of FI and of R 2 . For equilibrium the forces Fi, R, and P must 
intersect at one point which is evidently A, as P, the force due 
to the steam pressure, is taken to act along the center of the 
piston rod. On the crankshaft there is the force F\ from the 
connecting rod, and the force Q due to the weight lifted, and if 
there were no friction, their resultant would pass through their 
point of intersection B and also through the center of the crank- 
shaft. To allow for friction, however, R 2 must be tangent to 
the friction circle at the crankshaft and must touch the top of 
the latter, hence the position of R 2 is fixed. Thus the locations 
of the five forces, P, FI, Ri, R 2 and Q are known. 

Now draw the vector diagram, laying off CD *= P and drawing 
CE and DE parallel respectively to Ri and FI, which gives these 
two forces, next draw EF parallel to R 2 and DF parallel to Q 
which thus determines the magnitude of Q. 

If there were no friction, FI would be along the axis of the rod, 
and R i normal to the guides, both forces passing through AQ the 
center of the wristpin. Further, R 2 would pass through BQ the 
intersection of FI and Q, it would also pass through as shown 
dotted, so that the lines of action of all of the forces are known and 
the vector diagram CEoFoD may be drawn obtaining the resistance 
Qo = DFo, which could be overcome by the pressure P on the piston 
if there were no friction. The efficiency of the machine in this 

position is then r> = -*-, and may be found in a similar way for 

other positions. 

If desired, the value of the efficiency for a number of positions 
of the machine may be found and a curve plotted similar to a 
velocity diagram, Chapter III, from which the efficiency per 
cycle is obtained. 

In all illustrations the factor / is much exaggerated to make 
the constructions clear and in many actual cases the efficiency 
will be much higher than the cuts show. Where the efficiency 
is very close to unity, the method is not as reliable as for low 
efficiencies, but many of the machines have such high efficiency 
that such a construction as described herein is not necessary, 
nor is any substitute for it needed in such cases. 



1. In the engine crosshead, Fig. 110, if the friction factor is 0.05, what size 
is the friction angle? If the piston pressure is 5,000 pds., and the connect- 
ing rod is at 12 to the horizontal, what is the pressure in the rod and the 
efficiency of the crosshead, neglecting friction at the wristpin? 

2. Of two 12-in. journals one has a friction factor 0.002 and the other 0.003. 
What are the sizes of the friction circles? 

3. What would be the efficiency of the crank in Fig. 118 if the scale of the 
drawing is one-quarter and the pins are 1^ in. diameter? 

4. Determine the direction of the force in the side rod of a locomotive in 
various positions. 

6. A thrust bearing like Fig. 1 (6) has five collars, the mean bearing diam- 
eter of which is 10 in. If the shaft runs at 120 revolutions per minute and 
has a bearing pressure of 50 Ib. per square inch of area, find the power lost 
if the friction factor is 0.05. 

6. In the engine of Fig. 121, taking the scale of the drawing as one-six- 
teenth and the friction factor as 0.06, find the value of Q when P = 2,500 
pds. the diameters of the crank and wristpins being 3}^ and 3 in. respec- 

7. In a Scotch yoke, Fig. 6, the crank is 6 in. long and the pin 2 in. diam- 
eter, the slot being 3 in. wide. With a piston pressure of 500 pds., find 
the efficiency for each 45 crank angle, taking/ = 0.1. 



175. Methods of Governing. In all prime movers, which will 
be briefly called engines, there must be a continual balance be- 
tween the energy supplied to the engine by the working fluid and 
the energy delivered by the machine to some other which it is 
driving, e.g., a dynamo, lathe, etc., allowance being made for the 
friction of the prime mover. Thus, if the energy delivered by the 
working fluid (steam, water or gas) in a given time exceeds the 
sum of the energies delivered to the dynamo and the friction of 
the engine, then there will be some energy left to accelerate the 
latter, and it will go on increasing in speed, the friction also in- 
creasing till a balance is reached or the machine is destroyed. 
The opposite result happens if the energy coming in is insufficient, 
the result being that the machine will decrease in speed and 
may eventually stop. 

In all cases in actual practice, the output of an engine is con- 
tinually varying, because if a dynamo is being driven by it for 
lighting purposes the number of lights in use varies from time to 
time; the same is true if the engine drives a lathe or drill, the 
demands of these continually changing. 

The output thus varying very frequently, the energy put in 
by the working fluid must be varied in the same way if the desired 
balance is to be maintained, and hence if the prime mover is to 
run at constant speed some means of controlling the energy ad- 
mitted to it during a given time must be provided. 

Various methods are employed, such as adjusting the weight of 
fluid admitted, adjusting the energy admitted per pound of fluid, 
or doing both of these at one time, and this adjustment may be 
made by hand as in the locomotive or automobile, or it may be 
automatic as in the case of the stationary engine or the water 
turbine where the adjustment is made by a contrivance called a 

A governor may thus be defined as a device used in connection 
with prime movers for so adjusting the energy admitted with the 



working fluid that the speed of the prime mover will be constant 
under all conditions. The complete governor contains essentially 
two parts, the first part consisting of certain masses which rotate 
at a speed proportional to that of the prime mover, and the 
second part is a valve or similar device controlled by the part 
already described and operating directly on the working fluid. 

It is not the intention in the present chapter to discuss the 
valve or its mechanism, because the form of this is so varied as to 
demand a complete work on it alone, and further because its 
design depends to some extent on the principles of thermo- 
dynamics and hydraulics with which this book does not deal. 
This valve always works in such a way as to control the amount 
of energy entering the engine in a given time and this is usually 
done in one of the following ways: 

(a) By shutting off a part of the working fluid so as to admit 
a smaller weight of it per second. This method is used in many 
water wheels and gas engines and is the method adopted in the 
steam engine where the length of cutoff is varied as in high-speed 

(6) By not only altering the weight of fluid admitted, but by 
changing at the same time the amount of energy contained in 
each pound. This method is used in throttling engines of various 

(c) By employing combinations of the above methods in 
various ways, sometimes making the method (a) the most im- 
portant, sometimes the method (6) The combined methods 
are frequently used in gas engines and water turbines. 

The other part of the governor, that is the one containing the 
revolving masses driven at a speed proportional to that of the 
prime mover, will be dealt with in detail because of the nature 
of the problems it involves, and it will in future be briefly referred 
to as the governor. 

176. Types of Governors. Governors are of two general 
classes depending on the method of attaching them to the prime 
mover and also upon the disposition of the revolving masses, 
and the speed at which these masses revolve. The first type of 
governors, which is also the original type used by Watt on his 
engines, has been named the rotating-pendulum governor because 
the revolving masses are secured to the end of arms pivoted to the 
rotating axis somewhat similar to the method of construction of 
a clock pendulum, except that the clock pendulum swings in one 




plane, while the governor masses revolve. In this type there are 
three subdivisions : (a) gravity weighted, in which the centrifugal 
force due to the revolving masses or balls is largely balanced by 
gravity; (6) spring weighted, in which the same force is largely 
balanced by springs; and (c) combination governors in which 
both methods are used. Governors of this general class are 
usually mounted on a separate frame and driven by belt or gears 
from the engine, but they are, at times, made on a part of the 
main shaft. 

The second type is the inertia governor which is usually made 
on the engine power shaft, although it is occasionally mounted 
separately. The name is now principally used to designate a 
class of governor with its re- 
volving masses differently dis- 
tributed to the former class; 
its equilibrium depends on 
centrifugal force but during 
the changes in position the 
inertia of the masses plays a 
prominent part in producing Steam 
rapid adjustment. The name 
shaft-governor is also much 
used for this type. 

177. Revolving-pendulum 
Governor. Beginning with 
the revolving-pendulum type, 
an illustration of which is 
shown at Fig. 122 connected 
up to a steam engine, it is seen that it consists essentially of 
a spindle A, caused to revolve by means of two bevel gears 
B and C, the latter being driven in turn through a pulley D 
which is connected by a belt to the crankshaft of the engine; 
thus the spindle A will revolve at a speed proportional to that 
of the crankshaft of the engine. To this spindle at F two balls 
G are attached through the ball arms E, and these arms are 
connected by links J to the sleeve H, fastened to the rod R, 
which rod is free to move up and down inside the spindle A as 
directed by the movement of the balls and links. The sleeve 
H with its rod R is connected in some manner with the valve V, 
in this illustration a very direct connection being indicated, so 
that a movement of the sleeve will open or close the valve V. 

FIG. 122. Simple governor. 



The method of operation is almost self-evident; as the engine 
increases in speed the spindle A also increases proportionately 
and therefore there is an increased centrifugal force acting on 
the balls G causing them to move outward. As the balls move 
outward the sleeve H falls and closes the valve V so as to prevent 
as much steam from getting in and thus causing the speed of 
the engine to decrease, upon which the reverse series of opera- 
tions takes place and the valve opens again. It is, of course, the 
purpose of the device to find such a position for the valve V that 
it will just keep the engine running at uniform speed, by admit- 
ting just the right quantity of steam for this purpose. 

178. Theory of Governor. Several different forms of the 
governor are shown later in the present chapter and will be dis- 

FIG. 123. 

cussed subsequently, but it may be well to begin with the simplest 
form shown in Fig. 123, where the connection of the sleeve to 
the valve is not so direct as in Fig. 122 but must be made through 
suitable linkage. The left-hand figure shows a governor with the 
arms pivoted on the spindle, while the right-hand figure shows 
the pivots away from the spindle, and the same letters are used 
on both. Let the total weight of the two balls be w lb., each ball 

therefore weighing -^ lb., and let these be rotated in a circle of 


radius r ft., the spindle turning at n revolutions per minute corre- 


spending to co = -^=r radians per second. For the present, 
friction will be neglected. 


Three forces act upon each ball and determine its position of 
equilibrium. These are: (a) The attraction of gravity, which will 
act vertically downward and will therefore be parallel with the 
spindle in a governor where the spindle is vertical as in the illus- 


tration shown. The magnitude of this force is ^ pds. (6) The 


second force is due to the centrifugal effect and acts radially 
and at right angles to the spindle, its amount being - - . r. co 2 pds. 


(c) The third force is due to the pull of the ball arm, and will be 
in the direction of the line joining the center of gravity of the 
ball to the pivot on the spindle, which direction may be briefly 
called the direction of the ball arm. 

These three forces must be in equilibrium so that the vector 

triangle ABC may be drawn where AB = > BC o~ rc 2 anc * 

co must be such that AC is parallel to the arm. Now let D be 
the point at which the ball arm intersects the spindle and draw 
AE perpendicular to the spindle DE; then AE = r, the radius 
of rotation of the balls and the distance DE = h is called the 
height of the governor. 

The triangles DAE and ACB are similar and therefore: 

DE _ AB 
EA ~ BC 


h _ 2 
r ~~ w 

which gives 


h = ~2 

Thus, the height of the governor depends on the speed alone and 
not on the weight of the balls. The investigation assumes that 
the^ resistance offered at the sleeve is negligible as indeed is the 
case with many governors and gears, but allowance will be made, 
for this in problems discussed later. 

179. Defects of this Governor. Such a governor possesses 
several serious defects. In the first place, the sleeve must move 
in order that the valve may be operated, and this movement of 
the sleeve will evidently correspond with a change in the height 


and hence with a change in speed co. Thus, each position of the 
balls, corresponding to a given valve position, means a different 
speed of the governor and therefore of the engine ; this is what the 
governor tries to prevent, for its purpose is to keep the speed of 
the engine constant, although the valve may have to be opened 
various amounts corresponding to the load which the engine 
carries. This defect may be briefly expressed by saying that the 
governor is not isochronous, the meaning of isochronism being 
that the speed of the governor will not vary during the entire 
range of travel of the sleeve, or in other words the valve may be 
moved into any position to suit the load, and yet the engine and 
therefore the governor, will always run at the same speed. 

The second defect is that for any 
reasonable speed h is extremely small. 
To show this let the governor run at 
120 revolutions per minute so that co 

= -~Q- = 12.57 radians per second; 
then h = ~ = ,, ' r ^o = 0.2036ft. or 

2.44 in., a dimension which is so small, 
that if the balls were of any reasonable 
size, it would make the practical con- 
struction almost impossible. 
FIG. 124. Crossed-arm 180. Crossed -arm Governor. Now 

it is the desire of all builders to make 

their governors as nearly isochronous as is consistent with other 
desirable characteristics, which means that the height h must be 
constant, and to serve this end the crossed-arm governor shown 
in Fig. 124, has been built somewhat extensively. The propor- 
tions which will produce isochronism may be found mathematic- 
ally thus: 

Inspection of the figure shows that 

h = I cos 6 a cot 6. 

For isochronism h is to remain constant for changes in the 
angle 6 or 

-77 = = Z sin + a cosec 2 8. 

From which 

a = I sin 3 B 
h = I cos 3 

i - 3- "TV e/ 

h - (( , - t--- c LA 


and therefore a = Zsin 3 6 = h tan 3 Q. = 2 tan 2 0; which formulas 

give the relations between a, Z and 0, and it will be noticed that 
the weight w does not enter into the calculation any more than it 
does into the time of swing of the pendulum. 

As an example let the speed be w = 10 radians per second 
(corresponding to 97 revolutions per minute) and let 6 = 30. 
Then the formulas give a = 0.0618 ft. or 0.74 in., I = 0.495 ft. 
or 5.94 in. and the value of h corresponding to 6 = 30 is 0.322 
ft. With these proportions the value of h when 8 becomes 35 
will be 0.317 ft., a decrease of 1.56 per cent., corresponding to a 
change of speed of about 0.8 per cent. 

With a governor as shown at Fig. 123 and co = 10 as before, a 
change from 30 to 35 produces a change in speed of about 3 per 

It is possible to design a governor of this type which will 
maintain absolutely constant speed for all positions of the balls, 
and the reader may prove that for this it is only necessary to do 
away with the ball arms, and place the balls on a curved track 
of parabolic form, so that they will always remain on the surface 
of a paraboloid of revolution of which the spindle is the axis. 
In such a case, h and therefore u will remain constant. 

A perfectly isochronous governor, however, has the serious 
defect that it is unstable or has no definite position for a given 
speed, and thus the slightest disturbing force will cause the balls 
to move to one end or other of their extreme range and the gov- 
ernor will hunt for a position where it will finally come to rest. 
Such a condition of instability is not admissible in practice and 
designers always must sacrifice isochronism to some extent to 
the very necessary feature of stability, because the hunting of 
the balls in and out for their final position means that the valve 
is being opened and closed too much and hence the engine is 
changing its speed continually, or is racing. In the simple 
governor quoted in Sec. 178 it is evident that while it is not 
isochronous it is stable, for each position of the balls corresponds 
to a different but definite speed belonging to the corresponding 
value of the height h. 

181. Weighted or Porter Governor. In order to obviate 
these difficulties Charles T. Porter conceived the idea of plac- 
ing on the sleeve a heavy central weight, free to move up and 
down on the spindle and having its center of gravity on the 



axis of rotation. This modified governor is shown in Fig. 125, 
with the arms pivoted on the spindle, although sometimes the 
arms are crossed and when not crossed they are frequently sus- 

FIG. 125. Porter governor. 

pended by pins not on the spindle. In Fig. 126 a similar gover- 
nor is shown diagrammatically, with the pivots to one side. 
To study the conditions of equilibrium of such a governor 
find the image Q' of the point Q where the link Z 2 is attached to 



the central weight W. Then by the propositions of Chapter IX 
the half of the weight which acts at Q may be transferred to Q' , 
and let it be assumed that l\ and Z 2 are of equal length; then by 
taking moments of the weights and centrifugal force about 
the equation is 


-77- 21 1 sin 

w w 

-oh sin B -^- 
z zg 


From which it follows that 
h = 

FIG. 126. Porter or weighted governor. 

For example, let li = Z 2 = 9 in. or 0.75 ft., speed 194 revolutions' 
per minute for which w = 20 radians- per second, and let each ball 
weigh 4 lb., i.e., w = 8 lb., 6 = 45 and ai = a 2 = 0. Then 
by measuring from a drawing, or by computation, h is found to 
be 0.53 ft., and 




2W + w 

wh~ = 8 X 0.53 X 


W = 22.4 lb. 

= 52.8 lb. 

182. Advantages of Weighted Governors. The first advantage 
of such a governor is that the height h may be varied within 



wide limits at any given speed by a change in the central weight 
W, and thus the designer is left much freedom in proportioning 
the parts. In the numerical example above quoted the height 
would be 6.6 times as great as for an unweighted governor run- 
ning at the same speed, since - - = 6.6. 

Again the variation in height h corresponding to a given change 
in speed is much increased by the use of the central load, with the 
result that the sleeve will move through a certain height with 
smaller change in speed. Now the travel of the sleeve, or the 
lift as it is often called, is fixed by the valve and its mechanism, 
and the above statement means that for a given lift the variation 
in speed will be decreased, or the governor will become more 
sensitive. By sensitiveness is meant the proportional change 
of speed that occurs while the sleeve goes through its complete 
travel, the governor being most sensitive which has the least 

To prove this property let h', h, co' and co represent the heights 
and speeds corresponding to the highest and lowest positions 
of the sleeve. 

__ and 

W r W co" 



k' /C0\2 CO 

T~ = (~7) or ~~/ = 

But since h and In' are much greater in the weighted than in 
the unweighted governor, therefore , is more nearly unity in the 

former case. 

' 2 7 / 2 2 i,/ ft 

CO Al CO CO /i fl 

CO 2 = ~ h' CO 2 " h' 


co' co co' + co h' h 

co co h r 

Now usually co' and co do not differ very much, so that co' + o> = 
2co nearly, and therefore, 


- co h'- h 

The relation is evidently the sensitiveness of the governor 1 

and the smaller the ratio the more sensitive is the governor. 
For an isochronous governor 6co = 0. 

To compare the weighted and unweighted governors in regard 
to sensitiveness take the angular velocity co = 10 radians per 
second and let W = 60 Ib. and fjfa = 8 Ib. Let the change in 
height necessary to move the sleeve through its entire lif^Tbe 

(a) Unweighted Governor. For the data given h = 3.86 in., 
and, therefore, 2 

dh 0.5 

h ' 3.86 

= 0.129. 

Hence, 2 = 0.129 or = 0.064 or 6.4 per cent., so that the 


variation in speed will be 6.4 per cent. 

(6) Weighted Governor. For this governor 


- A o.ou 








h ~ 




2 = 0.008 giving = 0.004 or 0.4 per cent. 


the variation in speed being only 0.4 per cent. Such a gover- 
nor would therefore be very nearly isochronous. 

A third property of this weighted governor is that it is power - 

1 This may be simply shown by the calculus thus : 

, _ 2W + w g_ 

W ' 

and differentiating, 

2W -f w 8h n 8 a) 

8h = - 2W. So,. . . -r = 2 

w h w 


5co dh 

2 ~^ = " h 

where 5o> and 8h represent the small changes taking place in to and h. 

2 The negative sign appearing before dh on the preceding formula merely 
means that an increase in speed corresponds to a decrease in height. 


ful, that is to say, that as the central weight is very heavy the 
equilibrium of the device is very little affected by any slight dis- 
turbing force, such as that required to operate the valve gear or 
to overcome friction. Powerfulness is a very desirable feature, 
for it is well known in practice that the force required to operate 
the valve gear is not constant and therefore produces a variable 
effect on the governor mechanism, which, unless the governor is 
powerful, is sufficient to move the weights, causing hunting. 

The Porter governor thus enables the designer to make a very 
sensitive governor, of practical proportions and one which may 
be made as powerful as desired, so that it will not easily be 
disturbed by outside forces. 


183. A number of the results and properties of governors 
may be graphically represented by means of characteristic curves, 
and it will be convenient at this stage to explain these curves in 
connection with the Porter governor. Let Fig. 127 (a) represent 
the right-hand part of a Porter governor, the letters having the 
same significance as before. Choose a pair of axes, OC in the 
direction of the spindle and OA at right angles to the spindle, 
and let the centrifugal force on the ball be plotted vertically 
along OC, as against radii of rotation of the balls, which are 
plotted along OA, r and r% representing respectively the inner 
and outer limiting radii, the resulting figure will usually be a 
curved line somewhat similar to CiCCz in Fig. 127 (a). 

Let the angular velocities corresponding to the radii r\ and r 2 
be coi and co 2 radians per second respectively, and let co = 
J (^i + W2) represent the mean angular velocity to which the 
corresponding radius of rotation is r ft. 


W 2 

- TZ C02 



where the forces C, C\ and Cz are the total centrifugal forces 
acting on the two balls. The properties of this curve, which 
may be briefly called the C curve, may now be discussed. 



1. Condition for Isochronism. If the governor is to be iso- 
chronous then the angular velocity for all positions of the balls 
must be the same, that is co = coi = 402 and hence the centrifugal 
force depends only on the radius of rotation (see formulas above) 

Ci Cz C 

= = = a constant, 

7i 7 2 * 

a condition which is fulfilled by a C curve forming part of a 
straight line passing through 0. Thus any part of OC would 
satisfy this condition and the part ED corresponds to the radii 
TI and r 2 in the governor selected. 


FIG. 127. Characteristic curve. 

2. Condition for Stability. Although the curve ED will give 
an isochronous governor, it produces instability. The curve 
CiCCz indicates that the speeds are not the same for the various 
positions of the balls, and a little consideration will show that 
Ci corresponds to a lower speed and Cz to a higher speed than C. 
This is evident on examining the conditions at radius n, for the 
point E corresponds to the same speed as C, but since E and C\ 
are both taken at the same radius, and since the centrifugal force 
FE is greater than FC\ it is evident that the angular velocity cor 
corresponding to C\ is less than the angular velocity co correspond- 


ing to E. Thus a curve such as dCC 2 , which is steeper than the 
isochronous curve where they cross, indicates that the speed 
of the governor will increase when the balls move out, and it may 
similarly be shown that such a curve as Ci'CCz', which is flatter 
than the isochronous curve, shows that the speed of the governor 
decreases as the balls move out. 

Now an examination of these curves shows that the one CiCC 2 
belongs to a governor that is stable, for the reason that when the 
ball is at radius ri it has a definite speed and in order to make it 
move further out the centrifugal force must increase. But 
on account of the nature of the curve the centrifugal force must 
increase faster than the radius or the speed must increase as the 
ball moves out, and thus to each radius there is a corresponding 
speed. On the other hand, the curve Ci'CCz shows an entirely 
different state of affairs, for at the radius r\ the centrifugal force 
is greater than FE or the ball has a higher speed than co and thus 
as the ball moves out the speed will decrease. Any force that 
would disturb the governor would cause the ball to fly outward 
under the action of a resultant force Ci'E, and if it were at radius 
r 2 any disturbance would cause the ball to move inward. 

Another way of treating this is that for the curve CiCCz the 
energy of the ball due to the centrifugal force is increasing due 
both to the increase in r and in the speed, and as the weights 
W and w are being lifted, the forces balance one another and there 
is equilibrium; whereas with the curve Ci'CCz there is a decrease 
in speed and also in the energy of the balls while the weights are 
being lifted and the forces are therefore unbalanced and the 
governor is unstable. 

Thus, for stability the C curve must be steeper than the line 
joining any point on it to the origin 0. Sometimes governors 
have curves such as those shown at Fig. 127(6) and curve CiCCa 
indicates a stable governor, d'CC 2 ' an unstable governor, 
CiCCz partly stable and partly unstable and finally CYCCj 
partly unstable and partly stable. 

3. Sensitiveness. The shape of the curve is a measure of 
the sensitiveness of the governor. If S indicates the sensitive- 

Ci?2 Wl 

ness, then by definition S = - 

2 ~ 2 

a>2 coi _ yo?2 C01JICQ2 ~t~ ^L) _ 002 
co co(co2 ~T" ^i) 2co 



since co 2 + coi = 2co nearly. 

S = 

1 O>2 2 




2 == 

d = - 

= ror. 

Hence, by substituting in the formula for S, the result is 

w w 

rc 2 di 

. A* . rs* 

g ' g 


r 2 7*1 


~ 2 




Q _ " y v 

O c\ /^i ~ c\ S~i 

Referring now to Fig. 127 it is seen that 


- - " 


7 == tan ^ = OA 






fi _ 
J ~ 

- tan 0fi _ 1 rC 2 A - JgAi _ 1 (V? 
~ ^ ~ 2 DA ' 

2 DA 

Thus the C curve is also valuable in showing the sensitiveness 
of the governor. For an isochronous governor C 2 , B and D 
coincide and S = 0. Evidently the more stable the governor is 
the less sensitive it is, and in a general way an unstable governor 
is more sensitive than a stable one. At d, Fig. 127 (a), the stable 
governor is most nearly isochronous, arid evidently a fair degree 
of stability and sensitiveness could both be obtained in a governor 
having a reverse curve with point of inflexion near C, the part 
CC Z being concave to OA, the part CiC convex to OA. 

4. Powerfulness. The C curve also shows the powerfulness 
of the governor, since in this curve vertical distances represent 



the centrifugal forces acting on the balls, while horizontal distances 
represent the number of feet the balls move horizontally in the 
direction of the forces. Thus, an elementary area represents the 
product C.dr ft.-pds. and the whole area between the C curve and 
the axis OA gives the work done by the balls in moving over 
their entire range, and is therefore the work available to move 
the valve gear and raise the weights. The higher the curve is 
above OA the greater is the available work, and this clearly cor- 
responds to increased speed in a given governor. 

5. Friction. The effect of friction has been discussed in the 
previous chapter and need not be considered here. Some writers 
treat friction as the equivalent of an alteration to the central 
weight, and if this is done the effect is very well shown in Fig. 128 
where the C curve for the frictionless governor is shown at 

CiCC 2 . As the weight W is lifted 
4 the effect of friction when treated 
c 2 in this way is to increase W by 
e the friction/ with the result that 
the C curve is raised to 3-4, 
whereas when the weight W is 
falling the friction has the effect 
of decreasing the weight W and 
JS*" to lower the C curve to 5-6. The 
FIG. 128. effects of these changes are 

evident without discussion. 

184. Relative Effects of the Weights of the Balls and the 
Central Weight. For the purpose of further understanding the 
governor and also for the purpose of design, it is necessary to 
analyze the effects of the weights separately. Referring to Fig. 
129 and finding the phorograph by the principles of Chapter IV 
the image of D is at D' and taking moments about A, remember- 


ing that - may be transferred from D to its image D', 

MWb + M -MCh = 0. 

(Sec. 151, Chapter IX). Now let C w be the part of C neces- 
sary to support W and C w the corresponding quantity for w, so 

C w + C w = C where C = - 




That is 

b e 

W T and C w = w-r- 

n h 

The graphical construction is shown in Fig. 129. Draw JH 
and LG horizontally at distances below A to represent w and W 
respectively, then join AE, the line D'E being a vertical through 
D r . Then it may be easily shown that 

C w = AF and C w = AK. 


FIG. 129. Governor analysis. 

Making this construction for various positions and plotting 
for the complete travel of the balls the two curves are as drawn 
in Fig. 129. 

185. Example. The following dimensions are taken from an 
actual governor and refer to Fig. 129. ai = 0, a z = 1J^ in., AB = 
12K in., AM = 16 in. and BD = lOJ^ in -> while the travel 
of the sleeve is 2J^ in. and the point D is 15% in. below A 
when the sleeve is at the top of its travel. Each ball weighs 
15 Ib. so that w = 30 lb., also W = 124 Ib. 

Then drawing the governor mechanism in the upper, the mean 



and the lowest positions of the sleeve, the following table of 
results is obtained, since for the ball - = 00 ^, ~ = 0.933. 





r = e 


wl = cw 



w\ = C w 

Cw + C w = C 

W2 = -^ 


rev. per 

1. Upper. . 









2. Mean... 









3. Lower... 









The corresponding C curve and the two components C w and 
C w are plotted at Fig. 130 from which it is clear that the governor 

FIG. 130. 

is stable. From this curve it appears that the sensitiveness 

. ,/ 37 

is % X 222 = 0-^856 or 8.56 per cent., which checks very 

well with the speeds as shown in the last column, and which indi- 
cates a sensitiveness of 8.65 per cent. 

If it is desired to find the position of the balls for a speed of 
150 revolutions per minute, then o> = 15.7 radians per second 


and the force C = rco 2 = 0.933 X r X 246.5 = 230 r. Then 

draw the line for which the tangent is = 230 and where it cuts 

the C curve is the radius of the balls corresponding to this 

Assuming the mean height of the C curve to be 207 pds. the 
work done in the entire travel of the balls is 207 (0.98 0.83) = 
31 ft.-pds. 



2 = 10 in., and BM = 3 


186. Design of a Porter Governor. These curves may be con- 
veniently used in the design of a Porter governor to satisfy given 
conditions. Let it be required to design a governor of this type 
to run at a mean speed of 200 revolutions per minute with a 
possible variation of less than 5 per cent, either way for the ex- 
treme range. The sleeve is to have a travel of 2 in. and the 
governor is to have a powerfulness represented by 20 ft.-pds. 

From general experience select the dimensions a\, 2 , li, Z 2 
and BM in Fig. 129. Thus take 
in. ; also make i = a 2 = 1 in. 
Draw the governor in the central 
position of the sleeve with the 
arms at 90, as this angle gives 
greater uniformity than other 
angles, and measure the extreme 
radii and also that for the central origin 
position of the sleeve. The C 
curve may now be constructed 
and at Fig. 131 the three radii 
are marked, which are r\ = 9.5 
in., r = 10.22 in. and r 2 = 10.82 
in. Now the power of the 
governor is 20 ft.-pds., and divid- 
ingthisby r 2 7*1 = O.llft. gives 
the mean height of the C curve 
as 182 pds. Plot this at radius r 
making HG = 182 pds. and join 
to 0; it cuts r 2 at D. 

Now the sensitiveness is to be 
5 per cent., so that T and U are 
found such that DT = DU = 
0.05 X AD = 9.65 pds. Join 
T and U to 0, thus locating V and the resulting C curve will 
be VGT shown dotted. 

Next, since the centrifugal force GH = 182 pds. corresponds to 
a radius r = 10.22 in. and a speed of 200 revolutions, the weight 

w may be found from the formula C = rco 2 and gives w = 


15.75 Ib. By the use of such a diagram as Fig. 129 the three 
values of C w are measured for the three radii and the C w curve is 
drawn in Fig. 131, and then the values of C w are found. Thus, 







- 40 

- 20 

rj-9.5" r = 10.22" r 2 = 10.82" 

FIG. 131. Governor design. 



RV scales off as 153 pds. and hence C Wl = 153 - 13.6= 139.4 pds., 
and similarly the other values of CV are found, and from 
them W i = 104.6 pds., W = 107.6 pds. and W 2 = 110.6 pds. are 
obtained, as shown at Fig. 129. 

As a trial assume the mean of these values W 107.6 pds., as 
the value of the central weight and proceeding as in Sec. 185 find 
the three new values of C w and also of C = C w + C w and lay 
these off at the various radii giving the plain curve in Fig. 131. 
This will be found to correspond to a range in speeds from 192 

FIG. 132. Proell governor. 

to 207.5 revolutions per minute, and as this gives less than a 
5 per cent, variation either way from the mean speed of 200 
revolutions it would usually be satisfactory. If it is desired to 
have the exact value of 5 per cent., then it will be better to start 
with a little larger variation of say 6 per cent, and proceed as 

187. Proell Governor. The method already described may 
be applied to more complicated forms of governor with the same 
ease as is used in the Porter governor, the phorograph making 


these cases quite simple. As an illustration, the Proell governor 
is shown in Fig. 132 and is similarly lettered to Fig. 129, the 
difference between these governors being that in the Proell the 
ball is fastened to an extension of the lower arm DB instead of 
the upper arm AB as in the Porter governor. 

As before, AB is chosen as the link of reference and the* images 
found on it of the points D and M by the phorograph, Chapter IV. 
The force }^W is then transferred to D f and J^C and %w to M f 
from Chapter IX, but in computing C the radius is to be measured 
from the spindle to M and not to M f , since the former is the 
radius of rotation of the ball. The meanings of the letters will 
appear from the figure and by taking moments about A the same 
relation is found as in Sec. 184. The results for the complete 
travel of the balls is shown on the lower part of Fig. 132. 


188. Spring governors have been made in order to eliminate 
the central weight and to make possible the use of a nearly 
isochronous and yet sensitive and powerful governor. These 
governors always run at high speed and are sometimes mounted 
on the main engine shaft, but more frequently on a separate 

189. Analysis of Hartnell Governor. One form of this gover- 
nor, frequently ascribed to Hartnell of England, is shown in Fig. 
133 and the action of the governor may now be analyzed. Let 
the total weight of the two balls be w lb., as before, and let W 
denote the force on the ball arms at BB, due to the weight of the 
central spring and any additional weight of valve gear, etc. In 
this case W will remain constant as in the loaded governor. 
Now let F be the pressure produced at the points B by the 
spring, F clearly increasing as the spring is compressed due to 
the outward motion of the balls. 

In dealing with governors of this class it is best to use the mo- 
ments of the forces about the pin A in preference to the forces 
themselves, and hence in place of a C curve for this governor a 
moment or M curve will be plotted in its place, the radius of 
rotation of the balls being used as the horizontal axis. The 
symbols M , MF, M w and M w indicate, respectively, the mo- 
ments about A of the centrifugal force C, the spring force F, 
the weight of the balls w and the dead weight W along the spindle. 
Then M = M F + M w + M w 



or Ca cos 6 = Fb cos 6 + Wb cos 6 wa sin 6. 

The moment curves may be drawn and take the general shapes 
shown in Fig. 133 and similar statements may be made about 
these curves as about those for the Porter governor. If it is 
desired, the corresponding C curves may readily be drawn 
from the formula 


Ca cos 6 = Fb cos 6 + Wb cos wa sin 


C F + Cw 

p - _]_ iff w tan 

a a 

FIG. 133. Hartnell governor. 

and a graphical method for finding these values is easily devised. 
The curve for W is evidently a horizontal line since W, b and a 
are all constant, while that for w is a sloping line cutting the 
axis of r under the pin A and the C F curve may be found by 

190. Design of Spring. The data for the design of the spring 
may be worked out from the CF curve found as above. Evidently 

Cp = F~ or F = C F X r- = C F X a constant, and thus from 
the curve for C F it is possible to read forces F to a suitable scale. 



These forces F may now be plotted as at Fig. 134 which gives 
the values of F for the different radii of rotation. As the line 
EGL thus found is slightly curved, no spring could exactly fulfil 
the requirements, but by joining E and L and producing to H, a 
solution may be found which will fit two points, E and L, and will 
nearly satisfy other points. Draw EK horizontally; then LK 
represents the increase in pressure due to the spring while the 
balls move out EK in., and hence the spring must be such that 

r pds. will compress it 1 in., and further the force produced by 


the spring when the balls are in is EJ pds., that is the spring 
must be compressed through HJ in., for the inner position of 
the balls. 

In ordinary problems it is safe to assume for preliminary cal- 
culations that the effect of the weights W and w can be neglected 
and the spring may be designed to balance the centrifugal force 
alone. In completing the final computations the results may be 
modified to allow for these. In 
the diagrams here shown their 
effects have been very much ex- 
aggerated for clearness in the 

191. Governors with Hori- 
zontal Spindle. Spring gover- 
nors are powerful, as the complete 
computations in the next case 
will show, and are therefore well 
adapted to cases where the move- 

ment of the valve gear is difficult and unsteady. 

When such a governor is placed with horizontal spindle such 
as Fig. 135 the effects of the weights are balanced and the spring 
alone balances the centrifugal force. 

192. Belliss and Morcom Governor. One other governor of 
this general type may be discussed in concluding this section. 
It is a form of governor now much in use and the one shown in 
the illustration, Figs. 136 and 137, is used by Belliss and Morcom 
of Birmingham, England, in connection with their high-speed 
engine. The governor is attached to the crankshaft, and therefore 
the weights revolve in a vertical plane, so that their gravity effect 
is zero. There are two revolving weights W with their centers 
of gravity at G and these are pivoted to the spindle by pins A. 



FIG. 135. Governor with gravity effect neutralized. 

FIG. 136. Belliss and Morcom governor. 




Between the weights there are two springs S fastened to the 
former by means of pins at B. The balls operate the collar C, 
which slides along the spindle, thus operating the bell-crank lever 
DFV, which is pivoted to the engine frame at F and connected 
at V, by means of a vertical rod, to the throttle valve of the 
engine. There is an additional compensating spring S c with its 
right-hand end attached to the frame and its left-hand end con- 
nected to the bell-crank lever DFV at H t there being a hand 
wheel at this connection so that the tension in the spring may be 
changed within certain limits and thus the engine speed may be 
varied to some extent. This spring will easily allow the operator 
to run the engine at nearly 
5 per cent, above or below 

The diagrammatic sketch of 
the governor, shown in Fig. 
137, enables the different parts 
to be distinctly seen as well as 
the eolations of the various 
points. It will be noticed that 
this governor differs from all 
the others already described 
in that part of the centrifugal 
force is directly taken up by 
the springs S, while the forces 
acting on the sleeve are due to 
the dead weight of the valve 
V and its rod, and the slightly 
unbalanced steam pressure 
(for the valve is nearly 
balanced against steam pres- 
sure) on the valve, and in addition to these forces there is the 
pressure due to the spring S e . The governor is very efficiently 
oiled and it is found by actual experiment that the frictional effect 
may be practically neglected. 

In this case it will be advisable to draw the moment curve 
for the governor as well as the C curve and from the latter the 
usual information may be obtained. As this moment curve 
presents no difficulties it seems unnecessary to put the investi- 
gation in a mathematical form as the formulas become lengthy 
on account of the disposition of the parts. An actual case has 



FIG. 137. Belliss and Morcom 



been worked out and the results are given herewith and show the 
effects of the various parts of the governor. 

193. Numerical Example. The governor here selected, is 
attached to the crankshaft of an engine which has a normal mean 
speed of 525 revolutions per minute although the actual speed 
depends upon the load and the adjustment of the spring S c . 
The governor spindle also rotates at the same speed as the engine. 
The two springs S together require a total force of 112 pds. for 
each inch of extension, while the spring S c requires 220 pds. per 
inch of extension, the springs having been found on calibration 
to be extremely uniform. Each of the revolving masses has an 
effective weight of 10.516 Ib. and the radius of rotation of the 
center of gravity varies from 4.20 in. to 4.83 in. The other 
dimensions are: e = 4 in. radius, b = 3.2 in., c = 3.5 in., a = 
3.56 in., d = 10.31 in.,/ = 4.67 in., g = 5.31 in. 

The weight of the valve spindle, valve and parts together with 
the unbalanced steam pressure under full-load normal conditions 
is 20 Ib. 

The following table gives the results for the governor for four 
different radii of the center of gravity G, all the moments being 
expressed in inch-pound units, when reduced to the equivalent 
moment about the pivots A of the balls. 



of rota- 
tion of G, 

gal force, 

about A, 


of main 
about A, 

of spring 
S e about 
A, inch- 

due to 
about A, 

Sum of 
(D, (2) 
and (3), 

































In examining the table it will be observed that the sum of 
columns (1), (2) and (3) is always a little less than the moment 
due to the centrifugal force. As the results are all computed 
from measurements made on the engine during operation, there 
is possibly a slight error in the dimensions, and further the effect 
of centrifugal force on the springs S will make some difference. 



The results agree very well, however, and show that the calcula- 
tions agree with actual conditions. 

The results are plotted in Fig. 138, the left-hand part of the 
curves being dotted. The reason of this is that the observations 
below 508 revolutions per minute were taken when the 
engine was being controlled partly by the throttle valve, and 
do not therefore show the action of the governor fairly; the points 
are, however, useful in showing the tendency of the curves and 
represent actual positions of equilibrium of the governor. 

The effect of the weight of the valve and unbalanced steam 
pressure are almost negligible, so that the power of the governor 

sating 8pri2i_H- 

Valve Weight 

Radius r 
3.70 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 Inches 

FIG. 138. 

does not need to be large, but the spring S c produces an appre- 
ciable effect amounting to about 11 per cent, of the total at the 
highest speed. If the compensating spring S c were removed, 
the governor would run at a lower speed. 

Joining any point on the moment curve to the origin 0, as 
has been done on the figure, shows that the governor is stable. 

The sensitiveness and powerfulness may be found from the 
C curve shown at Fig. 139. At the radius 4.47 in. the centri- 
fugal force is 345.5 pds., and if the origin be joined to this point 
and the line produced it will cut the radius 4.83 in. at 373.5 pds., 
whereas the actual C is 410 pds. The sensitiveness then is 



o^o'rx = 0.0465 or 4.65 per cent. From the speeds 

^(^1(J -J- O/O.OJ 

/coo 508") 

the corresponding result would be 1x^500 _u 508") = 0.0442 or 4.42 

per cent., which agrees quite closely with the former value. 
The moment curves cannot be used directly for the determina- 
tion of the power of the governor because areas on the diagram 
do not represent work done. If the power is required, then the 
base must be altered either so as to represent equal angles passed 
through by the ball arm, or more simply by use of the C curve 
plotted on Fig. 139. It will be seen that the C curve differs 

^ Pounds 





3.7 3.8 3.9 4.0 4.1 

4.2 4.3 4.4 4.5 

FIG. 139. 

4.6 4.7 4.8 Inches 

very little in character from the moment curve. The power 
of the governor is only about 11.6 ft.-pds. 

The computations on this one governor will give a good general 
idea of the relative effects of the different parts in this style of 
governor, and also show that spring governors of this class 
possess some advantages. 



194. Reasons for Using this Type. The shaft governor was 
probably originally so named because it is usually secured to the 
crankshaft of an engine and runs therefore at the engine speed. 
In recent practice, however, certain spring governors, such as the 
Belliss and Morcom governor, are attached to the crankshaft 


and yet these scarcely come under the name of shaft governors. 
The term is more usually restricted to a governor in which the 
controlling forces differ to some extent from those already dis- 
cussed. This type of governor is not nearly so old as the others 
and was introduced into America mainly as an adjunct to the 
high-speed engine. 

On this continent builders of high (rotative) -speed engines 
have almost entirely governed them by the method first men- 
tioned at the beginning of this chapter, that is by varying the 
point of cut-off of the steam, and in order to do this they have 
usually changed the angle of advance and also the throw of 
the eccentric by means of a governor which caused the center 
of the eccentric to vary in position relative to the crank according 
to the load, the result will be a change in all the events of the 
stroke. The eccentric's position is usually directly controlled by 
the governor, and hence it is necessary to have a powerful gover- 
nor or else the force required to move the valve may cause very 
serious disturbances of the governor and render it useless. Again 
as the governor works directly on the eccentric, it is convenient 
to have it on the crankshaft. 

Governors of this class also possess another peculiarity In- 
those already described the pins about which the balls swung 
were in all cases perpendicular to the axis of rotation, so that the- 
balls moved out and in a plane passing through this axis. In 
the shaft governor, on the other hand, the axis of the pins is 
parallel with the axis of rotation and the weights move out and 
in in the plane in which they rotate. While this may at first 
appear to be a small matter, it is really the point which makes this 
class of governor distinct from the others and which brings into 
play inertia forces during adjustment that are absent in the 
other types. Such governors may be made to adjust themselves 
to their new positions very rapidly and are thus very valuable 
on machinery subjected to sudden and frequent changes of load. 

195. Description. One make of shaft governor is shown at 
Fig. 140, being made by the Robb Engineering Co., Amherst, 
Nova Scotia, and is similar to the Sweet governor. In this make 
there is only one rotating weight W, the centrifugal effect of 
which is partly counteracted by the flat leaf spring S, to 
which the ball is directly attached. The eccentric E is pivoted 
by the pin P to the flywheel, and an extension of the eccen- 
tric is attached by the link b to the ball W. The wheel 



rotating in the sense shown, causes the ball to try to 
move out on a radial line, which movement is resisted by the 
spring S. As the ball moves out, due to increased speed, the 
eccentric sheave swings about P, and thus the center of the eccen- 
tric will take up a position depending upon the speed. Two stops 
are provided to limit the extreme movement of the eccentric 
and ball. 

Other forms of governor are shown later at Fig. 143 and at 
Fig. 147, these having somewhat different dispositions of the 

FIG. 140. Robb inertia governor. 

Powerfulness in such governors is obtained by the use of 
heavy weights moving at high speed, for example in one governor 
the revolving weight is 80 Ib. and it revolves in a circle of over 
29 in. radius at 200 revolutions per minute, dimensions which 
should be compared with those in the governors already discussed. 

196. Conditions to be Fulfilled. The conditions to be ful- 
filled are quite similar to those in other spring governors so that 
only a brief discussion will be necessary, which may be illustrated 
in the following example. 

Let A, Fig. 141, represent a disc rotating about a center 
at n revolutions per minute, and let this disc have a weight w 



mounted on it so that it may move in and out along a radial line 
as indicated, and further let the motion radially be resisted by a 
spring S which is pivoted to the disc at E. Let the spring pull 
per foot of extension be $ pds., and let the weight be in equilibrium 
at distance r ft. from 0, the extension of the spring at this in- 


stant being a ft. The centrifugal force on the ball is C = rco 2 


pds. where co is the angular velocity of the disc in radians per 


second, and since co 

-AQ-, therefore 


= 0.000341 

where r is in feet. For the same position the spring pull will 
be Sa pds., so that for equilib- 
rium Sa = C or 

0.000341 wrn*, 

that is, 

S = 0.000341 w-n 2 . 

To make the meaning of 
this clear it will be well to 
take a numerical example, 
and let it be assumed that the 
weight w = 25 Ib. and the 
speed is 200 revolutions per 
minute. Three cases may be 

considered, according to whether r is equal to, greater than or 
less than a, and these will result as follows: 

1. r = a = 1ft., S = 0.000341 X 25 X j X 200 2 = 341 pds. 

2. r = lft., a = 0.57 ft., S = 0.000341 X25X^X 200 2 = 600 pds. 

3. r = 1 ft., a = 1.19 ft., S = 0.000341 X 25 X j^X 200 2 = 288 


So that, as the formula shows, the spring strength depends 
upon the relation of r to a. 

The resulting conditions when the ball is 10 in., 12 in. and 14 
in. respectively from the center of rotation with the three springs, 
are set down in the following table and in Fig. 142. 



Radius r, inches 

force at 200 

Spring pull 

S = 288 

S = 341 

S = 600 
















For the spring S = 288 pds. per foot of extension it is seen that 
at the smaller radius the spring pull is higher than the centrifugal 
force or the disc must run at a higher speed than 200 revolutions 
for equilibrium, while at the outer radius the spring pull is too lo\v 
and the speed must be below 200 revolutions for equilibrium, 




FIG. 142. 

that is the speed should decrease as the ball moves out. With 
spring S = 600 exactly the reverse is true, or the speed must 
increase as the ball moves out, while for spring S = 341 the speed 
will be constant for all positions of the ball. 

The spring S = 341 is properly designed and set for isochronism, 
but evidently there is no force holding the ball anywhere and the 
slightest push would send it oscillating along the scale, that is, 
it lacks stability. The spring S = 288 also gives an unstable 
arrangement for the reason that the centrifugal force increases 
and decreases faster than the spring pull, and thus if the ball 
happened to be 12 in. from the center and was disturbed it 
would instantly fly to the inner or outer extreme stop. However, 



spring S = 600 gives a stable arrangement, because whenever 
the ball is at say 12 in. from the center and any force pushes it 
away it immediately tries to return to this position, and will do 
so on account of the preponderating effect of the spring force 
acting upon it, unless there should be a change of speed forcing 
it to the new position, but to each speed there is a definitely fixed 
position of the ball. It is to be noticed that the curve for S = 
600, is always steeper than any line from it to the origin 0, which 
has been already given as a condition of stability (Sec. 183(2)). 

The gain in stability is, however, made at a sacrifice in 
sensitiveness. For the spring S = 600 the speed changes from 
184 to 211 revolutions per minute or the sensitiveness is 

1/^91 4- = 0-136 or 1^.6 per cent., while with spring 

FIG. 143. Buckeye and McEwen governors. 

S = 288 the range of speeds is from 198 to 203 revolutions per 
minute or the sensitiveness is 2.4 per cent. 

197. Analysis of the Governor. Having now discussed the 
conditions of stability and isochronism and the effect the design 
of the spring has on them, a complete analysis of the governor 
may be made. 

Two forms of governor are shown in Fig. 143 and these show a 
somewhat different disposition of the revolving weights. The 
one on the left is used by the Buckeye Engine Co. and has two 
revolving weights W connected by arms b to the pivots P. The 
centrifugal force is resisted by springs S attached to b and to the 
flywheel rim at K. The ends of the links b are connected at H 
to links attached to the eccentric E at C and the operation of 



the weights revolves E and changes the steam distribution. 
Auxiliary springs D oppose springs S at inner positions of weights 

The right-hand figure shows the McEwen governor having 
two unequal weights W\ and Wz cast on a single bar, the com- 
bined center of gravity being at G, and the pivot connection to the 
wheel is P. There is a single spring S attached to the weights at 
H and to the wheel at K. There is a dashpot at D attached to 
the wheel and to the weight at J5; this consists of a cylinder and 
piston, the latter being prevented from moving rapidly in the 
cylinder. The purpose of the dashpot is to prevent oscillations 

of the weight during adjustment and to keep it steady, but after 
adjustment has been made D has no effect on the conditions of 
equilibrium. In this governor a frictionless pin is provided at 
P by the use of a roller bearing. The valve rod is at E. 

A diagrammatic drawing of these two governors, which may 
be looked upon as fairly representative of this class, is given at 
Fig. 144, similar letters being used in both cases. 

Let the wheel revolve about A, Fig. 143, with angular velocity 
w radians per second and let F denote the spring pull when the 
center of gravity G is at radius r from A; further, let di and d z 
in. represent the shortest distances from the weight pivot P 
to the directions of r and S respectively. Then for equilibrium 


the moments about P due to the centrifugal force C and to the 
spring pull S must be equal if, for the present, the effect of 
gravity and of the forces required to move the valve are neglected. 
That is: 

Cdi = Sdz in.-pds. 

w . 

or rco 2 di = Sdz m.-pds. 

In such an arrangement as shown the effect of the forces re- 
quired to move the valves is frequently quite appreciable and 
is generally also variable, as is also the effect of friction and 
gravity, although usually gravity is relatively so small that it 
may be neglected. If it is desired to take these into account 

i = Sd 2 + moment due friction, valve motion and gravity. 

Denote the distance AP by a and the shortest distance from 
G to AP by x; thus a is constant but x depends on the position 
of the balls. From similar triangles it is evident that rd\ = ax 
and therefore 

w w . 

Thus the moment due to the centrifugal force is, for a given 
speed, variable only with x and hence the characteristics of the 
governor are very well shown on a curve 1 in which the base repre- 
sents values of x and vertical distances the centrifugal moments 

u z ax. 

Such a curve is shown below, Fig. 144, and the shape of the 
curve here represents a stable governor since it is steeper at all 
points than the line joining it to the origin 0. From this curve 
information may be had as to stability and sensitiveness, but the 
power of the governor cannot be determined without either 
placing the curve on a base which represents the angular swing 
of the balls about P or else by obtaining a C curve on an r base 
as in former cases. 

If co is constant, or the governor is isochronous, M varies 
directly with x or the moment curve is a straight line passing 
through the origin 0. 

Having obtained the M curve in this way the moment curves 

1 For more complete discussion of this method see TOLLE, " Die Regelung 
der Kraftmaschinen." 



about P corresponding to gravity, friction of the valves and parts 
and also those necessary to operate the valves are next found, 
these three curves also being plotted on the x base, and the 
difference between the sum of these three moments and the total 
centrifugal moment will give the moment which must be pro- 
vided by the spring which is Sd 2 in.-pds. From the curve giving 
Sd 2 the force S may be computed by dividing by <i 2 and these 
values of S are most conveniently plotted on a base of spring 
lengths, from which all information for the design of the spring 
may readily be obtained (see Sec. 190). 

In order that the relative values of the different quantities 
may be understood, Fig. 145 shows these curves for a Buckeye 

Eccentric Friction 

FIG. 145. 

Valve Friction 

governor, in which the gravity effect is balanced by using two 
revolving weights symmetrically located. Friction of the valve 
and eccentric and the moment required to move the valve are 
all shown and the curves show how closely the spring-moment 
and centrifugal-moment curves lie together. The curves are 
drawn from the table given by Trinks and Housom, in whose 1 
treatise all the details of computing the results is shown so as to 
be clearly understood. The governor has a powerf ulness of nearly 
600 ft.-pds. 


198. The inertia or shaft governor is particularly well adapted 
to rapid adjustment to new conditions and it is often made so 
1 THINKS and HOUSOM, "Shaft Governors." 



that it will move through its entire range in one revolution, which 
often means only a small fraction of a second. The rapidity of 
this adjustment depends almost entirely upon the distribution of 
the revolving weight and not nearly so much upon its magnitude. 
For a given position of the parts the only force acting is centrif- 
ugal force already discussed but during change of position the 
parts are being accelerated and forces due to this also come into 
play. Fig. 146 represents seven different arrangements of the 
weights; in five of these the weight is concentrated into a ball 
with center of gravity at G and hence with very small moment 
of inertia about G, so that the torque required to revolve such 
a weight at any moderate acceleration will not be great; the 
opposite is true of the two remaining cases, however, the weight 


being much elongated and having a large moment of inertia 
about G. 

Assuming a sudden increase of speed in all cases, then at (a) 
this only increases the pressure on the pin B because BG is 
normal to the radius AG, at (6) an increase in speed will produce 
a relatively large turning moment about the pin which is shown 
at A. Comparing (c) and (d) with (a) and (b) it is seen that the 
torque in the former cases is increased at (c) because in addition 
to the acceleration of G there is also an angular acceleration about 
G, whereas at (d) G is stationary and yet there is a decided torque 
due to its angular acceleration. At (e), (/) and (</) the sense of 
rotation is important and if an increase in speed occur in the 
first and last cases the accelerating forces assist in moving the 


weights out rapidly to their new positions, whereas at (/) the 
accelerating forces oppose the movement. 

Space prevents further discussion of this matter here, but it 
will appear that the accelerating forces may be adjusted in any 
desired way to produce rapid changes of position, the weights 
being first determined from principles already stated and the 
distribution of these depending on the inertia effects desired. 
Chapter XV will assist the reader in understanding these forces 
more definitely. 

FIG. 147. Rites governor. 

A form of governor made by Rites, in which the inertia forces 
play a prominent part during adjustment is shown at Fig. 147. 
The revolving weights are heavy and are set far apart, but their 
center of gravity G is fairly close to A so that the centrifugal 
moment is relatively small. In a governor for a 10 by 10-in. 
engine the weight W was over 120 Ib. and the two weight centers 
were 32 in. apart. 


1. Define a governor. What is the difference between the functions of a 
governor and a flywheel? 


2. What is the height of a simple governor running at 95 revolutions per 
minute ? 

3. What is meant by an isochronous governor? Is such a governor desir- 
able or not? Why? 

4. Explain fully the terms stability and powerfulness. 

5. Prove that in a governor where the balls move in a paraboloid of revo- 
lution, h is constant and the governor is isochronous. 

6. What are the advantages of the Porter governor? 

7. Using the data, n\ = 100, n 2 = 110, prove that -r = 2 

fl CO 

^ 8. Compare the sensitiveness of a simple and a Porter governor at 115 
revolutions per minute and with a sleeve travel of % in., taking W = 120 Ib. 
and w = 15 Ib. 

9. Analyze the following governor for sensitiveness and power (see Fig. 

n = 130, W = 110, w = 12, h = 12K, BM = 3^, h = 1(% i = 0, 
z = 2^, sleeve travel 2^ in. 

10. Design a Porter governor for a speed of 170 revolutions per minute 
with a speed variation of 5 percent, each way, travel 2^ in., power 35 ft.-pds. 

11. In a governor of the type of Fig. 133, a = 2. 1 in.; b = 0.75 in., dis- 
tance between pivots 2% in. inner radius of ball 1.6 in., weight per ball 1% 
Ib., travel Y in. and speed 250. Design the spring for 5 per cent, variation. 

12. What are the advantages of the shaft governor? Show how the dis- 
tribution of the weight affects the rapidity of adjustment. 


199. Nature of the Problem. The preceding chapter deals 
with governors which are used to prevent undue variations in 
speed of various classes of machinery, the governor usually con- 
trolling the supply of energy to the machine in a way to suit the 
work to be done and so as to keep the mean speed of the machine 
constant. The present chapter does not deal with this kind of a 
problem at all, but in the discussion herein, it is assumed that the 
mean speed of the machine is constant and that it is so controlled 
by a governor or other device as to remain so. 

Iri addition to the variations in the mean speed there are 
variations taking place during the cycle of the machine and which 
may cause just as much trouble as the other. Everyone is famil- 
iar with the small direct-acting pump, and knows that although 
such a pump may make 80 strokes per minute, for example, 
and keep this up with considerable regularity, yet the piston 
moves very much faster at certain times than others, and in fact 
this variation is so great that larger pumps are not constructed in 
such a simple way. With the larger pumps, on which a crank 
and flywheel are used, an observer frequently notices that, 
although the mean speed is perfectly constant, yet the flywheel 
speed during the revolution is very variable. Where a steam 
engine drives an air compressor, these variations are usually 
visible, at certain parts of the revolution the crankshaft almost 
coming to rest at times. These illustrations need not be multi- 
plied, but those quoted will suffice. The speed variations which 
occur in this way during the cycle are dealt with in this chapter. 

200. Cause of Speed Fluctuations. The flywheel of an engine 
or punch or other similar machine is used to store energy and to 
restore it to the machine according to the demands. Consider, 
for example, the steam engine ; there the energy supplied by the 
steam at different parts of the stroke is not constant, but varies 
from time to time; at the dead centers the piston is stationary and 
hence no energy is delivered by the working fluid, whereas when 



the piston has covered about one-third of its stroke, energy is being 
delivered by the steam to the piston at about its maximum rate, 
since the piston is moving at nearly its maximum speed and 
the steam pressure is also high, as cutoff has not usually 
taken place. Toward the end of the stroke the rate of delivery 
of the energy by the steam is small because the- steam pressure 
is low on account of expansion and the piston is moving at slow 
speed. During the return stroke the piston must supply energy 
to the steam in order to drive the latter out of the cylinder. 

Now the engine above referred to may be used to drive a pump 
or an air compressor or a generator or any other desired machine, 
but in order to illustrate the present matter it will be assumed to 
be connected to a turbine pump, since, in such a case, the pump 
offers a constant resisting torque on the crankshaft of the engine. 
The rate of delivering energy by the working fluid is variable, as 
has already been explained ; at the beginning of the revolution it 

Jerque Required 
V by Load 

O M f N' ^ ' R' S' ^A; 

FIG. 148. 

is much less than that required to drive the pump, a little further 
on it is much greater than that required, while further on again 
the steam has a deficiency of energy, and so on. 

At this point it will be well to refer again to Fig. 101 which 
has been reproduced in a modified form at Fig. 148 and shows in 
a very direct and clear way these important features. During 
the first part of the outstroke it is evident that the crank effort 
due to the steam pressure is less than that necessary to drive 
the load; this being the case until M is reached, at which point 
the effort due to the steam pressure is just equal to that necessary 
to drive the load; thus during the part OM f of the revolution the 
input to the engine being less than the output the energy of the 
links themselves must be drawn upon and must supply the work 
represented by OML. But the energy which may be obtained 
from the links will depend upon the mass and velocity of them, 
the energy being greater the larger the mass and the greater the 
velocity, the result is that if the energy of the links is decreased 



by drawing from them for any purpose, then since the mass of 
the links is fixed by construction, the only other thing which 
may happen is that the speed of the links must decrease. 

In engines the greater part of the weight in the moving parts is 
in the flywheel and hence, from what has been already said if 
energy is drawn from the links then the velocity of the flywheel 
will decrease and it will continue to decrease so long as energy is 
drawn from it. Thus during OM r the speed of the flywheel will 
fall continually but at a decreasing rate as M ' is approached, and 
at this point the wheel will have reached its minimum speed. 
Having passed M' the energy supplied by the steam is greater 
than that necessary to do the external work, and hence there is a 
balance left for the purpose of adding energy to the parts and 
speeding up the flywheel and other links, the energy available 
for this purpose in any position being that due to the height of 
the torque curve above the load line. In this way the speed of 
the parts will increase between M' and N' reaching a maximum 
for this period at N'. 

From N' to R f the speed will again decrease, first rapidly then 
more slowly, reaching a minimum again at R' and from R to S f , 
there is increasing speed with a maximum at S'. The flywheel 
and other parts will, under these conditions, be continually 
changing their speeds from minimum to maximum and vice versa, 
producing much unsteadiness in the motion during the revolu- 
tion. The magnitude of the unsteadiness will evidently depend 
upon the fluctuation in the crank-effort curve, if the latter curve 
has large variations then the unsteadiness will be increased; it 
will also depend on the weights of the parts. 

In the case of the punch the conditions are the reverse of the 
engine, for the rate of energy supplied by the belt is. nearly con- 
stant but that given out is variable. While the punch runs light, 
no energy is given out (neglecting friction), but when a hole is 
being punched the energy supplied by the belt is not sufficient, 
and the flywheel is drawn upon, with a corresponding decrease 
in its speed, to supply the extra energy, and then after the hole 
is punched, the belt gradually speeds the wheel up to normal 
again, after which another hole may be punched. To store up 
energy for such a purpose the flywheel has a large heavy rim 
running at high speed. 

It will thus be noticed that a flywheel, or other part serving 
the same purpose, is required if the supply of energy to the ma- 


chine, or the delivery of energy by the machine, i.e., the load, is 
variable; thus a flywheel is required on an engine driving a dyna- 
mo or a reciprocating pump, or a compressor, or a turbine pump; 
also a flywheel is necessary on a punch or on a sheet metal press. 
It is not, however, in general necessary to have a flywheel on a 
steam turbo-generator, or on a motor-driven turbine pumping 
set, or on a water turbine-driven generator set working with 
constant load, because in such cases the energy supplied is always 
equal to that given out. 

The present investigation is for the purpose of determining 
the variations or fluctuations in speed that may occur in a given 
machine, when the methods of supplying the energy and also of 
loading are known. Thus, in an engine-driven compressor, hoth 
the steam- and air-indicator diagrams are assumed known, as 
well as the dimensions and weights of the moving parts. 


201. Kinetic Energy of Bodies. In order to determine the 
speed fluctuations in a machine it is necessary, first of all, to 
find the kinetic energy of the machine itself in any given posi- 
tion and this will now be determined. 

If any body has plane motion at any instant, this motion 
may be divided into two parts: 

(a) A motion of translation of the body. 

(6) A motion of rotation of the body about its center of grav- 
ity. Let the weight of the body be w lb., then its mass will 

be m = , where g is the attraction due to gravity and is equal 


to 32.16 in pound, foot and second units, and let the body be 
moving in a plane, the velocity of its center of gravity at the 
instant being v ft. per second. Further, let the body be turning 
at the same instant at the rate of co radians per second, and assume 
that the moment of inertia of the body about its center of gravity 
is 7, the corresponding radius of gyration being k ft., so that 
I = mk\ 

Then it is shown in books on mechanics that the kinetic energy 
of the body is E = %mv 2 + ]4 /co 2 = ^mv 2 + i^ra/<; 2 a> 2 ft.-pds., 
and, hence, in order to find the kinetic energy of the body it is neces- 
sary to know its weight and the distribution of the latter because 
of its effect on k, and in addition the velocity of the center of 


gravity of the weight and also the angular velocity of the 

202. Application to Machines. Let Fig. 149 represent a 
mechanism with four links connected by four turning pairs, the 
links being a, b, c and d, of which the latter is fixed, and let I a , 
Ib and I c represent the moments of inertia of a, b and c respective- 
ly about their centers of gravity, the masses of the links being 
m a , nib and m c . Assuming that in this position the angular 

FIG. 149. 

velocity o> of the link a is known, it is required to find the corre- 
sponding kinetic energy of the machine. 

Find the images of P, Q, a, b, c, d and of G, H and N, the centers 
of gravity of a, b and c respectively, by means of the phorograph 
discussed in Chapter IV. Now if V G , V H and V N be used to rep- 
resent the linear velocities of G, H and N and also if co& and o> c be 
used to denote the angular velocities of the links b and c, it is 
at once known, from the phorograph (Sees. 66 and 68), that: 
VQ = OG'.u] V H = OH'.u and V N = ON'.u ft. per second, and 

6 1 c' 

fc>6 = "iT'w and co c = co radians per second, so that all the neces- 
sary linear and angular velocities are found from the drawing. 

203. Reduced Inertia of the Machine. The investigation wil) 
be confined to the determination of the kinetic energy of the 
link b, which will be designated by Eb, and having found this 
quantity the energy of the other links may be found by a similar 


process. Since for any body the kinetic energy at any instant 
is given by the formula: 

E b = y% mv 2 + M I"* ft.-pds. 
Therefore, E b = ^m b .v 2 H + M ^ b z ft.-pds. 

' v ' 2 

Now, JW = m b k b z w b 2 = m b k b 2 

Following the notation already adopted, it will be convenient 
'to write k' b for ^rk b , since the length j-k b is the length of the image 

of kb on the phorograph. The magnitude of k' b is found by draw- 
ing a line HT in any^ direction from H to represent kb and find- 
ing T' by drawing H'T' parallel to HT to meet TP produced in 
T' as indicated in Fig. 149; then H'T' is the corresponding value 
of fc' 6 . 

Hence E b = }$ mb-Vn* + % I b u b 2 

= y 2 m b [OH' 2 + k b ' 2 } ^ ft.-pds. 

Let the quantity in the square bracket be denoted by K b 2 ', 
then evidently K b 2 may be considered as the radius of gyration 
of a body, which if secured to the link a and having a mass ra& 
would have the same kinetic energy as the link 6 has at this 
instant. It is evidently a very simple matter to find K b graphi- 
cally since it is the hypotenuse of the right-angled triangle of 
which one side is OH' and the other k b \ this construction is shown 
in Fig. 149. 

Thus E b = M nibK b W ft.-pds, 

Similarly E a = ^m X a 2 co 2 ft.-pds. 

and E c = ^m c K c 2 u 2 ft.-pds. ; 

constructions for K a and K c are shown in the figure. 
For the whole machine the kinetic energy is 
E = E a + E b + E c 

+ /'* + /'el 


A study of these formulas and a comparison with the work 
just covered, shows that I' a is the moment of inertia of the mass 
with center of gravity at and rotating with angular velocity 
co which will have the same kinetic energy as the link a actually 
has; in other words, I r a may be looked upon as the reduced mo- 
ment of inertia of the link a, while similar meanings may be 
attached to /'& and I' c . Note that I' a and I a differ because the 
former is the inertia of the corresponding mass with center of 
gravity at 0, whereas I a is the moment of inertia about the 
center of gravity G of the actual link. The quantity J is, on 
the same basis, the reduced inertia of the entire machine, by 
which is meant that the kinetic energy of the machine is the 
same as if it were replaced by a single mass with center of gravity 
at 0, and having a moment of inertia J about 0, this mass rotat- 
ing at the angular speed of the primary link. It will be readily 
understood that J differs for each position of the machine and is 
also a function of the form and weight of the links. 

The foregoing method of reduction is of the greatest importance 
in solving the problems under consideration, because it makes it 
possible to reduce any machine, no matter how complex, down 

FIG. 150. 

to a single mass, rotating with known speed, about a fixed center, 
so that the kinetic energy of the machine is readily found from 
the drawing. 

204. Application to Reciprocating Engine. The method may 
be further illustrated in the common case of the reciprocating 
engine, which in addition to the turning pairs contains also a 
sliding pair. The mechanism is shown in Fig. 150 and the 
same notation is employed as was used in the previous case, and 
the only peculiarity about the mechanism is the treatment of 
the link c. 

The link c has a motion of translation only and therefore 
co c = and / c w c 2 = so that the kinetic energy of the link is 


E c = }4 m c - v c z = M m c OQ-' 2 o> 2 or 7 C ' = m c OQ' 2 since the 
point Q has the same linear velocity as all points in the link. 
The remainder of the machine is treated as before. 

Lack of space prevents further multiplication of these illus- 
trations, but it will be found that the method is easily applied 
to any machine and that the time required to work out the values 
of J for a complete cycle is not very great. 


205. Conditions Affecting Speed Variations. One of the most 
useful applications of the foregoing theory is to the determination 
of the proper weight of flywheel to suit given running conditions 
and to prevent undue fluctuations in speed of the main shaft of 
a prime mover. Usually the allowable speed variations are set 
by the machine which the engine or turbine or other motor is 
driving and these variations must be kept within very narrow 
limits in order to make the engine of value. When 1 a dynamo is 
being driven, for example, fluctuations in speed affect the lights, 
causing them to flicker and to become so annoying in certain cases 
that they are useless. The writer has seen a particularly bad 
case of this kind in a gas engine driven generator. If alternators 
are to be run in parallel the speed fluctuation must be very small 
to make the arrangement practicable. 

In many rolling mills motors are being used to drive the rolls 
and in such cases the rolls run light until a bar of metal is put in, 
and then the maximum work has to be done in rolling the bar. 
Thus, in such a case the load rises suddenly from zero to a maxi- 
mum and then falls off again suddenly to zero. Without some 
storage of energy this would probably cause damage to the motor 
and hence it is usual to attach a heavy flywheel somewhere 
between the motor and the rolls, this flywheel storing up energy 
as it is being accelerated after a bar has passed through the rolls, 
and again giving out part of its stored-up energy as the bar enters 
and passes through the rolls. The electrical conditions determine 
the allowable variation in speed, but when this is known, and also 
the work required to roll the bar and the torque which the motor 
is capable of exerting under given conditions, then it is necessary 
to be able to determine the proper weight of flywheel to keep the 
speed variation within the set limits. 

In the case of a punch already mentioned, the machine runs 


light for some time until a plate is pushed in suddenly and the 
full load is thrown on the punch. If power is being supplied by 
a belt a flywheel is also placed on the machine, usually on the 
shaft holding the belt pulley, this flywheel storing up energy while 
the machine is light and assisting the belt to drive the punch 
through the plate when a hole is being punched. The allowable 
percentage of slip of the belt is usually known and the wheel must 
be heavy enough to prevent this amount of slip being exceeded. 

The present discussion is devoted to the determination of 
the speed fluctuations with a given machine, and the investiga- 
tion will enable the designer to devise a machine that will keep 
these fluctuations within any desired limits, although the next 
chapter deals more particularly with this phase. 

206. Determination of Speed Variation in Given Machine. 
Let EI and E% be the kinetic energies, determined as already 
explained, of any machine at the beginning and end of a certain 
interval of time corresponding with a definite change of posi- 
tion of the parts. Then the gain in energy, E% EI, during the 
interval under consideration represents the difference between 
the energy supplied with the working fluid and the sum of the fric- 
tion of the machine and of the work done at the main shaft on 
some other machine or object during the same interval, because 
the kinetic energy of the machine can only change from instant 
to instant if the work done by the machine differs from the work 
done on it by the working fluid. In order to simplify the problem 
friction will be neglected, or assumed included in the output. 

Consideration will show that E 2 EI will be alternately posi- 
tive and negative, that is, during the cycle of the machine its 
kinetic energy will increase to a maximum and then fall again 
to a minimum and so on. As long as the kinetic energy is in- 
creasing the speed of the machine must also increase in general, 
so that the speed will be a maximum just where the kinetic energy 
begins to decrease, and conversely the speed will be a minimum 
just where the kinetic energy begins to increase again. But the 
kinetic energy of the machine will increase just so long as the 
energy put into the machine is greater than the work done by it 
in the same time; hence the maximum speed occurs at the end of 
any period in which the input to the machine exceeds its output 
and vice versa. 

The method of computing this speed fluctuation will now be 


It has already been shown that the kinetic energy of the machine 
is given by 

E = KJco 2 ft.-pds. 

from which there is obtained by differentiation 1 

5E = % {2J. co.5co + co 2 .5J| ft.-pds. 

5E -co 2 .5J 



where 6co is the change in speed in radians per second in the 
interval of time in which the gain of energy of the machine is 
5E and that in J is dJ. Of course, any of these changes may be 
positive or negative and they are not usually all of the same sign. 
The values of J and co used in the formula may, without sensible 
error, be taken as those at the beginning or end of the inter- 
val or as the average throughout the interval, the latter being 

207. Approximate Value of Speed Variation. The calculation 
is frequently simplified by making an approximation on the 
assumption that the variation in J may be neglected, i.e., that 
8J = fc0 The writer has not found that there is enough saving 
in time in the work involved to make this approximation worth 
while, but since it is often assumed, it is placed here for con- 
sideration and a slightly different method of deducing the 
resulting formula is given. Let Ei, Ez, coi and co 2 have the 
meanings already assigned, at the beginning and end of the 
interval of time and let the reduced moment / be considered 

1 To those not familiar with the calculus the following method may be of 

Let E, J and co be the values of the quantities at the beginning of the 
interval of time and E + dE, J + 8 J and co -}- 5co, the corresponding values 
at the end of the same interval. 

E = l^Jco 2 
E + 5E =%(J + 6J") (co + 5co) 2 

= J^C/co 8 + 2J.co5co + co 2 5j) 

where in the multiplication such terms as (5co) 2 , and Sj.Sco are neglected as 
being of the second order of small quantities. 
By subtraction, then, 

E + 8E-E=*dE = % {2J"co5w + co 2 5J"j as above. 




? - CO! 2 ) 

where co has been written for 1 , a substitution which causes 


little inaccuracy in practice. 
Therefore co 2 coi = 




000 -7-. 


This is the same result as would have been obtained from the 
former formula by making 5J = 0. 

208. Practical Application to the Engine. The meanings of 
the different quantities can best be explained by an example 

FIG. 151. 

which will now be worked out. The steam engine has been 
selected, because all the principles are involved and the method 
of selecting the data in this case may be rather more readily 
understood. The computations have all been made by the exact 
formula, which takes account of variations in J. 

Consider the double-acting engine, which is shown with the 
indicator diagrams in Fig. 151; it is required to find the change of 
speed of the crank while passing from A to B. Friction will be 

For simplicity, it will be assumed that the engine is driving a 


turbine pump which offers a uniform resisting turning moment 
and hence the work done by the engine during any interval is 
proportional to the crank angle passed through in the given 
interval. If the work done per revolution as computed from 
the diagram is W ft.-pds., then the work done by the engine during 

a __ n 

the interval from A to B will be QAn W ft.-pds. To make the 


case as definite as possible suppose that 2 0i = 18; then the 
work done by the engine will be Ho^ ft.-pds. 

209. Output and Input Work. Again, let AI and A 2 represent 
the areas in square inches of the head end and the crank end of 
the cylinder respectively, li and Z 2 being the lengths of the cor- 
responding indicator diagrams in inches. The stroke of the pis- 
ton is taken as L feet and the indicator diagrams are assumed 
drawn to scale s pds. per square inch = 1 in. in height. With 
these symbols the work represented by each square inch on the 

diagram is sAi >- ft.-pds. for the head-end and 8^27- ft.-pds. for the 
l>i L% 

crank-end diagram. 

Now suppose that during the crank's motion from A to B the 
area of the head-end diagram reckoned above the zero line is 
a\ sq. in., see Fig. 151, and the corresponding area for the crank- 
end diagram a 2 sq. in. Then the energy delivered to the engine 
by the steam during the interval is 

ij -- zzf- ft.-pds. 
LI L 2 

while the work done by the engine is 


Note that the work W is the total area of the two diagrams in 
square inches multiplied by their corresponding constants to 
bring the quantities to foot-pounds. 

Then the input work exceeds the output by 

L L W 


i y -- ^z-j -- ft.-pds. 

which amount of energy must be stored up in the moving parts 
during the interval. That is, the gain in energy during the 
period is 

E 2 EI = aisAi 7 -- a 2 sAz -, -- o~ ft.-pds. 

so that the gain in energy is thus known. 


Again, the method described earlier in the chapter enables the 
values of Ji and J 2 to be found and hence the value of J 2 J\. 
Substituting these in the formula, 

dE - Kco 2 5J 
5co = - 7 


the gain in angular velocity is readily found. The values are 
#2 - Ei = dE, J 2 - J l = 8J, MGA + Ji) = J and for co no 
error will result in practice by using the mean speed of rotation 
of the crank. 

During a complete revolution the values of 5co will sometimes 
be positive and sometimes negative, and in order that the engine 
may maintain a constant mean speed the algebraic sum of these 
must be zero. Should the algebraic sum for a revolution be 
positive, the conclusion would be that there is a gain in the mean 
speed during the revolution, that is the engine would be steadily 
gaining in speed, whereas it has been assumed that the governor 
prevents this. 

210. Numerical Example on Single -cylinder Engine. A nu- 
merical example taken from an actual engine will now be given. 

FIG. 152. 

The engine used in this computation had a cylinder 12>f 6 in. 
diameter with a piston rod 1% in. diameter and a stroke of 30 in. 
The connecting rod was 90 in. long, center to center, weighed 
175 Ib. and had a radius of gyration about its center of gravity 
of 31.2 in. The piston, crosshead and other reciprocating parts 
weighed 250 Ib., while the flywheel weighed 5,820 Ib. and had a 
moment of inertia about the shaft of 2,400, using pound and 
foot units. The mean speed of rotation was 86 revolutions per 

Using the notation employed in the earlier discussion, the data 
may be set down as follows: 

a = 1.25 ft., 6 = 7.5 ft., fa = 2.60 ft., I a = 2,400 pd.(ft.) 2 , 
m a = 181, m b = 5.44, m c = 7.78. 

* 27m 2 X T X 8 

Ine speed oy= ~^- -- = 9 radians per second. 


Using the above data the following quantities were measured 
directly from the drawing, Fig. 152: 
























From these the following quantities are obtained by computa- 



I'b =*= mb.Kb- 



J = 
















Zero Line 

FIG. 153. 

a g =.035 Sq. In. 

Thus, during the 18 under consideration there is a gain in 
the reduced inertia of 5.9, although as the complete table given 
later on shows, there is a loss in other parts of the revolution. 

The indicator diagrams for the engine are shown in Fig. 153 
and the areas corresponding to the crank motion considered are 



shown hatched and marked i and a 2 . These areas were meas- 
ured on the original diagrams which were drawn to 60- pd. scale, 
although these have been somewhat reduced in reproduction. 

Data for computations from the indicator diagrams are as 

Cylinder areas: Head end, AI = 114.28 sq. in. Crank end, 
A z = 111.52 sq. in. 

Diagram lengths: Head-end diagram, li = 3.55 in. Cranknend 
diagram, Z 2 = 3.58 in. 

Stroke of piston, L = 2.5 ft. 
Hence, each square inch on the diagrams represents 

L 25 

sAi r = 60 X 114.28 X 5-^ = 4,829 ft.-pds. for the head end, 

LI o.OO 



= 60 X 111.52 X 

= 4,673 ft.-pds. for the crank end. 

The original full-sized diagrams give ai = 0.550 sq. in. and 
a z = 0.035 sq. in., from which the corresponding work done will 

0.550 X 4 ; 829 = 2,656 ft.-pds. for the head end, 

0.035 X 4,673 = 163 ft.-pds. for the crank end. 
It is assumed that the engine is driving a turbine pump or 
electric generator which offers a constant resisting torque, so 

-I O -j 

that the corresponding work output is ^7: = ^ of the total work 

represented by the two diagrams, and is 1,079 ft.-pds. 
The quantities are set down in the table below. 


Diagram areas 

Work done on piston 

Work done 
by crank, 

Net work pro- 
ducing change of 
kinetic energy, 

Head 01, 
sq. in. 

Crank 02, 
sq. in. 












The total combined areas of the two diagrams represent 21,584 
ft.-pds., and since the speed was 86 revolutions per minute the 

indicated horsepower was QQ'QQQ X 86 = 56.2 hp. 

The quantity in the last column is the difference between 
2,493 and 1,079 and would evidently cause the machine to speed up. 


Then the work available for increasing the energy is 1,414 
ft.-pds. and this must represent the gain in kinetic energy of the 
machine, or 

5E = + 1,414 ft.-pds. 

The gain in angular velocity may now be computed. The 
average value of J is 

/ = K [2,410.9 + 2,416.8] = 2,413.8 
hence = 2,413.8 X 9 = 21,724.6 

and K " 2 .6J = MX 9 2 X 5.9 = 238.9 

dE - H " 2 5/ 

do) = 


_ 1,414 - 238.9 


= 0.0541 radians per second 

which is the gain in velocity during the period considered. Sim- 
ilarly the results may be obtained for other periods, and thus for 
the whole revolution. These results are set down in the table 
given on page 257. 

211. Speed-variation Diagram. The values of Sto thus ob- 
tained are then plotted on a straight-line base, Fig. 154, which 
has been divided into 20 equal parts to represent each 18 of 
crank angle. If it is assumed that the speed variation is small, 
as it always must be in engines, then no serious error will be made 
by assuming that these crank angles are passed through in equal 
times, and hence that the base of the diagram on which the values 
of 5 co are plotted is also a time base, equal distances along which 
represent equal intervals of time. 

If desired, the equal angle base may be corrected for the varia- 
tions in the velocity, using the values of 5co already found, so as 
to make the base exactly represent time intervals, but the author 
does not think it worth the labor and has made no correction 
of this kind on the diagram shown. 

Attention should here be drawn to the fact that the height of 
the original base used for plotting the speed-variation curve has 
to be chosen at random, but after the curve has been plotted, it is 
necessary to find a line on this diagram representing the mean 
speed of rotation, co = 9. This may be readily done by finding 
the area under the curve by a planimeter, or otherwise, and then 
locating the line co = 9 so that the positive and negative areas 




o o o o o 


o o o o d 

,-iiOi-Hr-(CMTt<l>OO'O-ciO<N r-l CO tfJ O 

I I I I I+ + + + +I I I I I 

P'S <u'.l3 
? 3 be a; 

^>T3 G 

+ I I I I I I 

I I I I I 


o a g a 





OOr^Tttt^lNOOcO^IN Ot^-<JiOOCOOl> l O<N 

* m 



I I I I+ + + + +I I I I I 







between this new line and the velocity-variation curve are equal. 
It is to be remembered that the computation gives the gain in 
velocity in each interval, and the result is plotted from the end 
of the curve, and not from the base line in each case. 

212. Angular-space Variation. Now since the space traversed 
is the product of the corresponding velocity and time, the angular- 
space variation, 56 in radians, is found by multiplying the value 
of d co by the time t in seconds required to turn the crank through 
the corresponding 18, that is 

56 = t.5u radians. 

But t.5u is evidently an area on the curve of angular-velocity 
variation, so that the angular space variation in radians up to 
any given crank angle, say 54, is simply the area under the 
angular-velocity variation curve up to this point, the area being 
taken with the mean angular velocity as a base and not with the 
original base line. In this case the area between the mean speed 
line, co = 9, and the speed-variation curve, from to 54, when 
reduced to proper units, represents 0.275 radian as plotted in 
the lower curve of Fig. 154. 

The upper curve shows that the minimum angular velocity 
was 8.922 radians per second; while the maximum was 9.063 
radians per second, a variation of 0.141 radian per second, or 
1.57 per cent. 

The lower curve shows the angular swing of the flywheel about 
its mean position, and shows that the total swing between the 
two extremes was 0.58, although the swing from the mean posi- 
tion would be only about one-half of this. 

The complete computations which have been given here in 
full for an engine, will, it is hoped, clearly illustrate the method 
of procedure to be followed in any case. The method is not as 
lengthy as would appear at first, and the results for an engine 
may be quickly obtained by the use of a slide rule and drafting 

In the case of engines, all moving parts have relatively high 
velocity, and it is generally advisable to take account of the 
variations in the reduced inertia, J. In other machines, such for 
example as a belt-driven punch, all parts are very slow-moving 
with the exception of the shaft carrying the belt pulleys and fly- 
wheel, and in such a case it is only necessary to take account of 
the inertia of the high-speed parts. Wherever the parts are of 


large size or weight or run at high speed, account must be taken 
of their effect on the machine. 

jg esBaioui o 


{3 8SBQJ09Q 

Frequently only the angular-velocity variation is required, but 
usually the space variation is also necessary, as in the case of 
alternators which are to work in parallel. 


213. Factors Affecting the Speed Fluctuations. A general dis- 
cussion has been given earlier in this chapter of the factors that 
affect the magnitude of the speed fluctuations in machinery and 
as an illustration here Fig. 155 has been drawn. This figure 
shows three speed-fluctuation curves for the engine just referred 
to, and for the same indicator diagrams as are shown in Fig. 153, 
but in each case the engine is used for a different purpose. The 
curve in the plain line is an exact copy of the upper curve in 
Fig. 154 and represents the fluctuations which occur when the 
engine is direct-coupled to an electric generator, the total fluc- 
tuation being 0.14 radian per second or about 1.57 per cent. 

The dotted curve corresponds to a water pump connected in 
tandem with the engine, a common enough arrangement, al- 
though the piston speed is rather too high for this class of work. 
The speed fluctuation here would be less than before, amounting 
to 0.123 radian or about 1.37 per cent., this being due to the 
fact that the unbalanced work is not so great in this class of re- 
sistance as in the generator. 

The broken line corresponds to an air-compressor cylinder in 
tandem with the steam cylinder and the resulting variation is 
0.305 radian per second or 3.38 per cent., which is over twice 
as much as the first case. 


1. A 12-in. round cast-iron disk 2 in. thick has a linear velocity of 88 ft. 
per second; find its kinetic energy. What would be its kinetic energy if it 
also revolved at 100 revolutions per minute? 

2. A straight steel rod 2 ft. long, 1^ in. diameter, rotates about an axis 
normal to its center line, and 6 in. from its end, at 50 revolutions per minute. 
What is its kinetic energy? 

3. Find the kinetic energy of a wheel 12 in. diameter, density 2. 14, at 500 
revolutions per minute. 

4. What is the kinetic energy of a cast-iron wheel 3 ft. diameter, 1% in. 
thick, rolling on the ground at 8 miles per hour? 

6. If the side rod of a locomotive is 5 ft. long and of uniform section 
2> by 5 in., with drivers 60 in. diameter, and a stroke of 24 in., find the 
kinetic energy of the rod in the upper and lower positions. 

6. Show how to find the kinetic energy of the tool sliding block of the 
Whitworth quick-return motion. 

7. Suppose the wheel in question 3 is a grinder used to sharpen a tool and 
that its speed is decreased in the process to 450 revolutions in 1 sec.; what is 
the change in kinetic energy? 

8. Plot the speed and angular velocity-variation curves for two engines like 
that discussed in the text with cranks at 90, only one flywheel being used. 

9. Repeat the above with cranks at 180. 



214. Purpose of Flywheels. In the preceding chapter a com- 
plete discussion has been given as to the causes of speed fluc- 
tuations in machinery and the method of determining the amount 
of such fluctuation. In many cases a certain machine is on hand 
and it is the province of the designer to find out whether it will 
satisfy certain conditions which are laid down. This being the 
case the problem is to be solved in the manner already discussed, 
that is, the speed fluctuation corresponding to the machine and 
its methods of loading are to be determined. 

Frequently, however, the converse problem is given, that is, 
it is required to design a machine which will conform to certain 
definite conditions; thus a steam engine may be required for 
driving a certain machine at a given mean speed but it is also 
stipulated that the variation in speed during a revolution must 
not exceed a certain amount. Or a motor may be required for 
driving the rolls in a rolling mill, the load in such a case varying 
so enormously, that, if not compensated for would cause great 
fluctuations in speed in the motor, which fluctuations might be so 
bad as to prevent the use of the motor for the purpose. In a 
punch or shear undue fluctuation in speed causes rapid destruc- 
tion of the belt. In all the above and similar cases these varia- 
tions must be kept within certain limits depending upon the 

In all machines certain dimensions are fixed by the work to 
be done and the conditions of loading, and are very little affected 
by the speed variations. Thus, the diameter of the piston of an 
engine depends upon the power, pressure, mean speed, etc., and 
having determined the diameter, the thickness and therefore the 
weight is fixed by the consideration of strength almost exclu- 
sively; the same thing is largely true regarding the crosshead, con- 
necting rod and other parts, the dimensions, weights and shapes 
being independent of the speed fluctuations. Similar statements 
may be made about the motor, its bearings, armature, etc., being 
fixed by the loading, and in a punch the size of gear teeth 
and other parts are also independent of the speed fluctuation. 



Each of these machines contains also a flywheel, the dimen- 
sions of which depend on the speed variations alone and not upon 
the power or pressures as do the other parts. The function of 
the flywheel is to limit these variations; thus on a given size 
and make of engine the weight of flywheel will vary greatly with 
the conditions of working; in some cases the wheel would be very 
heavy, while in other cases there might be none at all on the same 

Ordinarily the flywheel is made heavy and run with as high a 
rim speed as is deemed safe; in slow-revolving engines the 
diameter is generally large, while in higher-speed engines the 
diameter is smaller, as in automobile engines, etc. The present 
chapter is devoted to the method of determining the dimensions 
of flywheel necessary to keep the speed fluctuations in a given 
case within definitely fixed limits. 

Referring to Chapter XIII, Sec. 203, the kinetic energy of a 
machine is given by the equation E = %Ju 2 , where J is the re- 
duced inertia found as described therein. The method of 
obtaining E has also been fully explained; it depends upon the 
input and output of the machine, such, for example, as the indi- 
cator and load curves for a steam engine. E and J are thus 
assumed known and the above equation may then be solved for 

co, thus, j/2 o> 2 = - j.- 

215. General Discussion of the Method Used. In order that 
the matter may be most clearly presented it will be simplest to 
apply it to one particular machine and the one selected is the 
reciprocating engine, because it contains both turning and sliding 
elements and gives a fairly general treatment. In almost all 
machines there are certain parts which turn at uniform speed 
about a fixed center and which have a constant moment of in- 
ertia, such as the crank and flywheel in an engine, while other 
parts, such as the connecting rod, piston, etc., have a variable 
motion about moving centers and a correspondingly variable 
reduced moment of inertia; the table in the preceding chapter 
illustrates this. It will be convenient to use the symbol J a to 
represent the moment of inertia of the former parts, while J& 
represents that of the latter, and thus Ja is constant for all 
positions of the machine, and Jb is variable. The total reduced 
inertia of the machine is J = J a + Jb- Both of these quantities 



J and Jb are independent of the speed of rotation and depend 
only upon the mass and shape of the links, that is upon the rela- 
tive distribution of the masses about their centers of gravity. 

Suppose now that for any machine the values of J are plotted 
on a diagram along the ff-axis^tho nf which diagram 
represent the corresponding value of the energy \E this will give 
a diagram of the general shape shown at Fig. 156. where the curve 



\ 8 



FIG. 156. 

represents J for the corresponding value of E shown on the 
vertical line. 1 

Looking now at the figure KFGHK, it is evident from construc- 
tion that its width depends on the values of J at the instant and 
is thus independent of the speed. Also, the height of this figure 
depends on the difference between the work put into the machine 
and the work delivered by the machine during given intervals, 
that is, it will depend on such matters as the shapes of the indi- 
cator and load curves. The shape of the input work diagrams 
within certain limits depends on whether the machine is run by 
gas or steam, and on whether it is simple or compound, etc., but 
for a given engine this is also, generally speaking, independent of 
the speed : the load curve will, of course, depend on what is being 
driven, whether it is dynamo, compressor, etc. Thus the height 
of the figure is also independent of the speed. 

1 This form of diagram appears to be due to WITTENBAUER; see "Zeit- 
schrift des Vereines deutscher Ingenieure" for 1905. 


It will further be noted that the shape of the figure does not 
depend on J a , which is constant for a given machine, but only 
on the values of the variable Jbj hence the shape of this figure 
will be independent of the weight of the flywheel and speed, in so 
far as the input and load curves are independent of the speed, 
depending solely on the reciprocating masses, the connecting 
rod, the input-work diagrams and the load curves. 

Now draw from the two tangents, OF and OH, to KFGH, 
touching it at F and H respectively, then for OH the energy 

Ei = HH', and J l = OH' and (Sec. 203), JW = ^ = tan i, 

J i 

and since i is the least value such an angle can have it is evident 
that toi is the minimum speed of the engine. Similarly, E% = FF f 

and <7 2 = OF', and J^a>2 2 = -=- = tan .0:2 and hence, co2 would be 


the maximum speed of the engine, since oti is the maximum value 
of a. 

216. Dimensions of the Flywheel. Suppose now that it is 
required to find the dimensions of a flywheel necessary for a given 
engine which is to be used on a certain class of service, the mean 
speed of rotation being known. The class of service will fix the 
variations allowable and the mean speed ; in engines driving alter- 
nators for parallel operation the variation must be small, while 
in the driving of air compressors and plunger pumps very much 
larger variations are allowable. Thus, the class of service fixes 
the speed variation o>2 ~ coi radians per second, and the mean 

speed co = - ~ is fixed by the requirements of the output. 

Experience enables the indicator diagrams to be assumed with 
considerable accuracy and the load curve will again depend on 
what class of work is being done. 

The only part of the machine to be designed here is the fly- 
wheel, and as the other parts are known, and the indicator and 
load curves are assumed, the values of E and /& are found as 
explained in Chapter XIII and the E Jb curve is drawn in. 
In plotting this curve the actual value of E is not of importance, 
but any point may arbitrarily be selected as a starting point and 
then the values of 5E, or the change in E, and Jb will alone give 
the desired curve. Thus, in Fig. 156 the diagram KFGH has 
been so drawn and it is to be observed that the exact position 
of this figure with regard to the origin is unknown until J a is 


known, but it is J a that is sought. A little consideration will 
show, however, that an axis E'Oi may be selected and used as 
the axis for plotting J b , values of which may be laid off to the 

Further, any horizontal axis 0' Jb may be selected, and for 
any value of Jb a point may be arbitrarily selected to represent 
the corresponding value of E and the meaning of this point may 
be later determined. Having selected the first point, the remain- 
ing points are definitely fixed, since the change in E corresponding 
to each change in Jb is known. Thus, the curve may be found 
in any case without knowing J a or the speed, but the origin has 
its position entirely dependent upon both, and cannot be deter- 
mined without knowing them. Thus the correct position of the 
axes of E and J are as yet unknown, although their directions 
are fixed. 

Having settled on coi and co 2 , two lines may be drawn tangent 
to the figure at H and F and making the angles a\ and 2 respec- 
tively, with the direction 0' J&, where tan a\ = J^coi 2 and 
tan a 2 = M W 2 2 . The intersection of these two lines gives and 
hence the axis of E, so that the required moment of inertia of 
the wheel may be scaled from the figure, since J a = 00\. It 
should, however, be pointed out that if the position of the axis 
of E is known, and also the mean speed co, it is not possible to 
choose coi and co 2 at will, for the selection of either E or the speeds 
will determine the position of 0. In making a design it is usual 

to select oj and - , which give coi and co 2 , and from the 


chosen values to determine the position of and hence the axes of 
E and J. The mean speed o> corresponds with the angle a. 
Draw a line NMLR perpendicular to OJ, close to the E J 

T /? 

diagram but in any convenient position. Then ~ tan ai, 


jyn = tan a 2 and -^ = tan a, so that on some scale which may 

be found, LR represents coi 2 , or the square of the speed HI in 
revolutions per minute, NR represents n 2 2 and MR represents 
the square of the mean speed n all on the same scale. As in 
engines the difference between n\ and n 2 is never large it is fairly 
safe to assume 2n 2 = n z 2 + ni 2 or that M is midway between 
N and L. 

217. Coefficient of Speed Fluctuation. Using now 5 to denote 


the coefficient of speed fluctuation, then 5 is denned by the 

_ n 2 - HI 



= ^2 ni = n z ni _ 2 2 

n ~ 




25 = r ^^ 


But it has already been shown that 

i / o E\ 2-jrni 

j/2coi 2 = -y- = tan a\ and since coi = -^r- 
J i OU 


= 182.3 tan 

2X60 2 

2 X 60 2 

Similarly, n 2 2 = 182.3 tan 2 ; thus the speeds depend on a only. 
Since in Fig. 156 the base OR is common to the three triangles 
with vertices at N, M and L, it follows that 

RL = OR tan ai = OR X 

and RN = Cn 2 2 where C = 7^-5 in both cases. Further gen- 
erally, #M = Cn 2 . 

Then, referring to the formula for 25, which is 25 = - ~T~ 
this may be put into the following form : l 

RN _ RL 

nS - nS C C RN - RL NL 

25 = 

n 2 ^M RM ~ RM' 



Thus NL = 25 X #M. These are marked in Fig. 156. 

1 It is instructive to compare this investigation with the corresponding one 
for governors given in Sec. 183 and Fig. 127 a. 


In general, a* a\ is a small angle in practice, in which case 
M may be assumed midway between N and L without serious 
error, and on this assumption 

NM = ML = n 2 X 6. 

The foregoing investigation shows that the shape of the E J 
diagram has a very important effect on the best speed for a given 
flywheel and the best weight of flywheel at the given speed. 
Thus, Fig. 158 shows one form of this curve for an engine to be 
discussed later, while Fig. 160 shows two other forms of such 
curves for the same engine but different conditions of loading. 
With such a curve as that on the right of Fig. 160, the best 
speed condition will be obtained where the origin is located 
along the line through the long axis of the figure. In order to 
make this more clear, this figure is reproduced again on a re- 
duced scale at Fig. 157 and several positions of the origin are 
drawn in. This matter will now be discussed. 

218. Effects of Speed and Flywheel Weight. Two variables 
enter into the problem, namely the best speed and the most 
economical weight of flywheel. Now, the formula connecting 
the speed with the angle a is J^co 2 = tan a, Sec. 215, so that the 
speed depends upon the angle of alone, and for any origin along 
such a line as OF there is the same mean speed since a is constant 
for this line. To get the maximum and minimum speeds corre- 
sponding to this mean speed, tangents are drawn from to the 
figure giving the angles ai and /* 2 and hence coi and co 2 . A glance 
at Fig. 157 shows that the best speed corresponds to the line OF 
and that for any other origin such as Oi, which represents a 
lower mean speed, since for it a and hence tan a is smaller, there 
will be a greater difference between coi and co 2 in relation to co 
than there is for the origin at 0. A few cases have been drawn 
in, and it is seen that even for the case which represents a 
higher mean speed than the value of 5 will be increased; thus 
the best speed corresponds to the line OF and its value is found 
from ^co 2 = tan a. 

But the speed variations also depend on the weight of the 
flywheel and hence upon the value of^Tor the horizontal distance 
of the origin from the axis Q'E'. If the origin was at 0, there 
would be no flywheel at all but the speed variation taken from a 
scaled drawing, would be prohibitive as it is excessively large. 
For the position the inertia of the flywheel is represented by 



4 and the speed variations would be comparatively small, 
but if the origin is moved up along OF to 3 , the speed being the 
same as at 0, the variations will be increased very slightly, but 
the flywheel weight also shows a greater corresponding decrease. 
Similarly, Oi corresponding to the heaviest wheel, shows a varia- 
tion in excess of and nearly equal to 3 , and 2 with the same 

FIG. 157. Effect of speed and weight of flywheel. 

weight of wheel as at shows nearly double the variation that 

Thus, increasing the weight of the wheel may increase the 
speed variations if the speed is not the best one, and increas- 
ing the speed may produce the same result, but at the speed 
represented by OF, the heavier the wheel the smaller will be 
the variation, although the gain in steadiness is not nearly 
balanced by the extra weight of the wheel beyond a certain 
point. Frequently the operating conditions prevent the best 



speed being selected, and if this is so it is clear that the weight 
of the wheel must be neither too large nor too small. 

These results may be stated as follows: For a given machine 
and method of loading there is a certain readily obtained speed 
which corresponds to minimum speed variations, and for this 
best value the variations will decrease slowly as the weight of 
flywheel is increased. For a certain flywheel weight the speed 
variations will increase as the speed changes either way from the 
best speed, and an increase in the weight of the flywheel does 
not mean smaller fluctuation in speed unless the mean speed 
is suitable to this condition. 

219. Minimum Mean Speed.- The above results are not quite 
so evident nor so marked in a curve like Fig. 158 but the same 

Plain Line is for Outward Stroke 
Dotted Line is for Return Stroke 





M & 



5 10 

FIG. 158. Steam engine with generator or turbine pump load. 

conditions hold in this case also. The best speed is much more 
definitely fixed for an elongated E J curve and becomes less 
marked as the boundary of the curve comes most nearly to the 
form of a circle. The foregoing investigation further shows that 
no point of the E J curve can fall below the axis J, because 
if it should cut this axis, the machine would stop. The minimum 
mean speed at which the machine will run with a given flywheel 
will be found by making the axis J touch the bottom of the 
curve, and finding the corresponding mean speed; the minimum 



speed will, of course, be zero, since a 2 = 0. The minimum 
speed of operation may be readily computed for Figs. 158 and 
160 and it is at once seen that the right-hand diagram of Fig. 160 
corresponds to a larger minimum speed than any of the others, 
that is, when driving the air compressor the engine will stop at a 
higher mean speed than when driving the generator. 

220. Numerical Example of a Steam Engine. The principles 
already explained may be very well illustrated in the case of the 
steam engine used in the last chapter, which had a cylinder 12>{ 6 
in. diameter and 30 in. stroke and a mean speed of 87 revolutions 
per minute for which co = 9 radians per second. The form of 
indicator diagrams and loading are assumed as before and the 
engine drives a turbine pump which is assumed to offer constant 
resisting torque. The weight of the flywheel is required. 

Near the end of Chapter XIII is a table containing the values 
of J and 5E for equal parts of the whole revolution and for con- 
venience these results are set down in the table given herewith. 

TABLE OF VALUES OF J and E for 12% 6 BY 30-iN. ENGINE 

6, degrees 

J, total 

Jb = J - 2,400 

8E, foot-pounds 






- 233 




+ 1,387 




+ 1,414 




+ 699 




+ 186 












- 646 




- 858 








- 520 




+ 678 




+ 1,360 




+ 738 




+ 294 




- 33 








- 546 










Selecting the axes 0' E' and 0' J'&, Fig. 158, the corresponding 
E-J curve is readily plotted as follows: The table shows that 
when 6 = 0, J b = 3.2 and when = 18, J b = 5.4, the gain in 
energy which is negative, during this part of the revolution being 
6E = - 233 ft.-pds. Starting with J b = 3.2 and arbitrarily 
calling E at this point 1,000 ft.-pds. gives the first point on the 
diagram; the second point is found by remembering that when 
Jb has reached the value 5.4 the energy has decreased by 233 
ft.-pds., so that the point is located on the line J b = 5.4 and 233 
ft.-pds. below the first point. The third point is at J b = 10.9 
and 1,387 ft.-pds. above the second point and so on. 

Now draw on the diagram the line QM to represent the mean 
speed co = 9, its inclination to the axis of J b being a where tan a = 
J^co 2 = H X 9 2 = 40.5. The actual slope on the paper is readily 
found by noticing that the scales are so chosen that the same 
length on the vertical scale stands for 1,000 as is used on the hori- 

1 000 

zontal scale to represent 5, the ratio being ' = 200; then the 


40 5 

actual slope of QM on the paper is ^- = 0.2025, which enables 


the line to be drawn. This line may be placed quite accurately 
by making the perpendicular distance to it from the extreme 
lower point on the figure equal the perpendicular to it from F 
(see Figs. 156 and 158). Thus the position and direction of the 
mean speed line QM are known. 

Now suppose the conditions of operation require that the max- 
imum speed shall be 1.6 per cent, above the minimum speed, 
or that the coefficient of speed fluctuation shall be 1.6 per cent. 

COo 6)1 

Then, from Sec. 217, 5 = 0.016, that is 5 = - - = 0.016 


and the problem also states that the mean speed shall be co =9 

= - o . On comparing these two results it is found that 


co 2 = 9.072 and coi = 8.928. 

On substituting these two values in the equations for the 
angles, the results are tan ai = J^coi 2 = H X 79.709 = 39.854 
and tan 2 = H^2 2 - M X 82.301 = 41.150 which enables the 
two lines HL and AF to be drawn tangent to the figure at H 
and F and at angles <*i and <* 2 respectively to the axis of J b (on 
the paper the tangents of the slopes of these lines will be, for 

AF = 5 = a 2058 and for HL = = 0.1993). These 


lines are so nearly parallel that their distance apart vertically 
can be measured anywhere on the figure, and it has actually 
been measured along NML } the distance NL representing 
3,180 ft.-pds. 

Referring again to Fig. 156 it is seen that NR = OR tan 2 
and LR = OR tan ai and by combining these it maybe shown that 


OR = - . Substituting the results for this problem 

tan 2 tan on 



R -- 41.160 '-39.854 = 2 ' 453 -' + *-/.+ - 

Hence, the moment of inertia of the flywheel should be 2,430 
approximately, which gives the desired solution of the problem. 

221. Method of Finding Speed Fluctuation from E-J Dia- 
gram. The converse problem, that of finding the speed varia- 
tion corresponding to an assumed value of J a , has been solved in 
the previous chapter but the diagram may be used for this pur- 
pose also. Thus, let co = 9, the same mean ' speed as before, 


and J a = 2,000. Then, since J^co 2 = -j = tan a the value of 


E at M is (2,000 + 25) X % X 9 2 = 82,012. The points N 
and L will be practically unchanged and hence at N the value of 
E 2 is 82,012 + J(3,180) = 83,602 ft.-pds. and the value of 

83 602 
co 2 2 may be computed from the relation J^co 2 2 = ' , and in a 

similar way coi may be found and the corresponding speed varia- 

C0 2 

tion 5 = 


A somewhat simpler method may be used, however, by refer- 
ring to Fig. 156, from which it appears that NL = 2n 2 <5. Thus, 
2n 2 d is represented by 3,180 ft.-pds. and n 2 by 82,012 ft.-pds., from 
which the value of 5 is found to be 0.0175 which corresponds to 
a speed variation of 1.75 per cent. 

In order to show the effect of making various changes, let the 
speed of the engine be much increased to say 136 revolutions per 
minute for which co = 14.1, and let the speed variation be still 
limited to 1.6 per cent. The line QM will then take the position 
Q'M ' for which the tangent on the paper is % and the distance 
corresponding to LN measures 2,400 ft.-pds. On completing the 
computations the moment of inertia of the flywheel is found to 
be J = 740, that is to say that if the wheel remains of the same 



diameter it need be less than one-third of the weight required 
for the speed of 87 revolutions. 

The diagram Fig. 158 has been placed on the correct axis and 
is shown in Fig. 159 which gives an idea of the position of the 
origin for the value J a = 2,400 and w = 9. 

FIG. 159. 

222. Effect of Form of Load Curve on Weight and Speed. To 
show the effects of the form of load curve on this diagram and 
on the speed and weight of the flywheel, the curves shown in 
Fig. 160 have been drawn. The two diagrams shown there were 
made for the same engine and indicator diagrams as were used 
in Fig. 158, the sole difference is in the load applied to the engine. 
The left-hand diagram corresponds to a plunger pump connected 
in tandem with the steam cylinder, while the right-hand diagram 
is from an air compressor connected in tandem with the steam 
cylinder. The effect of the form of loading alone on the E J 
diagram is most marked and the air compressor especially pro- 
duces a most peculiar result, the best speed here being definitely 
fixed and being much higher than for either of the other cases, 
and if the machine is run at this speed it is clear that the weight 
of the flywheel is not very important so long as it is not extremely 

It is needless to say that the form of indicator diagram also 
produces a marked effect and both the input and output diagrams 
are necessary for the determination of the flywheel weight and 
the speed of the machine. The curves mentioned are sufficient 




to show that the weight of wheel and the best speed of operation 
depend on the kind of engine and also on the purpose for which 
it is used. It is frequently impossible, practically, to run an 
engine at the speed which gives greatest steadiness of motion 
and then the weight of wheel must be selected with care as out- 
lined in Sec. 218. 


20 ^ 

FIG. 160. Left-hand figure is for a plunger pump in tandem with steam 
engine; right-hand figure is for an air compressor in tandem with steam 

223. Numerical Example on Four-cycle Gas Engine. An illus- 
tration of the application to a gas engine of the four-cycle type 
is shown at Fig. 162, this being taken from an actual case of an 
engine direct-connected to an electric generator. The engine 
had a cylinder 14^ in. diameter and 22 in. stroke and was single- 
acting; the indicator diagram for it is shown at Fig. 161. The 
piston and other reciprocating parts weighed 360 lb., while the 
weight of the connecting rod was 332 lb., and its radius of gyra- 
tion about its center of gravity 1.97 ft., the latter point being 
24.3 in. from the center of the crankpin, and the length of the 
rod between centers was 55 in. 

There were two flywheels of a combined weight of 7,000 lb. 
and the combined moment of inertia of these and of the rotor 
of the generator was 1,600 (foot-pound units). 



The form of the E J diagram for this case is given in Fig. 162 
and differs materially in appearance from any of those yet shown, 
and the best speed is much more difficult to determine because 
of the shape of the diagram. The actual speed of the engine 

400 r 



o 1 - 


Zero JLine 

FIG. 161. 

'2~~~ 4 6 8 10 12 14 16 18 J $ 
FIG. 162. Gas engine driving dynamo. 

was 172 revolutions per minute and for this value the sloping 
lines on the diagram have been drawn. The mean-speed line 
would have an inclination to the axis of J b given by tan a = 

Ifa 2 = 162 and its slope on the paper would be 1 gr . A of this, 



since the vertical scale is 1,500 times the horizontal; thus the 


tangent of the actual slope is 1 = 0.108 and the lines are 


drawn with this inclination. 

The total height of the diagram is 31,200 ft.-pds. and using the 
value J a = 1,600, the mean value of E is 162 X 1,600 = 259,200 
ft.-pds. so that the speed variation is 

5 = Ji X 2QQ = 0.0602 or 6.02 per cent. 

The engine here described was installed to produce electric 
light and it is perfectly evident that it was entirely unsuited to 
its purpose as such a large speed variation is quite inadmissible. 
Owing to the peculiar shape of this diagram and the fact that the 
tangent points touch it on the left-hand side, it appears that the 
distance between them will not be materially changed by any 
reasonable change of slope of the lines, so that if the speed re- 
mains constant at 172 revolutions per minute the value of J a or 
the flywheel weight is inversely proportional to the speed varia- 
tion and flywheels of double the weight would reduce the fluc- 
tuation to about 3 per cent. 

A change in speed would bring an improvement in conditions 
and the results may readily be worked out. 


1. Show the effect of the following: (a) increase in flywheel weight, con- 
stant speed; (b) decrease under same conditions; (c) increase in speed with 
same flywheel; (d) increase and decrease in both weight and speed; all with 
reference to a gas engine. 

2. What would be the shape of the E-J diagram for a geared punch, 
neglecting the effect of the reciprocating head? 

3. If the connecting rod and piston of an engine are neglected, what would 
be the shape of the E-J curve? What would be its dimensions in the two 
examples of the chapter? 

4. What would be the best speed for the steam engine given in the text 
when driving the three different machines? At what mean speed would the 
engine stop in the three cases? 

6. What would be the best speed for the gas engine and at what mean 
speed would it stop? 

6. What flywheel weight would reduce the speed variation 5 per cent, for 
the steam engine? 

7. Examine the effect on the E-J diagram for the engine-driven com- 
pressor if a crank for operating the latter is set at 90 to the engine crank. 
What effect has this on the best speed? 



224. General Effects of Accelerations. It has become a prac- 
tice in modern machinery to operate it at as high a speed as possi- 
ble in order to increase its output. Where the machines con- 
tain parts that are not moving at a uniform speed, such as the 
connecting rod of an engine or the swinging jaw of a rock crusher, 
the variable nature of the motion requires alternate acceleration 
and retardation of these parts, to produce which forces are re- 
quired. These alternate accelerations and retardations cause 
vibrations in the machine and disturb its equilibrium; almost 
everyone is familiar with the vibrations in a motor boat with a 
single-cylinder engine, and many law-suits have resulted from 
the vibrations in buildings caused by machinery in shops and 
factories nearby. 

These vibrations are very largely due to the irregular motions 
of the parts and to the accelerating forces due to this, and the 
forces increase much more rapidly than the speed, so that with 
high-speed machinery the determination of these forces becomes 
of prime importance, and they are, indeed, also to be reckoned 
with in slow-speed machinery, as there are not a few cases of 
slow-running machines where the accelerating forces have caused 
such disturbances as to prevent the owners operating them. 

Again, in prime movers such as reciprocating engines of all 
classes, the effective turning moment on the crankshaft is much 
modified by the forces necessary to accelerate the parts; in some 
cases these forces are so great that the fluid pressure in the cylin- 
der will not overcome them and the flywheel has to be drawn 
upon for assistance. The troubles are particularly aggravated 
in engines of high rotative speed and appear in a most marked 
way in the high-speed steam engine and in the gasoline engines 
used in automobiles. 

The forces required to accelerate the valves of automobile 
engines may also be so great that the valve will not always re- 
main in contact with its cam but will alternately leave it and re- 
turn again, thus causing very noisy and unsatisfactory operation. 




Specific problems involving the considerations outlined above 
will be dealt with later but before such problems can be solved 
it will be necessary to devise a means of finding the accelerations 
of the parts in as convenient and simple a way as possible, and 
this will now be discussed. 

225. The Acceleration of Bodies. The general problem of accel- 
eration in space has not much application in machinery, so that the 
investigation will here be confined to a body moving in one plane, 
which will cover most practical cases. Let a body having weight 

w Ib. move in a plane at any instant; its mass will be m = and 


by the principle of the virtual center as outlined in Chapter III, 
its motion is equivalent at any instant to one of rotation about 
a point in the plane of motion, which point may be near or remote 
according to the nature of the motion; if the point is infinitely 
distant the body moves in a straight line, or has a motion of 

226. Normal and Tangential Acceleration. Let Fig. 163 repre- 

sent a body moving in the plane of 
the paper and let be its virtual 
center relative to the paper, being 
thus the point about which the body 
is turning at the instant. The body 
is also assumed to be turning with 
variable speed, but at the instant when 
it is passing through the position 
shown let its angular velocity be w 
radians per second. Any point P in 
the body will travel in a direction 
normal to OP, Sec. 34, in the sense 
indicated, as this corresponds with 
the sense of the angular velocity. 
This point P has accelerations in two 
directions : (a) Since the body is mov- 
ing about at variable angular velocity it will have an acceler- 
ation in the direction of its motion, that is, normal to OP] and 
(6) it will have an acceleration toward even if o> is constant, 
since the point is being forced to move in a circle instead of a 
straight line. The first of these may be called the tangential 
acceleration of the point since it is the acceleration of the point 
along a tangent to its path, while the second is its normal acceler- 


ation for similar reasons. Every point in a body rotating at a 
given instant has normal acceleration, no matter what kind of 
motion the body has, but it will only have tangential acceleration 
if the angular velocity of the body is variable. If the body has 
a motion of translation it can only have tangential acceleration. 
In Fig. 164 let OP be drawn separately, its length being r ft. 
and at the time 8t sec. later let OP be in the position OQ, the angle 
QOP being dd, so that the body has turned through the angle 50 
radians in dt sec. The angular velocity when in the position 
OP is co radians per second and in the position OQ is assumed to 
be co + 5co radians per second; thus the gain in angular velocity 
is 5 co radians per second in the time dt, or the angular acceleration 

of the body is a = radians per second per second. Now 


draw the corresponding velocity triangle as shown on the right 

FIG. 164. 

of Fig. 164, making SM normal to OP, equal to OP X o> = rco 
ft. per second and SN normal to OQ equal to OP(co + 5co) = 
r( + 5co) ft. per second, so that the gain in linear velocity in the 
time dt sec. is MN ft. per second and its components in the nor- 
mal and tangential directions are MR and RN ft. per second 

The normal gain in velocity of the point P in the time 8t is 
MR, so that its normal acceleration is 

MR rco50 50 

"N = -jr ^7- = rco = rco 2 ft. per second per second, 
ot ot ot 

and similarly the tangential acceleration is 

p r = ** = 


per second. 

NR SN-SR r(co + 5co) - rco 5co 

~ ~ =T=roL ft ' per second 


The sense of P T is determined by that of a while P N is always 
radially inward toward the center 0. Thus, the normal accelera- 
tion of the point is simply its instantaneous radius of rotation 
multiplied by the square of the angular velocity of its link while 
the tangential acceleration of the point is the radius of rotation 
of the point multiplied by the angular acceleration of its link. 
Where the link turns with uniform velocity a = and therefore 
P T = 0, but P N can only be zero if to is zero, which means that the 
link is at rest or has a motion of translation. In the latter case 
the link can only have tangential acceleration. 

227. Graphical Construction. Returning now to Fig. 163, the 
normal acceleration of P or P N is rco 2 toward 0, then take the 
length OP to represent this quantity, thus adopting the scale of 
a? 2 : 1 ; this is negative since the line OP represents the accelera- 
tion rco 2 in the direction and sense PO. Then the tangential 
acceleration P T will be represented by a line normal to OP, its 


length will be - 2 since the scale is co 2 : 1, and its sense is to the 


right, since the scale is negative, hence draw PP" = - ^ Now if 

be joined to P" then OP" = vector sum OP + PP" or OP" = 
P N + P T which will therefore represent the total acceleration 
of P, that is the total acceleration of P is P"0 X <o 2 in the direc- 
tion and sense P"0. It may very easily be shown that in order 
to find the acceleration of any other point R on this body at the 
given instant it will only be necessary to locate a point R" bear- 
ing the same relation to OP" that R does to OP, the acceleration 
of R, which is represented by OR" ', being #" 2 and its direction 
and sense R"0. The acceleration of R with reference to P is 

228. Application to Machines. The accelerations may now 
be found for machines and the first case considered will be as 
general as possible, the machine being one of four links with 
four turning pairs, Fig. 165. Let the angular velocity co and the 
angular acceleration a of the selected primary link a be known, it 
is required to find the angular accelerations of the other links as 
well as the linear accelerations of different points in them. From 
the phorograph, Chapter IV, the angular velocities of the links 

b and c are o>& = 7-00 and co c = to respectively, and these may 

U C 

readily be found. Further, if 05 and a c represent the, as yet 


unknown, angular accelerations in space of b and c respectively, 
and also if Q N and Q T represent respectively the normal and 
tangential accelerations of Q with regard to P, the point about 
which b is turning relative to a, and R N and R T have the same 
significance as regards R relative to Q, then the previous para- 
graph enables the following relations to be established: 

PAT = aco 2 ; P T = act] Q N = frcob 2 ; QT = bab', RN = c 2 and R T 

= COi c . 

Using the principle of vector addition the total acceleration of 
R with regard to is the vector sum of the accelerations of R 

FIG. 165. 

with regard to Q, of Q with regard to P and of P with regard to 0. 
But as R and are stationary, the total acceleration of R with 
regard to is zero, hence, the sum of the above three accelera- 
tions is zero, or 

RT + RN + Q T + QN + PT + PN = 0, 

that is, the vector polygon made up with these accelerations as 
its sides must close, or if the polygon be started at it will end 
at also. 

Now the point P" may be located according to the method 
previously given, and in order to locate Q", giving the total 
acceleration of Q, proceed from P" to by means of the vectors 
QN + QT + RN + RT. The direction and sense of both Qv 
and R N are known, they are respectively QP and RQ, further, the 
direction, but not the sense of Q T and of R T is known, in each case 


it is normal to the link itself, or Q T is normal to b and R T is normal 
to c. 

In order to represent the results graphically they may be put 
into the following convenient form : 

Q N = fe W6 2 = &> 2 : - X a, 2 



v/ .2 

~ A to 

and remembering that the scale for the diagram is w 2 :!, draw 

QN b' 2 R N c' 2 

P" A = f = -j- and follow it with AB = * = , the negative 

sign having been taken into account by the sense in which these 
are drawn. The polygon from B to may now be completed 
by adding the vectors Q T and R T , and as the directions of these 
are known, the process is evidently to draw from the line OC 
in the direction R T , that is normal to c, and from B the line BC 
normal to 6, which is in the direction of Q T , these lines inter- 
secting at the point C. Then it is evident that BC represents 
Q T on the scale co 2 to 1, and that OC represents R T on the same 
scale, so that in the diagram OPP"AECQ"0 it follows that 
OP = P Nj PP" = P T , P"A = Q N , AB = R N , BC = Q T and CO 
= R T , all on the scale co 2 to 1. Complete the parallelogram 
CBAQ"; then OP" = P N + PT, P"Q" = Q + QT and Q"0 = 
RN + RT, and therefore, the vector triangle OP"Q!'R" gives the 
vector acceleration diagram of all the links on the machine. 

229. Acceleration of Points. The linear acceleration of any 
point such as G on b is readily shown to be represented by OG" 
and to be equal to G"0.u 2 , where the point G" divides P"Q" in 
the same way that G divides PQ, the direction and sense of the 
acceleration of G is G"0. Similarly, the acceleration of H in 
c is H"0.u 2 in magnitude, direction and sense where H" divides 
OQ"(R"Q") in the same way as H divides RQ. In this way the 
linear acceleration of any point on a machine may be directly 

Angular Accelerations of the Links. The angular accelera- 
tions of the links may be found as follows. Since Q T = AQ" X co 2 

= ba b , then ba b = - AQ".u* or - a b = AQ" X ^- so that the 


length AQ" represents a b , the angular acceleration of the link 


6, and similarly CO represents the angular acceleration <x c of 

c or <x c = CO X The sense of these angular accelerations 

may be found by noticing the way one turns to them in going 
from the corresponding normal acceleration line; thus, in going 
from PN to P T one turns to the right, in going from Q N (P"A) to 
Q T (AQ") the turn is to the left and hence o: fe is in opposite sense 
to a, and by a similar process of reasoning a c is in the same sense 
as a. Thus, in the position shown in the diagram, Fig. 165, the 
angular velocities are increasing for the links a and c, and that 
of the link b is also increasing since both a& and w& are in opposite 
sense to a and co. 

It will be found that the method described may be applied to 
any machine no matter how complicated, and with comparative 
ease. The construction resembles the phorograph of Chapter 
IV, which it employs, and hence this latter chapter must be care- 

6' 2 
fully read. Simple graphical methods for finding -=--, etc., may 

be made up, one of which is shown in the applications given 


230. The real object of determining the accelerations of points 
and links in a machine is for the purpose of finding the forces 
which must be applied on the machine parts in order to produce 
these accelerations and also to learn the disturbing effects pro- 
duced in the machine if the accelerations of the parts are not 
balanced in some way. The investigation of these disturbing 
effects will now be undertaken, the first matter dealt with being 
the forces which must be applied to the links to produce the 
changing velocities. 

It is shown in books on dynamics, that if a body having plane 

motion, has a weight w Ib. or mass m = and an acceleration of 


its center of gravity of / ft. per second per second, then the force 
necessary to produce this acceleration is mf pds., and this force 
must act through the center of gravity and in the direction of the 
acceleration /. In many cases the body also rotates with variable 
angular velocity, or with angular acceleration, in which case a 
torque must act on the body in any position to produce this 
variable rotary motion, and if the body has a moment of inertia 



I about its center of gravity and angular acceleration a radians 
per second per second this torque must have a magnitude of 
I X a ft.-pds. Let the mass of the link be so distributed that its 
radius of gyration about the center of gravity is k; then I = 
mk 2 and the torque is mk 2 a. For proof of this the reader is 
referred to books on dynamics. 

To take a specific case let a machine with four links be selected, 
as illustrated in Fig. 166, and let the vector acceleration diagram 

FIG. 166. Disturbing forces due to mass of rod. 

OP"Q"0, as well as the phorograph OP'Q'O be found, as already 
explained; it is required to find the force which must be exerted 
on any link such as b to produce the motion which it has in the 
given position. Let G be the center of gravity of the link and 
let its weight be Wb lb. and its moment of inertia about G be 
represented by Ib in feet and pound units; then 7& = mbkb 2 where 

mb = and kb is the radius of gyration about the point G. From 


the vector diagram it is assumed that the angular acceleration 
otb has been found; also the acceleration of G, which is G"0 X co 2 . 
To produce the acceleration of G a force must act through it 
of amount F = m X G"0 X co 2 in the direction and sense G"0, 
while to produce the angular acceleration a torque T must act 
on the link of amount T = /&o& = m^k^a*. The torque T may 


be produced by a couple consisting of two parallel forces ac-ting 
in opposite sense and at proper distance apart, and these forces 
may have any desired magnitude so long as their distance apart 
is adjusted to suit. For convenience let each of the forces be 
selected equal to F; then the distance x ft. between them will 
be found from the relation T = Fx. 

Now, as this couple may act in any position on the link b let 
it be so placed that one of the forces passes through G and the 
two forces have the same direction as the acceleration of G. 
Further, let the force passing through G be the one which acts 
in opposite sense to the accelerating force F', this is shown on 
Fig. 166. Now the accelerating force F and one of the forces 
F composing the couple act through G and balance one another 
and thus the accelerating force and the couple producing the 
torque reduce to a single force F whose magnitude is ra&.G"0.a> 2 , 
whose direction and sense are the same as the acceleration of the 
center of gravity G of b, and which acts at a distance x from G, 
determined by the relation T = Fx } and on that side of G which 
makes the torque act in the same sense as the angular accelera- 
tion a&. 

The distance x of the force F from G may be found as follows : 


Since Q T = ba b = Q"A X co 2 , Fig. 165, then a b = Q"A X ~, because 


the line AQ" represents QT on a scale co 2 : 1. 

Also T = I b a b = m b k b 2 X^V- X co 2 


and F = m b .G"0.u\ 


T _ 

therefore *- -^r^ -^-where -^- is a 

constant, so that x = const. X ^7777 which ratio can readily be 

Or U 

found for any position of the mechanism. This gives the line of 
action of the single force F and, having found the position of the 
force, let M be its point of intersection with the axis of link b. 
Now find M' the image o f M and move the force from M to its 
image M'] then the turning moment necessary on the link a to ac- 
celerate the link b is Fh, where h is the shortest distance from 
to the direction of F, Fig. 166. 



This completes the problem, giving the force acting on the link 
and also the turning moment at the link a necessary to produce 
this force. The same construction may be applied to each of 
the other links such as a and c and thus the turning moment on a 
necessary to accelerate the links may be found as well as the 
necessary force on each link itself. 


231. The results just obtained may be used to find the bending 
moment produced in any link at any instant due to its inertia. 






, dm 


b c 


3 ( 

; ( 

FIG. 167. Bending forces on rod due to its inertia. 

Any part such as the connecting rod of an engine is subject to 
stresses due to the transmission of the pressure from the piston 
to the crankpin, but in addition to this the rod is continually 
being accelerated and retarded, these changes of velocity pro- 
ducing bending stresses in the rod, and these latter stresses may 
now be determined. 

To make the case as general as possible, let OPQR, Fig. 167, 
represent a machine for which the vector acceleration diagram is 
OP"Q"0, it is required to find the bending moment in the rod b 
due to its inertia. Lay off at each point on b the acceleration of 
that point; thus make PA\, GCi, QBi, etc., equal and parallel re- 
spectively to OP", OG", QO", etc., obtaining in this way the 
curve AiCiBi. 

Now resolve the accelerations at each point in b into two parts, 


one normal to b and the other parallel to the link. Thus PA is 
the acceleration of P normal to 6, and GC and QB are the corre- 
sponding accelerations for the points G and Q respectively. In 
this way a second curve ACB may be drawn, and the perpen- 
dicular to b drawn from any point in it to the line ACB represents 
the acceleration at the given point in b in the direction normal 
to the axis of the latter, the scale in all cases being co 2 : 1. 
Thus the acceleration of P normal to b is AP.w 2 , and so for other 

Now let the rod be placed as shown on the right-hand side of 
Fig. 167 with the acceleration curve ACB above it to scale. 
Imagine the rod divided up into equal short lengths one of which 

is shown at D, having a weight 5w Ib. and mass dm = , and 

let the normal acceleration at this point be represented by DE. 
Should the rod be of uniform section throughout its length all 
the small masses like dm will be equal since all will be of the same 
weight dw, but if the rod is larger at the left-hand end than at 
the right-hand end, then the values of dm will decrease in going 
along from P to Q. Now tlie force due to the acceleration of 
the small mass is equal to dm multiplied by the acceleration 
corresponding to DE and this force may be set off along DE 
above D. Proceeding in this way for the entire length of the 
rod gives the dotted curve as shown which may be looked upon as 
the load curve for the rod due to its acceleration. From this 
load curve the bending moments and stresses in the rod may be 
determined by the well-known methods used in statics. 

For a rod of uniform cross-section throughout the acceleration 
curve ACB will also be a load curve to a properly selected scale, 
but with the ordinary rods of varying section the work is rather 
longer. In carrying it out, the designer usually soon finds out 
by experience the position of the mechanism which corresponds 
to the highest position of the acceleration curve ACB, and the 
accelerations being the maximum for this position the rod is 
designed to suit them. A very few trials enable this position 
to be quickly found for any mechanism with which one is not 

The process must, of course, be carried out on the drafting 

232. To Find the Accelerations of the Various Parts of a Rock 
Crusher. In order to get a clearer grasp of the principles in- 



volved, a few applications will be made, the first case being that 
of the rock crusher shown in Fig. 168. The mechanism of the 
crusher is shown on the left and has not been drawn closely to 
scale as the construction is more clear lor the proportions shown. 
A crank OP is driven at uniform speed by a belt pulley on the 
shaft and to this crank is attached the long connecting rod PQ. 
The swinging jaw of the crusher is pivoted to the frame at T 
and connected to PQ by the rod SQ, while another rod QR is 
pivoted to the frame at R. As OP revolves Q swings in an arc 
of a circle about R } giving the jaw a swinging motion about T 


FIG. 168. Rock crusher. 

and crushing between the jaw and the frame any rocks falling 
into the space. In large crushers the jaw is very heavy and its 
variable velocity, or acceleration, sometimes sets up very serious 
vibrations in buildings in which it is placed. 1 

The acceleration diagram is shown on the right and there is 
also drawn the upper end of the rod b and the whole of the crank 
a. It is to be noted that the actual mechanism may be drawn 
to as small a scale as desired and the diagram to the right to as 
large a scale as is necessary, because in the phorograph and the 
acceleration diagram only the directions of the links are required 
and these may be easily obtained from the small scale drawing 

1 See article by PROF. O. P. HOOD in American Machinist, Nov. 26, 1908. 


shown. The phorograph of the mechanism and acceleration dia- 
gram should give no difficulty because the mechanism is simply 
a combination of two four link mechanisms, OPQR and RQST 
exactly similar to that shown in Fig. 165 and already dealt with. 
The crank OP has been chosen as the primary link. 

The crank OP is assumed to turn at uniform speed of co radians 
per second. For the phorograph, OQ' parallel to RQ meeting 6 
produced gives Q' and P'Q' = V and OQ' = c'; further OS' par- 
allel to ST meeting Q'S' parallel to QS gives S' and Q'S' = e' 
while S'O gives /'. The points R' and T r lie at 0. 

For the acceleration diagram P" lies at P since a is assumed to 
run at uniform speed; then, following the method already de- 


scribed in Sec. 228, lay off P"A = -j- and AB parallel to c and 

c / 2 

of length AB = , and finish the diagram by making BC per- 

pendicular to 6 and OC perpendicular to c, these intersecting 
at C. Complete the parallelogram ABCQ" and join P"Q" and 
Q"0; then in the acceleration diagram OP" = a", P"Q" = V 
and Q"0 = c" which gives the vector acceleration diagram for 
the part OPQR. Then starting at Q", which gives the accelera- 

c / 2 
tion of Q on the vector diagram, draw Q"D = and parallel 

T' S /2 
to e; this is followed by DE parallel to TS and of length ~~- = 


' , and the vector diagram is closed by drawing EF perpendicular 

to c to meet OF perpendicular to'/ in F. On completing the par- 
allelogram DEFS", the point S" is found and then S" is joined 
to and to Q". The line S"Q" represents e on the acceleration 
diagram while OS" = f represents / on the same figure. 

The length OS" represents the acceleration of S on a scale of 
co 2 : 1 and the acceleration of any other point on / is found by 
locating on OS" or R"S" a point similarly situated to the desired 
point on ST. If the angular acceleration of the jaw is required v 
it may be found as described at Sec. 229 and evidently is a/ = 

~ FOX T 

Calling G the center of gravity of the jaw / and locating G" 
in the same way with regard to S"T" that G is located with regard 




to ST, the acceleration of G is G"0 X co 2 and the force required 
to cause this acceleration and therefore shaking the machine is 

parallel to G"0 and is equal to G"0 X co 2 X 

weight of jaw 


Or the torque required for the purpose is // X a/ where // is the 
moment of inertia of / with regard to G. 

233. Application to the Engine. This construction and the 
determination of the accelerations and forces has a very useful 
application in the case of the reciprocating engine and this ma- 
chine will now be taken up. Fig. 169 represents an engine in 

FIG. 169. 

which is the crankshaft, P the crankpin and Q the wristpin, the 
block c representing the crosshead, piston and piston rod. Let 
the crank turn with angular velocity co radians per second and 
have an acceleration a in the sense shown, and let G be the center 
of gravity of the connecting rod b. To get the vector accelera- 
tion diagram find P" exactly as in the former construction, OP 
representing the acceleration 2 and PP" the acceleration aa, 
both on the scale -co 2 to 1. 

Now the motion of Q is one of sliding and thus Q has only 
tangential acceleration, or acceleration in the direction of sliding, 
in this case QS, the sense being determined later. Hence, the 
total acceleration of Q must be represented by a line through O 
in the direction, QS therefore Q" lies on a line through the center 
of the crankshaft, and the diagram is reduced to a simpler form 


than in the more general case. Having found P", draw P" 'A 


parallel to 6, of length -j-> to represent Q N , and also draw AQ", 

normal to P"A, to meet the line Q"0, which is parallel to QS, 
in Q". Then will AQ" represent the value of the angular accel- 


eration of the rod 6. Since bab = Q" 2 or a& = Q"A.j- } and 

since AQ" lies on the same side of P"A that PP" does of OP, 
therefore a& is in the same sense as a] thus since co& is opposite to 
o>, the angular velocity of the rod is decreasing, or the rod is being 

The acceleration of the center of gravity of b is represented by 
OG" and is equal to(r" 2 , and similarly the acceleration of the 
end Q of the rod is represented by OQ" and is equal to <2"0.o> 2 , 
this being also the acceleration of the piston. 

It will be observed that all of these accelerations increase as 
the square of the number of revolutions per minute of the crank- 
shaft, so that while in slow-speed engines the inertia forces may 
not produce any very serious troubles, yet in high-speed engines 
they are very important and in the case of such engines as are 
used on automobiles, which run at as high speeds as 1,500 revo- 
lutions per minute, these accelerations are very large and the 
forces necessary to produce them cause considerable disturb- 
ances. Take the piston for example, the force required to move 
it will depend on the product of its weight and its acceleration, 
so that if an engine ran normally at 750 revolutions per minute 
and then it was afterward decided to speed it up to 1,500 revo- 
lutions per minute, the force required to move the piston in any 
position in the latter case would be four times as great as in the 
former case. 

234. Approximate Construction. In the actual case of the 
engine, the calculations may be very much simplified owing to 
certain limitations which are imposed on all designs of engines 
driving other machinery, these limitations being briefly that the 
variations in velocity of the flywheel must be comparatively 
small, that is, the angular acceleration of the flywheel must not 
be great, and in fact, on engines the flywheels are made so heavy 
that a. cannot be large. 

To get a definite idea on this subject a case was worked out for 
a 10 by 10-in. steam engine, running at 310 revolutions per min- 
ute, and the maximum angular acceleration of the crank was 



found to be slightly less than 7 radians per second per second. 
For this case the normal acceleration of P is rco 2 = ^{ 2 X 1,100 = 
458 ft. per second per second, while the tangential acceleration 

is ra = y~ X 7 = 5.8 ft. per second per second, which is very 

small compared with 458 ft. per second per second, so that on 
any ordinary drawing the point P" would be very close to P. 
Thus without serious error ra. may be neglected compared with 
rcc 2 and hence P" is at P. 

With the foregoing modification for the engine, the complete 
acceleration diagram is shown at Fig. 170, the length PA repre- 


FIG. 170. Piston acceleration. 

sen ting -j- and AQ" is normal to b, thus P"Q" is the acceleration 

diagram for the connecting rod and OQ " represents the accelera- 
tion of the piston on the scale -co 2 to 1. Two cases are shown: 
(a) for the ordinary construction; and (&) for the offset cylinder. 
The acceleration of any such point as G is found by finding G", 
making the line GG" parallel to QQ", the acceleration then is 
G" 2 . 

It should be noticed that the greater the ratio of 6 to a, that 
is the longer the connecting rod for a given crank radius, the 
more nearly will the point A approach to P because the distance 


PA represents the ratio -v- and this steadily decreases as b in- 

creases, and at the same time AQ" becomes more nearly vertical. 
In the extreme case of an infinitely long rod, carried out practi- 
cally as shown at Fig. 6, the point A coincides with P and 


is vertical and then the acceleration of the piston which is OQ" 
is simply the projection of a on the line of the piston travel or 
the acceleration Q"0 X co 2 = a . cos 6 X co 2 where is the crank 
angle POQ". 

235. Piston Acceleration at Certain Points. Taking the more 
common form of the mechanism shown at Fig. 170 (a) the num- 
erical values of the acceleration of the piston may be found in a 
few special cases. When the crank is vertical, b f is zero and there- 
fore A is at P vertically above 0, so that when AQ" is drawn, 
Q" lies to the left of showing that the piston has negative ac- 
celeration or is being retarded. For this position a circle of diam- 
eter QQ" will pass through P and therefore Q"0 X OQ = OP 2 or 

OP 2 a 2 

Q"0 = ~7^ = jr^ and the acceleration of the piston is 


o> 2 X /ry~^^2 ^' P er secon d P er second. 

fr' 2 a 2 

At both the dead centers b f = a hence P"A = -7- = ~r> so 

6 6 

a 2 

that for the head end, Q"0 = a + -r- and the piston has its maxi- 
mum acceleration at this point, which is (a + ~r ] <o 2 toward O, 
while for the crank end, Q"O = a -- -r and the acceleration is 

(a j-\ co 2 toward 0, or the piston is being retarded. 

Example. Let an engine with 7-in. stroke and a connecting 
rod 18 in. long run at 525 revolutions per minute. Then a = 

3/^ 18 

-TTT = 0.29 ft., b = -r~o = 1.5 ft., and co = 55 radians per second, 
-i-^j \.2i 

At the head end the acceleration of the piston would be: 

(o> 2 \ I 29 2 \ 

a+ -T-) co 2 = (0.29 + -4-=-) X 55 2 = 931 ft. per second per 
/ \ l.O / 


At the crank end the acceleration would be: 

(a - ~) co 2 = (o.29 - -J-R-) X 55 2 = 623 ft. per second per 


At the time when the crank is vertical the result is: 

X 55 2 = 173 ft. per second per 



The angular acceleration of the connecting rod, being deter- 
mined by the length AQ", is zero at each of the dead points but 
when the crank is vertical it has nearly its maximum value; the 

formula for it is Q"A X y. When the crank is vertical a dia- 
gram will show that 



and the acceleration is 

Vb 2 - a 2 

For the engine already examined, when in this position, 

[0 29 2 1 
= 55 2 = 596 radians per second per second. 
Vl.5 2 - 0.29 2 J 

236. In the approximate method already described, in which 

FIG. 171. 

the angular acceleration of the crankshaft is neglected and P" 
is assumed to coincide with P, it will be noticed that the length 


P"A = -T-, is laid off along the connecting rod, the length P'Q' 


representing &', and PQ the length b, and then AQ" is drawn per- 
pendicular to PQ. This may be carried out by a very simple 
graphical method as follows: With center P and radius P'Q' 
= b' describe a circle, Fig. 171, also describe a second circle, 
having the connecting rod b as its diameter, cutting the first 
circle at M and N, and join MN. Where MN cuts b locates 
the point A and where it cuts the line through in the direction 
of motion of Q gives Q". 

The proof is that PMQ being the angle in a semicircle is a right 
angle also the chord MN is normal to PQ and is bisected at A. 
Then in the circle MPNQ there are two chords PQ and MN inter- 
secting at A, and hence from geometry it is known that: 

PA.AQ = MA.AN = MA 2 


PA(PQ - PA) = MP 2 - PA 2 = b' - PA 2 . 

Multiplying out the left-hand side and cancelling 

PA.PQ = b' 2 
that is 

PQ~ b 

which proves the construction. 


In order to accelerate or retard the various parts of the engine, 
some torque must be required or will be produced at the crank- 
shaft, and a study of this will now be taken up in detail. 

237. (a) The effect produced by the piston. 

By the construction already described the acceleration of the 
piston is readily found and it will be seen that Q" lies first on the 
cylinder side of and then on the opposite side. When Q" lies 
between and Q, Fig. 172, then the acceleration of the piston is 
Q"O X co 2 , and the acceleration of the piston is in the same sense 
as the motion of the piston, or the piston is being accelerated. 
Conversely, when Q" lies on QO produced the acceleration being 
in the opposite sense to the motion of the piston, the latter is 
being retarded. These statements apply to the motion of the 
piston from right to left, when the sense of motion of the 
piston reverses the remarks about the acceleration must also 
be changed. If now the accelerations for the different 



piston positions on the forward stroke be plotted, the diagram 
EJH will be obtained, Fig. 172, where the part of the diagram 
EJ represents positive accelerations of the piston, and the part 
JH negative accelerations, or retardations. The corresponding 
diagram for the return stroke of the piston is omitted to avoid 


FIG. 172. Acceleration of piston on forward stroke. 

Let the combined weight of the piston, piston rod and cross- 
head be w c pounds, the corresponding mass being m c = , and 


let / represent the acceleration of the piston at any instant ; then 
the force P c necessary to produce this acceleration is P c = ra c ./. 

FIG. 173. Modification of diagram due to inertia of piston. 

This force will be positive if / is positive and vice versa, that is, if / 
is positive a force must be exerted on the piston in its direction of 
motion and if it is negative the force must be opposed to the 
motion. In the first case energy must be supplied by the flywheel, 
or steam, or gas pressure, to speed up the piston, whereas, in the 
latter case, energy will be given up to the flywheel due to the 


decreasing velocity of the piston. Since no net energy is received 
during the operation, therefore, the work done on the piston 
in accelerating it must be equal to that done by the piston while 
it is being retarded. 

Two methods are employed for finding the turning effect of this 

force, P c ] the first is to reduce it to an equivalent amount per 


square inch of piston area by the formula p c = -r where A is 


the area of the piston, and then to correct the corresponding pres- 
sures shown by the indicator diagram by this amount. In this 
way a reduced indicator diagram for each end is found, as shown 
for a steam engine in Fig. 173, where the dotted diagram is the 
reduced diagram found by subtracting the quantity p c from the 
upper line on each diagram. The remaining area is the part 
effective in producing a turning moment on the crankshaft. 

The second method is to find directly the turning effect neces- 
sary on the crankshaft to overcome the force P c , and from the 
principles of the phorograph this torque is evidently T c = 
PC X OQ' = m c X / X OQ'. In the position shown in Fig. 
172, P e would act as shown, and a torque acting in the same 
sense as the motion of a would have to be applied. 

The first method is very instructive in that it shows that the 
force necessary to accelerate the piston at the beginning of the 
stroke in very high-speed engines may be greater than that pro- 
duced by the steam or gas pressure, and hence, that in such cases 
the connecting rod may be in tension at the beginning of the 
stroke, but, of course, before the stroke has very much proceeded 
it is in compression again. This change in the condition of stress 
in the rod frequently causes " pounding" due to the slight slack- 
ness allowed at the various pins. 

238. (6) The Effect Produced by the Connecting Rod. 
This effect is rather more difficult to deal with on account of 
the nature of the motion of the rod. The resultant force acting 
may, however, be found by the method described earlier in the 
chapter, Sec. 230, but in the case of the engine, the construction 
may be much simplified, and on account of the importance of the 
problem the simpler method will be described here. It consists 
in dividing the rod up into two equivalent concentrated masses, 
one at the crosshead pin the other at a point to be determined. 

Referring to Fig. 174, the rod is represented on the acceleration 
diagram by P"Q" and the acceleration of any point on it or the 



angular acceleration of the rod may be found by processes 
already explained. Let h be the moment of inertia of the rod 
about its center of gravity, k b being the corresponding radius of 
gyration and nib the mass, so that 7& = ra&fcb 2 , and let the center 
of gravity G lie on PQ at distance TI from Q. Instead of consider- 
ing the actual rod it is possible to substitute for it two concen- 
trated masses mi and m 2 , which, if properly placed, and if of 
proper weight, will have the same inertia and weight as the 

original rod. Let these masses be nil and m 2 where m\. = 

and m 2 = 'in which w\ and w 2 are the weights of the masses in 
pounds. Further, let mass m\ be concentrated at Q, it is required 

FIG. 174. 

to find the weights Wi and w 2 and the position of the weight w 2 . 
Let r 2 be the distance from the center of gravity of the rod to 
mass m 2 . 

These masses are determined by the following three conditions : 

1. The sum of the weights of the two masses must be equal 
to the weight of the rod, that is, Wi + w 2 = Wb, or nil + m 2 = nib. 

2. The two masses m\ and m 2 , must have their combined 
center of gravity in the same place as before; therefore, m\r\ = 
m 2 r 2 . 

3. The two masses must have the same moment of inertia 
about their combined center of gravity G as the original rod has 
about the same point; hence 

-\- m 2 r 2 2 = 



For convenience these are assembled here: 
nil -\- m 2 = mb 

= m 2 r 2 



Solving these gives: 

mi = m b X ~ and w 2 = m b X - 

7*1 -r 7*2 r i ~r r 2 

7 2 ^ 2 

and rir 2 = fc & 2 or r 2 = 

Thus, for the purposes of this problem the whole rod may be 
replaced by the two masses mi and w 2 placed as shown in Fig. 174. 
The one mass mi merely has the same effect as an increase in the 
weight of the piston and the method of finding the force required 
to accelerate it has already been described. Turning then to the 
mass ra 2 , which is at a fixed distance r 2 from G] the center of grav- 
ity of m 2 is K and the acceleration of K is evidently K"0 X co 2 , 
K"K being parallel to G"G. The direction of the force acting 
on w 2 is the same as that of the acceleration of its center of gravity 
and is therefore parallel to K /f O, and the magnitude of this force 
is w 2 X K"0 X co 2 . The force acts through K, its line of action 
being KL parallel to K"0. 

The whole rod may now be replaced by the two masses m\ 
and w 2 . The force. acting on the former is m\ X Q"0 X co 2 
through Q parallel to Q"0, that is, this force is in the direction of 
motion of Q and passes through L on Q"0. The force on the 
mass ra 2 is ra 2 X K fr O X co 2 , which also passes through L, so 
that the resultant force F acting on the rod must also pass through 
L. Thus the construction just described gives a convenient 
graphical method for locating one point L on the line of action 
of the resultant force F acting on the connecting rod. 

Having found the point L the direction of the force F has been 
already shown to be parallel to G"0 and its magnitude is 
m b X G"0 X co 2 . Let F intersect the axis of the rod at H, find 
the image H f of H, and transfer F to H 1 '. The moment required 
to produce the acceleration of the rod is then Fh. 

A number of trials on different forms and proportions of en- 
gines have shown that the point L remains in the same position 
for all crank angles, and hence if this is determined once for a 
given engine it will be only necessary to determine G"0 for the 
different crank positions; as this enables the magnitude and 
direction of F to be found and its position is fixed by the 
point L. 

239. Net Turning Moment on Crankshaft. For the position 
of the machine shown in Fig. 174, let P be the total pressure on 


the piston due to the gas or steam pressure; then the net turning 
moment acting on the crankshaft is 

P X OQ' - [m c X Q"0 X co 2 X OQ f + m b X G"0 X w 2 X h] 

after allowance has been made for the inertia of the piston 
and connecting rod. This turning moment will produce an ac- 
celeration or retardation of the flywheel according as it exceeds 
or is less than the torque necessary to deliver the output. 

All of these quantities have been determined for the complete 
revolution of a steam engine and the results are given and dis- 
cussed at the end of the present chapter. 

240. The Forces Acting at the Bearings. The methods de- 
scribed enable the pressures acting on the bearings due to the 
inertia forces to be easily determined, and this problem is left for 
the reader to solve for himself. 

In high-speed machinery the pressures on the bearings due 
to the inertia of the parts may become very great indeed and all 
care is taken by designers to decrease them. Thus, in automobile 
engines, some of which attain as high a speed as 3,000 revolutions 
per minute, or over, during test conditions, the rods are made as 
light as possible and the pistons are made of aluminum alloy in 
order to decrease their weight. In one of the recent automobile 
engines of 3-in. bore and 5-in. stroke the piston weighs 17 oz. 
and the force necessary to accelerate the piston at the end of the 
stroke and at a speed of 3,000 revolutions per minute is over 800 
pds., corresponding to an average pressure of over 110 pds. per 
square inch on the piston and the effect of the connecting rod 
would increase this approximately 50 per cent; thus during the 
suction stroke the tension in the rod is over 1,200 pds. at the head- 
end dead center and the compressive stress in the rod is much 
less than that corresponding to the gas pressure. At the crank- 
end dead center the accelerating force is also high, though less 
than at the head end, and here also the rod is in compression due 
to the inertia forces. If the gas pressure alone were considered, 
the rod would be in compression in all but the suction stroke. 

241. Computation on an Actual Steam Engine. In order 
that the methods may be clearly understood an example is worked 
out here of an engine running at 525 revolutions per minute, 
and of the vertical, cross-compound type with cranks at 180, 
and developing 125 hp. at full load. Both cylinders are 7 in. 
stroke and 11 in. and 15 J^ in. diameter for the high- and low- 



pressure sides respectively. The weight of each set of recip- 
rocating parts including piston, piston rod and crosshead is 
161 lb., while the connecting rod weighs 47 Ib. has a length 
between centers of 18 in. and its radius of gyration about its cen- 

FIG. 175. 

ter of gravity is 7.56 in., the latter point being located 13.3 in. 
from the center of the wristpin. 

From the above data o> = 55 radians per second, 


5, m b = 

= 1.46 and 7* = - = 0.63 ft. 

Alsor 1 = ^- = 1.11 ft., r 2 = 


1.46 X 

0.63 2 

1.11 + 0.36 

0.36 ft. and 

- = 0.35 while m z = 1.11. 







tal Torque E< 




-Torque Requ 

red Ifor 







\ 90 1 

)8 1 

6 144 1 



\ 2 

Crank Angles 
38 306 324 342 36 


8 3 

6 5 

4 1 



B R d ol 


80 1 

J8 2 

(> 2, 

i4 2, 

U 2" 




g ^ 
















FIG. 176. Effect of connecting rod and piston. 

The construction for the crank angle 36 is shown in Fig. 175 
with all dimensions marked on, and the complete results for the 
entire revolution for one side of the engine are set down on the 
accompanying table, all of the quantities being tabulated. The 
point L for this engine is located 0.44 in. from and on the 
cylinder side of it. The table shows that at the head end a 
force of 5,262 lb. would be required to accelerate each piston 
which corresponds to a mean pressure for the high-pressure 



side of 55 Ib. per square inch., in other words if the net steam 
pressure fell below 55 Ib. at this point the high-pressure rod 
would be in tension instead of compression. 

The disturbing effect of the connecting rod is much less marked 
as the table shows, but in accurate calculations cannot be neg- 
lected. The combined effect of the two as shown in the last 
column is quite decided. 

In order that the results may be more clearly understood 
they have been plotted in Fig. 176, which shows the turning 
moment at the crankshaft required to move the piston and the 
crosshead separately, and also the combined effort required for 
both. The turning effort required for the rod is not quite one- 
twelfth that required for the piston. 





Piston, crosshead, etc. 

Connecting rod 

Total turn- 
ing moment 






G -rt 


at crank to 







move all 


^ ' 

*O N 

-^ --j 






c 2 



G v ^4 











+ 1 


fe * 





PH * 




+ 1,053 

+ 5,262 









+ 526 





+ 43 




























+ 142 

+ 711 







+ 22 

































































. 711 







+ 702 

+ 3,509 







+ 20 








































+ 181 



+ 265 
































































The relative effects of these turning moments is shown more 
clearly at Fig. 177 in which separate curves are drawn for the 
high- and low-pressure sides. The dotted curves in both cases 
show the torque due to the steam pressure found as in Chapter X, 
while the broken lines show the torque required to accelerate the 
parts and the curves in solid lines indicate the net resultant 
torque acting on the crankshaft. The reader will be at once 





* 800 

I 400 







? hPr 

essure Cyl 




. / 

t \ 






Torque Due 
^\ to Steam 








il > 

' \ 

\ i 













l \ 



270 \ 


\ > 






rque Kequirt 
ccelerate Pa 

d to 










.^x-Kesulta'ut To 









to St 

: Due, 








\ ' 














270 \ 









S s 

^^Torque Require 
Atcelenite Pa 

d to 





FIG. 177. Torque diagrams allowing for inertia of parts. 

struck with the modification produced by the inertia of the parts, 
but it must always be kept in mind that this only modifies the 
result but produces no net change, as the energy used up in 
accelerating the masses for one part of the revolution is returned 
when the masses are retarded later on in the cycle of the ma- 
chine. That these forces must be reckoned with, especially in 
high-speed machinery, is very evident. 

242. The Gnome Motor. One further illustration of the 
principles stated here may be given in the Gnome motor, which 
has had much application in aeroplane work. The general form 
of the motor has already been shown in Fig. 12 in the early 



part of this book and it has been explained that the mechanism 
is exactly the same as in the ordinary reciprocating engine 
except that the crank is fixed and the connecting rod, cylinder and 
other parts make complete revolutions. The mechanism is shown 
in Fig. 178 in which a is the cylinder and parts secured to it, 6 is 
the connecting rod and c is the piston, and power is delivered 
from- the rotating link a, which is assumed to turn at constant 
speed of co radians per second. 

At the wristpin two letters are placed, Q on the rod b and P 
on a directly below Q, and thus as the revolution proceeds P 

FiG. 178. Gnome motor. 

moves in and out along a; the motion of Q relative to P must, in 
the nature of the case, be one of sliding in the direction of a. 
The phorograph is obtained in a similar way to that for the 
Whitworth quick-return motion, Fig. 38, the only difference here 
being that the cylinder link a turns at uniform speed while in 
the former case the connecting rod did so. The image Q' lies 
on P'Q' normal to a and on R'Q' through parallel to b. 

To find the acceleration diagram the plan followed in Sec. 
228 is employed. Thus, the acceleration of R relative to Q added 
to that of Q relative to P and that of P relative to must 
be zero. Using the notation of Sec. 228 it follows that R T + 
RN + QT + QN + PT + PN = 0; and since the only acceleration 


which Q can have relative to P is tangential, it follows that Q N = 
0. Again, since a turns at a uniform speed, the value of P T is 
also zero. Hence, the result is R T + RN + QT + PN = 0. 

Now, adopting the scale of co 2 : 1, the acceleration P N = aco 2 
is represented by OP" = a and it has also been shown in Sec. 228 


that R N = -j- X co 2 , so that P" B is laid off along b and equal to 

6' 2 

-T-, and thus P"# will represent R N . The vector diagram is closed 

by R T and Q T , the former perpendicular to 6 and the latter parallel 
to a, so that BC perpendicular to b represents R T and hence CR" 
= Q T - It will make a more correct vector diagram to lay off 
p"Q" = CR" and make Q"A and AR" equal respectively to P"B 
and BC. Then OP" = P*, P"Q" = Q r , Q"A = #, AR = R T 
and Q".R" represents the rod b vectorially on the acceleration 
diagram, G" corresponding to its center of gravity G. The accel- 
eration of the center of gravity G of b is G"0 X co 2 and the an- 


gular acceleration of the rod is R"A X -j- as given in Sec. 229. 

The pull on 6 due to the centrifugal effect of the piston is Q"0 

weight of position . 

X co 2 X - ^n~n ~~ ln the direction of a. 

The resultant force F on the rod 6 may be found as in Sec. 230 
and is in the direction G"0, that is, parallel to b. Its position 
is shown on the figure and the pressure between the piston and 
cylinder due to this force is readily found knowing the value and 
position of F. 


1. A weight of 10 Ib. is attached by a rod 15 in. long to a shaft rotating at 
100 revolutions per minute; find the acceleration of the weight and the ten- 
sion in the rod. 

2. If the shaft in question 1 increases in speed to 120 revolutions per 
minute in 40 sec., find the tangential acceleration of the weight and also its 
total acceleration. 

3. A railroad train weighing 400 tons is brought to rest from 50 miles per 
hour in 1 mile. Find the average rate of retardation and the mean resistance 

4. At each end of the stroke the velocity of a piston is zero; how is its 
acceleration a maximum? 

6. Weigh and measure the parts of an automobile engine and compute 
the maximum acceleration of the parts and the piston pressure necessary to 
produce it. 


6. Find the bending stresses in the connecting rod of the same engine, due 
to inertia, when the crank and rod are at right angles. 

7. Divide the rod in question 6 up into its equivalent masses, locating one 
at the wristpin. 

8. Make a complete determination for an automobile engine of the result- 
ing torque diagram due to the indicator diagram and inertia of parts. 


243. General Discussion on Balancing. In all machines 
the parts have relative motion, as discussed in Chapter I. Some 
of the parts move at a uniform rate of speed, such as a crankshaft 
or belt-wheel or flywheel, while other parts, such as the piston, 
or shear blade or connecting rod, have variable motion. The 
motion of any of these parts may cause the machine to vibrate 
and to unduly shake its foundation or the building or vehicle 
in which it is used. It is also true that the annoyance caused by 
this vibration may be out of all proportion to the vibration 
itself, the results being so marked in some cases as to disturb 
buildings many blocks away from the place where the machine 
is. This disturbance is frequently of a very serious nature, 
sometimes forcing the abandonment of the faulty machine alto- 
gether; therefore the cause of vibration in machinery is worthy of 
careful examination. 

It is not possible in the present treatise to discuss the general 
question of vibrations, as the matter is too extensive, but it 
may be stated that one of the most common causes is lack of 
balance in different parts of the machine and the present chapter 
is devoted entirely to the problem of balancing. Where any 
of the links in a machine undergo acceleration forces are set up 
in the machine tending to shake it, and unless these forces are 
balanced, vibrations of a more or less serious nature will occur, 
but balancing need only be applied where accelerations of the 
parts occur. 

It must be borne in mind, however, that the accelerations are 
not confined solely to such parts as the piston or the connecting 
rod which have a variable motion, but the particles compos- 
ing any mass which is rotating with uniform velocity about 
a fixed center also have acceleration 1 and may throw the 
machine out of balance, because, as explained in Sec. 226, 

1 In connection with this the first part of Chapter XV should be read 
over again. 



a mass has acceleration along its path when its velocity is 
changing, and also acceleration normal to its curved path 
even when its velocity is constant. In discussing the sub- 
ject it is most convenient to divide the problem up into two 
parts, dealing first with links which rotate about a fixed center 
and second with those which have a different motion, in all 
cases plane motion being assumed. 

244. Balancing a Single Mass. Let a weight of w Ib. which has 


a mass m = ~ rotate about a shaft with a fixed center, at a fixed 


radius r ft., and let the radius have a uniform angular velocity 
of co radians per second. Then, referring to Sec. 226 this mass 
will have no acceleration along its path since co is assumed con- 
stant, but it will have an acceleration toward the axis of rotation 
of rco 2 ft. per second per second, and hence a radial force of 


amount rco 2 = mrco 2 pds. must be 


applied to it to maintain it at the 
given radius r. This force must be 
applied by the shaft to which the 
weight is attached, and as the 
weight revolves there will be a pull 
on the shaft, always in the radial 
direction of the weight, and this 
pull will thus produce an unbalanced 
FIG. 179. force on the shaft, which must be 

balanced if vibration is to be avoided. 

Let Fig. 179 represent the weight under consideration in one 
of its positions; then if vibration is to be prevented another 
weight Wi must be attached to the same shaft so that its accelera- 
tion will be always in the same direction but in opposite sense to 
that of w, and this is possible only if w\ is placed at some radius 
7*1 and diametrically opposite to w. Clearly, the relation be- 

tween the two weights and radii is given by rco 2 = rico 2 or 
rw = riWi, since is common to both sides, from which the 

product riWi is found, and having arbitrarily selected one of 
these quantities such as r if the value of Wi is easily determined. 



If the two weights are placed as explained there will be no resul- 
tant pull on the shaft during rotation, and hence no vibration; 
in other words the shaft with its weights is balanced. 

It sometimes happens that the construction prevents the 
placing of the balancing mass directly opposite to the weight w, 
as for example in the case of the crankpin of an engine, and then 
the balancing weights must be divided between two planes 
which are usually on opposite sides of the disturbing mass, 
although they may be on the same side of it if desired. Let Fig. 
180 represent the crankshaft of an engine, and let the crankpin 
correspond to an unbalanced weight w Ib. at radius r. The planes 
A and B are those in which it is possible to place counterbalance 
weights and the magnitude and position of the weights are 

FIG. 180. Crank-shaft balancing. 

required. Let the weights be w\ and w z Ib. and their radii of 
rotation be r\ and r 2 respectively; then clearly the vector sum 

Let all the 

' wr) - - 

0, or 

+ w z r 2 = wr. 

masses be in the plane containing the axis of the shaft and the 
radius r. 

Now it is not sufficient to have the relation between the 
masses and radii determined by the formula w\r\ + w 2 r z = wr 
alone, because this condition only means that the shaft will 
be in static equilibrium, or will be balanced if the shaft is sup- 
ported at rest on horizontal knife edges. When the shaft re- 
volves, however, there may be a tendency for it to "tilt" in the 
plane containing its axis and the radii of the three weights, and 
this can only be avoided by making the sum of the moments of 


CO 2 

the quantities r X w X about an axis through the shaft normal 

to the last-mentioned plane, equal to zero. 

For convenience, select the axis in the plane in which wi re- 
volves, and let a and a 2 be the respective distances of the planes 
of rotation of w and Wz from the axis; then the moment equation 

(wra WzTzdz) = or wra = 

Combining this relation with the former one 


/- a \ a 

= wr(l -- and w<>r<> = w r 

O,2/ 0, 


so that wtfi and W 2 r 2 are readily determined. 

As an example let w = 10 lb., r = 2 in., a = 4 in. and a^ = 10 

in.; then w 2 r 2 = ^f2> an 4 if r z be taken as 4 in. w 2 = = % 


= 2 lb., since the radii are to be in feet. Further, the value of 


) = 10 X TO! T7^ = 1 f rom which if 

I l-^l 1" I 



be arbitrarily chosen as 4 lb., it will have to revolve at a radius 
of y ft. or 3 in. from the shaft center. In this way the two 
weights are found in the selected planes which will balance the 

245. Balancing Any Number of Rotating Masses Located on 
Different Planes Normal to a Shaft Revolving at Uniform 
Speed. Let there be any number of masses, say four, of weights 
Wi, 102, Ws and w, rotating at respective radii ri, r 2 , r 3 and r 4 on a 
shaft with fixed axis and which is turning at w radians per second, 
the whole being as shown at Fig. 181. It is required to balance 
the arrangement. 

As before, this may be done by the use of two additional 
weights revolving with the shaft and located in two planes of 
revolution which may be arbitrarily selected; these are shown in 
the figure, the one containing the point 0, and the other at A t 
and the quantities ai, a 2 , a s , a* and a 5 represent the distances 
of the several planes of revolution from 0. 



It is convenient to use the left-hand plane, or that through 0, 
as the plane of reference and, in fact, the reference plane must 
always contain one of the unknown masses, and it will be evident 

that if the masses are balanced the vector sum - X r X co 2 must 


be zero. Further, the vector sum of the tilting moments 


w X r X a X of the various masses in planes containing the 

masses and the shaft must also be zero; otherwise, although the 
system may be in equilibrium when at rest, it will not be so while 
it is in motion. Now, since o> 2 and g are the same for all the 


[ I ] 


: i i 

03 > 

a 4 - 

2 "' *1 

FIG. 181. Balancing revolving masses. 

masses, therefore, the above equations may be reduced to the 
form: (1) vector sum of the products w X r must be zero; and (2) 
vector sum of the products wra must be zero. Since the first of 
these is the condition to be observed if the shaft is stationary, 
it may be called the static condition, while the second is the 
dynamic condition coming into play only when the shaft is 

Now the tilting moment w X r X a has a tendency to tilt the 
shaft in the plane containing r and the shaft, and it will be most 
convenient to represent it by a vector parallel to the trace of 


this plane on the plane of revolution, or what is the same thing, 
by a vector parallel with the radius r itself, and a similar method 
will be used with other tilting moments. Two balancing weights 
will be required, w at an arbitrarily selected radius r in the normal 
plane through 0, and w^ at a selected radius r$ in the normal 
plane through A. 

Now from the static condition the vector sum 

wr + WiTi + wtfz + w s r 3 + w^r* + w^r^ = 

where w and w b are unknown, and these cannot yet be found 
because the directions of the radii r and r 5 are not known. Again, 
since the reference plane passes through 0, tilting moments about 
O must balance, or 

w 3 r 3 a 3 + w^r^a^ + w b r b a^ = 

and here the only unknown is w- r^a^ which may therefore be 
determined. The vector polygon for finding this quantity is 
shown at (a) in Fig. 181 and on dividing by a& the value 
of iy 5 r 5 is given. The force polygon shown at (&) may now be 
completed, and the only other unknown w X r found, and thus 
the magnitude and positions of the balancing weights w and w$ 
may be found. The construction gives the value of the prod- 
ucts wr and w$r$ so that either w or r may be selected as 
desired and the remaining factor is easily computed. 

By a method similar to the above, therefore, any number of 
rotating masses in any positions may be balanced by two weights 
in arbitrarily selected planes. Many examples of this kind 
occur in practice, one of the most common being in locomotives 
(see Sec. 253), where the balancing weights must be placed in the 
driving wheels and yet the disturbing masses are in other planes. 

246. Numerical Example on Balancing Revolving Masses. 
Let there be any four masses of weights Wi 10 lb., w z = 6 Ib. 
w s = 8 lb. and u' 4 = 12 lb., rotating at radii r\ 6 in., r^ = 8 in., 
r 3 = 9 in. and r = 4 in. in planes located as shown on Fig. 12.8 
It is required to balance the system by two w'eights in the planse 

through the points and A respectively. 


The data of the problem give w\ri = 10 X TO = 5, W^TZ = 
6 X TO = 4, w 3 r 3 = 8 X TO = 6 and w^t = 12 X T^ = 4, and 




= 5 X TO = 2.5, 

"I O 

= 4 X o = 3.33, 

-j r 

= 6 X = 7.5 and 


= 4 X = 6. 

The first thing is to draw the tilting-couple vector polygon as 
shown on the left of Fig. 182 and the only unknown here is w b r b a$ 
which may thus be found and scales off as 4.65. Dividing by a& 


= -= = 1 ft. gives w b r$ = 4.65 and the direction of r 5 is also given 

as parallel to the vector 

Next draw the vector diagram for the products w.r as shown 
on the right of Fig. 182, the only unknown being the product wr 

FIG. 182. 

for the plane through 0. From the polygon this scales off 
as 2.9 and the direction of r is parallel to the vector in the 

In this way the products wr and w^r b are known in magnitude 
and direction, and then, on assuming the radii, the weights are 
easily found. This has been done in the diagram. It is advis- 
able to check the work by choosing a reference plane somewhere 
between and A and making the calculations again. 


247. The Balancing of Reciprocating and Swinging Masses. 

The discussion in the preceding sections shows that it is always 



possible to balance any number of rotating masses by means 
of two properly placed weights in any two desired planes "of 
revolution, and the method of determining these weights has 
been fully explained. The present and following sections deal 
with a much more difficult problem, that of balancing masses 
which do not revolve in a circle, but have either a motion of 
translation at variable speed, such as the piston of an engine or 
else a swinging motion such as that of a connecting rod or of the 
jaw of a rock crusher or other similar part. Such problems 
not only present much difficulty, but their exact solution is 
usually impossible and all that can generally be done is to parti- 
tially balance the parts and so minimize the disturbing effects. 
248. Balancing Reciprocating Parts Having Simple Harmonic 
Motion. The first case considered is that of the machine shown 

FIG. 183. 

in Fig. 6 somewhat in detail and a diagrammatic view of which 
is given in Fig. 183. The crank a is assumed to revolve with 
uniform angular velocity co radians' per second, the piston c hav- 
ing reciprocating motion, and it has been shown in Sec. 234 that 
the acceleration of c is, at any instant, equal to the projection of 
a upon the direction of c multiplied by co 2 , or the acceleration of 
the piston is OA X co 2 = a cos 6 X co 2 . The force necessary to 

produce this acceleration of the piston then is F = X a cos 6 X co 2 


where w is the weight of the piston, and it is this force F 
which must be applied to give a balance. A little consideration 
will show that this force F is constant in direction, always co- 
inciding with the direction of motion of c, but it is variable in 
magnitude, since the latter depends on the crank angle 0. 



Suppose now that at P is placed a weight w exactly equal to the 
weight of the reciprocating parts; then the centrifugal force act- 
ing radially is a X co 2 and the resolved part of this in the direction 

of motion of c is clearly X a cos 6 X o> 2 , that is to say, the 


horizontal resolved part of the force produced by the weight w at 
P is the same as that due to the motion of the piston. It there- 
fore follows that if at PI, located on PO produced so that OPi 
= OP, there is placed a concentrated weight of w lb., the hori- 
zontal component of the force produced by it will balance the 
reciprocating masses; the vertical component, however, of the 
force due to w at PI is still unbalanced and will cause vibrations 
vertically. Thus, the only effect produced by the weight w at 
PI is to change the horizontal shaking forces due to c into vertical 
forces, and the machine still has the unbalanced vertical forces. 

FIG. 184. 

Frequently in machinery there is no real objection to this 
vertical disturbing force, because it may be taken up by the 
foundation of the machine, but in portable machines, such as loco- 
motives or fire engines or automobiles, it may cause trouble 
also. It is seen, however, that complete balance is not ob- 
tained in this way, that is, a single revolving mass cannot be 
made to balance a reciprocating mass. 

There is only one way in which such a mass can be completely 
balanced and that is by duplication of the machine. Thus, if it 
were possible to use PI as a crank and place a second piston, as 
shown in Fig. 184, the masses would be completely balanced. 
If the second machine cannot be placed in the same plane nor- 
mal to the shaft as the first, then balance could be obtained 
by dividing it into two parts each having reciprocating weights 



"2 and moving in planes equidistant from the plane of the first 


When the reciprocating mass moves in such a way that its 
position may be represented by such a relation as a cos 9 it is 
said to have simple harmonic motion and its acceleration may 
always be represented by the formula a cos 6 X co 2 . Balancing 
problems connected with this kind of motion are problems in 
primary balancing and are applicable to cases where the connect- 
ing rod is very long, giving approximate results in such cases, and 
exact results in cases where the rod is infinitely long, and in the 
case shown in Fig. 183, just discussed. 

One method in which revolving weights may be used to produce 
exact balance in the case of a part having simple harmonic 


Geared to 

Crank Shaft 

Ratio 1:1 

(JL of Engine 

FIG. 185. Engine balancing primary balance. 

motion is shown in Fig. 185, where the two weights %w are 
equal and revolve at the speed of the crank and in opposite 
sense to one another, their combined weight being equal to the 
weight w of the reciprocating parts. Evidently here the vertical 
components of the two weights balance one another, leaving their 
horizontal components free to balance the reciprocating parts. 
Taking the combined effective weights as equal to that of the 
reciprocating parts, then they must rotate at a radius equal to 
that of the crank, and must be 180 from the latter when it is 
on the dead center. 

249. Reciprocating Parts Operated by Short Connecting Rod. 
The general construction adopted in practice for moving 
reciprocating parts differs from Fig. 183 in that the rod imparting 



the motion is not so long that the parts move with simple har- 
monic motion, and in the usual proportions adopted in engines 
the variation is quite marked, for the rods are never longer than 
six times the crank radius and are often as small as four and one- 
half times this radius. 

The method to be adopted in such cases is to determine the 
acceleration of the reciprocating parts and to plot it for each of 
the crank angles as described in the preceding chapter. To 
illustrate this, suppose it is required to balance the reciprocating 
parts in the engine examined in Sec. 241 ; then the accelerations 
of these are found and set down as shown in the table belonging 
to this case. The accelerations shown in the third column have 








Accelerations -Ft. per Sec t per Sec, 


















8 5 

G 'i 


\ 10^ ^ 

2G 1 

14 1( 

>2 IS 


38 2 





*3Qb 324 342 3(X 
Crank lAngles 








. -* 


FIG. 186. 

been plotted in the plain line A on Fig. 186. This curve must 
now be broken up into its corresponding harmonic components, 
and it is usual to assume that these are in phase at the inner 
dead center with the original curve. The dotted line B repre- 
sents a simple harmonic or sine curve in phase with the plain 
curve and having maximum height of J( 1,053 + 711) = 882, 
which is the mean value for the true curve heights at and 
at 180 crank angles. The difference between these two curves 
has been plotted in the broken line curve C and will be found on 
examination to be almost a true sine curve, in fact, it differs 
so little from a sine curve that it would be impossible to distin- 
guish between them on the scale of this drawing. 1 

It will be observed that the curve C is also in phase at the 

1 See Appendix A for mathematical proof of these statements. 


inner dead center with the curve A but has twice the frequency 
and maximum height on the drawing of 171 ft. per second per 

second. It will also be found' that 171 is ~ X 882 or - X 882 


= 171. 

The reciprocating parts of this engine could also be balanced 
in the manner shown in Fig. 185, but it would require that in- 
stead of one pair of weights, two pairs should be used; one pair 
rotating at the speed of the crank and 180 from it at the dead 
centers, and another pair in phase at the inner dead center with 
the first but rotating at double the speed of the crank. The 
weight rotating at the speed of the crankshaft should be the 
same as that of the piston, namely 161 Ib. (m = 5), if placed at 
at 3^2 in. radius, while the weight making twice the speed (co 
= 110) of the crank might also be placed at a radius of 3^ m -> 

. , 5 X 171 X 32.2 
in which case it would weigh Q - - = 7.76 Ib. In 

jj X (110)' 

order that these weights could rotate without interference they 
might have to be divided and separated axially, in which case 
the two halves of the same weight would have to be placed equi- 
distant from the plane of motion of the connecting rod. 

It is needless to say that the arrangement sketched above is 
too complicated to be used to any extent except in the most 
urgent cases, where some serious disturbance results. Counter- 
weights attached directly to the crankshaft are sometimes used, 
but at best these can only balance the forces corresponding to 
the curve B and always produce a lifting effect on the engine. 
The reader must note that the above method takes no account 
of the weight of the connecting rod, which will be considered later. 

If the method already described cannot be used, then the only 
other method is by duplication of the parts and this will be de- 
scribed at a later stage. 

Where the acceleration of the reciprocating masses cannot be 
represented by a simple harmonic curve, but must have a second 
harmonic of twice the frequency, superposed on it, the problem 
is one of secondary balancing, so called because of the latter 

250. Balancing Masses Having Any General Form of Plane 
Motion. It is impossible, in general, to balance masses moving 
in a more or less irregular way, such, for example, as the jaw of 


the rock crusher shown in Fig. 168, or the connecting rod of an 
engine. The general method, however, is to plot the curve of 
accelerations for the center of gravity of the mass, using crank 
angles as a base, and if the resulting curve at all approximates 
to a simple harmonic curve a weight may be attached to the 
crank, as already described, which will roughly balance the 
forces, or will at least reduce them very greatly. The magnitude 
of the weight and its position will be found by drawing a simple 
harmonic, curve which approaches most nearly to the actual 
acceleration curve. As the accelerations are not all in the same 
direction, the most correct way is to plot two curves giving the 
resolved parts of the acceleration in two planes and balance each 
separately, but usually an approximate result is all that is desired, 
and, as the shaking forces are mainly in one plane, the resolved 
part in this plane alone is all that is usually balanced. 

In engines, the method of balancing the connecting rod is 
somewhat different to that outlined above. The usual plan is 
to divide the rod up into two equivalent masses in the manner 
described in Sec. 238, one of the masses mi being assumed as 
located at the wristpin and the location of the other mass m z 
is found as described in the section referred to. In this way the 
one mass mi may be regarded simply as an addition to the weight 
of the reciprocating masses and balancing of it effected as de- 
scribed in the last or following sections. The other mass ra 2 , 
however, gives trouble and cannot, as a matter of fact, be exactly 
balanced at all, so that there is still an unbalanced mass. 

Consideration of a number of practical cases shows that w 2 
in many steam engines lies close to the center of the crankpin. 
The engine discussed in Sec. 241, for instance, has the mass mi 
concentrated at the wristpin and the mass m 2 will be only 0.36 in. 
away from the crankpin; for long-stroke engines, however, the 
mass m 2 may be some distance from the crankpin and in such 
cases the method described below will not give good results. In 
automobile engines the usual practice is to make the crank end 
of the rod very much heavier and the crankpin larger than the 
same quantities at the piston end and hence the first statement 
of this paragraph is not true. In one rod examined the length 
between centers was 12 in., and the center of gravity 3.03 in. 
from the crankpin center; the weight of the rod was 2.28 Ib. and 
selecting the mass mi at the wristpin its weight would be 0.43 Ib. 
and the remaining weight would be 1.85 Ib. concentrated 0.92 in. 


from the crankpin and on the wristpin side of it, so that consider- 
able error might result by assuming the latter mass at the crank- 
pin center. 

The fact that the mass w 2 does not fall exactly at the crankpin 
has been already explained in Sec. 241, and in the engine there 
discussed the resultant force on the rod passes through L, slightly 
to the right of the crankshaft, instead of passing through this 
center, as it would do if the mass w 2 fell at the crankpin. If an 
approximation is to be used, and it appears to be the only thing 
to do under existing conditions, m 2 may be assumed to lie at the 
crankpin, and thus the rod is divided into two masses; one, mi 
concentrated at the wristpin and balanced along with the recip- 
rocating masses, and the other, m 2 , concentrated at the crank- 
pin and balanced along with the rotating masses. 

It should be pointed out in passing, that the method of divid- 
ing the rod according to the first of two equations of Sec. 238, 
that is, so that their combined center of gravity lies at the true 
center of gravity of the rod, to the neglect of the third equation, 
leads to errors in some rods. Much more reliable results are 
obtained by finding wii and w 2 according to the three equations in 
Sec. 238 and the examples of Sec. 241, except that m 2 is assumed 
to be at the crankpin center. Dividing .the mass m so that its 
components mi and w 2 have their center of gravity coinciding 
with that of the actual rod will usually give fairly good results, if 
the diameters of the crankpin and wristpin do not differ unduly. 

251. Balancing Reciprocating Masses by Duplication of Parts. 
Owing to the complex construction involved when the recipro- 
cating masses are balanced by rotating weights, such a plan is 
rarely used, the more common method being to balance the recip- 
rocating masses by other reciprocating masses. The method 
may best be illustrated in its application to engines, and indeed 
this is where it finds most common use, automobile engines being 
a notable example. 

For a single-cylinder engine the disturbing forces due to the 
reciprocating masses are proportional to the ordinates to such 
a curve as A, Fig. 186, or what is the same thing to the sum of 
the ordinates to the curves B and (7. Suppose now a second en- 
gine, an exact duplicate of the first, was attached to the same 
shaft as the former engine and let the cranks be set 180 apart 
as at Fig. 187 (a), then it is at once evident that there will be a 
tilting moment normal to the shaft in the plane passing through 



the axis of the shaft and containing the reciprocating masses, 
and further, a study of Fig. 186 will show that while the ordinates 
to the two curves B belonging to these machines neutralize, still 
the two curves C are additive and there is unbalancing due to the 
forces corresponding to curves C. In Fig. 187 are shown at (6) 
and (c) two other arrangements of two engines, both of which 
eliminate the tilting moments; in the arrangement (6) the cranks 
are at 180 and the unbalanced forces are completely eliminated, 
producing perfect balance, whereas at (c) the sum of the crank 
angles for the two opposing engines is 180 and the forces cor- 
responding to curve B are balanced, while those corresponding 
to C are again unbalanced and additive, so that there is still an 
unbalanced force. 1 Since the disturbing forces are in the direc- 

O) CO 

FIG. 187. Different arrangements of engines. 

tion of motion of the pistons, nothing would be gained in this 
respect by making the cylinder directions different in the two 

If a three-cylinder engine is made, with the cylinders side by 
side and cranks set at 120 as in Fig. 188, an examination by the 
aid of Fig. 186 will show that the arrangement gives approxi- 
mately complete balance, since the sum of ordinates to the curves 
B and C for the three will always be zero, but there is still a 
tilting moment normal to the axis of the crankshaft which is 
unavoidable. Four cylinders side by side on the same shaft, 
with the two outside cranks set together and the two inner ones 
also together and set 180 from the other, does away with the 
tilting moment of Fig. 187 (a) but still leaves unbalanced forces 
proportional to the ordinates to the curves C. A six-cylinder 
arrangement with cylinders set side by side and made of two 

1 In order to get a clear grasp of these ideas the reader is advised to make 
several separate tracings of the curves B and C, Fig. 186, and to shift these 
along relatively to one another so as to see for himself that the statements 
made are correct. 


parts exactly like Fig. 188, but with the two center cranks parallel 
gives complete balance with the approximations used here. 

In what has been stated above the reader must be careful to 
remember that the rod has been divided into two equivalent 
masses and the discussion deals only with the balancing of the 
reciprocating part of the rod and the other reciprocating masses. 
The part acting with the rotating masses must also be balanced, 
usually by the use of a balancing weight or weights on the crank- 
shaft according to the method already described in Sec. 245. It 
is to be further understood that certain approximations have been 
introduced with regard to the division of the connecting rod, and 
also with regard to the breaking up of the actual acceleration 
curve for the reciprocating masses into two simple harmonic 
curves, one having twice the frequency of the other. Such a 
division is a fairly close approximation, but is not exact. 

FIG. 188. 

The shape of the acceleration curves A, Fig. 186, and its com- 
ponents B and C, depend only upon the ratio of the crank radius 
to the connecting-rod length, and also upon the angular velocity 

co. For the same value of T the curves will have the same shape 

for all engines, and the acceleration scale can always be readily 
determined by remembering that at crank angle zero the accelera- 
tion is f a + -j- } co 2 ft. per second per second. These curves also 

represent the tilting moment to a certain scale since the mo- 
ment is the accelerating force multiplied by the constant distance 
from the reference plane. 

The chapter will be concluded by working out a few practical 

252. Determination of Crank Angles for Balancing a Four- 
Cylinder Engine. An engine with four cylinders side by side 
and of equal stroke, is to have the reciprocating parts balanced by 
setting the crank angles and adjusting the weights of one of the 
pistons. It is required to find the proper setting and weight, 
motion of the piston being assumed simple harmonic. The 



dimensions of the engine and all the reciprocating parts but one 
set are given. 

Let Fig. 189 represent the crankshaft and let w 2 , WB, w 4 repre- 
sent the known weights of three of the pistons, etc., together 
with the part of the connecting rod taken to act with each of 
them as found in Sec. 250. It is required to find the remain- 
ing weight Wi and the crank angles. 

Choose the reference plane through Wi, and all values of r are 
the same; also the weights may be transferred to the respective 
crankpins, Sec. 248, as harmonic motion is assumed. Draw the 
wra triangle with sides of lengths W 2 ra 2) W 3 ra 3 and W 4 ra 4 which 
gives the directions of the three cranks 2, 3, and 4. Next draw 
the wr polygon, from which WIT is found, and thus Wi, and the 

Moments wra Forces wr 

FIG. 189. Balancing a four crank engine. 

corresponding crank angle. The part of the rods acting at the 
respective crankpins, as well as the weight of the latter, must be 
balanced by weights determined as in Sec. 245. The four recip- 
rocating weights operated by cranks set at the angles found 
will be balanced, however, if harmonic motion is assumed. 

Example. Let w 2 = 250 lb., w z = 220 lb. and iv* '= 200 lb., 
r = 6 in. and the distance between cylinders as shown. Then 
WzTz = 125, WsTa = 110, i# 4 r 4 = 100, WzTtfig = 250, w^fya^ = 
550 and i(; 4 r 4 a 4 = 700. The solution is shown on Fig. 189 
which gives the crank angles and weight Wi = 216 lb. The 
rotating weights would have to be independently balanced. 

253. Balancing of Locomotives. In two-cylinder locomotives 
the cranks are at 90, and the balance weights must be in the 



driving wheels. In order to avoid undue vertical forces it is 
usual to balance only a part of the reciprocating masses, usually 
about two-thirds, by means of weights in the driving wheels, 
and these balancing weights are also so placed as to compensate 
for the weights of the cranks. Treating the motion of the piston 
as simple harmonic, this problem gives no difficulty. 

Example. Let a locomotive be proportioned as shown on Fig. 
190. The piston stroke is 2 ft. and the weight of the revolving 
masses is equivalent to 620 Ib. attached to the crankpin. The 
reciprocating masses are assumed to have harmonic motion and 
to weigh 550 Ib. and only 60 per cent, of these latter masses are 
to be balanced, so that weight at the crankpin corresponding 
to both of these will be 620 + 0.60 X 550 = 950 Ib. for each side. 

Moments WrcC 

1 1 


i 3 




11 , 

FIG. 190. Locomotive balancing. 

The reference plane for the tilting moments must always pass 
through one of the unknown masses, and the plane is here taken 
through the wheel 2. Note that the crank 1 being on opposite 
side of the reference plane to wheel 3 and crank 4, the sense of 
the moment vector must be opposite to what it would be if it 
were in the position 4. The crank 1 is thus drawn from the 
shaft in opposite sense to the vector Wiriai. 

A diagrammatic plan of the locomotive in Fig. 190 gives the 
moment arms of the masses and the values of the corresponding 
moments are plotted on the right. Thus Wir^i = 950 X 1 X 1.12 
= 1,069 and w 4^0,4, = 950 X 1 X 6.04 = 5,740 and from the 
vector diagram W 3 r 3 a 3 scales off as 5,840. Since a* = 4.92 ft., 
w 3 r 3 = 1,187. The force polygon may now be drawn with 
sides 950, 950 and 1,187 parallel to the moment vectors and then 
wtfz is scaled off as 1,187. Selecting suitable radii r 2 and r 3 
give the weights w 2 and w z and the end view of the wheels and 


axle on the right shows how the weights would be placed in 
accordance with the results. 

The above treatment deals only with primary disturbing forces, 
only part of which are balanced, and further, it is to be noticed 
that there will be considerable variation in rail pressure, which 
might, with some designs, lift the wheel slightly from the tracks 
at each revolution, a very bad condition where it occurs. 

254. Engines Used for Motor Cycles and Other Work. In 
recent years engines have been constructed having more than 
one cylinder, with the axes of all the cylinders in one plane nor- 
mal to the crankshaft. Frequently, in such engines, all the 
connecting rods are attached to a single crankpin, and any 
number of cylinders may be used, although with more than five, 
or seven cylinders at the outside, there is generally difficulty in 
making the actual construction. The example, shown in Fig. 

FIG. 191. Motor cycle engine. 

191, represents a two-cylinder engine with lines 90 apart and is 
a construction often used in motorcycles, in which case a vertical 
line passes upward through 0, midway between the cylinders; 
in these motorcycle engines the angle between the cylinders is 
frequently less than 90. The same setting has been in use for 
many years with steam engines of large size, in which case one 
of the cylinders is vertical, the other horizontal. 

When a similar construction is used for more than two cylin- 
ders, the latter are usually evenly spaced; thus with three cylin- 
ders the angle between them is 120. 

These constructions introduce a number of difficult problems 
in balancing, which can only be touched on here, and the method 
of treatment discussed. The motorcycle engine of Fig. 191 


with cylinders at 90 will alone be considered, and it will be 
assumed that both sets of moving parts are identical and that 
the weight of each piston together with the part of the rod that 
may be treated as a reciprocating mass is w Ib. In the following 
discussion only the reciprocating masses are considered, and the 
part of the rods that may be treated as masses rotating with the 
crankpin, and also the crankpin and shaft are balanced indepen- 
dently ; as the determination of the latter balance weights offers 
no difficulty the matter is not taken into account. 

It has already been shown in Sec. 249 (and in Appendix A) that 
the acceleration of the piston may be represented by the sum of 
two harmonic curves, one of the frequency of the crank rotation, 
and another one of twice this frequency; these are shown in Fig. 
186 in the curves B and C. It is further explained in Sec. 249 

that the maximum ordinate to the curve C is 7 times that to the 


curve B. 

The discussion of Sees. 248 and 249 should also make it clear 
that if a weight w be secured at S, in Fig. 191, the component of 
the force produced by this weight in the direction OX will balance 
the primary component of the acceleration of the piston Q, that 
is, it will balance the accelerations corresponding to the ordinates 
to the curve B, Fig. 186. There is still unbalanced the accelera- 
tions corresponding to the curve C and also the vertical compo- 
nents of the force produced by the revolving weight w at S, these 
latter being in the direction OY. A very little consideration will 
show that the latter forces are exactly balanced by the recipro- 
cating mass at R, or that the weight w at S produces complete 
primary balance. 

The forces due to the ordinates to the curve C for the piston 
Q could be balanced by another weight, in phase with S when is 
zero, but rotating at double the angular speed of OS. If this 

weight is at the crank radius, its magnitude should be (^j X r 
times w, since its angular speed is double that of OS, and also the 
maximum height of the curve C is r times that of B, Sec. 249. 

The difficulty is that this latter weight has a component in the di- 
rection of OF which will not be balanced by the forces corres- 
ponding to the curve C for the piston R. It is not hard to see 
this latter point, for when becomes 90 the fast running weight 


should coincide with S to balance the forces of the piston R, 
whereas if it coincided with w when is zero it will be exactly 
opposite to it when 6 = 90. 

With such an arrangement as that shown there is, then, per- 
fect primary balance but the secondary balance cannot be made 
at the same time. If the secondary balance weight coincides 
with w when 6 is zero, then the unbalanced force in the direction 
OF will be a maximum, and if Qis exactly balanced the maximum 
unbalanced force in the direction OF is twice that corresponding 

to the ordinates to the curve C, or is 2 X X 7 X aco 2 . Owing 

Q o 

to the difficult construction involved in putting in the secondary 
balance weight, the latter is not used, and then the maximum un- 
balanced force may readily be shown to be \2 X X r- X aco 2 pds. 

The use of the curves like Fig. 186 will enable the reader 
to prove the correctness of the above results without difficulty. 

In dealing with all engines of this type, no matter what the 
number or distribution of the cylinders, the primary and second- 
ary revolving masses are always to be found, and by combining 
each of these separately for all the cylinders the primary and 
secondary disturbing forces may be found, and the former 
always balanced by a revolving weight on the crank, but the lat- 
ter can be balanced only in some cases where it is possible to 
make the reciprocating masses balance one another. Thus, a 
six-cylinder engine of this type with cylinders at equal angles 
may be shown to be in perfect balance. 


1. What are the main causes of vibrations in rotating and moving parts of 
machinery? What is meant by balancing? 

2. The chapter deals only with the balancing of forces due to the masses; 
why are not the fluid pressures considered? 

3. If an engine ran at the same speed would there be any different arrange- 
ment for balancing whether the crank was rotated by a motor or operated by 
steam or gas? Why? 

4. Two masses weighing 12 and 18 Ib. at radii of 20 and 24 in. and inclined 
at 90 to one another revolve in the same plane. Find the position arid size 
of the single balancing weight. 

5. If the weights in question 4 revolve in planes 10 in. apart, find the 
weights in two other planes 15 and 12 in. outside the former weights, which 
will balance them. 


6. Examine the case of V-type of engine similar to Fig. 191 with the 
cylinders at 60 and see if there are unbalanced forces and how much. 

7. A gas engine 15 in. stroke has two flywheels with the crank between 
them, one being 18 in. and the other 24 in. from the crank. The equivalent 
rotating weight is 200 Ib. at the crankpin, while the reciprocating weight is 
250 Ib. Find the weights required in the flywheels to balance all of these 

8. In a four-crank engine the cylinders are all equally spaced and the 
reciprocating weights for three of the engines are 300, 400 and 500 Ib. Find 
the weight of the fourth set and the crank angles for balance. 


Approximate Analytical Method of Computing the Acceleration of the 
Piston of an Engine 

The graphical solution of the same problem is given in Sec. 236. 

Let Fig. 192 represent the engine and let be the crank angle 
reckoned from the head-end dead center, and further let x denote 
the displacement of the piston corresponding to the motion of the 
crank through angle 6. Taking o> to represent the angular ve- 
locity of the crank, and t the time required to pass through the 
angle 0, then = ut. 

FIG. 192. 

When the crank is in the position shown, the velocity of the 

piston is and its acceleration is -j-. 
at dt 

Examination of the figure shows that: 


x = a + b a cos b cos <f> 
x = a cos + b cos"0 (b + a). 

/ la 2 & 

Now cos0 = \1 sin 2 < =\/l :rr sin 2 0; since sin <f> = - sin 6. 

\ b 2 b 

a 2 
Further, since sin 2 is generally small compared with unity 

the value of 

B is equal 1 - p^ sin 2 approximately. 

(It is in making this assumption that the approximation is intro- 
duced and for most cases the error is not serious.) 


x = a cos 

~i sin2 e ] - 



or x = a cos ut sin 2 ut a 

dx a 2 

therefore = aco sin co co sin at cos 
d/ 6 

= aco sin co - co sin 2co 

d 2 z a 2 

and = aco 2 cos co co 2 cos 2co 
d/ 2 b 

a 2 

= aco- cos co 2 cos 20. 

Therefore the acceleration is 

a 2 
f = aco 2 cos -j- co 2 cos 20 

= aco 

I cos T cos 201 . 

Since x is negative and the acceleration is also negative, the 
latter is toward the crankshaft, in the same sense as x. 

The above expression will be found to be exactly correct at the 
two dead centers and nearly correct ,at other points. It shows 
that the acceleration curve for the piston is composed of two 
simple harmonic curves starting in phase, the latter of which has 

twice the frequency of the other, and an amplitude of y times the 

former's value. This has been found to be the case in the curves 
plotted from the table at the end of Chapter XV and shown on 
Fig. 186, and the error due to the factor neglected is found very 
small in this case. 

In the case shown in Fig. 186, co is 55 radians per second and 


a = -^~ = 0.292 ft., so that the value of aco 2 cos at crank angle 

. = is aco 2 = 0.292 X (55) 2 = 882 ft. per second per second, 
and this is the maximum height of the first curve. At the same 

angle = the value of aco 2 X g cos 20 = 882 X ~ = 171 ft. 

per second per second, which gives the maximum height of the 
curve of double frequency. These values are the same as those 
scaled from Fig. 186. 


Experimental Method of Finding the Moment of Inertia of 
Any Body 

For the convenience of those using this book the experimental 
method of finding the moment of inertia and radius of gyration 
of a body about its center of gravity is given herewith. 

Suppose it is desired to find these quantities for the connecting 
rod shown in Fig. 193. Take the plane of the paper as the plane 
of motion of the rod. Balance the rod carefully across a knife 
edge placed parallel with the plane of motion of the rod, and 
the center of gravity G will be directly above the knife edge. 

FIG. 193. Inertia of rod. 

Next secure a knife edge in a wall or other support so that its 
edge is exactly horizontal and hang the rod on it with the knife 
edge through one of the pin holes and let it swing freely like a 
pendulum. By means of a stop watch find exactly the time 
required to swing from one extreme position to the other; this 
can be most accurately found by taking the time required to do 
this say, 100 times. Let t sec. be the time for the complete 

Next measure the distance h feet from the knife edge to the 
center of gravity, and also weigh the rod and get its exact weight 
w Ib. 

Then it is shown in books on mechanics that 

/2 ni\ 


in foot and pound units, gives the moment of inertia of the rod 
abtout its center of gravity. 



As an example, an experiment was made on an automobile 
rod of 12-in. centers and weighing 2 Ib. 4>^ oz. or 2.281 Ib. The 
crank and wristpins were respectively 2 in. and ! ^{Q in. diameter, 
and when placed sideways on a knife edge it was found to bal- 
ance at a point 3.03 in. from the center of the crankpin. The 
rod was first hung on a knife edge projecting through the crank- 
pin end, so that h = 4.03 in. or 0.336 ft., and it was found that 
it took 94% sec. to make 200 swings; thus 

94 6 
t = = 0.473 sec. 

(0 473V 2 281 

Then I = '^ X 2.281 X 0.336 - X (0.336) 2 

= 0.00937 in foot and pound units. 

When suspended from the wristpin end, it is evident that h = 
9.315 in. or 0.776 ft., while t was found to be 0.539 sec. giving the 

I = ftii)t X 2.281 X 0.776 - 0.0709 X (0.776) 2 
= 0.00940. 

The average of these is 0.0094 which may be taken as the 
moment of inertia about the center of gravity. The square of 
the radius of gyration about the same point is 

/x 0.0094X32.16 


or k = 0.36 ft. 


Absolute motion, 26 
Acceleration in machinery, Chap. 
XV, 277 

angular, 282 

bending moment due to, 287 

connecting rod, 292, 297, 301 

effect m crank effort, 295-297 
on i arts, 283 

force lequired to produce, 284 

forces at bearings due to, 300 
due to, 283 

general effects of, 277 

graphical construction, 280 

in bodies, 278 

in engine, 290, 291, 300 

in rock crusher, 287 

links, 282 

normal, 278 

piston, 291-297, 300 

piston-formula, 329 

points, 282 

stresses in parts from, 286 

tangential, 278 

total, 280 

vibrations due to, 277 
Addendum line, 76, 83 

circle, 76, 83 
Adjustment of governors, rapidity 

of, 236 
Angle of approach, 77 

friction, 183 

obliquity, 79 

recess, 77 
Angular acceleration, 282 

space variation, 258 

velocity, 35, 38, 39, 57 
Annular gears, 80 
Approach, arc of, 77 

angle of, 77 

Arc of approach, 77 

contact, 77 

recess, 77 
Automobile differential gear, 133 

gear box, 115 
Available energy variations, 164 


Balancing, Chap. XVI, 307 

connecting rod, 320 

crank angles for, 322 

disturbing forces, 315 

duplication of parts for, 320 

four-crank engine, 322 

general discussion, 307 

locomotives, 323 

masses, general, 318 

motor cycle, 325 

multi-cylinder engines, 320, etc. 

primary, 316 

reciprocating masses, 314, 320 

rotating masses, 308 

short connecting rod, 316 

secondary, 318 

swinging masses, 313 

weights, 316 
Base circle, 77 
Beam engine, 154 
Bearing, 15 

Belliss and Morcom governor, 223- 

curves of, 227 

Bending moment due to accelera- 
tion, 287 
Bevel gears, Chap. VI, 90 

cone distance for, 93 

teeth of, 92 
Bicycle, 6 

Brown and Sharpe gear system, 84 
Buckeye governor, 233 

curves for, 236 




Cams, Chap. VIII, 136 

gas engine, 143 

general solution of problem, 144 

kinds of, 137 

purpose of, 136 

shear, 141 

stamp mill, 137 

uniform velocity, 140 
Center, fixed, 30 

instantaneous, 28 

permanent, 30 

pressure, 190 

theorem of three, 30 

virtual, 28 
Chain closure, 11 

compound, 17 

double slider crank, 22 

inversion of, 17 

kinematic, 16 

simple, 17 

slider crank, 18 
Characteristic curve for governor, 


Chuck, elliptic, 22 
Circle, base, 77 

describing, 72 

friction, 191 

pitch, 70 
Circular or circumferential pitch, 77, 


Clearance of gear teeth, 83 
Cleveland drill, 129 
Clock train of gears, 114 
Closure, chain, 11 

force, 11 
Coefficient of friction, 180 

speed fluctuation, 265 
Collars, 10 
Compound chain, 16 

engine, 170 

gear train, 110 
Cone, back, distance, 93 

pitch, 92 
Connecting rod, 4 

acceleration of, 292 

balancing of, 320 

friction of, 193 

Connecting rod, velocity of, 61 
Constrained motion, 6, 10 
Contact, arc of, in gears, 77 

line of, skew bevel gears, 90 

path of, in gears, 77 
Continued fractions, application, 122 
Cotter design, 184 
Coupling, Oldham's, 22 
Crank, 4 

Crank angles for balancing, 322 
Crank effort, 152, 165 

diagrams, Chap. X, 166 

effect of acceleration on, 295, 


of connecting rod, 297 
of piston, 295 
Crankpin, 4 

Crossed arm governor, 206 
Crosshead, friction in, 183 
Crusher, rock, 158 
Cut teeth in gears, 83 
Cycloidal curves, 73, 80 

teeth, 72 

how drawn, 74 
Cylinders, pitch, 70 

Dedendum line, 83 
Describing circle, 72, 75 
Diagrams of crank effort, 166 

E-J (energy-inertia), 262 

indicator, 167, etc. 

motion, Chap. IV, 49 

polar, 45 

straight base, 45 

torque, 166, 169 

vector, phorograph, 58 

velocity, uses of, 49 

Chap. Ill, 35 
Diametral pitch, 84 
Differential gear, automobile, 133 
Direction of motion, 29, 30, 32 
Discharge, pump, 46 
Divisions of machine study, 8 
Drill with planetary gear, 129 
Drives, forms of, 68 




Efficiency, engine, 196 

governor, 194 

maqhine, Chap. XI, 177 

mechanical, 177 

shaper, 186 
Effort, crank, 152, 165 
E-J (energy-inertia) diagram, 262 

for steam engine, 270 
Elements, 11 
Elliptic chuck, 22 
Energy available, 164 

kinetic of bodies, 243 
of engine, 246 
of machine, 244 

producing speed variations, 251 
Engine, acceleration in, 290, 291, 300 

beam, 154 

compound, 170 
' crank effort in steam, 166, etc. 

diagram of speed variation, 250, 
252, 255 

efficiency, 196 

energy, kinetic, 246 

internal combustion, 171 

multi-cylinder, 321, etc. 

oscillating, 25 

proper flywheel for steam, 270, 


for gas, 274 
Epicyclic gear train, 110, 124, etc. 

ratio, 110, 125 
Epicycloidal curve, 73 
Equilibrium of machines, 150 

static, 8 
External forces, 149 

Face of gear, width of, 84 

tooth, 84 

Factor, friction, 181 
Feather, 11 
Fixed center, 30 
Fluctuations of speed in machinery, 

Chap. XIII, 240 
approximate determination of, 

Fluctuations of speed, cause of, 242 

conditions affecting, 247, 260 

diagram of, 255 

energy, effect on, 240, 247, 251 

in any machine, 248 

in engine, 250, 252 

nature of, 240 

Flywheels, weight o f, Chap. XIV, 

best speed, 267 

effect of power on, 261 
of load on, 261 

gas engine, 274 

general discussion, 262 

given engine, 270 

minimum mean speed, 269 

purpose, 261 

speed, effect on, 264, 267 
Follower for cam, 138 
Forces in machines, Chap. IX, 149 

accelerating, 283, 300 
effect on bearings, 300 
general effects, 283 

causing vibrations, 314 

closure, 17 

external, 149 

in machine, 151 

in shear, 152 
Ford transmission, 131 
Forms of drives, 68 
Frame, 3, 6 
Friction, 178 

angle of, 183 

circle, 191 

coefficient of, 180 

crosshead, 183 

factor, 181 

in connecting rod, 193 

in cotter, 184 

in governor, 194, 216 

laws of, 180 

sliding pairs, 181 

turning pairs, 189, 192 

Gas engine cam, 143 
crank effort, 172 
flywheel, 274 



Gears and gearing, Chap. V and VI, 

68, 90 
Gears, annular, 80 

bevel, Chap. VI, 90, 94 

Brown and Sharpe system, 84 

conditions to be fulfilled in, 71 
when used, 68 

diameter of, 69 

examples, 85 

face, 84 

hunting tooth, 123 

hyperboloidal, 90, 94 

interference of teeth, 81 

internal, 80 

methods of making, 83 

path of contact of teeth, 73 

proportions of teeth, 84 

sets of, 79 

sizes of, 69 

spiral, Chap. VI, 90 

spur, 68 

systems, discussion on, 87 

toothed, Chap. V, 68 

types of, 90 

Gears for nonparallel shafts, 91 
Gears, mitre, 91 

screw, 90, 102 

skew bevel, 90, 93 

spiral, 90, 102 
Gears, worm, 90, etc. 

construction, 104 

ratio of, 103 

screw, 106, 107 
Gearing, trains of, Chap. VII, 110 

compound, 110 

definition of, 110 

epicyclic, 110, 124, etc. 

kinds of, 110 

planetary, see Epicyclic. 

reverted, 110 

Gleason spiral bevel gears, 93 
Gnome motor, 20 

acceleration in, 303 
Governors, Chap. XII, 201 

Belliss and Morcom, 223 
curves for, 227 

Buckeye, 233 

characteristic curves, 212 

crossed arm, 206 

Governors, definition, 201 

design, 219 

efficiency of, 194 

friction in, 194, 216 

Hartnell, 221 

design of spring for, 222 

height, 205 

horizontal spindle, 223 

inertia, 228, 229 
properties of, 229 

isochronism, 206, 213, 230 

McEwen, 233 

pendulum, theory of, 203, 204, 

Porter, 207 

powerfulness, 211, 212, 215 

Proell, 159, 220 

Rites, 238 

Robb, 230 

sensitiveness, 210, 214 

spring, 221 
design of, 222 

stability, 207, 213 

types of, 201, 202 

weighted, see Porter governor. 
Governing, methods of, 201 
Graphical representation, see Matter 


Hartnell governor, 221 
Height of governor, 205 
Helical motion, 9 

teeth, 87 

uses of, 88 

Hendey-Norton lathe, 119 
Higher pair, 14 
Hunting in governors, 207 

tooth gears, 123 
Hyperboloidal gears, 91, 94 

pitch surfaces of, 95, 101, 108 

teeth of, 102 
Hypocycloidal curve, 73 


Idler, 113 
Image, 53, 55 



Image, angular velocity from, 57 

copy of link, 59 

how found, 55 

of point, 53 

Inertia-energy (E-J) diagram, 262 
Inertia governor, 203, 228, etc. 

analysis of, 233 

distribution of weight in, 237 

isochronism in, 234 

moment curves for, 234 

properties of, 229 

stability, 234, 235 

work done by, 236 
Inertia of body, 331 

parts, Chap. XVI, 307 

reduced, for machine, 244 
Input work, 176, 251 
Instantaneous center, 28, 32, 35 
Interference of gear teeth, 81 
Internal gears, 80 
Inversion of chain, 18 
Involute curves, method of drawing, 
78, 79 

teeth, 78 

Isochronism in governors, 206, 213, 

Jack, lifting, 185 
Joy valve gear, 41 

velocity of valve, 41, 66 


Kinematic chain, 16 
Kinematics of machinery, 8 
Kinetic energy of bodies, 243 

of engine, 246 

of machine, 244 

Lathe, 4; 116, etc. 

Hendey-Norton, 119 
screw cutting, 116 

Line of contact in gears, 90 

Linear velocities, 29, 35 

Link, 16 

Link, acceleration of, 282 
angular velocity of, 68 
kinetic energy of, 243 
motion, Stephenson, 63 
primary, 66 
reference, 49 
velocity of, 36, 37, 40 
general propositions, 38 

Load, effect on flywheel weight, 261, 

Locomotive balancing, 323 

Lower pair, 14 


Machine, definition, 7 

design, 8 

efficiency of, Chap. XI, 176 

equilibrium of, 150 

forces in, Chap. IX, 150 

general discussion, 3 

imperfect, 8 

kinetic energy of, 244 

nature of, 3 

parts of, 5 

purpose of, 7 

reduced inertia of, 244 

simple, 17 

Machinery, fluctuations of speed in 
Chap. XIII, 240 

cause of speed variations, 240 

effect of load, etc., 240, 247 

kinematics of, 8 
McEwen governor, 233 
Mechanical efficiency, 177 
Mechanism, 17, 55 
Mitre gears, 91 
Module, 85 
Motion, absolute, 26 

constrained, 6 

diagram, Chap. IV, 49 

direction of, 29, 30 

helical, 9 

in machines, Chap. II, 24 

plane, 4, 9, 24 

quick-return, 20, 62 

relative, 5, 25, 26 

propositions on, 26, 27 

screw, 9 



Motion, sliding, 11 

spheric, 9 

standard of comparison, 25 

translation, 3 

turning, 10 

Motor cycle balancing, 325 
Multi-cylinder engines, 320, 322, 325, 


Normal acceleration, 278 
Numerical examples; see Special 

Obliquity, angle of, 79 
Oldham's coupling, 22 
Oscillating engine, 19 
Output work, 176, 251 

Pair, 3 

friction in, 167, 170 

higher, 14 

lower, 14 

sliding, 12, 33 
friction in, 181 

turning, 11 

friction in, 189 
Permanent center, 30 
Phorograph, 50, 52, 58, Chap. IV 

for mechanism, 55 

forces in machines by, 155 

principles of, 50-53 

property of, 66 

vector velocity diagram, 58 
Pinion, 69 
Piston, 3 

acceleration of, 291-296, 329 

velocity, 44, 61 

Pitch, circular or circumferential 77, 

circle, 70 

cones, 92 

cylinder, 70 

diametral, 84 

Pitch, normal, 101 

point, 84 

surfaces, 95, 100 
Plane motion, 4, 9, 24 
Planetary gear train, see Epicyclic 

examples, 126 

purpose, 124 

ratio, 125 
Point, acceleration of, 282 

image of, 53 

of gear tooth, 84 
Polar diagram, 45 

Porter governor, see also Weighted 

advantages, 209 

description, 207 

design of, 219 

height, 209 

lift, 210 

sensitiveness, 210 

Power, effect on flywheel weight, 273 
Powerfulness in governors, 211, 215 
Pressure, center of, 190 
Primary balancing, 316 
Primary link, 56 
Proell governor, 159, 220 
Pump discharge, 46 

Quick-return motion, 20, 62, 186 
Whitworth, 20, 62 


Racing in governors, 207 

Rack, 80 

Rapidity of adjustment in governors, 

Recess, angle of, 77 

arc of, 77 

Reciprocating masses, balancing of, 
314, 320 

balancing by duplication, 320 
Reduced inertia, 244 
Reeves valve gear, 64 
Reference link, 49, see Primary link. 
Relative motion, 5, 25, 26 



Relative velocities, 38, 40 
Resistant parts of machines, 5 
Reverted gear train, 110 
Rigid parts of machines, 5 
Rites governor, 238 
Riveters, toggle joint, 161 
Robb governor, 230 
Rock crusher, 158 

acceleration in, 287 
Rocker arm valve gear, 60 
Root circle, 76, 83 

of teeth, 84 
Rotating masses, balancing of, 308 

pendulum governor, 202, etc. 
defects of, 205 
theory of, 204 
Rotation, sense of, in gears, 113 

sense of, for links, 57 

Screw gears, 90, 102 

motion, 9 

Secondary balancing, 319 
Sensitiveness in governors, 210, 214 
Sets of gears, 79 

Shaft governor, 203, 229; see also 
Inertia governor. 

properties of, 228 
Shaper, efficiency of, 187 
Shear, cam, 141 

forces in, 157 
Simple chain, 17 
Skew bevel gears, 90, 94 

pitch surfaces of, 95 
Slider-crank chain, 18 

double chain, 22 
Sliding motion, 11 

friction in, 181 

pair, 12, 33 

Slipping of gear teeth, 76 
Space variation, angular, 258 
Speed fluctuations' in machines, 
Chap. XIII, 240 

approximate determination, 249 

cause of, 240 

coefficient of, 265 

conditions affecting, 247, 260 

diagram of, 255 

Speed fluctuations, effect of load on, 
240, 247 

energy causing, 251 

in engine, 250, 252 

in any machine, 248 

minimum, mean, 269 

nature of, 240 

Speed of fly-wheel, best, 267 
Spheric motion, 9 
Spiral bevel teeth', 90, 93 

gears, Chapter VI, 90 
Spring governor, 221 
Spur gears, 68 

Stability in governors, 196, 207, 213 
Stamp-mill cam, 137 
Static equilibrium, 8, 150 
Stephenson link motion, 63 
Stresses due to acceleration, 286 
Stroke, 4 
Stub teeth, 79, 85 
Sun and planet motion, 128 
Swinging masses, balancing of, 313 

Tangential acceleration, 278 
Teeth, cut, 83 

cycloidal, 72, 74 

drawing of, 74, 78 

face, 84 

flank, 84 

helical, 87 

hyperboloidal, 117 

interference, 81 

involute, 77, 79 

of gear wheels, 76 

parts of, 84 

path of contact, 73, 78 

point, 84 

profiles of, 73 

root, 83 

slipping, amount of, 76 

spiral bevel, 93 

stub, 79, 85 

worm and worm-wheel, 103, 104 
Theorem of three centers, 30 
Threads, cutting in lathe, 122 
Three-throw pump, 47 
Toggle-joint riveters, 161 



Toothed gearing, Chap. V, 68 
Torque on crankshaft, 165, etc. 

effect of acceleration on, 299, 301 
Total acceleration of point, 280 
Trains of gearing, Chap. VI, 90 

automobile gear box, 116, 133 

change gears, 117 

clock, 114 

definition of, 110 

epicyclic, 124 

examples, 114 

formula for ratio, 112 

lathe, 116 

planetary, 124 
ratio of, 112 

rotation, sense of, 113 
Translation, motion of, 3 
Transmission, Ford automobile, 131 
Triplex block, Weston, 128 
Turning motion, 10 

moment, Chap. X, 164 

pair, 11 

friction in, 189 


Uniform velocity cam, 140 
Unstable governor, 207 

Valve gears, Joy, 41, 66 
Reeves, 64 
rocker arm, 60 

Velocity, diagram, Chap. Ill, 35 

graphical representation, 43 

linear, 29, 35 
of points, 35 
relative, 38, 40 

piston, 44 
Velocities, angular, 35, 37, 39 

how expressed, 39 
Vibration due to acceleration, 277 
Virtual center, 28, 32, 35 


Watt sun and planet motion. 128 
Weight of fly-wheels, Chap. XIV, 

effect of speed, etc., on, 267 
Weighted governor, 207 

advantages of, 209 

effect of weights, 216 

height, 209 

lift, 210 

Weights for balancing, 316 
Weston triplex block, 128 
Wheel teeth, 76 
Whitworth quick-return motion, 20, 

Work, 7 

in governors, 235 

input and output, 176 
Worm, 102 

gear, 102 

teeth of, 105 
Wrist pin, 4 




APR 4 1337 

19 1.937 

MAR ~4 1923 

JUN 1019 

OCT 28 t93 

SEP 9 1942 



JUN 21