GIFT OF
MICHAEL REESE
SLIDE-VALVES.
A BOOK FOR PRACTICAL MEN
+
ON THE
PRINCIPLES AND METHODS OF DESIGN;
WITH
AN EXPLANATION OF THE PRINCIPLES
OF SHAFT-GOVERNORS.
BY
C. W. MAcCORD, JR., M.E.
FIRST EDITION.
FIRST THOUSAND.
NEW YORK:
JOHN WILEY & SONS.
LONDON: CHAPMAN & HALL, LIMITED.
1897.
Copyright, 1897,
BY
C. W. MAcCORD, JR.
ROBERT DRUMMOND. KLECTROTYPER AND PRINTER, NEW YORK.
PREFACE.
A SERIES of articles entitled "Valve-Gears" was pub-
lished by the author in Power during 1895 and 1896. The
object of these papers was to put the principles of design
of slide-valves in a practical form for practical men. This
book contains the subject-matter of those papers entirely
revised and rearranged, with a complete new set of cuts.
C. W. MACCORD JR.
AUBURN, N. Y.,
July 4, 1897.
JT*Se^ ~ "-'"^r^A^
iii
CONTENTS.
CHAPTER I.
PAGE
General Principles I
CHAPTER II.
Crank, Connecting-Rod, and Eccentric-Rod 17
CHAPTER III.
Valve Diagrams 27
CHAPTER IV.
Steam and Exhaust Ports Bridges Steam-Pipes 44
CHAPTER V.
General Problems in Valve Design 54
CHAPTER VI.
Rockers and Bell-Cranks 62
CHAPTER VII.
Design of a Plain D-Valve 68
CHAPTER VIII.
Valve and Eccentric Rods, Steam-Chest , 78
v
vi CONTENTS.
CHAPTER IX.
PAGE
Design of an Allen or " Trick" Valve 82
CHAPTER X.
Design of a Double-Ported Valve 9
CHAPTER XI.
Valve-Setting 99
CHAPTER XII.
Shaft-GovernorsGeneral Principles and Types 108
CHAPTER XIII.
Shaft-GovernorsAnalysis of Principles 124
SLIDE-VALVES.
CHAPTER I.
GENERAL PRINCIPLES.
FlG. I shows a cylinder which is entirely empty, except
for the piston shown; the latter being free to slide in the
cylinder under the influence of any force. Suppose, for
PlG.l
example, that steam were admitted to the space A at the
left. The natural result would be that the piston would
move over to the right-hand end, as shown in Fig. 2. Now,
FIG. 2
in order to bring the piston back to the position shown in
Fig. I, a force must be applied to the right of the piston.
2 SLIDE-VALVES.
Suppose steam were admitted there. This would not suffice
to move the piston, because the space A is filled with steam
and the pressures on the two sides of the piston would
balance each other. It therefore follows that the mere
admission of steam to the ends of the cylinder in alternation
is not enough to cause the piston to reciprocate, or move
backward and forward ; but, in addition, some means must be
provided for emptying or exhausting the steam from the
cylinder as soon as it has done its work. If this exhausting
is accomplished, the steam-pressure will then move the piston
in the desired way.
In Fig. 3 is represented the simplest contrivance designed
to regulate the admission and exhaust of steam from the
FIG. 3
cylinder. The space S is called the " steam-chest," and is
in direct communication with the boiler, so that it is at all
times full of steam at boiler-pressure. The passages P and
P' lead to the right- and left-hand ends, respectively, of the
cylinder, and are called " steam-ports " or, more often,
"porf-s." The piece Fis the' 'valve," and it is free to slide
in a horizontal direction only, or parallel with the " line of
GENERAL PRINCIPLES. 5
stroke" of the piston. The surface of the main casting in
which the ports terminate is called the " valve-seat," and
the lower surface of the valve, which is in contact with the
seat, is called the " valve-face." The valve is moved by
means of the "va'lve-rod," VR, which passes through a
" stuffing-box " at the left of the steam-chest; the object of
the stuffing-box being to prevent the leakage of steam. The
piston, A, with its "piston-rod," does not require any partic-
ular description. The passage E in the valve-seat is called
the "exhaust-port," and it is in direct communication with
the atmosphere at all times.
The valve as shown is separated entirely frcm the "valve-
gear" which operates it. This is done because a clearer
understanding of the subject can be had by first studying
thoroughly the movements of the valve and then considering
the gear required to produce these movements.
The operation of this valve is as follows:
In Fig. 3 the valve just covers both ports, and the steam
cannot reach either end of the cylinder. Then the valve
FIG. 4
moves slightly to the left, assuming the position shown in
Fig. 4, when the right-hand port is uncovered, and the steam
4 SLIDE-VALVES.
naturally follows the course indicated by the curved arrows,
thus getting on the right of the piston and forcing it over to
the left, as shown by the straight arrow. When the piston
reaches the middle of its stroke the valve is at the extreme
left of its stroke, or travel, and begins to move in the oppo-
sition direction; this being shown in Fig. 5. When the
piston reaches the extreme left-hand position the valve is
FIG. 5
again " central or in the middle position, and just covers
both ports, as shown in Fig. 6. The valve continues to
move to the right, thus admitting steam to the left-hand end
of the cylinder, as shown in Fig. 7. At the same time, it
will be noticed, the cavity E' in the valve is over the right-
hand steam-port, thus permitting steam to escape from the
right-hand side of the piston through the exhaust-port E, as
shown by the arrows. The exhaust-port leads to the atmos-
phere. The piston now reverses its stroke, and when in the
middle of its stroke the valve is over at the extreme right-
hand end of its travel, as shown in Fig. 8. The piston con-
tinues to move to the right, while the valve reverses and
moves to the left, and the original condition, shown in
GENERAL PRINCIPLES.
5
Fig. 3, is once more attained. The relative movements of
the piston and valve are then, repeated. When the valve is
FIG. 6
again over at the left, as shown in Figs. 4 and 5, the steam
at the left of the piston is exhausted through P' and .
v it
FIG. 7
It should be at once apparent that the motion of the valve
bears a fixed relation to the motion of the piston, and should
SLIDE-VALVES.
be derived from it in some way. When the valve is in its
central position the piston is at either one end or the other of
FIG. 8
its stroke. That is, the valve is either a half-stroke ahead of
or behind the piston.
In every engine the reciprocating or "to-and-fro " motion
of the piston is converted into a rotary motion of the fly-
wheel. Fig. 9 shows the simplest possible contrivance for
effecting this conversion; and at the same time it shows the
simplest form of apparatus for deriving the motion of the
valve from the stroke of the piston. The center of the
crank-shaft, on which the fly-wheel is keyed, is at <9, and the
crank-pin is at C. The crank itself is omitted for the sake of
clearness. The piston-rod, R, terminates in a slotted cross-
head, in which the crank-pin fits and is free to slide. With
this gear, as the piston moves to and fro, the crank rotates,
and the crank-pin slides up and down in the slotted cross-head.
It will be noticed that when the positions of the piston-rod,
crank-pin, and shaft are as shown in the figure that is, when
their centers are all in the same straight line, there is no
rotative effort on the crank-pin, the entire pressure on the
GENERAL PRINCIPLES. 7
piston being expended in an attempt to crush the crank-pin
or shaft. At the other end of the stroke a similar state of
affairs occurs. These two points are called the "dead-
points" or " dead-centers." If an engine, when stopped,
gets on either dead-center, it is necessary to pry it off before
FIG. 9
it can be started. When the engine is on the center the
piston is at the end of its stroke.
Another pin is secured to the shaft just 90, or a quarter
circumference, away from C, as shown at E. If, therefore,
the valve be driven by E, it will always be a half-stroke away
from the piston. The pin E is called the "eccentric-pin,"
or the " eccentric," and another slotted cross-head is em-
ployed to communicate the motion of E to the valve-rod,
VR, through the "rocker," or " rock-shaft," L. This
rocker is pivoted at >, which is some point on the frame of
the engine; and, in the case shown, the arms are equal, so
that the motion of the valve is the same in amount as if the
valve-rod were fastened to the end of the eccentric-rod ; but
it is opposite in direction, or the rocker reverses the move-
8 SLIDE-VALVES.
ment of the valve. A rocker is not an indispensable adjunct,
but it is used when the line of travel of the valve is above or
below the line of stroke of the engine. The type shown in
the cut is only one of many, some of which will be described
later, and is taken on account of its simplicity.
This device answers the two requirements mentioned;
for the motion of the valve is obviously derived from the
stroke of the piston, and, as the angle between the crank and
the eccentric is 90, it is evident that when the piston is at
one end of its stroke, as it must be in Fig. 9, the rocker L is
vertical and the valve is consequently in the middle position.
The distance OE from the center of the crank-shaft to the
center of the eccentric-pin is called the "throw" of the
eccentric, or simply the " eccentricity," and a momentary
inspection of the figure will show that the travel of the valve
is equal to twice the eccentricity, or to the diameter of the
circle in which the eccentric-pin travels. Similarly, the
length of the crank, OC, is often called the "throw " of the
crank, and it is equal to one-half the stroke of the engine.
It will be noticed that the travel of the valve is much less
than that of the piston. It is made so in order to decrease
the power required to move the valve; for the friction
between the valve and the valve-seat varies directly with the
distance through which the valve has to be moved.
This type of valve-gear has two serious drawbacks: The
steam is not used expansively, and there is no "compres-
sion " or cushioning of the steam at the end of the stroke to
take up the momentum of the moving parts, to avoid
" pounding." Both of these faults are due to the fact that
steam is admitted to the cylinder from the moment that the
piston leaves one end of the cylinder until it reaches the other
end, or, in other words, "steam follows full stroke," a fact
which should be plain from the foregoing text. The exhaust
is also open during the entire stroke.
If steam were "cut off" from the cylinder some time
GENERAL PRINCIPLES. 9
before the piston reached the end of its stroke, the expansive
force of the steam would, in combination with the reduction
of pressure in front of the piston due to the exhaust, carry
the piston along to the end of the stroke. There is great
gain in economy in so using steam expansively, but it lies
without the province of this work to enter upon its proof.
Similarly, if the port were closed to exhaust a short time
before the piston reached the end of its stroke, a certain vol-
ume of steam would be imprisoned and compressed by the
advancing piston, thus acting as a cushion to overcome the
momentum of the moving parts.
Steam may be cut off from the cylinder at any desired
point in the stroke of the piston by adding to the outside of
the valve a piece equal to the distance that the valve has
moved from its central position when the piston is at that
point of the stroke. This is shown in Fig. 10, which is a
center-line sketch of the valve-gear shown in Fig. 9, with the
addition of the valve and ports at the end of the valve-rod.
Suppose, for example, that it is desired to cut off at three-
fourths stroke that is, when the piston has completed three
10 SLID E- VA L VES.
fourths of its travel toward one end of the cylinder the port
shall be closed to steam.
FIRST. Find the position of the crank-pin when the piston
is at the point where cut-off is required. This is done by
describing, about O as a center, a circle with a radius equal
to the length of the crank ; then take TF equal to | of A T,
and draw a perpendicular to A T from F until it cuts the
crank-circle at C, the required crank-pin position ; and OC is
the " crank-line," or line representing the position of the
center-line of the crank. This is true because, with the
slotted connection, the horizontal movement of the crank-pin
is the same as that of the piston.
SECOND. Locate the eccentric-pin position corresponding to
the determined crank-pin position. This is done by drawing
OE, the eccentric line, perpendicular to OC, because the
angle between the crank and eccentric is 90, and describing
a circle about O as a center, and with a radius equal to the
eccentricity. This circle cuts OE at E, which is the required
position of the eccentric-pin.
THIRD. Locate the position of the end of the eccentric-rod.
To do this, it is only necessary to drop a perpendicular ED
from E to the line of stroke.
FOURTH. Draw in the rest of the figure. This is simply
done by taking the dimensions of the valve-rod, eccentric-rod,
rocker, etc., from Fig. 9.
Now, the "displacement of the valve," or the distance it
has moved from its central position, is OR or /, and it is evi-
dent that the valve must move that distance to the right
before the right-hand port is closed. If a piece of the length
/ is added to the outside of the valve, as shown by the dotted
lines, the port will be closed with the gear in the position
indicated, and the desired result will be obtained for the
right-hand end of the stroke.
The angle which the crank-line at any time makes with
the line of stroke is called the ''crank-angle "; and with the
GENERAL PRINCIPLES. II
slotted connection it is evident that for the same piston
position on the forward and return strokes the crank-angles
are equal. Consequently the valve displacements are equal,
and, taking the case just solved, it is only necessary to
lengthen the valve on the other end by the amount / in order
to secure cut-off at the same point on the return stroke.
The end of the 'cylinder nearest the crank is called the
" crank end," the other being called the " head end." The
stroke in which the piston approaches the crank-shaft is called
the " forward stroke"; and the stroke in which it recedes
from the crank-shaft is called the "return stroke."
The length /which is added is called the "outside lap "
or " steam-lap," because it is on the outside of the valve and
controls the admission of live steam. It is defined as the
amount which the valve projects over the outside edge of the
steam-port when the valve is in its central position, as shown
in Fig. ii. It is very evident that if the valve originally just
I, I, = outside laps.
i, t, = inside laps.
6, b,= bridges. PIG. 11
e, = exhaust port.
p, p,=steam ports.
covered the port when in its central position, a piece of length
/ would project just that amount over the edge of the port.
The compression of steam may be secured by adding a
length /' to the inside of the valve, the value of i being
determined in the same manner that /was found; that is. by
making it equal to the distance which the valve would have
to travel in order to close the port, with the piston in the
1 2 SLIDE- VA L VES.
required position. This addition is called the "inside lap'*
or "exhaust-lap," for obvious reasons. It maybe defined
as the amount which the valve projects over the inside edge
of the port, with the valve central, as in Fig. n.
The partition which divides the exhaust from the steam-
port is called the bridge. The inside lap therefore rests on
the bridge.
A difficulty due to the introduction of outside lap is at
once apparent. The admission of steam is made late by just
the amount /. That is, the valve must move a distance /
from its middle position before the port is opened to steam,
or the eccentric must turn through the corresponding angle.
This may be remedied by changing the angle between the
crank and eccentric, moving the eccentric ahead that is, in
the direction in which the engine is to run an angle corre-
sponding to Z\ That is, in Fig. 10, through an angle EOP
making the eccentric position OE*. This will make the
admission occur sooner, but it will also make the cut-off
occur earlier, so that the result will be different from the
three-quarter cut-off desired, as shown in Fig. 12, where E*
is the eccentric position with the port closed, and OC the
corresponding crank-position. It is not necessary to show
how this three-quarter cut-off could be obtained, as the
general method has been indicated, and this type of gear is
only useful as a mode of explanation. By increasing the angle
between the crank and eccentric, the opening of the port to
exhaust will be made earlier that is, it will occur before the
end of the stroke, in the simple valve with outside lap but no
inside lap; and the compression will occur before the end of
the return stroke. The inside lap, if added, need be much
smaller than the outside lap. Early exhaust opening, pro-
vided it does not occur earlier than seven-eighths stroke, does
no harm, as it simply serves to reduce the pressure against
which the piston must work on its return stroke.
The angle between the crank and the eccentric has now
GENERAL PRINCIPLES. 13
been changed from 90 to COE* by the advance of the eccen-
tric, and the amount of the change is called the "angular
advance"; and angular advance is therefore defined as the
FIG. 12
difference between 90 and the angle between the crank and
eccentric.
The rocker affects very materially the angle between the
crank and eccentric. With the device shown, the horizontal
motion of the upper arm is directly opposite to that of the
lower arm. If, in the simple device shown in Fig. 9, the
eccentric-rod were fastened direct to the valve-rod, and the
rocker in consequence eliminated, it would be necessary to
move the eccentric-pin around to E' ', which is 180 away from
E, in order to secure the same movement of the valve. If,
with the eccentric in this new position, outside lap were
added to the valve, it would be necessary to move the eccen-
tric ahead in the direction in which the engine is to run to
secure the admission of steam at the beginning of the stroke,
just as before. But in this case it would tend to increase
the angle between the crank and eccentric, while with the
rocker it resulted in a decrease of the angle. In the case of
14 SLIDE- VA L VES.
direct connection the eccentric precedes the crank, while with
the reversing rocker the eccentric follows. The definition of
angular advance may therefore be stated as follows:
li Angular advance" is the difference between the angle
between the crank and eccentric arid 90.
When the right angle is the greater the angle becomes
the "angle of follow," or the eccentric follows the crank.
When the right angle is the less, the angle becomes the
"angle of advance," or the eccentric precedes the crank.
This affords a ready means of determining whether a
rocker is used.
With this type of valve-gear the valve displacement for a
given piston position is found by drawing the perpendicular
from the eccentric-pin to the line of stroke, as ER, Fig. 10.
The displacement from the end of its travel is RS, and from
the central position is OR. The latter is the one usually
employed. It must be remembered that, on the contrary,
"piston displacement " is the amount the piston has moved
from the beginning of its stroke, and is usually expressed by
the fraction of the stroke completed.
It has been pointed out that setting the eccentric ahead
makes all the events of the stroke earlier. This may be car-
ried to such an extreme as to reverse the direction of rotation
of the crank. This may be explained by the use of Fig. 13.
Let CO be the crank and OE the eccentric, with the
engine on the head-end dead-center, running in the direction
shown by the arrow. The relative positions of crank and
eccentric are taken from Fig. 12. At this stage the valve
would commence to move to the left, opening the head-end
port. But suppose that the eccentric-pin were suddenly
slipped around to E 1 , the angle COE 1 being equal to COE.
The valve displacement would remain the same, so that the
right-hand port would be just covered. But if the engine
continues to move in the direction of the arrow, the valve
will move to the right, with the eccentric at E\ thus opening"
GENERAL PRINCIP
the right-hand or head-end port to exhaust and the crank-end
port to steam. The difference between this motion of the
valve and that obtained with the eccentric at E is shown in
the figure. The piston is moving toward the left in both
cases; but with the head-end port exhausting and the crank-
FIG. 13
ECCENTRIC AT E.
end port admitting steam, as shown by the upper valve, there
soon comes a time. when the pressure at the left of the piston
overcomes the momentum of the moving parts and forces
the piston to reverse its movement. This carries the crank in
the opposite direction. The engine will continue to run in
this reversed direction, as E 1 is as much behind the crank in
the new motion as E was behind the old.
It will be noticed that in the first case, with the eccentric
at E y the crank goes above the line of stroke on the forward
stroke. The engine is said to ''run over." When reversed,
the crank goes below the line of stroke on the forward stroke,
and the engine "runs under."
The results obtained from Fig. 13 may be stated as
follows :
1 6 SLIDE- VA L VES.
To reverse an engine, it is only necessary to slip the eccen-
tric-pin around on the shaft until it is on the other side of the
crank-pin, making the angular distance between the crank
and eccentric the same that it was at first.
In this connection it is to be noted that it is not correct
to slip the eccentric around 1 80 on the shaft, until it is
directly opposite its original position. This condition is
shown by OE\ and it is at once apparent that that will make
all the events of the stroke late by the angle E'OE\ Of
course, if the angle between the crank and eccentric were
90, it would be correct to slip it around 180, and that is
the only case in which it would be right.
CHAPTER II.
CRANKS, CONNECTING-RODS, AND ECCENTRICS.
THE slotted cross-head form of connection is but seldom
used, on account of certain mechanical difficulties which it is
unnecessary to describe here. The commonest means of
converting the reciprocating motion of the piston into the
rotary motion of the crank is shown in Fig. 14. The piston-
FIG. 14
rod, B, terminates in the cross head, H, to which it is rigidly
secured, the cross-head sliding between the guides G G. The
crank-shaft, S, carries the crank or pitman, M, on which is
the crank-pin, C. The connecting-rod, R, joins the crank-
pin with the wrist-pin, W, of the cross-head, the connecting-
rod being provided with bearings in which these pins fit, so
that the rod may alter its position and still keep the distance
between the pins sensibly constant, provisions being made in
various ways for taking up the wear.
Very frequently in fact always the eccentricity is less
than the radius of the crank-shaft. If, therefore, the slotted
17
18
SLIDE-VALVES.
form of connection to the valve-rod were employed, it would
be necessary either to place the eccentric-pin at the end of
the shaft, which is done in some engines, or to reduce the
diameter of the shaft at the point from which the motion of
the valve is derived, as shown in Fig. 15. Here the shaft
FIG. 15
center is <9, and the crank-shaft is cut away to form the pin
C, on the same principle that center-crank engines are built.
The eccentricity is OE, the center of the pin being at E\
and motion would be imparted to the valve by a rod attached
to C. To avoid these difficulties, and also to obtain
adjustability, the connection shown in Fig. 16 is employed.
FIG. 16
Here the eccentric-pin is expanded until it surrounds the
shaft. This is done by taking a disk of greater diameter than
the shaft, as in Fig. 17, where the center of the disk is E. and
boring in it a hole the diameter of which is equal to that of
CRANKS, CONNECTING-RODS, AND ECCENTRICS. 19
the crank-shaft. The hole and the disk are not concentric,
the center of the former being at O. The distance OE is the
eccentricity, or throw of the eccentric.
The disk may then be keyed to the shaft
in any desired position, thus obtaining
the power to adjust the angular advance.
Returning to Fig. 16, B is the disk, now
called the eccentric-sheave, and OE is the
eccentricity. A ring, T, called the eccen-
tric-strap, surrounds the eccentric-sheave,
and is free to turn on it. The strap is secured to the eccen-
tric-rod, N t and the latter communicates the motion of the
eccentric to the valve, either by a rocker, or by direct attach-
ment to the valve-rod. The motion of the valve thus secured
is exactly the same as if the eccentric-rod were attached to an
eccentric-pin at E, as shown by the center-lines.
The use of the crank and connecting-rod introduces am
irregularity due to the "angularity of the rod." With the
slotted-cross-head form of connection, it will be remembered,
the displacement of the piston from the end of its stroke is
exactly equal to the horizontal displacement of the crank-pin;
and, therefore, when the crank-pin is in its middle position
the piston is at the center of its stroke; and this is true on
both the forward and return strokes. The velocity of the
crank-pin is assumed to be uniform, and hence the piston
always takes the same amount of time to complete a half-
stroke. Both strokes being alike, it follows that for a given
piston displacement from the end of the stroke, the angle
made by the crank-line with the line of stroke is the same at
both ends. This is shown in Fig. 18, in which XZ is the line
of stroke, AB is the length of the stroke, and the circle whose
center is O, and whose diameter MN is equal to AB, is the
path of the crank-pin. Now if BP is the piston displacement
on the forward stroke, the crank-pin position, C, correspond-
ing to this is found by taking PD equal to BN, and drawing
20
SLIDE-VALVES.
CD perpendicular to the line of stroke until it cuts the circle
at C. Then is CO the crank position, and CON is the crank
angle corresponding to the given piston displacement on the
forward stroke. Now let AP', equal to BP, be the piston
displacement on the return stroke. Take P'D f , equal to AM,
FIG. 18
or, which is the same, BN or PD, and locate C' by drawing
D ' C' perpendicular to the line of stroke until it cuts the circle
at C'. Then is C'O the crank position and C'OM is the
crank angle corresponding to the given piston displacement.
As MD' is equal to ND, both being equal to the piston dis-
placement, the angles C'OM and CON are equal.
With the crank and connecting-rod arrangement none of
these things are true. This is shown in Figs. 19 and 20, which
are lettered similarly to Fig. 18. The length of the connect-
ing-rod is BN With the piston in its middle position on the
forward stroke, as at P, Fig. 19, the crank- pin position is C,
found by taking P as a center and striking an arc, with the
length of the rod as a radius, until it cuts the circle at C.
Then is CO the crank position for mid-stroke, forward, of the
piston, and CON is the crank angle. During the first half of
the piston stroke the crank-pin travels from N to C\ and
CRANKS, CONNECTING-RODS, AND ECCENTRICS.
21
during the last half the crank-pin moves from C to M. With
a uniform velocity of the crank-pin, which is assumed, the
piston travels slower during the last half of the forward stroke
FIG. 19
than during the first half. On the return stroke, with the
piston at P, the crank position is C'O, and the crank angle is
C'OM, which is evidently greater than CON- It is also
FIG. 20
c'
evident that the piston velocity is less during the first half of
the return stroke than during the last half; for the crank-pin,
moving at a uniform rate, takes longer to pass over MC' than
it does to cover C'N.
22 SLIDE- VA L VES.
In Fig. 20 the piston-displacements are the same as in
Fig. 1 8, but the crank-pin positions are found by striking
arcs about P and P' as centers, the radius in each case being
the length of the connecting-rod. The crank angles C'OM
and CON are evidently unequal, C'OM being the greater.
The foregoing may be summed up as follows:
With the ordinary crank and connecting-rod the crank
angles corresponding to equal piston displacements on the for-
ward and return strokes are unequal, that on the return stroke
being the greater.
With the ordinary crank and connecting-rod, assuming tJie
velocity of the crank-pin to be uniform, the piston velocity is
not uniform, being less during the crank- end half of each stroke
than during the head-end half.
The variation between the crank angles on the forward and
return strokes depends on the length of the connecting-rod as
compared with that of the crank; the longer the connecting-
rod, the less will be the difference between the crank angles
for equal piston displacements on the forward and return
strokes. When the rod is infinitely longer than the crank,
there is no difference between the angles; and hence the
slotted-cross-head form of connection is usually referred to as
the " infinite connecting-rod." With the ordinary types of
engines, where the rod is from four to ten times the length
of the crank, the difference must be considered, and affects
the amounts of lap required on the ends of the valve to secure
cut-off at the same point in each stroke, making the laps
unequal. This is because the angle between the crank and
eccentric is fixed, and the valve displacement varies with the
<crank-angle; hence if the crank-angles are unequal, the valve
displacements are unequal; and the lap required to cut off at
any point in the stroke being equal to the valve displacement
at that point, it follows that unequal crank-angles require
unequal laps.
With the eccentric the case is different, as the eccentricity
CRANKS, CONNECTING-RODS, AND ECCENTRICS. 2$
is so much less than the length of the eccentric-rod that the
effect of angularity may be neglected. This is because the
eccentricity is much less than the throw of the crank, while
the eccentric-rod is equal to or longer than the connecting-
rod; hence the ratio of the eccentric-rod to the eccentricity
is much greater than the ratio of the connecting-rod to the
crank. The valve displacement is therefore reckoned as equal
to the eccentric displacement.
It is, therefore, necessary to determine the crank positions
corresponding to the piston positions at which cut-off is
desired, when designing a valve for any engine; or, when
studying the action of a valve already in use, to determine the
piston position corresponding to a given crank position. The
method of Figs. 19 and 20 may be employed, but it has the
disadvantage of requiring to be drawn on a large scale, and
of having the lines intersect at acute angles, making it hard
to obtain good results. A better method is shown in Fig. 21.
This method is a modification of the original method of
M. Marcel Deprez, and the author acknowledges his indebted-
ness to the Practical Engineer.
Lay off the straight line X Y of indefinite length, and on
it take any convenient point, as O, for a center; and about
O describe a circle with a radius equal to the length of the
crank. This gives the crank-circle, whose diameter is MN.
Lay off the distance OO' , equal to the square of the crank
length, divided by four times the length of the rod, or
where R = length of connecting-rod;
C = length of crank.
About O' as a center, and with a radius equal to the length
of the crank, describe an auxiliary crank-circle, whose
diameter is M'N'. The point O f should always be between
24 SLIDE- VA L VES.
the center of the crank-shaft, O, and the cylinder; that is, in
the case shown, the cylinder is at the right.
Now determine the crank-pin position corresponding to
the given piston position on the auxiliary crank-circle, as if
the slotted cross-head were used. For example, if the piston
is at \ stroke, take N'D equal to i of N'M', and draw DC
FIG. 21
perpendicular to XY until it cuts the auxiliary circle at c.
Then c is the crank-pin position on the auxiliary circle.
Next, from c, draw a line to O. Then the part of this
line cO included within the crank circle is the crank position
corresponding to the given piston position; that is, CO is the
required crank position. This diagram shows very clearly the
CRANKS, CONNECTING-RODS, AND ECCENTRICS. 2$
difference between the finite and infinite rods. For cO' is
the crank position corresponding to the given piston position
with the infinite rod, and the crank angle cO ' N' is evidently
different from CON.
For the return stroke the process is similar. Take M'D'
equal to the given piston displacement. Draw D 1 ' c' ', per-
pendicular to the line of stroke until it cuts the auxiliary
crank-circle at c 1 '. Join c' and (9, and produce Oc' until it
strikes the crank-circle at C' . Then is OC 1 the crank position
corresponding to the given piston displacement.
A reversal of this method will give the piston displace-
ment corresponding to any given crank position. Draw the
given crank position, and determine its intersection with the
auxiliary circle, producing the given crank-line, if necessary.
From this intersection drop a perpendicular upon the line of
stroke. The distance from the foot of this perpendicular to
the dead-center of the auxiliary crank-circle gives the piston
displacement.
It will be seen at once that the distance OO' varies with
the ratio of the connecting-rod to the crank, and that the
difference between the crank angles depends on the length of
00'.
To save calculation, Table I has been prepared, which
gives the values of OO' in terms of the crank length. For
example, if the stroke length is 36 inches and the connecting-
rod is 5 feet 3 inches long, the distance OO' is i-g^- inches, if
laid out full-size, found as follows: The crank is one half of
the stroke, or 18 inches, and, the connecting-rod being 5 feet
3 inches or 63 inches length, the ratio of the connecting-rod
to the crank is 63 ^- 18 = 3^. From Table I it is found
that for this ratio the distance OO is equal to the crank
length divided by 14 or multiplied by .0714. Using the
latter gives 18 X .0714 = 1.2852 inches. From Table II
it is found that the nearest fraction of an inch to .2852 is
26 SLIDE- VA L VES.
.28125, corresponding to -/$ of an inch. OO' is therefore
made i 9 ^- inches.
If the diagram is to be made on any pther scale, OO' is
reduced or enlarged in proportion. At half-size it would be
.6426 or 41..
CHAPTER III.
VALVE DIAGRAMS-GENERAL PRINCIPLES.
IN designing a new valve, or in studying the action of an
old one, it is necessary to know the valve displacement corre-
sponding to a given crank position. This may be found by
making a center-line sketch of the gear employed, but this
requires too much space, is not sufficiently accurate, and is
too long and tedious.
There are various diagrams designed to show at a glance
the relation between the crank position and the valve dis-
placement, all having certain good points. The one which
will be used in this work is that developed by Dr. Gustave
Zeuner.
CASE I.
That of a simple valve, with neither outside nor inside lap;
angle between crank and eccentric 90. Such a valve is
shown in Figs. 3 to 8.
Draw the two lines JTFand VZ, Fig. 22, at right angles,
intersecting at O. This point O represents the center of the
crank-shaft, and VZ represents the line of stroke. About
O as a center, and with a radius equal to the length of the
crank, describe the crank-circle as shown. Take OE on X Y
equal to one-half the eccentricity, and describe the valve-circle
shown. Locate E' in the same way, and draw the lower
valve circle.
Now, having this figure, the valve displacement for any
crank position is equal to that portion of the crank line which
27
28
SLIDE-VALVES.
is included in the valve-circle. For example, when the crank
is in the position 0i, the valve displacement is Oj.
This diagram shows the action of the valve to be just as
described in Chapter I. Suppose the engine to be running
over, as shown by the arrow; then the action of the valve OR
the head-end is as follows: With the crank at A, the crank
line is A O, and the valve displacement is zero, as AO does
not cut either valve-circle. When the crank moves ahead to
I, the valve displacement is Oj. When the crank reaches 2
VALVE DIAGRAMS GENERAL PRINCIPLES. 29
at right angles to line of stroke, the valve is displaced the
distance Ov, which is the eccentricity, or half travel. While
the crank travels from A to 2 the port is gradually opening.
From 2 to R the port gradually closes, until at R it is closed.
The valve now commences to reverse its direction, as shown
by the fact that the crank line, as it moves on, cuts the lower
valve circle. The valve is at the other end of its travel when
the crank-pin is at 3, as shown by the fact that the valve
displacement is Os, equal to the eccentricity. When the
crank reaches OA, the valve is again central.
With this type of valve the port-opening is equal to the
valve displacement. Therefore, during the entire forward
stroke, or while the crank-pin moves from A to R, the head-
end port is open to admission, and during the entire return
stroke, or while the crank-pin goes from C to A, the head-end
port is open to exhaust, as shown in the figure.
The action of the valve on the crank end is directly the
reverse; that is, the valve displacement which opens the head-
end port closes the port at the crank end. The events of the
strokes are therefore read from- opposite valve circles, as
shown in Fig. 23. With the crank at A' the valve is central,
but as it moves on the crank-end port is opened to admit
steam, by the amount of the valve displacement, until, with
the crank-pin at 3, the crank-end port is wide open. From
there on to R! the port is being closed, until at R the valve
is central. Admission and exhaust are therefore reversed.
CASE II.
Simple D-valve, with outside lap; angle between crank
and eccentric 90.
The only difference between such a valve as this and that
of Case I is that the port-opening for admission is not equal
to the valve displacement, but is less by the amount of outside
lap, as pointed out in Chapter I. The exhaust remains
SLIDE-VALVES.
unchanged. The diagram for this case is shown in Figs. 24
and 25.
In order to find the port-opening for a given crank posi-
tion, it is only necessary to subtract the amount of the lap
from the valve displacement; and this can be done by striking
an arc about O as a center and with a radius OL equal to the
outside lap. This being done, the port-opening is given by
the following
RULE: The port- opening for any given crank position is
equal to that portion of the crank-line which is included
VALVE DIAGRAMS GENERAL PRINCIPLES. 31
between the lap and valve-circles. For example, with the
crank-pin at 2, Fig. 24, the port-opening is Lv.
Fig. 24 represents the diagram for the head end. Events
on the forward stroke are read above the line of stroke;
events on the return stroke are read below it, the direction of
rotation of the engine being shown by the arrow. With the
crank at T there is no displacement of the valve. When it
gets to A, the displacement is just equal to the lap, and any
further movement of the crank will open the head-end port
OF THE
UNIVERSITY
<>A
SLIDE-VALVES.
to steam. Therefore admission begins at A ; and it lasts
until the port is again closed, which will be when the crank
is at C, when the valve displacement is equal to the outside
lap. This gives the two following rules:
"Admission" begins when the crank line passes through
the first intersection of the valve and outside-lap circles. OA
passes through Oa.
"Cut-off" occurs when the crank line passes through the
second intersection of the valve and lap circles.
VALVE DIAGRAMS GENERAL PRINCIPLES. 35
In Fig. 24 the lap-circle is only drawn above the line of
stroke. This is because the outside lap does not affect the
exhaust, and the exhaust and admission are read from opposite
valve-circles.
The exhaust opens at R, and closes at T that is, it lasts
during the entire stroke, just as in Case I.
From C to R no steam is admitted, and the steam
admitted up to cut-off expands.
From T to A the cylinder is empty, and the engine runs
only by virtue of the momentum of the fly-wheel.
Fig. 25 shows the diagram for the crank-end of the
cylinder, the valve being supposed to have the same outside
lap on both ends.
CASE III.
Simple D-valve, both outside and inside lap; angle
between the crank and eccentric 90.
This case is shown in Figs. 26 and 27, the former being
for the head-end, and the latter for the crank-end. The valve
is supposed to have the same amount of inside lap at both
ends. The valve travel, crank throw, and outside lap are the
same as in the preceding problem.
The inside lap affects only the exhaust; and as its effect
is to lessen the amount that the port is open to exhaust by
the amount of lap, allowances can be made for it by striking-
an arc about O as a center, and with a radius OI equal to the
inside lap, in a manner similar to that employed for the out-
side lap. This gives as a rule:
The port-opening to exhaust corresponding to any given
crank position is equal to that portion of the crank-line which
is included between the valve-circle and inside lap-circle. The
exhaust-opening and steam-opening are read from different
valve circles, it must be remembered.
Fig. 26 is the same as Fig. 24 as far as the facts relating
34
{SLIDE-VALVES.
to admission and cut-off are concerned. The exhaust does
not open until the valve displacement is equal to the inside
lap, or until the crank is at R. It remains open while the
cranK passes from R to T, when the valve displacement is
again equal to the inside lap. From which these two rules:
"Release" or exhaust -opening, occurs when the crank-line
passes through the first intersection of the valve and inside lap
circles. OR passes through r.
" Compression," or exhaust-closure, occurs when the crank
VALVE DIAGRAMS GENERAL PRINCIPLES.
35
line passes through the second intersection of the valve and in-
side-lap circles. OT passes through /.
From C to M expansion occurs; and a peculiarity of this
type of valve is that from M to R the steam is compressed.
FIG. 27
This is evident because the piston reverses the direction of
its motion, and the exhaust does not open until the piston
has passed over a portion of its return stroke. In the same
way, compression occurs from T to N, and from N to A this
compressed steam expands.
SLIDE-VALVES.
CASE IV.
Simple D-valve, with both inside and outside laps, with
the angular advance sufficient to open the port to steam at
the beginning of the stroke.
The diagrams for this case are shown in Figs. 28 and 29,
<o-
ADMISSION
to'
FIG. 28
Fig. 28 being for the head-end. These figures, as in the
previous cases, are lettered alike, the sole difference being that
VALVE DIAGRAMS GENERAL PRINCIPLES.
37
the letters are primed in the crank-end diagram. The nota-
tion shown and used thus far will be used throughout the
work.
In order to make the valve-circle pass through the inter-
section of the outside lap-circle and the valve-circle, it is
I
o
o
V I
ADMISSION
L'l
2 ~~^+5
.V
I
Y
FIG. 29
necessary to incline the line DD' joining E and /?', as shown
in the cuts. The angle DOX is equal to the angular advance ;
and it makes no difference whether the eccentric leads or
follows the crank that is, whether the angle between the
38 SLID E- VA L VES.
crank and eccentric is 90 + the angular advance, or 90
the angular advance the \mzDD' is always inclined as shown,
with the upper end, D, brought over toward the cylinder.
The angular advance will take care of itself when it comes to
the valve-setting. Nor does it make any difference whether
the cylinder is to the right or left of the crank-shaft as it is
located in the diagram. The diagrams are for the head-end
and crank-end, which disposes of that question. Neither
does it make any difference whether the engine runs over or
under. This is one of the beauties of the diagram that it
fits any and all circumstances.
The dimensions of the valve being taken the same as in
the preceding case, it will be seen at once that the diagram
shows the effect of angular advance to be, as stated in
Chapter I, that of making all the events of the stroke earlier.
While the crank is traveling from 5 to 4 the valve moves
in one direction ; and while the crank travels from 4 to 5 the
motion of the valve is reversed. This will help to explain
why the facts relating to exhaust and admission are not read
from the same valve circles. For, referring to Figs. 23 to
28, it will be seen that when the valve leaves its central
position and moves to the left, it uncovers the head-end port
to steam, and continues to leave it open until it is again
central on its travel to the right. That is, the steam-opening
is affected by the last half of the left-hand travel of the valve
^
and the first half of the right-hand travel. Then, in all the
valve diagrams, Ov represents the last half of the travel in
one direction, and vO represents the first half of the travel in
the other direction. The exhaust on the head-end, Figs. 3
to 8, is affected in the opposite way, that is, by the first half
of the left-hand travel and the last half of the right-hand
travel, a-nd is therefore read from the other valve-circle, where
Os is travel in one direction, and sO is travel in the other.
VALVE DIAGRAMS GENERAL PRINCIPLES.
39
CASE V.
Simple D-valve, both inside and outside laps, and having
the angular advance sufficient to o^pen the port before the
beginning of the stroke.
This case, which is shown in Figs. 30 and 31, simply
EIG. 30
amounts to moving the eccentric a little farther ahead.
Admission occurs when the crank is at A. The amount de,
40
SLIDE-VALVES.
which the port is opened when the piston begins its stroke,
is called the " steam-lead," or simply " lead." The angle
ZOA is called the "angle of lead," or "lead angle.'
FIG. 31
Similarly, fg is the exhaust-lead, and ROM is the exhaust-
lead angle.
The "lead" angle is the angle between the crank line at
Admission and the line of stroke. ZOA is the lead angle.
This case is the general one. The instructions for drawing
it may be summed up as follows:
VALVE DIAGRAMS GENERAL PRINCIPLES. 4!
Draw the two lines XY and VZ at right angles, intersect-
ing at (9, which represents the center of the crank-shaft, VZ
representing the line of stroke. About O as a center, and
with a radius OM equal to the crank, describe the crank-circle
whose diameter is MN, which is equal to the stroke of the
engine. Lay off the angle XOD equal to the angular advance.
Draw DOD'. Lay off OE equal to half of the eccentricity,
and with E as a center and a radius OE describe the upper
valve-circle. Locate E' and describe the lower valve-circle
in the same way. (It is neither necessary nor advisable to
use the same scale for the valve and crank-circles. The
crank-circle is only employed to determine the inclination of
the crank to the line of stroke, and need be drawn only on a
small scale.) About O as a center, and with a radius OL equal
to the outside lap, describe the outside lap circle. About O
as a center, and with a radius Of equal to the inside lap,
describe the inside lap circle.
The diagram is then read in accordance with the rules
given.
EFFECT OF CHANGING VARIOUS DIMENSIONS OF THE VALVE,
AS SHOWN BY THE DIAGRAM, WHEN THERE IS
ANGULAR ADVANCE.
Angular Advance. The effect of changing this is shown
in Figs. 28 and 30, or 29 and 31. If the angular advance is
increased, its effect is to make all the events of the stroke
earlier. If it is decreased, all the events occur sooner.
Refer to Table III.
Valve Travel. If this is increased, cut-off and compres-
sion come later and admission and release occur earlier. If
it is decreased, cut-off and compression occur earlier and
admission and release come later. Refer to Fig. 32 and
Table III.
SLIDE-VALVES.
Outside Lap. Increasing this makes admission later and
cut-off earlier, prolongs the expansion and compression,
decreases the lead, and leaves the exhaust unchanged. De-
creasing the outside lap increases the lead, makes cut-off later
F
FIG. 32
and admission earlier, shortens expansion and compression,
and does not affect the exhaust. Refer to Fig. 33 and
Table III.
Inside Lap. Increasing this does not affect lead, admission,
or cut-off, but it makes release later and compression earlier,
*&
L^
OF THB
UNIVERSITY
^^ .. "" ~ ; ^
VA L VE D I A GRA MS GENERA L^Sf&ef&fES.
43
thus prolonging expansion and compression. Decreasing the
inside lap makes release earlier and compression later, thus
\o
V
V
ADMISSION
FIG. 33
shortening both compression and expansion. See Fig. 33
and Table III.
CHAPTER IV.
DIMENSIONS OF PORTS, STEAM-PIPES, AND BRIDGES.
A STUDY of the valve diagrams will show at once that the
dimensions of the valve are influenced very strongly by the
point of cut-off. Next in importance to this, as a controlling
influence, is the width of the steam-port. This latter is not
assumed arbitrarily, but is calculated with a reasonable degree
of accuracy.
When steam passes through an opening at a greater speed
than 6000 or 8000 feet per minute it is choked, throttled, or
(< wire- drawn " ; that is, its pressure after passing through the
opening is less than the pressure urging it through. For this
reason the steam must not be forced to pass through the
ports at a greater speed than 6000 feet per minute. Now, in
order to fill the cylinder up to cut-off with steam at boiler-
pressure, it is necessary that the volume of steam admitted
shall be equal to the volume swept through by the piston.
The volume of the cylinder up to cut-off is of course equal
to the area of the piston multiplied by the distance through
which the piston has traveled. If the piston area is one square
foot, and the travel up to cut-off is 2 feet, then 1X2 = 2
cubic feet of steam must be admitted while the piston is
moving through those two feet. The amount of steam to be
supplied can also be found by multiplying the area of the
piston in square feet by the piston speed in feet per minute,
and multiplying this result by the time in minutes that the
port is open. This may be expressed in a formula as follows:
C= A X SX T, . . . . . . (i)
44
DIMENSIONS OF PORTS, STEAM-PIPES, AND BRIDGES. 45
where C '= volume in cubic feet up to cut-off;
A = piston area in square feet;
5 = piston speed in feet per minute;
T time in minutes that port is open.
For example, if the piston area is I square foot, the pis-
ton speed 400 feet per minute, and the port is open ^^ of a
minute, then
C i X 400 X - - = 2 cubic feet.
200
Now, this steam must be admitted through the port, which
is open for the same length of time. The volume of steam
thus admitted is equal to the velocity of the steam in feet per
minute multiplied by the time the port is open, and this
result then multiplied by the area of the port in square feet.
Expressed in a formula this is
C= VX PX T, ...... (2)
where C = volume admitted in cubic feet;
V =. velocity of steam in feet per minute;
T= time in minutes that port is open.
For example, if the velocity of the entering steam is 6000
feet per minute, the port area ^ of a square foot, and the
port is open ^.J^ of a minute, then
C = 6000 X X - = 2 cubic feet.
The results given by formulas (i) and (2) must be equal,
of course, so that
A X SX T= FXPX T,
or, as T is on both sides, it may be dropped, giving
A X 5= VXP. . . . . ..-,. (3)
46 SLIDE- VA L VES.
In designing a valve, the area of the piston is known, and
the velocity of the steam is assumed in accordance with the
statement made above. The piston speed is found by multi-
plying the length of the stroke by twice the number of
revolutions per minute; because the piston makes two strokes
for every revolution. That is,
....... (4)
where N = number of revolutions per minute;
L length of stroke in feet.
Now, from formula (3),
or
piston area X piston speed
port area -- ; r- --- - f - .
velocity of steam
This may be expressed in the following:
RULE.
To Find the Port Area for a Given Engine : Multiply the
piston area in square feet by the piston speed in feet per minute ',
and divide the product by the velocity of the steam, in feet per
minute. The result will be the area of the port in square feet.
For example, take an engine 18 X 24 that is, one in
which the cylinder is 18 inches in diameter and 24 inches
stroke, running at 200 revolutions per minute, and assume
the velocity of the steam to be 6000 feet per minute. To find
the port area required. The piston speed is, according to
formula (4),
24
2 X r X ioo = 400 feet.
The 24 is divided by 12 to reduce the stroke to feet. The
area of an 1 8-inch circle is 254.47 square inches, from Table V
at the end of the book, or 254.47 -f- 144 1.766 square feet.
DIMENSIONS OF PORTS, STEAM-PIPES, AND BRIDGES. 47
Now, applying the rule, 1.766 X 400 = 706.4, and 706.4 -f-
6000 = .1174 square feet, or 16.91 square inches.
The work of this is renderedve ry light by using Table IV,
which contains the values of
piston speed
velocity of steam
for various velocities of steam, and for piston speeds from 100
to 1200 feet per minute, advancing by 25. It is only neces-
sary to multiply the figure given in the table by the piston
area to get the port area. Take the example just worked.
Under 6000, in the column headed " Port Area, Piston Area
as Unity," and opposite 400 feet piston speed, will be found
.067. Multiplying the area of piston, 254.47 square inches,
by .067, gives 17.049 square inches. The difference between
this result and the one obtained by the longer method is very
slight, only .13 of a square inch, and is due to the use of
decimals.
The length of the port should be made, as nearly as possi-
ble, equal to the diameter of the cylinder, and the width is
of course found by dividing the area found as above by the
length. If the length of the port is nine tenths of the
diameter of the cylinder, the width may be found by multi-
plying the diameter of the piston by the figure given in
Table i V. For example, in the engine just considered, if the
port length is .9 of the cylinder diameter, the port width is
found by mutiplying 18 (the diameter of the cylinder) by
.058, which is the figure opposite the piston speed, 400, and
under " 6000, Width of Port." This gives 18 X .058 =
1.024 inches as the width of the port.
If the velocity of the steam were assumed as 4000, the
multiplier would be .086. If the velocity were 8000 and the
piston speed 300, the multiplier would be .033.
After the steam has been expanded in the cylinder it is at
a lower pressure than when it was admitted, and will travel
4 SLID E- VA L VES.
at a lower rate of speed. The exhaust-ports should therefore
be made with a greater area than the steam-ports. In order
to allow for this it is usual to assume that the velocity at
entrance is 6000 feet per minute, and at exhaust is 4000 feet
per minute. These velocities are exceeded in certain cases,
such as locomotives, and the table has been extended to cover
these excesses; but throughout this work the figures just
mentioned will be employed.
The area of the exhaust-ports may be found by applying
the rule given, or the table may be employed.
Of course when the same port is used for admission and
exhaust, as in the case of the common slide-valve, the port
width must be made that corresponding to the lower velocity,
in order that the exhaust may be free and unimpeded.
The diameter of the steam- and exhaust-pipes for a given
engine may be determined by the following rule:
To Find Diameter of Steam- and Exhaust-pipes : Multiply
the diameter of the piston by the square root of the port
area; the latter being expressed as a fraction of the piston
area.
This rule is obtained as follows:
The piston area is taken as unity. As explained before,
the port area is a certain fraction of the piston area. Repre-
sent this fraction by F. The areas of two circles are to each
other as the squares of their diameters, so that
I : F : : D* : d\
where I = piston area;
F= port area, as a fraction of piston area, given in
column of table headed " Port Area";
D = diameter of piston;
d = diameter of pipe.
This formula gives
DIMENSIONS OF PORTS, STEAM-PIPES, AND BRIDGES. 49
In order to save the trouble of the extraction of the square
root, a column in Table IV, headed " Diameter of Steam-
Pipe," has been provided, which contains the values of VF
corresponding to different piston speeds. This column is used
in the same manner as the others. For instance, for the
engine just discussed, 18 inches diameter and 400 feet piston-
speed, the steam-pipe diameter is 18 X .258 = 4.82 inches;
the figure .258 being under " 6000, Diameter of Steam-Pipe "
and opposite 400. The exhaust-pipe diameter is found from
he table by assuming the lower velocity. In this case it
would be 18 X .316 = 5.688 inches.
Suppose an engine designed for a slide-valve having a port
width E, Fig. 34, equal to that required for free exhaust.
If the port is opened its full width, the valve assumes the
position shown in the figure when at the end of its travel. It
is only necessary, however, to open the steam-port sufficiently
to allow a free admission; that is, the valve assumes the posi-
tion shown in Fig. 35, A being the width required for
unchoked admission. By designing a valve to open the port
only the smaller distance, the travel can be decreased by the
difference between E and A. This will not interfere with a
free opening for exhaust, as this event is controlled by the
inner edge of the valve; and the lap there can be so modified
as to give the desired opening. As shown in Fig. 35, the
exhaust is fully open.
When the valve has reached its extreme position, one port
is open to its greatest extent to admit steam, and the other
one is opened as fully as possible to exhaust. The amount
that the port is open to steam in this position is called the
" maximum port-opening."
The maximum port-opening \ made equal to the width of
port necessary for free admission of steam.
The maximum opening for exhaust is equal to the width of
the port.
When the travel of the valve is so great that the valve
5 SLIDE-VALVES.
assumes the position shown in Fig. 36 when in its extreme
position, the amount x that it runs over the edge is called the
overtravel.
In order to fix these principles, as well as to illustrate the
use of the tables again, another example will be worked out.
GIVEN. An engine 20 X 30, running at 150 revolutions.
Length of ports, 19 inches.
REQUIRED, (i) Width of port. (2) Diameter of steam-
pipe. (3) Diameter of exhaust-pipe. (4) Maximum port-
opening.
150 X 20 X 2
SOLUTION, (i) Piston speed = 500 feet per
minute. See formula (4). In Table I, under "4000, Area
of Port" and opposite 500, is .125. The area of a 2O-inch
circle is 314. 16 square inches. 314.16 X .125 = 39.27 square
DIMENSIONS OF PORTS, STEAM-PIPES, AND BRIDGES. 5
inches. 39.27 -f- 19 = 2.668. Nearest sixteenth,
Width of port = 2\\.
(2) In Table I, under " 6000, Diameter of Steam-Pipe "
and opposite 500, is .288. Then the diameter of steam-pipe
should be 20 X .288 = 5.76 or $ inches.
(3) In Table I, under "4000, Diameter of Steam-Pipe"
FIG. 36
and opposite 500, is .353. Then the diameter of the exhaust-
pipe should be 20 X -353 7.06 or 7 T V inches.
(4) The velocity of the steam at exhaust being 4000, while
that at exhaust is 6000, it follows that the port need only be
opened | of its width to secure free admission. This is
equal to f. The maximum port-opening is therefore f X 2|
= 1.79 inches. It is therefore made iff- inches.
The same result would be obtained by using the table.
That part of the cylinder-casting which divides the steam-
and exhaust-ports is called the " bridge." B, Fig. 36, is the
width of the bridge. This width must be sufficient so that if
the valve has overtravel that is, if the end of the valve runs
beyond the edge of the port, as shown in the figure the
steam- and exhaust-ports will not be in communication. If
the greatest opening of the port does not exceed the width
of the port, it is sufficient to make the bridge width equal to
the thickness of the cylinder-casting, which makes it compara-
tively easy to obtain a good casting; but if the valve has
5 2 SLIDE- VA L VES.
overtravel greater than the thickness of the cylinder-walls, it
is necessary to increase the thickness of the bridge ; which
may be summed up in the following rule:
The Width of tJie Bridge is made equal to the thickness of
the cylinder-wall when the valve has either no overtravel or
an overtravel less than the thickness of the cylinder-walls.
When the overtravel exceeds this amount, the bridge thick-
ness is found by adding J inch to the port-opening and sub-
tracting the width of the port from the sum. The remainder
is the thickness of the bridge.
Fig. 36 shows the valve in its extreme left-hand position.
The exhaust is then passing through the port P and the open-
ing O. These two openings must be equal in order that the
exhaust shall not be cramped; for it was determined that the
width P is necessary for free exhaust from the cylinder, and
the same width is evidently necessary at the exhaust, in
order not to increase the velocity of steam at exhaust to such
an extent as to increase the back pressure. The valve shown
in the figure has no inside lap; and as the valve has moved a
distance E, equal to one half the valve travel from its middle
position, the distance D, which is the width of the exhaust-
port, must be equal to E + O B\ but O is equal to P, so
that
D = E + P - B.
If the valve has inside lap, the width of the exhaust-port
must be increased by the amount of the inside lap.
To Determine the Width of the Exhaust-port : i . When
the valve has no inside lap, add together the width of the
steam-port and one half the valve travel. From the sum sub-
tract the width of the bridge. The remainder is the width of
the exhaust-port. D = E -f- P B.
2 . When the valve has inside lap, add together the width of
the steam-port and the eccentricity or half-travel of the valve.
From the sum thus obtained subtract the width of the bridge,
DIMENSIONS OF PORTS, STEAM-PIPES, AND BRIDGES. 53
and to the remainder add the amount of the inside lap. The
sum is the width of the exhaust-port. Z> E -{- P B I.
3. When the inside lap is negative that is, when the valve
allows the port to be open to exhaust when the valve is in its
middle position add together the half -travel or eccentricity and
the width of the steam-port. Then add together the amount of
the negative lap or inside clearance and the thickness of the
bridge. Subtract this latter sum from the first, and the re-
mainder is the zvidth of the exhaust-port. D E -\- P B /.
CHAPTER V.
VALVE DIAGRAMS GENERAL PROBLEMS OF DESIGN.
THERE are several different problems which may present
themselves for solution, circumstances beyond the designer's
control having fixed certain dimensions. But the following
problems will cover the probable cases very completely.
PROBLEM I.
GIVEN. Eccentricity, point of cut-off, angle of advance,
and point of compression.
REQUIRED. Lap, exhaust-lap, lead, exhaust-lead, and
greatest possible openings of port to both admission and
exhaust.
SOLUTION. The solution of this problem is shown in Fig.
37. Draw XY and VZ at right angles, intersecting at O.
Draw DOD f , making the angle DOX equal to the given angle
of advance. Find CO, the crank position corresponding to
the given point of cut-off by either of the two methods pre-
viously described. It is best to make the determinations of
crank positions on a separate sheet, and transfer them to the
sheet on which the solution is being made. Then lay off ov
equal to the eccentricity or half-travel. Find E, the middle
point of ov, and about E as a center, and with a radius equal
to EO, describe the upper valve circle. In a similar way find
R' and draw the lower valve circle.
Now the upper valve circle intersects oc at c. Then,
according to Chapter III, oc must be the outside lap. Then
54
VALVE-DIAGRAMSGENERAL PROBLEMS OF DESIGN. 55
draw in the lap circle about O as a center and with a radius
equal to oc. This will cut VZ at d, and the distance de" from
d to e> the intersection L the valve circle with VZ, is the
lead. The lap circle cuts the valve circle at a, and therefore
OA, drawn through the point a is the crank-position at
admission. The angle ZOA is the lead angle.
Now draw in OT, the crank-position corresponding to
the given point of compression. This intersects the lower
valve circle at /, and Ot is therefore the amount of the inside
lap. The inside lap-circle is then drawn in about O as a
center and with Ot as a radius. Its intersection with the
5& SLIDE- VA L VES.
valve-circle at r determines OR, the crank-line at release.
The exhaust lead isj^, and the exhaust lead angle is VOR.
PROBLEM II.
GIVEN. The lap, point of cut-off, and lead.
REQUIRED. The valve travel and angle of advance.
SOLUTION. Fig. 38 shows one solution of this problem.
The lines XY and VZ are drawn at right angles, intersecting
FIG. 38
at O. The crank-position at cut-off is OC, as given. Then
the lap-circle is drawn in about O as a center and with a
radius equal to the given lap. Its intersection c with OC is
a point through which the valve-circle must pass. O is
another point on the valve-circle. Another point is ^, found
by laying off de equal to the given lead. These three points
VALVE-DIAGRAMSGENERAL PROBLEMS OF DESIGN. 57
serve to determine the center of the valve-circle, as follows:
Find m, the middle point of OC, and draw my perpendicular
to Oc. Find n, the middle point of Oe, and draw nx perpen-
dicular to Oe. Then nx and my intersect at E, which is the
center required. Then draw in the valve-circle about E as a
center and with EO as a radius. Draw DEO, and its intersec-
tion v with the valve-circle determines Ov, which is the
eccentricity. The valve-travel is twice the eccentricity.
Another solution is shown in Fig. 39. Having found the
points c and ^, draw ev perpendicular to VZ. Draw cv per-
FIG. 39
pendicular to OC, intersecting ev in v. Then draw DvO.
Then is vO equal to the valve travel. If the middle point,
E, of vO is located, and > circle of radius EO drawn about E
as a center, it will pass through c and e as well as O, showing
the construction to be correct.
SLIDE-VALVES.
PROBLEM III.
GIVEN. Cut-off, angle of lead, width of port, and over-
travel.
FIG. 40
REQUIRED. Eccentricity, lap, lead, and angle of advance.
SOLUTION. Draw XY and VZ, Fig. 40, at right angles to
VALVE-DIAGRAMSGENERAL PROBLEMS OF DESIGN. $$
each other. Draw OA, the crank position corresponding to
the given lead angle; that is, make ZOA equal to the given
angle. Draw OC, the crank-line at cut-off, as given. Now,
the center of the valve-circle must lie somewhere on the line
bisecting the angle formed by these two crank-lines. There-
fore draw DOD' ', bisecting the angle CO A, and the angle
XOD is the angular advance.
Take any convenient radius, as E 'O, and describe a trial
valve-circle, intersecting OD at i/. Then Oc' is the corre-
sponding lap, and by drawing in this trial lap-circle it will be
found that the greatest possible opening of the port is L'v 1 ',
with this lap and valve-travel. The greatest possible opening
of the port is evidently equal to the width of the port plus
the overtravel, and as both of these are given in the problem,
the maximum port-opening is known. It is hardly probable,
though possible, that the distance L'v' will equal the given
amount on the first trial. That being the case, proceed by
drawing v'g in any convenient direction, and take v'x equal
to the required greatest port-opening that is, equal to the
width of the port plus the overtravel. Then join L' and x,
and from O draw Og parallel to L' x, until it cuts v'g in g.
Then is v'g equal to the required eccentricity. Next lay off
Ov on OD, equal to v'g. Bisect Ov in E, and about E as a
center and with a radius OE draw in the valve-circle which
will cut OC at Cj thus determining the lap, Oc; and by draw-
ing the lap-circle the lead de is determined.
The maximum port-opening with this travel and lap is
evidently vL\ and if the construction is accurate, vL will be
found to be equal to v'x.
PROBLEM IV.
GIVEN. Point of cut-off, lead, and greatest possible port-
opening.
REQUIRED. Lap, valve-travel, and angle of advance.
6o
SLIDE-VALVES.
SOLUTION. This is shown in Fig. 41. Draw JTFand VZ
at right angles, intersecting at O. Draw OC t the crank- line
at cut-off, as given. Then produce OC to ;, making Om
equal to the given lead. This should be drawn on a large
scale, as the lead is usually small, seldom exceeding T 3 ^- of an
inch. A scale of from two to four times full size should be
employed.
Next take ms on mC, equal to the given maximum port-
opening. Through s draw nst parallel to the line of stroke
VZj making st equal to sm as shown by the arc mt. Then
join O and t, and on Ot take Oh equal to Om as shown by the
VALVE-DIAGRAMSGENERAL PROBLEMS OF DESIGN. 6 1
arc mh. Next about O as a center, and with a radius equal to
On, describe an indefinite arc. Through h draw gh parallel
to VZ. Then gh will cut the arc just drawn in g. The arc
cuts Ot in k, and this arc kn is next picked up in the compass
and laid off from g to D 1 '. Then draw D'O and produce it
indefinitely, as to D. Then is the center of the valve circle
on D'OD, and DOX is the angle of advance.
The problem now reduces to:
GIVEN. Cut-off, lead, angle of advance, and greatest port-
opening.
REQUIRED. Eccentricity.
This is the same as Problem III, and the solution is made
in accordance with the directions given for that. The draw-
ing, Fig. 41, covers this solution, and is lettered the same as
Fig. 40 to enable the construction to be followed throughout,
if desired.
The proof of these solutions is in one or two cases rather
mathematical, and is therefore omitted, as it is entirely un-
necessary to understand this proof in using the diagrams.
'In order to fix these principles, it is not sufficient to rest
content with merely reading the foregoing text, but it is
necessary to solve numerous examples under each problem.
Other problems will probably suggest themselves to the
mind, but they can either be reduced to the preceding ones,
or can be solved by the application of the general principles
of the diagram as outlined in Chapter III.
CHAPTER VI.
ROCKERS AND BELL-CRANKS.
IT has been said that the motion of the eccentric is very
often imparted to the valve through a rocker which joins the
end of the eccentric- rod to the end of the valve-stem.
This is a necessary appliance when the path of the valve
is not in the same straight line, that is, when the line of
valve-travel is parallel to the line of stroke, but is above or
below it, or on one side of it.
Figs. 42 and 43 show two forms of this rocker, the prin-
cipal difference between them being that with the plain
rocker, Fig. 42, where the pivot is at the bottom, the eccen-
A V
FIG. 42
trie-rod at the top, and the valve-stem attached somewhere
between, the valve motion is the same in direction as that of
the eccentric, and therefore the eccentric precedes the crank,
62
ROCKERS AND BELL-CRANKS. 63
as shown in the drawing; while with the bell-crank shown in
Fig. 43, with the pivot at the angle, the motion of the valve
is opposite in direction to that of the eccentric, arid therefore
the eccentric follows the crank. In both cases the amount of
motion of the valve depends upon the relative lengths of the
FIG. 43
rocker-arms to which the valve-stem and eccentric-rods are
attached.
The method of laying out these rockers will be understood
by reference to the figures, which are lettered alike, as the
same method is pursued in both cases.
Let VZ be the line of stroke, and O the center of the
shaft. Then take O as a center, and with a radius OE equal
to the eccentricity describe the eccentric-circle, as shown.
Let the direction of rotation be as represented by the arrow.
Then draw inn at the proper distance from VZ and parallel to
64 SLIDE- VA L VES.
it, to represent the line of travel of the valve-stem, and let X
be the position of the end of the valve-stem when the valve
is in its middle position. Lay off XL and XL ', each equal to
the outside lap of the valve. Then take L and L' as centers,
and with radii equal to the length of the longer arm of the
rocker describe the two arcs vs and xy, intersecting at R.
Then is R the pivot of the rock-shaft. Then take R as a
center, and a radius equal to the other arm of the rocker, and
draw the arc st, making it of indefinite length. From O draw
O Y tangent to the arc st, determining the point of tangency,
Y y by drawing RY perpendicular to OY, the point of intersec-
tion being the required point. Next, from O, draw OE
perpendicular to OY. Then OE represents the eccentric, and
by joining E and Y the length EY is determined, which is
the length of the eccentric-rod.
Peculiarities of the design may make it an object to make
the eccentric-rod of a certain length. This may be done by
shifting the pivot of the rocker until the desired length is
obtained. If the distance through which the pivot is to be so
moved is short, proceed as follows:
Through Y draw a line YG parallel to the line of stroke
VZ. With E as a center, and with a radius Ea equal to the
given length of the eccentric rod, describe an arc ah, intersect-
ing YG in//. Then move the whole arrangement through
the distance YH, changing the length of the valve-stem by
that amount. The same construction would hold good if the
rod were to be shortened instead of lengthened.
With this construction the action of the valve is not made
irregular at either admission or cut-off. The angle between
the crank and eccentric is changed, but this is a matter which
affects the valve-setting alone. The amount of the change
may be found by laying off ZOC equal to the angle of advance
as found by the valve diagram, and then OC represents the
position of the crank with the valve in its middle position, and
COE is the angle between the crank and eccentric.
ROCKERS AND BELL-CRANKS. 6$
Sometimes the valve-face is not parallel to the line of
stroke. This is a combination seldom met, as the lack of
parallelism renders it a difficult matter to face off the valve-
seat and bore the cylinder. Therefore an extended descrip-
tion of this type will be omitted. There is one special case,
however, which it is well to bear in mind, and that is when the
center line of the valve-stem is inclined to the line of stroke
in such a manner that it would, if produced, pass through the
center of the shaft. This changes the angle of advance by
just the angle between the two lines the line of stroke and
the center line of the valve-stem. If the change is but slight,
it may be neglected. In any case, it is a matter which affects
the valve-setting alone.
In both Figs. 42 and 43 the rocker-arms are unequal, the
shorter arm being f of the larger one, the resultant travel of
the valve being the same as if the eccentricity were of OE+
In designing and laying out the valve, it is treated as if it
were actuated by the larger eccentric. That is to say, no
allowance is made in construction until it comes time to lay
out the eccentric, when that is made with, in this case, f of
the throw which would be given it if it were either connected
direct or with an equal-armed rocker.
The valve diagrams given in Chapters III and V show that
if the cut-off is equalized on the two ends that is, if made
to occur at the same point on each stroke by making the out-
side laps unequal, the leads will be unequal; and if the leads
are kept equal, the cut-offs will be unequal. It is possible,
however, by the use of a bell-crank designed in a certain way,
to equalize the cut-offs and have the leads equal or nearly so.
This bell-crank is designed as follows: Determine the outside
lap, angle of advance, and eccentricity necessary to secure
the required cut-off and lead on one end of the valve. It
makes no difference which end of the cylinder is selected, as
the difference between the laps would be compensated for by
the difference between the resulting bell-cranks. Then pro-
66
SLIDE-VALVES.
ceed to construct the diagram shown in Fig. 44, which should
be drawn on as large a scale as possible in order to render the
work accurate.
Draw VZ to represent the line of stroke, and let O be the
center of the crank-shaft. About O as a center, and with
/,' T VALVE STEM
a radius equal to the length of the crank, describe the
crank-circle. Then describe the eccentric-circle about O as a
center and with a radius Oc equal to the eccentricity. Next
find the crank positions corresponding to the given point of
cut-off on the forward and return strokes respectively. These
are OC and OC' , and may be found by either of the two
methods given in Chapter II. The first method is the one
adopted here, because the method of Deprez, requiring the
use of the auxiliary crank-circle, would complicate the draw-
iftg too much at the right-hand end. Next draw in OA and
OA', the crank-positions at admission on the forward and
return strokes respectively. These positions are determined
by taking the crank-position for admission from the end of
the cylinder for which the lap, etc., were determined, and
making the other lead angle the same. For instance, if OA
was found, then the angle ZOA' is made equal to VOA. The
next step is to locate the eccentric-positions corresponding to
these four crank-positions. This is done by laying off from
each crank-position the angle between the crank and eccentric.
That is, COc, C'Oc'AOa, and A' Oa' are each equal to the
angle between the crank and eccentric.
Now, it must be remembered that when admission takes
ROCKERS AND BELL-CRANKS. 6?
place, the valve is just opening the port; and when cut-off
occurs, the valve is just closing the port. Consequently the
valve is in the same position at admission and cut-off, but the
motions are in opposite directions. That is to say, with the
eccentric-rod end at either a or c the valve must be in the
same position. Then with a and c as centers, and with radii
equal to the length of the eccentric-rod, describe arcs inter-
secting at 5. Then take the same radius, and with a' and
c' as centers describe two more arcs which intersect at t.
Join s and /, and find the middle point x. At x draw xR,
perpendicular to ts. Then if a line be drawn perpendicular
to the line of stroke, as, for example, RT, it will determine
the angle xR T between the two arms of the bell-crank. The
lengths of these two arms must be such as to give the required
movement of the valve, and are determined as follows.
On xR lay off xm equal to xs, and from m draw the line
mn perpendicular to the line of travel of the valve-stem, and
make mn equal to the lap of the valve. Draw xT from x
through n until it meets the line of travel of the valve-stem
at T; and from T draw TR, also perpendicular to the line of
valve-stem travel until it meets xR at R. Then will TR and
Rx be the lengths required for the two arms of the bell-
cranks, while R will be the point of suspension. The valve-
stem is fastened to T, and the eccentric-rod to x. The
general appearance is shown by the heavy lines.
In a similar manner the exhaust and compression can be
equalized, obtaining points similar to / and s\ and if these
points should happen to lie in such positions that an arc pass-
ing through them would also pass through s and /, it would
be possible to design a bell-crank which would equalize all the
events of the stroke. As an ordinary thing this coincidence
is impossible, so it may be taken as an accepted fact that with
a plain D valve it is impossible to equalize all the events of
the stroke. This is one of the greatest drawbacks to this type
of valve.
CHAPTER VII.
DESIGN OF A PLAIN D-VALVE.
IF the preceding chapters have been thoroughly mastered,
the student is in possession of sufficient information to design
a plain slide-valve. In order to fix clearly the various steps
and their order, a numerical example will be worked out
completely.
Let the engine be 24 X 24, running at 135 revolutions per
minute. Cut-off to be at $ stroke, compression at T *L stroke,
and the lead -J of an inch. The connecting-rod is five times
the length of the crank. Length of port 22 inches.
ORDER IN WHICH THE VARIOUS DIMENSIONS ARE TO BE
DETERMINED.
T 3 "V 2 A. y 2
1. Piston Speed. This is - - = 540 feet per
minute.
2. Area of Steam-port. This is determined by reference
to Table III, assuming that the velocity of steam at exhaust
is 4000 feet per minute. Under * * 4000, Area of Port ' ' and
opposite 550, which is the nearest to the given piston speed,
is .138. The piston being 24 inches in diameter, its area is
452.39 square inches, and the port area is therefore
452.39 X .138 = 62.430 square inches.
3. Width of Steam-port. This is the area just found
divided by the length given, or
62.430 -r- 22 = 2.837 inches.
68
DESIGN OF A PLAIN D-VALVE. 69
The nearest sixteenth is 2-J-J, from Table II, which is there-
fore the width required.
4. Maximum Port-opening. This will be the amount
which would allow the entering steam to have a velocity of
6000 feet per minute. It may be found by taking the num-
ber from Table IV, multiplying it by the piston area, and
then dividing it by the port length; but it is easier to multiply
the result given in (3) by f , as 4000 is f of 6000. This gives
2.837 X f 1.891 inches,
and, from Table II, the maximum port-opening is made i-J
inches.
5. Outside Lap, Angle of Advance, Valve-Travel. These
must be such as to produce the required point of cut-off with
the given amount of lead, and the port-opening found in (4).
This amounts to an example in Problem IV, given in
Chapter V.
First determine the crank-position corresponding to the
given point of cut-off on either end the head-end, for
example. This is done, as shown in Fig. 45, by the method
given in Chapter II. For this purpose a sheet of paper is
required which will allow the drawing to be made on a large
enough scale to measure accurately. The crank circle being
24 inches in diameter, a sheet of paper fourteen inches square
will be large enough for a half-size drawing. Then draw the
two lines XY and VZ at right angles, locating O, which will
represent the center of the shaft. Then locate O', the
center of the auxiliary crank-circle. The connecting-rod
being five times as long as the crank, OO' will be ^ of the
crank, as shown in Table I. That is, OO' will be
= - = .6 of an inch,
20 5
and on the half-size drawing it will be .3 of an inch. Then
SLIDE-VALVES.
draw in the two circles shown, the dotted circle being the
auxiliary crank-circle. Lay off vn equal to -J- of vz, as the
cut-off is to be at -J stroke. Draw nc perpendicular to vz,
cutting the auxiliary circle at c, and join O and c, producing
it until it meets the crank circle at C. Then OC is the crank
position at cut-off on the head end. (Refer to Chapter II
for a complete explanation of this method.)
Now, having located the crank-position at cut-off, take
another sheet of paper, and proceed as in Problem IV.
Draw XY and VZ at right angles, Fig. 46. Their intersec-
tion represents the center of the shaft. Draw OC, the crank-
DESIGN OF A PLAIN D -VALVE. ? I
position at cut-off, as found in Fig. i. Produce OC to m,
making Om equal to the given lead, j- of an inch. This
drawing should be made at least twice full size, in which case
a sheet of paper 12 inches square will be large enough. Next,
on mC take ms equal to I-J inches, the given maximum port-
opening. Through s draw nst parallel to the line of stroke
VZ, making st equal to sm or I-J inches, as shown by the arc
mt. Then join O and /, and on Ot take Oh equal to Om or
J- inch, as shown by the arc mk. Then about O as a center,
and with a radius equal to On, describe an indefinite arc.
Through h draw gh parallel to VZ. Then gh will cut the
indefinite arc just drawn in g. The arc cuts Ot in k. Then
pick up the arc kn in the compass or dividers, and lay it off
from g to D'\ that is, kn equals gD 1 . Then join D' and O,
and produce D' O to any point, as D. Then is DOD' the
center line of the valve circles, and DOX is the angle of
7 2 SLIDE-VALVES.
advance. Now take any convenient radius, as E'O, and
describe the trial valve-circle, intersecting OD at v' . Then
Oc' is the corresponding lap, and by drawing in this trial lap-
circle it will be found that the greatest possible port-opening
is L'v' with this lap and valve-travel. If this is not equal to
i-J inches, the given amount and it probably will not be,
draw v'y in any convenient direction, and take v'x equal to i-J
inches, the required maximum port-opening. Then join L'
and x, and from O draw Oy parallel to L ' x, until it cuts v'y
in y. Then is v'y the required eccentricity. Lay off Ov
equal to v'y, and draw in the valve-circle with this diameter,
which will be found to be 2-J inches. This valve-circle will
cut OC, the crank-position at cut-off, at c, and Oc, which is
ij 1 ^ inches, is the outside lap on the head-end. The lead is
de, and it will be found to measure \ of an inch.
6. Inside Lap, Head- End. This could be found on the
same sheet, but by this time that drawing has become too
complicated and full of lines to permit of any close determina-
tions. Take another sheet of paper, and proceed to draw the
full diagram for the head-end, as shown in Fig. 47. This
drawing may be made on a smaller scale than the preceding
one, if desired, but it is not desirable to make it any less than
actual size. Lay off the things already determined on this
sheet that is, draw XY, VZ, DOD' , OC, the upper valve-
circle whose center is E, and the outside lap-circle. Now
turn back to Fig. 45, and locate thereon the crank-position
O T, the crank-position corresponding to the given point of
compression on the head end. (Compression occurs on the
return stroke, and the crank is therefore below the line of
stroke when it occurs.) Transfer OT to Fig. 47, where it will
intersect the lower lap circle at /, thus determining Ot, the
inside lap on the head end, which will be found to measure -J
an inch. Release and admission will take place, as shown, at
the crank-positions OA and OR.
7. Inside and Oiitside Laps on Crank-End. These can be
DESIGN OF A PLAIN- D-VALVE.
73
V
74 SLIDE- VA L VES.
found very readily on the last drawing, but for purposes of
explanation they are determined separately in Fig. 48, which
is an exact reproduction of Fig. 47 as far as the angle of
advance and the valve-circles are concerned. Then pass back
to Fig. 45 and determine OC' and OT', the crank-positions
corresponding to cut-off and compression on the crank-end.
Then transfer them to Fig. 48, and thus determine Oc' , the
outside lap, to be -J^ inch, and Of , the inside lap, which is |-
of an inch.
Of course in making these determinations, all the crank-
positions required will be first located on Fig. 45 and marked
for identification, and then transferred to Figs. 46 and 47 as
needed. Fig. 48 will be dispensed with.
8. Minimum Width of Bridge. In this case the valve has
no overtravel, and the bridge width should be made equal to
the thickness of the cylinder-walls, which are assumed in this
case to be one inch.
9. Width of Exhaust-port. This is determined by the
method given in the preceding chapter. There are three
different rules given there, and in addition to that there is a
difference between the two ends of the valve, thus giving rise
to the question as to which is the proper one to be used.
Always employ the one which gives the greatest result, thus
being sure that the port is plenty wide enough. In this case
the third rule is the one to be taken, giving
Half-travel + port width bridge -f inside lap = exhaust-port width.
2* + 2 T\ ' + t 5 T V
The results obtained on the preceding text may be col-
lected as follows:
Eccentricity 2 J-
Valve travel ( 2 X eccentricity) 5 j-
Head end, outside lap i-^
" inside lap J-
DESIGN OF A PLAIN D-VALVE.
FIG. 48
SLIDE-VALVES.
Crank-end, outside lap
' ' inside lap
Exhaust-port width
The next thing in order is
TO LAY OUT THE VALVE.
This is a comparatively simple matter, but one in which
many beginners go astray, owing to the fact that the valve is
not the same on both ends, and consequently cannot be laid
out very well from a center line.
Draw the line FZ, Fig. 49, to represent the valve-seat.
Then start to lay out the valve from either end the crank
end being taken in this case. Start at any point, as A, and
lay out the valve as follows:
Make AB = outside lap, crank-end I T 3 F inches.
BC = width of port = 2^\ "
CE = " ll bridge "= I "
EP exhaust
FH= ll " bridge = I
HI li " port =
//= outside lap, head-end =
DESIGN OF A PLAIN D-VALVE. 77
Then draw in the section of the ports as shown, making
the thickness of the seat one inch, the same as the cylinder-
walls and bridge. Next,
Make CD = inside lap, crank end = inch.
HG = inside lap, head end ^ "
Then draw in the valve proper, making it thick enough to
withstand the pressure to which it is subjected. The height
of the cavity in the valve need not be very great, the general
proportions shown in the cut being very good.
Laying out the valve-seat first and the valve afterwards
avoids confusion.
CHAPTER VIII.
LENGTH OF VALVE-CHEST, VALVE-STEM, AND ECCEN-
TRIC-ROD.
IN addition to the dimensions of the valve and seat as
determined in the preceding chapter, there are three more to
be found. These are:
1. The length of the valve-chest.
2. The length of the valve-stem.
3. The length of the eccentric-rod.
These will be considered in the order given.
LENGTH OF VALVE-CHEST.
This must be at least long enough to allow the full
travel of the valve. That is, it must permit the valve to go
to the two extremes of its travel as shown in Figs. 50 and 51.
The minimum length of the chest may then be measured
directly from the drawing, which represents the valve in its
78
LENGTH OF VALVE-CHEST, VALVE-STEM, ETC. Jg
middle position, as in Fig. 49, by laying off the half-travel at
each end of the valve, and measuring between the marks
thus obtained. Or if it is desired to obtain the length of
valve-chest necessary without drawing the figure, it can be
done by applying the following rule:
Add together the outside laps at the two ends of the valve >
twice t/ie width of the steam-port, twice the thickness of the
bridge, the 'cvidth of the exhaust-port and the valve-travel.
This gives the minimum inside length of the steam-chest ', and it
should be increased by at least one inch.
LENGTH OF VALVE-STEM.
One of the most common methods of connecting the valve
and valve-rod or stem is shown in Fig. 52. This consists of
a collar which fits closely around the valve, and may or may
not be otherwise secured to it. The collar is tapped at B, and
the valve-stem screwed in. The length of the valve-rod, VR,
is measured from the inside of the collar to the point at
which the stem is secured to the eccentric-rod, or, if a rocker
is employed, to that. The length of the valve-stem must be
sufficient to permit its reaching from the valve when in its
extreme position, as shown in Fig. 51, to the point where
8c
SLIDE-VALVES.
it joins either the eccentric-rod or the rocker-arm. This
gives a direct means of obtaining the proper length from the
drawing. The valve shown in the drawing has overtravel.
LENGTH OF THE ECCENTRIC-ROD.
Figs. 42, 43, and 44 in the sixth chapter show the
determination of this length when a rocker or bell-crank is
Fia. 52
employed. When neither is used, the length of the eccentric-
rod may be found by this rule:
Find the distance from the center of the exhaust-port to the
center of the shaft. This may be done from the drawing.
From that distance subtract the distance from the center of the
exhaust-port to the end of the valve -stem, when the valve is in
its middle position. The remainder is the required length of
the eccentric-rod.
This is not strictly true, but it is near enough, the differ-
ence between the length thus obtained and the correct length
being negligible, as will be apparent from Fig. 53. O is the
LENGTH OF VALVE-CHEST, VALV.
81
center of the shaft, P is the center of the exhaust-port, and
PV is the length of the valve-stem. The length given by the
FIG. 53
rule is OV, whereas the correct length would be EP. But,
for example, suppose OE were 3 inches and O V were 5 feet
or 60 inches. Then
= 1/3600 -|- 9
= 1/3609
= 60.07 inches.
The correct length is 60.07 inches, then, against 60
obtained by the rule.
The rules for the lengths of valve-stem and eccentric-roxl
are given in such a way as to permit either one to be assumed
and the other one found. The location of the point of sus-
pension of a rocker, if one is employed, usually determines
both of these lengths.
CHAPTER IX.
DESIGN OF AN ALLEN OR "TRICK" VALVE.
THE difficulty with the ordinary slide-valve is that it can-
not be used for early cut-offs, on account of the large outside
laps and travel required to produce the desired result. This
renders the friction of the valve a very considerable item,
and it is unusual to find a plain slide-valve used to secure cut-
offs much earlier than five-eighths of the stroke, three-quarters
being usually the limit for which they are employed.
There are, however, several valves constructed on the
same general principle, which permit an early cut-off to be
attained with moderate lap and travel. Prominent among
these is the Allen, or " Trick," valve, shown in Figs. 54, 55,
and 56, which is so constructed as to give double the amount
of port-opening as a plain slide with the same amount of
travel, or the same port-opening with one half the travel.
This advantage is secured by means of the passage X in the
valve. This passage is used for admission only.
While this valve is designed to take the place of a slide-
valve, it cannot be used on the same seat, but requires the
face-plate shown between the slide-valve seat and the face of
the Allen valve. Its operation will be best understood by
reference to the figures.
The figures show the valve in its progress from one
extreme of its travel to the other, and the arrows show the
course of the steam. Fig. 54 represents the extreme left-
hand position, Fig. 55 shows the middle position, and in Fig.
56 is shown the extreme right-hand position. With the
82
DESIGN OF AN ALLEN OR " TRICK" VALVE. 83
valve out at the left the piston is over toward the right, and
steam must be admitted to the right-hand end of the cylinder
through the port P. This is open by the amount BE,
Fig. 54, and the steam enters through that space just as it
would with an ordinary D-valve. But, in addition, the other
end of the valve has moved beyond the edge of the face-plate,
putting the passage X in communication with the steam-chest,
through the opening C 'D' r , and steam is therefore admitted to
the right-hand end of the cylinder through CD. The total
opening of the port is therefore the combined amount of CD
and BE. If CD and BE are equal, the total port-opening is
84 SLIDE- VA L VES.
2BE-, and as BE is the amount that the port would be open
with a plain slide-valve, it is evident that the Allen valve
secures double the port-opening with the same travel. The
action at the other end of the valve is the same.
-Another point which must be borne in mind in regard to
this valve is that it cannot be used to advantage when the
cut-off is much later than one-half stroke.
In order to bring out clearly one or two points, an actual
valve will be designed.
PROBLEM. To lay out an Allen valve and seat for a
12 X 24 engine, running at 150 revolutions per minute, con-
necting-rod four times the crank, cut-off on both ends at 10
inches, compression at 22 inches, lead on head-end -J inch.
Ports to be 1 1 inches long.
ORDER IN WHICH THE DIMENSIONS ARE DETERMINED,
I. Piston Speed. This is, by the rule given before,
24
2 X 150 X = 600 feet per minute.
2. Area of Port. From Table IV this is found to be
.15 of the area of the piston, if the velocity of steam at
exhaust is 4000 feet per minute. The area of the 12-inch
piston is 113.10 inches, so that the port area must be
113.10 x .15 = !6.95 inches.
3. Width of Port. This is the area divided by the length*
16.95 -*- ii 1.54 inches,
or from Table II, I T V inches.
DESIGN OF AN ALLEN OR " TRICK" VALVE. 8$
4. Maximum Port-Opening. The velocity of steam at
admission being assumed at 6000, the port-opening required is
4000
X iA = %
and therefore, according to Table II, it is made I T \ inches.
5 . Outside Lap, A ngle of Advance, and Valve- Travel. This
means the outside lap, angle of advance, and valve-travel
necessary to give the required point of cut-off on the head-
end for an ordinary D- valve, but with one-half the given port-
opening. The Allen valve will double this half, giving the
required amount.
First determine the crank-position corresponding to cut-
off on the head-end. This operation has been fully explained
in previous chapters, and will not be touched on again.
Having located this crank-position, the example then
becomes a case of Problem IV, Chapter V. The diagram is
given in Fig. 57, and the dimensions determined therefrom
are:
Valve travel ................... 4^ inches.
Eccentricity (= half-travel) ...... 2j
Outside lap, head-end ........ .... if "
6. Inside Lap, Head-End. This is determined from the
diagram, Fig. 58, where T is the crank-position with the
piston at 22 inches. The inside lap is, as explained in
Chapter III, that part of the crank-position included between
the valve-circle and the point O of the diagram. The inside
lap and outside lap are read from different valve-circles. It
will be noticed that the line O T does not cut the lower valve-
circle below the line of stroke, but above it. The inside lap
is therefore negative ; that is, the valve has inside clearance.
While this is a possible construction, it is not at all desirable,
as there would evidently be a time in the stroke when the two
ends of the cylinder would be in direct communication with
86
SLIDE- VAL VES.
each other and with the exhaust. This will be plain on refer-
ence to Fig. 55, where the dotted lines x x' show the construc-
tion of a valve with inside clearance at both ends. If only one
end has this negative inside lap, the blow-through of steam
would not be in evidence with the valve in its middle posi-
tion, but will appear later, unless the other end has a con-
FIG. 57
siderable amount of inside lap, at least equal to the inside
clearance of the first end.
The inside lap on the head-end will therefore be made zero.
7. Inside and Outside Laps, Crank- End. These are found
from the diagram, in the manner already explained. The
outside lap is Oc' ', Fig. 58, which is I T 5 ^- inches. The inside
lap is again negative, as shown by OT' cutting the upper
valve-circle below the line of stroke. The inside lap will
therefore be made zero.
DESIGN OF AN ALLEN OR " TRICK" VALVE. 7
For the rest of the dimensions the description will refer to
the valve in its middle position, as in Fig. 55.
8. The Distances DE and D' E' . These are equal and may
be assumed, and in this case will be made -J inch.
9. Width of Port, MN and M' N' . These being the open-
ings in the cylinder-face, they are made equal to the port
width determined, or I T 9 F inches each.
10. Width of Passage in Valve, or CD and C'D'. These
are each made equal to one-half the port-opening. The port-
opening being i^, the half is ii|-, and CD and C'D' will be
made the next larger sixteenth, or T 9 .
These last three are all the dimensions that are the same
on both ends of the valve.
1 1. BE is made equal to the outside lap on the head-end,
assuming the cylinder to be at the right. BE = if inches.
88 SLIDE- VA L VES.
12. B ' E' is made equal to the outside lap on the crank-
end, or I T 5 F inches.
13. AC. This is equal to the outside lap, head-end, ij
inches.
14. A'C'. This is equal to the outside lap on the crank-
end, i T 5 g- inches.
15. BC. This is made equal to the outside lap, head-end
one-half the port-opening DE. This gives if T 9 ^ -J
= i-jig- inches.
16. B'C'. This is made equal to the outside lap, crank
end one-half the port-opening D'E'. This gives I T 5 T 9 ^
-J -| inches.
No allowance is made for the difference between the port-
openings at the two ends of the cylinder, because the crank-
end opening, as shown by the diagram, is greater than that
of the head-end. Therefore, if the dimensions be made large
enough to permit the required opening on the head-end, the
crank-end opening will be ample with the same allowance.
The next thing is to describe the method of
LAYING OUT THE FALSE SEAT AA'.
17. Thickness. This may be made anything desirable.
1 8. AE. This is made equal to the outside lap, head-end
-\- one half the port-opening -f- DE\ or, in this case, if -f- -$
+ i= 2^ inches.
19. A'E'. This is calculated in a similar manner: Out-
side lap, crank-end + one half of the port-opening + D'E',
which gives I T ^ + yV 4~ t ~ 2 mcri es.
20. EH. This made equal to the total port-opening -f-
BC, or i-jig- + I T V = 21 inches.
21. E'H 1 '. In a similar way this is made equal to the
total port-opening + B'C', which gives i^ + = i-j-^.
22. Width of Bridge. This is made equal to the half-
travel (EH or E'H f ) inside lap -)- J inch, for a 'minimum.
DESIGN OF AN ALLEN OR "TRICK" VALVE. 89
EH or E'H' is used, according to which gives the greater
value. The inside lap to be used is determined in the same
way. E'H' is the value for this case, and the inside lap is
zero at both ends, so that the rule gives 2j iJ-J -|- J ||
inches. This is the minimum thickness of the bridge, and if
it is any greater than the thickness of the rest of the casting,
the bridge must be thickened to the required amount. If not,
the bridge may be made of the same thickness as the rest of
the cylinder-casting. This will be done in this case, and the
bridges HL and H'L' will each be made one inch.
23. Width of Exhaust-port. This is made at least equal,
to the half-travel -\- width of steam port bridge -f- inside
lap; and the width will then be, in this case, at least 2j -[" JyV
i -j- o = 2 T 5 ff ; and it may be anything greater.
All the dimensions of the valve, valve-seat, and false seat
are now determined, and the valve may be laid out in its
middle position, as shown in Fig. 55, beginning at either end.
CHAPTER X.
DESIGN OF A DOUBLE-PORTED VALVE.
THE valve shown in Figs. 59-63, known as the " double-
ported " valve, is another type of valve used to replace a
D valve when it is desired to secure an early cut-off; and it
does this in the same general manner as the Allen valve, by
securing the same port-opening as a D-valve with one-half the
travel, or double the opening with the same travel. This-
enables the same point of cut-off to be secured with a smaller
amount of outside lap. The lap and travel being reduced, it
follows that the friction of the valve is lessened. From Figs.
60 and 61 it will be seen that the steam enters the cylinder
beneath the outer edges of the valve, and that the action of
the outer shell is therefore similar to that of a plain D-valve.
In addition to this means of admission, passages 5 and S' are
provided which run all the way through the valve, and arc
therefore open to live steam from the steam-chest at all times.
These passages have ports in the bottom, as shown at GH
and G'H'. There are corresponding ports IL and I ' L' in the
valve-seat, and steam is thus admitted through these secondary
ports as shown by the arrows in Figs. 60 and 61. If the
valve as rhown in Fig. 60 is in its extreme left-hand position,
the port opening to steam is the sum of G'H' and A'C' .
The opening A'C 1 is that which would be obtained by a plain
D-valve, and therefore, if the valve is so constructed that
A ' C' and G'H' are equal, the total port-opening is double that
which would be obtained by a plain slide-valve.
The exhaust passes through the passage X y as shown by
90
DESIGN OF A DOUBLE-PORTED VALVE. 9 1
92 SLIDE-VALVES.
the arrows; and this passage gives the valve a certain simi-
larity to the Allen valve; but here the similarity ends, for it
will be remembered that with the Allen valve the passage is
used for the admission alone, while here it is only for the
exhaust. .
The action of this valve is therefore the same as would be
obtained by two slide-valves, one within the other. The
difference in construction is that the exhaust-passage, X" , of
the inner valve is thrown open to the exhaust-passage, X, of
the outer valve. It will be noticed that 5 and S' are used
for admission only. They are not of the same size through-
out, but are greater at the sides, where they open to the
steam-chest, as shown clearly in Fig. 63, in which the left half
FIG'. 63
of the figure shows a transverse section through the middle of
the valve, and the right-hand half is a section through the
center of the passage 5. This figure also shows the general
shape of the passage X, which is divided into two parts by the
central web. Fig. 62 shows a section through the center of
the valve, Figs. 59, 60, and 61 being some distance to one
side.
This type of valve is more often applied to vertical engines
than to any others, and is designed to give equal cut-offs, for
the following reasons. With equal cut-offs secured by equal
laps the leads are unequal, and the greatest lead is at the
crank-end of the cylinder, as shown by the diagrams in
Chapter III. The crank-end of the cylinder being the lower
DESIGN OF A DOUBLE-PORTED VALVE. 93
end, the extra amount of lead is useful in cushioning the
piston and piston-rod, etc., which must, from the nature of
the construction, have a greater force on the down than on the
up stroke. Another reason, or, rather, one in the same line,
for not designing the valve to have equal leads, is that that
would give a longer cut-off on the crank-end, which is the end
where the weight of the moving parts tends to increase their
motion rather than retard it.
This valve, unlike the Allen, cannot be used on the same
valve-seat as a plain D valve by the interposition of a false
seat, but requires the valve-seat to be laid out with reference
to the double-ported valve. It is therefore impossible to take
off a D-valve from an engine and put a double-ported valve
in place.
To show the process of designing a valve, an actual case
will be worked out.
PROBLEM.
GIVEN. A cylinder 27 X 24 inches; ports 24 inches long;
revolutions per minute, 140; width of bridge, if inches;
inside lap, both ends, o; cut-off, both ends, stroke; con-
necting-rod, 72 inches; velocity of admission, 6000 feet per
minute; velocity of exhaust, 4000 feet per minute; lead,
head end, T V inch.
REQUIRED. Outside lap, both ends; travel of valve; lead,
crank-end; maximum port-opening, both ends; and to lay
out the valve in its middle position as shown in Fig. 59.
ORDER IN WHICH THE DIMENSIONS ARE DETERMINED.
I. Piston Speed. This is
140 X 2 X -- = $60 feet per minute.
94 SLIDE-VALVES.
2. Area of Steam- Port. The nearest piston speed, to the
given 560, in Table IV is 550, and the port area is given as
.138 times the piston area for a velocity of 4000 feet per
minute. The piston area from Table V is 572.56 square
inches, so that the port area is
572.56 X .138 = 79-0 1 -
3. Width of Steam-Port. The ports being 24 inches long,
the width must be
79.01 -7-24= 3.3.
This is, however, the total width of port at one end of the
cylinder, and will be taken as 3^ inches. There are two ports
at each end of the cylinder, so that the width of each one is
made 3^ -r- 2 = if inches. This determines I'L', C'F' at the
head-end, and IL and CF at the crank-end, each being made
if inches. The width of the main port that is, after the
two ports join must be made 3^ inches.
4. Maximum Port -Opening on the Head-End. This is the
opening which will give the entering steam the required
velocity, and is found as follows:
velocity of exhaust
Port-opening = port width X \ rr~ t j : : .
velocity of admission
This gives:
"Pi-M-4-_rvt-=riinrr T V
6000
. 4000
Port-opening = if X %- - = i-
This is the opening of each port, and the total opening at one
end of the cylinder is therefore 2 X I T \ = 2 f
The same result would be obtained by using Table IV to
find the area required for free admission.
5 . Valve- Travel, Inside and Outside Laps on Both Ends, and
Maximum Port-opening and Lead on Crank- End. These are
determined precisely as for the Allen valve, and the explana-
DESIGN OF A DOUBLE-PORTED VALVE. 95
tion given of Fig. 57 will answer for this, the numerical values
only being changed. The results obtained are as follows:
Outside lap, head end \\ inches.
Outside lap, crank end i "
Lead, crank end f " ,
Maximum port-opening, crank end .... i-J- "
Valve travel 5 "
6. Width of Bridge. The minimum width allowable is
Half-travel width of one port inside lap -f- J inch.
2\ if o + J = i inch.
This is less than the if inches given in the problem.
This rule is only applied to see whether the width given is
sufficient to prevent blowing through. The result obtained
shows that the if inches, which is the thickness of the
cylinder-casting, is ample.
7. Width of GH. This is made equal to the port-opening
for that end of the cylinder, and is therefore i inches. GH
is at the crank end.
8. Width of G'H' . Make this equal to the port-opening
at its end of the cylinder, or I T 3 ^ inches, G'H' being at the
head end.
9. Width EG. This may be made any convenient figure,
and in this case it will be one inch.
10. Width E'G'.~ This is made equal to EG.
11. Width of Exhaust-Port. This must be made great
enough so that when the valve is in either of its extreme
positions, as shown in Figs. 60 and 61, the opening to exhaust
is so great that the outflowing steam will not have a greater
velocity than 4000 feet per minute. The total width of steam-
ports at one end is that found to be necessary for this velocity,
and, as the exhaust-port takes care of the exhaust from both
9 SLIDE- VA L VES.
of these ports, the free opening of the exhaust MK 1 ', Fig. 60,
or KM', Fig. 61, must be equal to the total width of the
ports at one end of the cylinder. The exhaust-port, MM', is,
then made of a width equal to
Half-travel + inside lap bridge -\- port width at one end.
2 + o if + 3i = 4i inches.
12. Width FI. The edge E must not travel beyond F if
it is desired to have a full opening during exhaust. That
gives the value of FI as
Outside lap, crank-end -f- GH -\- EG -\- half-travel.
I + i + I + 2j =: 6 inches.
13. Width FT . This is obtained in a similar manner to
FI\ or it is equal to
Outside lap, head-end + G' H ' + E'G' + half-travel.
it + i A + i + 2* =
The edge E may be allowed to overtravel very slightly
the point F, in case it is an object to shorten the valve.
Doing so prevents the exhaust from being wide open all the
time, but the choking only occurs during a small part of the
stroke. When this overtravel is allowed, FI and FT are to
be shortened by the amount allowed. No allowance is made
in this case.
14. Width DE. This is made equal to
Half-travel, inside lap, crank-end amount that E overtravels F.
2j o 0=2% inches.
15. Width D'E'. This is made equal to
Half-travel inside lap, head end amount E' overtravels^.
2^0 ^0=2% inches.
DESIGN OF A DOUBLE-PORTED VALVE. 97
The inside laps being equal, it happens that DE and D'E'
are equal. If the inside laps are unequal, the values obtained
by the rules would be unequal; but both DE and D'E' would
be made equal to the larger value obtained, keeping the valve
symmetrical in this respect.
16. Width BC. This must be so long that D will not
overtravel B; for if it did, the steam-chest would be in direct
communication with the exhaust-passage X, allowing steam
to blow straight through the valve. Its width must therefore
be at least equal to
Half-travel width of one port inside lap, crank-end + J inch.
2\ if O -f- 4 l i nc h-
17. Width B'C . This is determined, from the same con-
siderations that govern the dimension BC, to be
Half-travel width of one port inside lap, head-end + \ inch,
2^ if o + J = I inch.
The next thing in order is
TO LAY OUT THE VALVE.
1 8. Tabulate the Dimensions. The dimensions of the valve
and seat should be tabulated separately, in the order in which
they are to be laid down, beginning at either end. In this
case the start will be made from the crank end. The letters
apply to Fig. 59.
Valve Seat.
BC. I inch obtained from 15
CF. if " " " 3
FI. 6 " " " ii
IL. if " " " 3
LM. if " " " 6
SLID E- VA L VES.
MM'. 4^ inch ....... . ---- obtained from 10
M'L'. if " ............ " " 6
LT. if " ............ " " 3
I'F'. 5H " ............ " " l2
F'C. if " ............ " " 3
B'C. i " ............ " " 16
Lay this out on any convenient scale.
The Valve.
AC. I inch outside lap ........ obtained from 5
DF. o " inside lap . ....... given
DE. 2\ " ....... obtained " 13
EG. i " ........ " " 8
GH. i " ........ " <4 6
J/A'. i outside lap ....... " " 5
LK. o " inside lap ...... . given
K'L'. o " " " ........ "
I'H'. ij " outside lap ..... ... obtained from 5
H'G'. I T 8 75- " port opening ...... " " 7
G ' E' ' . i " ........ obtained from 9
E'D'. 2% *> ........ " " 14
F'D'. o " inside lap ....... , given
ij " outside lap ........ obtained from 5
Reference to Fig. 63 will show that the exhaust-passage
XX, Fig. 59, is divided into two portions, X' and X" , when
it passes over either of the steam-passages, as 5. Care must
be taken to make the valve of so great a height that the sum
of the areas of these two portions shall be at least as great as
the total area of the ports at one end of the cylinder in order
to maintain the proper velocity of exhaust.
CHAPTER XL
VALVE-SETTING.
HAVING designed a valve and having it in place on the
engine, the next thing to be considered is the proper method
of setting it in order to secure the best results. The method
to be employed will depend on the design of the valve
whether it is intended to secure equal cut-offs on the two
ends, or whether it was made to give equal leads; or, again,
whether the cut-offs and leads were equalized, according to
the method given in Chapter VI.
In any case the first step is to put the engine on the center,
by which is meant that the piston is at the end of its stroke
and the crank and connecting-rod are in the same straight
line. This operation must be performed very precisely, as
little variation in the position of the crank from the line of
stroke or piston from the end of the stroke will make a large
difference in the position of the valve. This is because the
eccentric is at or near its middle position at the time the crank
is on the center. The crank-pin is moving vertically while
the eccentric is moving horizontally at that instant, so that a
small angular movement of the eccentric will make a great
difference in the position of the valve. The crank, however,
is moving vertically, and a comparatively large angular move-
ment of that will only move the cross-head a little ways from
the end of its travel. Consequently the dead-center must
be located very exactly- or the valve-setting will be thrown
out.
99
IOO
SLIDE-VALVES.
TO PUT AN ENGINE ON THE CENTER
The engine is put on the center by moving the cross-head
a measured distance on each side of its extreme travel and
FIG. 64
measuring the amount of the movement of the fly-wheel by
means of marks on the rim. This distance is bisected, and
FIG. 65
the middle point determines the true center. The practical
method of doing this is shown in Figs. 64, 65, and 66.
First. Turn the engine in the direction in which it is to
run until the cross-head is nearly at the end of its stroke, as
shown in Fig. 64.
VA L VE-SE TTING.
101
Second. With the cross-head in this position, take a piece
of chalk and make a mark across the cross-head and guides, as
Fief. 66
shown at A, Fig. 64. This mark may be put anywhere on
the cross-head, as it serves only as a reference-mark.
Third. Fix an upright pointer as near as possible to the
face of the fly-wheel, as shown at B, and mark the height
reached by its end, as shown at C.
Fourth. Turn the engine, still in the direction in which it
is to run, until the mark on the cross-head again comes even
with the mark on the guides, as shown in Fig. 65.
Fifth. Make another chalk-mark on the fly-wheel opposite
the end of the upright, as shown at C' , Fig. 65. The posi-
tion of the first mark after this movement is then as shown
at C. The distance CC' then represents the amount that the
fly-wheel has moved, while the cross-head has moved from A
out to the dead-center and back again to A. That is, the
cross-head movement has been twice the distance from A to
the dead-center. Consequently, if the wheel revolved until
the pointer comes midway between C and C f , the engine will
be on the dead-center.
Sixth. Find the point midway between C and C' . This
is D, Fig. 65.
-1 02 SLID E- VA L VES.
Seventh. Turn the engine, still in the direction in which it
is to run, until the mark D comes opposite the end of the
pointer, as shown in Fig. 66. The engine is then on the
dead-center.
Eighth. With the engine on the dead-center, make another
chalk-mark on the guides opposite the one on the cross-head,
as at E, Fig. 66. This is to serve as a reference-mark, and
the first mark is then to be erased.
The dead-center at the other end is found in the same
manner.
When putting the engine on the center, or when perform-
ing any of the other operations of valve-setting, the engine
must always be turned in the direction in which it is to run.
This is because any lost motion, back-lash, or play in the
moving parts will then affect the valve in setting precisely as
under running conditions. The wheel must never be turned
beyond the required point and then back to it, as the lost
motion would allow a considerable movement of the fly-wheel
to take place without a corresponding motion of the cross-
head ; and the valve-setting would then be thrown put of true.
TO SET A VALVE FOR EQUAL LEADS.
First Method.
First. Put the engine on the dead-center at one end of its
stroke, using the method just described.
Second. Give the eccentric as nearly as possible the
proper amount of angular advance, as determined from the
valve diagrams, taking care that the amount given shall be
more, rather than less, than the required amount.
Third. Adjust the length of the valve-stem or eccentric-
rod until the lead at the end of the stroke for which the
adjustment is then being made is equal to the required
amount.
Fourth. Turn the engine to the other dead-center.
VA L VE-SE T TING. 1 03
Fifth. Measure the lead at that end. If the measure-
ment is the same as that at the first end, the eccentric is in its
proper place, aad it only remains to secure it there; but the
chances are that the leads will be unequal, and in that case
the next step is the
Sixth. Correct half the difference in the leads by changing
the length of the valve-stem.
Seventh. Correct the remaining half of the difference by
moving the eccentric. It will be apparent at once, from the
nature of the difference, in which direction the eccentric is to
be moved.
Eiglith. Turn back to the other center and measure the
lead. If it is the required amount, the setting is complete.
If not, repeat operations Sixth and Seventh until the leads are
equal.
Ninth. Secure the eccentric in place.
When a valve-gear has a rocker, the latter is usually de-
signed to swing to an equal angle on each side of the perpen-
dicular, and in any case the length of the valve-stem must be
such that the rocker will move as designed. This being so,
it is evident that in performing the sixth and seventh opera-
tions those of correcting the variation in leads at the two
ends of the cylinder the length of the valve-stem must be
changed very little, if any, making the sixth operation very
small, and the greater part of the adjustment must be made
in the seventh operation, that of changing the position of the
eccentric.
Second Method.
This method is a convenient one when it is difficult to
turn tlu engine over. It is applicable only to that class of
valves having harmonic motion. Harmonic motion is such
as that of the foot of a perpendicular from the crank-pin upon
the line of stroke when the motion of the crank-pin is uniform.
This is the case in an ordinary engine, and an ordinary slide-
1 04 SLID E- VA L VES.
valve has harmonic motion. Among the valve-gears in which
the valve does not have harmonic motion may be mentioned
a slide-valve having equal lead and the cut-offs equalized by
means of a rocker or bell-crank lever, and the link motion and
radial gears.
When the motion is harmonic, the maximum port-open-
ings will be equal when the leads are equal.
First. Loosen the eccentric on the shaft.
Second. Turn the eccentric until it gives the maximum
port-opening, first at one end and then at the other.
Third. If the maximum port-openings are not equal and
the chances are that this will be the case make them so, by
changing the length of the valve-stem by half the difference,
thus adjusting the length of the valve-stem.
Fourth. Put the engine on the center. This is the only
time that it is necessary to perform this operation.
Fifth. Turn the eccentric to give the proper lead, thus
adjusting the angle of advance.
Sixth. Secure the eccentric in place.*
TO SET A VALVE FOR EQUAL CUT-OFFS.
First. Put the engine on one dead-center say the head-
end.
Second. Give the eccentric, as nearly as can be judged,
the angle of advance determined by the valve-diagram; tak-
ing care that if there is any difference, it shall be in excess of
the proper amount rather than less.
Third. Give the valve the correct amount of lead, as
nearly as possible.
Fourth. Move the engine in the direction in which it is to
run until cut-off occurs.
Fifth. Measure the distance that the cross-head has moved,
*The author wishes to acknowledge his indedbteness to Peabody on
Valve-gears for this method.
VAL VE-SETTING. 10$
up to this point, from the end of the stroke. This is found
by means of the chalk-marks on the guides and cross-head.
Sixth. Turn the engine in the direction in which it is to
run until cut-off occurs on the return stroke.
Seventh. Measure the travel of the cross-head from the
beginning of the return stroke up to cut-off. If this is the
same as on the forward stroke, the valve is set correctly, and
the eccentric should then be secured in place. But it is
hardly probable that this result will be secured on the first
trial; and in that case the next operation is the
Eighth. Correct the difference in cut-offs by changing the
length of the valve-stem. If the cut-off is earlier on the
crank-end or return stroke, the valve-stem should be length-
ened. If it is earlier on the head-end or forward stroke, the
valve-stem should be shortened.
Ninth. Put the engine on the head-end center again, and
adjust the lead by moving the eccentric.
Tenth. Test the cut-offs and see whether or not they are
equal. If they are, the adjustment is finished. If not, repeat
the eighth operation until they are equal, and then
Eleventh. Secure the eccentric in place.
As pointed out in Chapter III, designing a valve for equal
cut-offs will make the leads unequal. At the time the valve
is designed the lead can be measured at each end of the
cylinder, and the* operation of setting the valve can be per-
formed by using the first method given above for equal leads,
except that the leads, instead of being made equal, are made
equal to the required amounts. In addition, the travel of the
cross-head from the beginning of the stroke up to cut-off
must be determined and the setting completed by the eighth
and ninth operations just given.
. By means of any of these methods, it is assured that the
action of the valve shall be just as intended at admission or
cut-off. This also assures the fact that any irregularity of
action, or error of design, caused by neglecting the angularity
io6
SLIDE-VALVES.
of the eccentric-rod, will not affect the valve while either
opening or closing, but will make itself felt while the port is
either opened or closed; which is of no particular conse-
quence.
Now, having set the valve, it is of importance to make a
distinct set of reference-marks on the eccentric, so that it will
be possible to set it in place again very readily if it slips, thus
avoiding vexatious delays, or, perhaps, the necessity of going
over the whole ground cf valve-setting again.
When any one of the preceding methods of valve-setting
is employed, it is necessary to remove the cover of the steam-
chest in order to observe the action of the valve, measure
leads, etc.
TO SET A VALVE WITH THE CHEST-COVER ON.
After having once set the valve for equal leads, with the
cover off, it is possible to arrange things so that thereafter the
valve can be set with the chest-cover on, and, if absolutely
necessary, with steam on. This method is illustrated in
Figs. 67 and 68.
After having set the valve, the engine is placed on the
center, and a reference-mark made on the valve-stem some-
where outside of the stuffing-box, as shown at A, Fig. 67.
Then another mark is made somewhere on the chest-cover, as
at B. Next a piece of heavy wire is taken, and a tram or
spanner, T, Fig. 67, is made, of such length that it will reach
from A to B. Then the engine is put on the other center,
which operation will change the distance of the mark on the
valve-stem from the mark on the chest-cover to AB, Fig. 68.
Another spanner, T' , is then made which will reach that dis-
tance. Then, when it is again necessary to set the valve, the
operation can be performed by employing the first method for
equal leads, using the trams to determine the equality of the
leads, as follows:
VAL VE-SE TTING.
107
First. Put the engine on one center, say the head-end.
Second. Give the eccentric the proper amount of angular
advance, as nearly as possible, making it too great rather
than too little.
Third. Adjust the length of the valve-rod until the lead
at the head-end is the proper amount; that is, until the tram
T reaches from A to B.
Fourth. Turn the engine to the other center.
Fifth. Measure the crank-end lead. This is done by using
the tram T' . Remember that when it reaches from A to B,
FIG. 67
as in Fig. 68, the leads are equal. When it reaches to any
other point, the lead is changed by the distance from A to
the point which the tram marks on the valve-stem. If the
tram spans over to the crank side of A, the lead is too great
at the crank-end. If the tram reaches to a point on the
cylinder side of A, the lead at the crank-end is too little.
The rest of the operations are the same as given for equal
leads.
A variation of this method consists in using but one tram,
and making two marks on the valve-stem, at the points
reached by the tram with the engine on the two -enters.
The method of setting is then the same. This is frequently
used on locomotives.
CHAPTER XII.
SHAFT-GOVERNORS. GENERAL PRINCIPLES AND TYPES.
IT was shown in the first chapter that in order to reverse
the direction of rotation of an engine it is necessary to move
the eccentric around the shaft past the crank until it makes
the same angle with it on the opposite side that it did in its
original position. That is, if the arrangement of crank and
eccentric shown in Fig. 69 will cause the engine to run over,
1C
0'
o
1C'
\
\
\
S
FIG. 69
FIG. 70
as shown by the arrow, the arrangement shown in Fig. 70
will reverse the engine, the angle C'O'E' being equal to the
angle COE. With this arrangement no rocker-arm is em-
ployed between the eccentric-rod and the eccentric.
Figs. 71 and 72 show the arrangement when a rocker-arm
100
SHA FT- G O VERNORS.
I0 9
is employed. The angle CO' E' is equal to the angle COE,
just as before, and the direction of rotation has been reversed,
as shown by the arrows.
In both cases the angle through which the eccentric has
been turned to effect the reversal is equal to 180 twice the
angle of advance.
V
E
FIG. 71
rr FIG. 72
There are numerous means by which the eccentric can be
moved around the shaft ; and they may be divided into two
general classes:
First Class. Movable eccentrics with which the engine
must be stopped in order to effect the reversal.
Second Class. Movable eccentrics with which the reversal
can be effected while the engine is in motion.
FIRST CLASS.
This is by far the simpler class, and, for reasons which
should be very obvious, it is but seldom employed, although
it was used on early stationary and locomotive engines.
Figs. 73 and 74 illustrate an eccentric of this type. The shaft
carries a disk which is forged or cast on it, or fastened in
HO SLIDE- VA L VES.
place. This disk is slotted in an arc of a circle about the
center of the shaft O. The eccentric is loose on the shaft and
is slotted in a similar manner. A bolt is used to fasten the
two together. With the eccentric in one extreme position,
such as the one shown in Fig. 73, the angle of advance is
XOD. The arc of the slot is made of such a length that
when the eccentric is slipped around so that it is bolted in its
other extreme position, as shown in Fig. 74, the angle of
advance is YOD' ', equal to XOD of Fig. 73. In both figures
the path of the eccentric center is the dotted circle shown.
That this change in the angle of advance will result in a
reversal of the engine should be understood from Chapter I.
Figs. 75 and 76, the valve diagrams corresponding to Figs.
73 and 74 respectively, will aid in the comprehension of the
fact that the events of the stroke occur at the same points in
both cases. The angles of advance are laid off on opposite
sides of the vertical in the two figures for the sake of em-
phasis, but, as explained in Chapter III, this is not necessary.
Fig. 77 shows another type of this class. The eccentric-
slot is replaced by a pin which projects through the slot in
the disk, and is secured in place by a jam-nut on the opposite
side. The same results will be secured by this type as with
the one shown in Figs. 73 and 74, and the diagrams 75 and
76 apply to it as well.
SECOND CLASS.
This is the more important class, possessing many points
of advantage over the first class, and is the one which is
employed on all or nearly all high-speed engines, where they
are used as "shaft-governors."
The fundamental principle in this class is to have the
eccentric slotted to clear the shaft, and, by moving the eccen-
tric across the shaft, change the angle of advance as required.
Eccentrics of this class are of either one of two kinds:
SHA FT-GO VERNORS.
Ill
112
SLIDE-VALVES.
SHA FT-GO VERNORS. 1 1 $
I. "Swinging" Eccentrics; in which the eccentric is
pivoted at some point which is secured to the shaft and rotates
with it. In that case the eccentric swings across the shaft
in the arc of a circle, and the slot is therefore curved.
2. " Shifting" Eccentrics; which move squarely across the
shaft, and the slot is of course straight.
The first question to be considered is the effect of moving
the eccentric across the shaft and holding it in any given
position; the means by which this movement is effected being
reserved for later discussion.
114
SLIDE-VALVES.
SWINGING ECCENTRICS.
Figs. 78 and 79 show the two extreme positions of an
eccentric of this class. It is pivoted at the point P, which is
usually located on the arm of a small fly-wheel called the
* 'governor- wheel " or "spider," which is securely keyed to
the shaft. When in its upper position, Fig. 78, the angle
between the crank and the eccentric is COE] and when in its
lower position it is C 'O 'E' ', Fig. 79. The shaded area in each
figure is the shaft, and the path of the eccentric center is
shown by the dotted circle.
The manner of laying out an eccentric in this type is
.shown in Fig. 80. The point /*, about which the eccentric
is to swing, is generally fixed, and the center O of the shaft is
also fixed. Then draw the line VZ through O and C, and
through O draw XY perpendicular to VZ. The angle of
advance having been determined, lay off XOD equal to that
angle. Then describe the circle whose center is O and whose
o
radius, OE, is equal to the eccentricity. This will cut the
line OD at E, which is the center of the eccentric. Now
about P as a center describe an arc passing through E,
Again with P as a center describe an arc passing through the
center of the shaft O. Now E' , which is where the eccentric
SHA FT- G O VERNORS. 1 1 5
center must be in order to secure a reversal of the engine,
will of course lie on the arc through E. Then the eccentric
center must be shifted through the arc EE' to secure the
reversal. The next thing is to determine the length of slot
required to allow this movement, and the smallest diameter
of eccentric. Therefore from e, the point where the arc EE'
cuts the line VZ, lay off the arc ee' equal to EE' . Draw e'P,
which cuts the arc drawn- through O at O'. Now the slot
must be drawn in as shown. That is, the lower center is O,
and the radius is just enough larger than the radius of the
shaft to permit it to clear. The upper center is O' , and the
radius is of course the same as the lower. The rest of the
slot outline is made up of two arcs having P as their center.
It is obvious that the center line of the crank must coincide
with VZ, and the spider or containing-wheel of the governor
must be keyed to the shaft so that this result will be obtained.
Now suppose the eccentric to have swung around so far
that the center is at I, Fig. 80. What is the result ? The
angular advance has been increased to XO\, and the eccen-
tricity has been decreased to Oi. (In order to have the
eccentricity remain the same, it would be necessary to have
the eccentric move around O as a center.) Now, the effect
of increasing the angular advance is to make the events of the
stroke all occur earlier. Decreasing the travel makes the
admission later, cut-off earlier, release earlier, and compres-
sion earlier, as shown in Table III. The combined effect of
the two changes can best be understood by constructing the
diagram. Fig. 81 shows the diagrams for five positions of the
eccentric center in Fig. 80. Diagrams E and E' correspond
to the extreme positions of the eccentric, or, as they are called,
*' full-gear forward" and "full-gear backward/' Diagram 2
shows the steam distribution when the eccentric center is at
2, Fig. 80, midway between its two extremes; that is, when
the eccentric is in "mid-gear," Diagrams I and 3 show the
SLIDE-VALVES.
results obtained with the eccentric center at I and 3 respect-
ively, Fig. 80.
It will be noticed that the cut-off varies more than the
admission or lead, because both changes of angular advance
and eccentricity affect the cut-off in the same way, making
it earlier; while the admission is made later by the decreased
travel, and earlier by the increased angular advance. These
two opposite changes in the lead may neutralize each other,
but only when the center, P, about which the eccentric
swings is removed to infinity that is, when the arc through
which it swings becomes a straight line, and the eccentric
moves straight across the shaft. The valve diagrams show
that the release varies less than the compression, with tlu
swinging eccentric.
SHAFT-GO VERNORS.
117
SLIDE-VALVES.
SHIFTING ECCENTRICS.
This is the type in which the lead is constant, as stated in
the preceding paragraph. It differs from the swinging eccen-
tric more in degree than in kind. Figs. 82 and 83 show one
Y
FIG. 82
of this kind in its two extreme positions. It will be noticed
that the cut-off has a greater range of variation in this than
in the swinging eccentric, because the valve travel has a
greater range. This will be clearly understood by reference
to Fig. 80, where the straight line from E to E' shows the
path which would be followed by the center of a shifting
eccentric designed to secure the same results as the swinging
eccentric. The. eccentricity being the distance from the
center of the shaft to the center of the eccentric, the variation
is evidently the greater with the shifting eccentric.
The next thing is to demonstrate that these variations in
the events of the stroke can be made to serve a useful pur-
pose.
Suppose an engine be running at full load ; the eccentric
SHA FT- G O VERNORS. 1 1 9
being fixed in place. Then suppose that a large part of
the load is suddenly thrown off. If the steam-pressure
remains the same, the engine will begin to speed up; and if
enough load has been thrown off, the increase in speed may
be sufficient to result in a bursting fly-wheel.
Now suppose that the engine is fitted with a movable
eccentric, and that when the full load is on the eccentric
is in full gear forward, giving the latest admission and latest
cut-off. Again, suppose that the load is thrown off, just as
before, but at the same time the eccentric is moved in toward
the 1 shaft. What will happen ? Under the lighter load the
engine will tend to speed up, but the earlier admission, due
to the increased angular advance, will cushion the piston,
tending to decrease the speed of the piston; and the cut-off
will come earlier in the stroke, also tending to reduce the
piston speed by shortening the length of time the moving
force is applied. In other words, the period of admission,
while remaining the same, is divided more nearly evenly
between the two strokes.
Now, with the cut-off made earlier, the engine will not
develop so much power; and the change may be just suffi-
cient to adapt the engine to the lighter load. A light load
at given speed requires less power than a heavy load at the
same speed.
Figs. 84 to 87, inclusive, illustrate various forms of shaft-
governors. The general principle of their construction is as
follows :
A small wheel or spider is keyed on the shaft, as stated
in the previous chapter. This wheel carries weights which
are pivoted on the arms or ring, so that under the influence
of centrifugal force they fly outward, this motion being
opposed by springs. The weights are linked to the eccen-
tric, so that a motion of the weights will cause a correspond-
ing motion of the eccentric. Now, if the engine speed
remains constant, the centrifugal force remains constant, and
120
SLIDE-VALVES.
FIG. 84
FIG. 85
SHAFT-GO VERNORS.
121
the springs will be stretched a certain amount, and the
eccentric will be held in one position. If, now, the load be
lessened, the engine will speed up momentarily, increasing
FIG. 86
FIG. 87
the centrifugal force, thus throwing the weights outward and
moving the eccentric across the shaft, shortening the cut-off
and adapting the power to the load. If the load is increased,
122 SLIDE-VAL VES.
the engine will slow down, decreasing the centrifugal force so
that the springs will pull the eccentric back, lengthening the
cut-off. . In this way the events of the stroke are regulated so
as to keep the speed of rotation sensibly constant at all loads.
Figs. 84 and 85 show two positions of the Westinghouse
governor, Fig. 84 representing the governor when at rest or
running below its normal speed; that is, at its latest cut-off.
Fig. 85 shows the governor when in the position correspond-
ing to its earliest cut-off; that is, when the engine has
speeded up considerably above the desired point and the
governor is acting to bring it down to the proper point.
The governor-weights B and B are pivoted to the wheel
A, which revolves with the shaft M at b and b. The upper
weight carries a link, PT, one end of which, P, is pivoted to
the governor-weight, and the other end, 7", is pivoted to the
eccentric casting. The weights are connected by the link ee.
The springs D and D are fastened to the weights and to the
wheel or disk, as shown; thus resisting any tendency of the
weights to fly outward. Under an increase of speed the
centrifugal force increases, and the springs are extended, thus
allowing the eccentric to move across the shaft.
Fig. 86 shows the Straight-Line governor, which is con-
tained within the fly-wheel. The center of the shaft is at O.
One of the fly-wheel arms, N, has a pivot on which the
weight-lever, WNM, is hung. The end, M, of the lever is
connected to the eccentric casting at Fand to the spring at
Z-, by the link MVL. The spring is secured to a boss, (9, on
the fly-wheel rim, as shown, and the eccentric is of the swing-
ing type, being pivoted at 5. The whole arrrangement
occupies the position shown when the engine is at rest; it
will be noticed that the eccentric is in its extreme position,
thus securing the greatest cut-off, as required when the engine
starts up. When the engine starts up the action of the cen-
trifugal force will force the weight outward, thus moving the
eccentric across the shaft.
Fig. 87 represents the Buckeye governor. When the
engine is at rest the springs F F hold the weights A A against
the inner stops. When the engine starts up the weights tend
to fly outward ; and when a certain rotative speed is reached
they move away from the stops, thus stretching the springs.
The eccentric being connected to the weights by the rods BB,
it is moved around the shaft, thus causing an earlier cut-off.
When a point of cut-off corresponding to the load has been
established, the speed ceases to increase, and the spring pull
and the centrifugal force will balance each other as long as
there are no changes in the load on the engine. If the load
is increased, or the pressure of the steam is reduced, the
speed is reduced momentarily, and a later cut-off is estab-
lished by the springs drawing the arms toward the shaft and
changing the position of the eccentric, thus bringing the
engine back to speed.
CHAPTER XIII.
SHAFT-GOVERNORSANALYSIS.
THE first principle of shaft-governors that varying the
events of the stroke properly will cause the engine to run at
the same speed under all loads having been established, the
next point in order for consideration is an analysis of the
action of the springs and weights whereby the movement of
the eccentric is accomplished. The method pursued in this
will be one drawn from a valuable monograph entitled "The
Mechanics of the Shaft-Governor," by Prof. Barr, published
in the Sibley Journal of Engineering, 1 896.
First, reduce the springs and weights to the simplest form,
shown in Fig. 88. Here there is but one weight, represented
by the ball W^ which slides on the radial rod R. The length
of this rod is such that when the ball is in the extreme inner
position its center coincides with the center of the containing-
wheel or spider G. The movement of the ball is resisted by
the spring 5, which is fastened to the rim of the spider. The
line of travel of the spring and ball is a diameter of the circle.
In this investigation friction is neglected.
First, suppose that the ball is moved outward by any force
from its central position until it occupies a position W. In
that case the spring must be extended a certain amount, equal
to the distance from W^ to W, or the distance the ball is
moved; and it will then exert a pull on the ball, tending to
restore it to its original position. The amount of this inward
pull is found as follows. "Spring strength " is the term used
to designate the force in pounds required to extend or com-
124
SHAF7'-GO VERNORSANAL YSIS.
I2 5
press a spring one inch. Thus, a 2O-pound spring means one
which requires a pull of 20 pounds to extend it one inch; a
6o-pound spring requires a pull of 60 pounds to accomplish
the same result. The total amount of extension or compres-
FIG. 88
sion is directly proportional to the force exerted upon the
spring. Thus, a weight of 120 pounds resting on the 20-
pound spring would compress it
120 -f- 20 = 6 inches,
or, if directly suspended from it, would lengthen it the same
amount. The same weight would only lengthen or shorten
a 6o-pound spring
1 20 -T- 60 = 2 inches.
The weight or force required to produce any given change in
the length is found by multiplying the spring strength by the
126 SLIDE-VAL VES.
amount of the change in inches. For example, to lengthen
a 3O-pound spring J inch would require a force of
30 X \ = 15 pounds.
The force exerted by a spring is equal to the force exerted in
compressing or extending it to the length it has. The
3O-pound spring just mentioned would exert a force of 15
pounds if compressed or extended. "To every action there
is an equal and opposite reaction.'*
Again, suppose the wheel to revolve with the ball at W.
There will be a centrifugal force set up which will tend to
throw the ball outward; this force will act radially, and as the
ball is free to slide along the rod it may be considered that
the centrifugal force acts along that line. The amount of this
force depends upon the weight of the ball, the rotative speed
of the wheel, and the distance from the ball to the center of
the wheel. Expressed in a formula, it is
" C= .00002 84 WN*R,* ''"./. . . (i)
where C = centrifugal force in pounds;
W= weight of ball in pounds;
N = revolutions per minute;
R = radius, or distance from the center of the wheel
to the center of the ball, in inches.
This may be stated in a rule as follows:
To find the centrifugal force in pounds exerted by a weight:
Multiply together the weight in pounds, the square of the num-
ber of revolutions per minute, and the radius, or distance in
inches from the center about which the weight revolves to the
center of gravity of the weight. Multiply this product by the
constant .0000284, and the result is the centrifugal force. (In
the case of a ball, the center of gravity is the center of the
ball.)
For example, suppose that the ball at W weighs 20
SHA FT- GO VERNORSANA L YSIS. 1 2 7
pounds, is 20 inches from the center of the wheel, and that
the latter is running at 150 revolutions per minute. The
centrifugal force is 255.60 pounds, found by the above rule as
follows: The square of the number of revolutions per minute
is 150 X 150 22,500. Then the centrifugal force is
C= 20 X 22500 X 20 X .0000284 = 255.60 pounds.
Formula (i) may be transposed to give the value of N
when the other factors are known, giving
(2)
This expressed in the form of a rule is as follows:
To find the number of revolutions per minute which a ball
of a known weight, at a known radius, must make to exert a
given centrifugal force : Multiply the weight in pounds by the
radius in inches, and divide the centrifugal force by this
product. Then extract the square root of the quotient, and the
figure is the number of revolutions per minute.
For example, if a weight of 20 pounds is rotating about
a center 20 inches distant, and exerting an outward pull of
255.60 pounds on the link connecting it with that center, it
is making 150 revolutions per minute; because, applying
formula (2) or its corresponding rule,
20 X 20 = 400
255.60 -r- 400 = .639
^^639= .7994
187.7 x .7994 = 150-04
Now, if the ball when running at any given speed is held
stationary, it is obvious that at that moment the centrifugal
force of the ball and the spring pull must be equal to each
other. That is, taking the 2O-pound weight which was just
figured to have 255.60 pounds of centrifugal force urging it
128 SLIDE- VA L VES.
outward, the inward spring pull must be 255.60 pounds in
order to hold the ball at W t . It is obvious that this amount
of spring pull could be secured by using a spring of any
strength and stretching it the required amount; or the stretch
of the spring could be assumed and the strength calculated.
It is better to assume the spring strength, however, because
this affects the closeness of regulation, as will be explained
later.
Suppose, then, that the spring is a 4<D-pound. Then to
hold the ball at W the extension necessary is
255.60 -^ 40 = 6.39 inches.
This extension will not permit the ball, when the tension is
relieved, to move into the center of the containing-wheel;
because the ball is 20 inches out from the center, so that,
with the tension removed and the spring collapsed, the ball
would still be
20 6.39 = 13.61 inches
away from the center. The spring pull with the ball that
distance out from the center would be zero, and therefore,
to maintain a balance between the centrifugal force and the
spring pull, the centrifugal force must be zero. This can
only happen when the radius at which the weight revolves, or
the rotative speed, is zero. But as the radius is 13.61, the
rotative speed must be zero.
Now suppose that the ball moves inward to W z , which is
4 inches from W, or 16 inches from the center. In that case
the spring extension has been decreased to 6.39 4 = 2.39
inches, and the spring pull to
2.39 X 40 = 95.6 pounds.
Then the speed at which the wheel must revolve to main-
tain the ball at W^ , the revolutions per minute, must be such
SHA FT- GO VERNORSA NA L YSIS. 1 29
as to produce a centrifugal force of 95.6 pounds. Formula
(2) or Rule II gives the value of N as 113 07.
95-6
2O X 2O
= 187.7 I/. 239
= 113-67.
That is, the governor-ball will not move to W^ until the
speed has decreased to 113.67 revolutions per minute.
Next, find the effect when the ball is on the other side of
W, say at W^ , which is 4 inches farther out than W y or 24
inches from the shaft center. The spring extension at that
place is
6-39 + 4= 10.39,
and the spring pull is
10.39 X 40 = 4I5-6.
In order to have the centrifugal force equal to the spring
pull, the revolutions per minute must increase to 175.52,
because, from formula (2),
- :8 7 . 7 ' 4 ' 5 ' 6
20 X 24
= 187.7 ^.86583
= 187.7 x .9351
= 175.52.
That is, with the engine and governor-wheel running at
175.52 revolutions per minute the ball would be held in
balance at W 19 24 inches from the center of the shaft.
Then if W 9 and W^ are the extreme inner and outer posi-
tions of the governor-ball, the extreme variation in speed is
from 113.67 revolutions per minute to 175.52. If the ball in
I 3 SLIDE- VA L VES.
moving from W^ to W l moves the governor from its earliest
to its latest cut-off, the engine will vary
175.52 - 113.67 = 61.85
revolutions per minute in passing from light to full load. The
normal speed of the engine being taken at 150, the variation is
61.85
- - = 41.23 percent.
Prof. Barr has devised a diagram which is of great service
in showing clearly and at a glance the relation between the
spring pull and centrifugal force. Fig. 89 shows it as applied
to the case just discussed.
Draw the line W Q W l , and let the distances on it from W ,
on any convenient scale, represent distances from the center
of the shaft, or radii, at which the ball revolves. The scale
adopted here is two small squares to the inch. Then let dis-
tances perpendicular to W W^ represent the forces acting on
the ball. For example, it was shown that with the ball at
W lt the spring pull, no matter what the speed, is 415.6
pounds. Then lay off W 1 S 1 to represent this amount. The
scale chosen in the figure is 20 pounds for one small square;
and it will generally be found necessary to employ a much
smaller scale for the forces than the distances in order to bring
the drawing within reasonable limits. Next, lay off W^S^
equal to 95.6, which was found to be the spring pull with the
ball at W^ , or 16 inches from the center. Join 5, and 5 2 , and
produce the line 5,5, until it meets W W, at 5 . It will be
found that S is 13.61 inches from W . This is as it should
be; for it was shown that with the ball at that distance from
the center the spring pull is zero. The line 5 5,, represent-
ing the spring pull, is straight, because the pull varies directly
as the extension of the spring. The spring pull at any point
is then found by drawing a perpendicular from the point on
W^Wi , and measuring the length included between W^W^ and
SHA FT-GO VERNORSA NA L YSIS. 1 3 l
the line of spring pull; thus with the ball at W, 20 inches
from the center, the spring pull is WS, which is 255.6 pounds.
In other words, horizontal distances from 5 , as S^W^
S^W, S^W^ , represent the spring extension.
Now to consider the centrifugal force. It is desirable to
have the engine run at a constant speed, and the centrifugal
force will be considered on that basis. There must be some
radius at which the spring pull and the centrifugal force at
that speed will balance. For example, it was found that with
the engine running at 150 revolutions, the spring pull and the
centrifugal force balance with the ball at W, 20 inches out
from the center. Now, with a constant speed, the centrifugal
force varies directly as the radius, as shown by formula (i).
With the ball at the center of the wheel -with the radius
zero the centrifugal force is zero. Therefore the lines of
spring pull and centrifugal force at constant speed coincide at
20 inches from W^ and W n C l is the line of centrifugal force,
found by drawing a straight line from W through C. The
centrifugal force at any other radius is found by drawing the
132 SLIDE-VALVES.
perpendicular from the end of the radius to the line of centrif-
ugal force. Thus at W^ the centrifugal force is W l C l , equal
to 304.9 pounds, and with the ball at W^ the centrifugal force
is 202.5 pounds, measured by W^S^.
Now this diagram shows very clearly that if the speed be
maintained constant, the centrifugal force and the spring pull
will not be equal at different radii. They are equal at W
only; beyond that point the spring pull is greater, and below
it the centrifugal force is the larger. This shows that the
governor is possessed of a considerable amount of stability;
that is, it is but little liable to derangement by outside forces,
such as the drag of the valve. Suppose, for example, that
the engine is running at 150 revolutions. Then the ball
balances at W t and if the speed be maintained constant, a
force equal to S 1 C l must be applied to move the ball out to
W l and balance it there. For, if the ball is at W t , the cen-
trifugal force urging it outward at 150 revolutions is Wf^
and the spring pull drawing it inwards is W^S^. The differ-
ence between these two is C 1 S 1 , equal to
415.6 304.9 = 1 10.7 pounds,
which is the force necessary to send the ball out to W l and
hold it there. Of course, if a lesser force were applied, it
would move the ball somewhat, but it would not force it out
all the way.
The action of this type of governor, with a very strong
spring, may be summed up as follows: It does not give very
close regulation, the example chosen showing a variation of
from 175 to 113 revolutions, but the stability, or resistance to
external forces, is very great. This stability increases as the
spring strength increases, which shows clearly on the diagram,
the line ss being the spring pull for the stronger spring. This
is because the point at which the spring pull is zero is at a
greater distance from W , and the line of spring pull, passing
through 5, makes a greater angle with the line of centrifugal
SHA FT -GO VERN ORSA NA L YSIS.
133
force. Hence the distance between the two lines is greater
at any given point; and as this distance measures the force
required to displace the ball, the stability is greater.
Next, consider this form of governor having in a spring
of such strength that when there is no spring extension the
ball will be at the center of the spider. That is, with the ball
out anywhere on the rod the spring extension is equal to the
radius at which the ball revolves. The effect of this spring
can be best understood by reference to Fig. 90, which is the
Barr diagram for this case. Take the same numerical exam-
ple as before, a 2O-pound ball running at 150 revolutions at a
radius of 20 inches. The centrifugal force is 255.6 pounds as
determined before, and the spring strength must be
255.6 -v- 20 = 12.78 pounds
to bring the weight to the center of the spider when the
spring is collapsed, or has no tension on it. This figure is
drawn on the same scale as the preceding one, and the line of
centrifugal force at constant speed is the same as before; but
the line of spring pull is decidedly different. At W the
spring pull is equal to the centrifugal force. At W it is
zero. Therefore the straight line joining these two points,
134 SLWE-VAL VES.
which represents the spring pull, coincides with the line of
centrifugal force.
What does this show ? It shows that the governor regu-
lates very closely, for the spring pull is equal to the centrif-
ugal force at all positions. With the ball out at W l the
speed would be 150 revolutions, and with the ball at W^ the
speed would be the same. But in obtaining this closeness of
regulation the stability has been sacrificed. The distance
between the lines of spring pull and centrifugal force measures
the resistance to deranging forces; and, as this distance is
zero at all points, the slightest force would move the ball
over its entire length of travel. The slightest increase of
speed would send the ball out to its extreme outer position,
and the slightest decrease would bring it in to the center of
the wheel. A governor having this property is said to be
isochronous.
Fig. 91 shows the Barr diagram for the remaining case of
the elementary governor, that in which the spring is too weak
for isochronous action. The same figures are used as before
weight of ball, 20 pounds, revolutions 150, spring pull and
centrifugal force balancing at a 2O-inch radius. Assume the
spring strength as 8 pounds. Then, the centrifugal force with
W at 20 inches being 255.6 pounds, the spring extension
must be
255.6 -7- 8 = 31.95 inches.
That is, the spring extension is
31.95 20= 11.95 inches
greater than the radius at which the ball revolves. This
locates the point S Q over at the left of W^ , as shown in the
figure. Joining S and S, which is located the same as in the
previous cases, gives the line of spring pull. Joining W and
5 gives the line of centrifugal force at constant speed. This
case is the reverse of the first one, shown in Fig. 89. Beyond
SHA FT-GO VERNORSA NA L YSIS.
135
1 36 SLIDE- VA L VES.
W the centrifugal force is greater than the spring pull, and
below W the spring pull is the greater of the two, thus show-
ing that this form of governor is decidedly unstable. Sup-
pose, for example, that this engine is running at its normal
speed of 150 revolutions with the ball at W. Then suppose
a part of the load to be suddenly thrown off. As a natural
consequence the engine will speed up, and the increased cen-
trifugal force will throw the ball out ; and as the centrifugal
force increases faster than the spring pull, the ball will go
clear out to its extreme position, thereby making the cut-off
much earlier than is necessary for the change in the load.
This early cut-off of course results in a reduction of speed.
If the outer limit is W^ where the spring extension is
(31.95 +4)= 35-95 inches,
and the spring pull is
35-95 X 80 = 284.60 pounds,
the ball will balance there until the speed is reduced below
the point where the centrifugal force equals 2 84. 6 pounds, or,
from formula (2),
N =
20 X 35-95
= I87.71/.3964
= 187.7 X .629
= 1 1 8. 06 revolutions per minute.
As soon as the speed is reduced below this point the ball
begins to move inward; and as the centrifugal force decreases
faster than the spring pull, it will continue to move in until it
reaches its inner limit. This will result in a later cut-off than
is necessary to bring the engine back to speed, and the result
is that the speed will grow greater than is necessary to hold
SHA FT-GO VERNORSA NA L YSIS. 1 3 7
the ball at W^. If this inner position is, as before, at W 9 ,
which is 16 inches from the center of the wheel, the spring
extension at that point will be
16 + 11.95 = 27.95 inches,
and the spring pull will be
27.98 X 8 = 223.60.
The revolutions required to produce this amount of centrif-
ugal force with the ball out 16 inches are
= I87 V
223.6
20 X 16
= 187/7^.69875
= 187.7 X .836
= 156.92.
That is, the engine must speed up to about 157 revolutions
before the ball will move out again. When it does start, it
will go out to the outer end again, and it will keep on "hunt-
ing " or "racing " up and down, the speed meanwhile varying
from 118 to 157 revolutions, or
(157- 118)
^jp- - ; = 26 per cent.
The same argument holds true for any other deranging
force, such as the weight of the valve pulling on the valve-
rod, so that this form of governor is evidently very unstable.
These three cases stability, isochronism, and unstability
cover the ground completely. The first two qualities are
greatly to be desired, but it will be seen at once from the
preceding text that they cannot be obtained at the same
time. Stability can be secured easily enough by using a
strong spring, but this renders isochronism out of the ques-
1 3 8 SLID E- VA L VES.
tion. In the real governor, friction in the moving parts
renders the governor stable, but at the same time it destroys
the desired isochronism. Good results have been secured by
reducing the external forces acting on the governor to a
minimum and then making the governor as nearly isochronous
as possible. It was shown that a perfectly isochronous
governor is a possibility when the spring pull and centrifugal
force are considered to be the only forces acting on the
weight. But when gravity is considered it will be found that
the perfectly frictionless governor will race between its outer
and inner limits under a steady load. This is due to the fact
that when the ball is above the shaft, gravity tends to draw
it in, and when the ball is below the shaft, gravity tends to
draw it out, or away from the center. The average cut-off
obtained by this action would be the one suited to the load,
but would be alternately too early and too late. This could,
of course, be obviated by making the friction of the governor
enough to prevent the racing, but this is a poor plan, as the
desired isochronism is thereby rendered impossible, because,
when the ball moves outward to compensate for a reduction
of the load or for a reduction of the steam-pressure, the
centrifugal force must overcome the friction in addition to the
spring pull. The best method of overcoming the gravity
distortion is to balance the governor; that is, to employ two
weights, or their equivalent, one on each side, the result
being that the gravity effects on the two will be opposed to
each other and will therefore be negligible.
The same argument applies to the disturbing force of the
valve; it could be compensated for by friction of the governor,
but it is not desirable to so arrange it, on account of the
destruction of isochronism. With an ordinary slide valve
this pull would be very great, because the steam of a pressure
equal to or but little less than the boiler-pressure bears
directly upon the valve, forcing it against the seat, while the
pressure underneath the valve acting upward against this is
SHA FT- G O VERNORSA NA L YSIS.
'39
only that of the exhaust steam which is expanded to a con-
siderably lower pressure. This unbalanced downward pressure
increases greatly the effort required to move the valve.
When this difference of pressure is obviated or greatly
reduced, the valve is said to be balanced.
The simplest form of balanced valve is the piston-
rfi
FIG. 92.
valve, shown in section in Fig. 92. The valve is cylindrical,
the shape being similar to that which would be obtained by
rotating a plain D valve about its valve-stem. Consequently
the valve-seat must be cylindrical to fit the valve-seat. The
valve may be arranged with double pistons, such as in Fig.
92, or the pistons may be single, as in Fig. 93. By referring
to either figure, it will be seen that the steajgi==f^s*t^ must
140
SLIDE-VALVES.
be equal on the two end faces of the valve ; and as the valve
is in contact with the seat throughout its entire length, there
is no possibility of any other pressure than friction getting at
the valve sidewise. The objections to this form of valve are
leakage and wear. The piston cannot fit tight in its bore,
because unequal expansion would cause the piston to bind in
the bore when the latter is cold and the pistons hot. This
necessary difference in size would result in leakage if not
FIG. 93.
guarded against. Spring rings are used to effect the steam-
tight joint, and these rings cause friction and wear. Some-
times no rings are used, but in that case great care must be
taken to keep the temperature of the piston and the bore the
same. This is done by steam-jackets, as shown in Fig. 93.
Another system of balanced valves employs pressure-
plates. Here a flat plate is used which is secured in the
steam-chest, and which receives the unbalanced steam-press-
ure, while the valve slide sunder the plate. It will be seen
that the principle is practically the same as that of the pistor -
SHAFT-GOVERNORSANAL YSIS.
141
valve, with the difference that with the pressure-plates the
valve can be made flat and steam-tight at the top and bottom
only, where it touches the pressure-plate and seat.
There are three classes of pressure-plates fixed, adjust-
able, and flexible. Fig. 94 shows one of the fixed type.
Here the plate is bolted to the bottom of the steam-chest, and
the length of the plate is such that the valve never projects
FIG. 94.
beyond it. The exposed ends of the valve, being of equal
area, balance each other, and the pressure under the valve
through the ports is balanced by recesses of equal area under
the hood and over the valve.
In some cases a spring ring or similar arrangement is fitted
to the top of the valve in such a way as to make a continuous
contact between the plate and the valve, thus preventing any
steam from acting on the top of the valve. Fig. 95 shows
such a valve, where the rings// are inserted in the top or
142
SLIDE-VALVES.
back of the valve and are pressed upward against the plate A
by means of springs. The space between the plate and the
back of the valve is open to the exhaust through the opening
//, as shown. This balances nearly all the top of the valve,
and the pressure on the remaining portion is sufficient to pre-
vent leakage.
An example of the second type is shown in Fig. 96. In
this design the pressure-plate is supported on an inclined
plane, the plate being made to slope at the same angle, as
shown by the dotted line in the figure. By means of the
adjustable handle E, the movable plate can be adjusted to
FIG. 95.
secure any desired amount of pressure on the back of the
valve. The valve also illustrates another type of double-
ported valve.
The third system, that of flexible pressure-plates, consists
of a flexible plate of steel or other elastic metal, which is so
arranged as to allow the pressure upon it to force it down
upon the valve, but only with force enough to prevent leak-
age between the valve and the plate. An example of this
type is shown in Fig. 97.
By the use of such valves as these the friction is greatly
reduced, and the pull on the governor becomes a minimum.
Another disturbing element in the action of a governor is
the inertia of the valve. This comes into play at each end of
the valve travel, when it is necessary to reverse the direction
. G O VERNORSA NA L YSIS.
143
144
SLIDE-VALVES.
of motion of the valve, thus giving a sudden pull. One
method of obviating or compensating this deranging force
consists of employing a dash-pot. This consists of a piston
fitting loosely in a cylinder containing oil. If the piston is
moved slowly and steadily, the oil offers little or no resistance
to the motion, as it will readily slip past the circumference of
the piston. But if the piston be moved suddenly, the oil
will be unable to flow by quickly enough to permit the piston
to move rapidly, and will therefore bank up and impede the
progress of the piston. A piston-rod passes through a
FIG. 97.
stuffing-box on the dash-pot, and is attached to the governor-
weight, the dash-pot itself being fastened to and revolving
with the governor-wheel or spider. This dash-pot corrects
the distortion due to the inertia of the valve, and in no way
impairs the action of the governor. Another result is obtained
by the employment of this contrivance. When the weight
starts to change its position under the influence of a change
of speed it starts with a velocity which, owing to its inertia,
would carry it beyond the proper place were it not for the
retarding influence of the dash-pot. The dash-pot then
serves as a preventive of too sudden a motion in either direc-
tion.
SHAFT-GO VERNORSANAL YSIS. 145
The last-mentioned deranging force or disturbing element
the inertia of the governor-weight may be turned from a
hindrance to a help, making it perform the functions of a
dash-pot. This is explained as follows by Mr. E. J. Arm-
strong, in a paper read before the American Society of
Mechanical Engineers, and forming a part of Vol. XI. of the
Transactions : When any governor is engaged in its task of
controlling the engine, the weight travels at a variable rate
of speed, resulting from its revolving in a circle of variable
size. This change in velocity is often quite considerable,
depending of course upon the amount of radial movement
and the rotative speed. To take an example from the
Straight-Line engine: a fly-weight is i6 inches from the
center of the shaft when in, and 2oJ inches when out, mak-
ing, at 220 revolutions per minute, a difference of 194 feet 4
inches per minute. Whenever the fly-weight takes a new
position it must change its speed must move either faster or
slower, be accelerated or retarded and must absorb or give
out power somewhere. This resistance to a change in velocity
acts at right angles to a radial line drawn through the center
of gravity of the weight, and if the weight were pivoted so as
to move radially, as in Fig. 98, the only result would be to
increase the pressure on the pivot. If the fly-weight were so
pivoted as to move at an angle to a radial line, as in Fig. 99,
so that in its outward movement it goes toward the way the
wheel rotates, then the outward movement of the fly-weight
will be opposed, to some extent, by this resistance to accel-
eration, depending on the angle which the line of movement
forms to a radial line or, to put it in another way upon the
length of the lever-arm AB y CA being the line of resistance,
and B the fly-weight pivot. When the weight moves toward
the shaft, the action is the same. It has to part with some
of its momentum, and so hangs back, as in its outward move-
ment, thus making, similarly to a dash-pot, a resistance to
movement in both directions, which can only be overcome
146
SLIDE-VALVES.
FIG. 98
FIG. 99
SHAFT-GO VERNORSANAL YSIS.
147
quickly by a great force, or by a small one moving slowly;
this resistance increases with the velocity of the weight, which
is all that is accomplished by the ordinary dash-pot.
To return to the analysis of the governor proper. Fig.
100 represents a slight modification of the elementary type.
The change consists in putting a stop, T, on the radial rod,
for the purpose of preventing the ball from traveling in to
the center of the wheel. With the wheel at rest it is possible
FIG. 100
to maintain the ball in the position shown in the figure by
either one of two methods: First, by the employment of a
spring of any strength, but of such length that it will hold
the ball in that position without any extension that is, the
governor will be of the stable type, such as explained before,
and whose action is illustrated in Fig. 89; and, second, by
shortening such a spring, and then extending it by the amount
148 SLIDE-VALVES.
that it is shortened, thus getting an inward pull on the ball.
This pull is called the initial tension, because it is the tension
existing in the spring while the wheel is at rest, or before any
centrifugal force is developed to extend the spring. The
amount of the initial tension of the spring may be expressed
in two ways the amount, in inches, that the spring is
shortened, or the pull, in pounds, on the ball. Suppose for
example that a lo-inch spring of 40 pounds strength will hold
the ball at the stop without any extension. If, then, this
spring is shortened to 7 inches and then stretched out to cover
the 10, the pull required will be
(10 7)40 = 120 pounds,
and if the end is fastened to the ball, it will exert that pull on
it. The amount of the initial tension in pounds is therefore
equal to the amount that the spring is extended when the
wheel is at rest, and the ball against the stop, multiplied by
the spring strength. If the initial tension is expressed by the
inches that the spring is shortened, the spring strength must
also be specified in order to render it exact. In the practical
governors it is usual to supply some device, such as a screw-
thread, by which the extension of the spring can be altered
at will, and any desired amount of initial tension secured.
If it is desired to produce an isochronous governor, it is
very evident that the initial tension must be such that if the
stop on the radial rod were removed, the ball would move
inward to the center of the retaining wheel.
If the initial tension is less than that required to bring the
ball in to the center of the wheel with the stop removed, the
result will be the same as if a strong spring were employed ;
that is, the governor will be stable.
If the initial tension is too great for isochronism that is,
if the ball would, on removal of the stop, go in beyond the
center the result obtained would be the same as if a weak
spring were employed, as discussed and shown in Fig. 91.
SHA FT-GO VERNORSA NA L YSIS. 1 49
This initial tension produces a slight variation in the
operation of the governor. It will be remembered that with
the elementary isochronous governor the weight starts out as
soon as the engine is started; but with initial tension the
engine must continue to speed up until it is going so fast that
the centrifugal force of the ball exceeds the initial tension.
For example, if the initial tension is 120 pounds, and if the
ball, weighing 20 pounds, is held by the top at the point 6
inches from the center of the wheel, the engine will speed up
until the centrifugal force is greater than 120 pounds; that is,
from from the rule on page 127, until the revolutions per
minute are greater than
120 .
= 187.7 Vi
= 187.7.
The previous argument being mastered, it is next in order
to discuss the simple type of shaft-governor shown in Fig.
101. Here the weight C is secured to the lever AC, the latter
being pivoted at A on an arm of the containing-wheel. The
spring is fastened to the lever at B, and to the rim of the con-
taining-wheel at D. Now, when the wheel revolves, a certain
centrifugal force is induced which tends to throw the weight
out, that is, to turn the lever about A. This is resisted by
the spring as before, but with a decided modification. The
spring pull and the centrifugal force are not directly opposed
to each other, but act on the lever AC at different distances
from A. The distance AB is the spring leverage, and the
distance A C is the weight leverage. It is at once apparent
that in order to secure a balance between the opposing forces
at any given speed, it must be true that spring pull X spring
leverage = centrifugal force X weight leverage. Or, putting
this in symbols,
5 X L = C X L',
1 5 O SLIDE- VA L VES.
where 5 is the spring pull;
L is the spring leverage;
C is the centrifugal force;
L' is the weight leverage.
Another difference between this and the elementary type
is occasioned by the fact that the spring and weight are not
FIG. 101
directly opposed to each other; that is, the spring extension
is not equal to the weight displacement. The spring exten-
sion is directly proportional to its leverage, and is found by
multiplying the displacement of the weight by the spring
leverage, and dividing by the weight leverage. For example,
suppose the weight leverage to be 8 inches, while the spring
leverage is but 4 inches, and that the weight has moved out
SHA FT-GO VERNORSA NA L YSIS. 1 5 1
6 inches from its inner position under the influence of centrif-
ugal force. Then the spring extension is equal to
6X4
- = 3 inches.
That is, if the spring has half the leverage of the weight,
it has half the extension, and so on. It must be borne in
mind that this extension is from the inner position of the
weight. If a stop is used and initial tension is supplied, the
total spring extension is equal to that above found, and which
may be called centrifugal extension, increased by the initial
tension in inches.
This form of governor possesses certain obvious advan-
tages. When it is desired to secure a greater spring pull, it
is not necessary to substitute a stronger spring. The same
effect may be secured by fastening the spring at a greater dis-
tance from A, thereby securing a greater spring leverage, and
at the same time increasing any existing initial tension, or, if
none exist, adding some.
The practical governors have weights of various shapes,
but the principle remains the same. The weight is supposed
to be concentrated at its center of gravity, and the weight
leverage is the distance from this center of gravity to the
pivot, irrespective of the form of the connecting link. Of
course the weight of the lever must be taken into considera-
tion in finding the center of gravity. The radius at which the
centrifugal force is found is the distance from the center of
gravity to the center of the containing-wheel.
The foregoing text leads to the following general rules,
which are applicable to all governors built on the above
principle; and this type of governor is by far the most
common.
1 5 2 SLIDE- VA L VES.
TO INCREASE THE SPEED.
It may be desirable to increase the speed at which the
engine is to run. In that case, increase the spring tension,
observing the following limitations: If the increase of spring
tension is sufficient to produce isochronism, the governor will
race as shown before. The increase must be kept below this
limit, and as the governor is usually set near this limit, the
above adjustment is applicable only to small changes of speed.
For larger changes the governor-weight may be decreased.
This will produce an increase of speed, because the centrifugal
force will be reduced and it will require a greater number of
revolutions per minute to develop sufficient force to move the
weight across its extreme travel. Or the increase may be
secured by moving the weight in toward the pivot about
which it revolves. This lessens the weight leverage, and con-
sequently a greater number of revolutions per minute are
required to develop the same centrifugal force as before. An
increase of the spring leverage will produce an increase of
speed, because, the spring pull being thereby increased, a
greater centrifugal force is required to stretch it, and this can
only be secured by an increase in the number of revolutions
per minute.
TO DECREASE THE SPEED.
The adjustments here are necessarily the opposite of those
cited above. For small changes, reduce the initial tension.
For larger changes, the weight may be increased; or the
weight may be moved out from the pivot, increasing the
weight leverage; or the spring leverage may be reduced.
TO REDUCE THE VARIATION OF SPEED.
That is, to produce isochronism. Increase the spring ten-
sion or diminish the spring leverage.
All of these changes are subject to one limitation that
SHAFT-GO VERNORSANAL YSIS. 1 5 3
of performance. It is impossible to set a governor to produce
exact results by theory alone. The adjustment must be
made as nearly right as may be, and then the engine must be
run. If the engine runs as calculated, all well and good. If
not, the adjustment must be carried out on the lines indicated
above until the engine runs smoothly and regulates closely.
154
SLIDE-VALVES.
TABLE I.
VALUES OF O(y, FIG. 2O, FOR VARIOUS RATIOS OF CONNECTING-ROD TO CRANK.
Ratio
6>0' = Crank.
Ratio
OO' = Crank.
Ratio
OO' = Crank.
of
/-*_
of
f- _
of
/-. _
Con-
nect'g-
rod to
Divided
Multiplied
Con-
necting-
rod to
Divided
Multiplied
Con-
nect'g-
rod to
Divided
Multiplied
Crank.
by
by
Crank.
by
by
Crank.
by
by
I
4
.2500
4
16
.0625
7
28
.0356
.1
6
.2000
.1667
3
17
.0588
.0556
7\
29
30
0345
0333
If
7
.1429
4f
19
.0526
7l
31
.0323
2
8
.1250
5
20
.0500
8
32
0313
H
9
.1119
If
21
.0476
8 i
33
.0303
4
10
.lOOO
sl
22
0455
8if
34
.0294
2f
ii
.0909
5t
23
0435
8f
35
.0286
3
12
0833
6
24
.0417
9 t
36
.0278
3?
13
.0769
63-
25
.0400
37
.0270
3*
14
.0714
6
26
0385
9^
38
.0263
3f
15
.0667
6f
27
.0370
9t
39
.0256
SHA FT- G O VERNORSA NA L YSIS.
TABLE II.
DECIMAL EQUIVALENTS OF FRACTIONS OF AN INCH.
1/64
.015625
17/64
.265625
33/64
515625
49/64
.765625
1/32
03135
9/32
.28125
17/32
.53125
25/32
.78125
3/64
.046875 ;
19/64
.296875
35/64
.546875
51/64
.796875
1/16
.0625
5/i6
.3125
9/16
5625
13/16
.8125
5/64
.078125
21/64
.328125
37/64
.578125
53/64
.828125
3/32
09375
17/32
34375
19/32
59375
27/32
.84375
7/64
.109375
23/64
.359375
39/64
.609375
55/64
859375
1/8
.125
3/8
.375
5/8
.625
7/8
.875
9/64
. 140625 !
25/64
.390625
41/64
.640625
57/64
.890625
5/32
.15625
13/32
.40625
21/32
.65625
29/32
.90625
11/64
.171875
27/64
.421875
43/64
.671875
59/64
.921875
3/i6
1875
7/16
4375
11/16
.6875
15/16
9375
13/64
.203125 i
29/64
.453125
45/64
.703125
61/64
.953125
7/32
.21875
15/32
.46875
23/32
.71875
31/32
.96875
15/64
.234375
31/64
.484375
47/64
734375
63/64
.984375
i/4
.25
r/2
5
3/4
75
i
.i
I 5 6
SLIDE-VALVES.
TABLE III.
EFFECT OF CHANGING OUTSIDE AND INSIDE LAPS, TRAVEL AND ANGULAR
ADVANCE. (THURSTON.)
CHANGE.
ADMISSION.
EXPANSION.
EXHAUST.
COMPRESSION.
Increase
Outside Lap
begins later,
ceases sooner
occurs earlier,
continues longer
unchanged
unchanged
Decrease
Outside Lap
begins earlier,
ceases later
begins later,
period shortened
unchanged
unchanged
Increase
Inside Lap
unchanged
begins as before,
continues longer
begins later,
ceases earlier
begins sooner,
contin. longer
Decrease
Inside Lap
unchanged
begins as before,
period shortened
begins earlier,
ceases later
begins later,
period short'd
Increase
Travel
begins sooner,
ceases later
begins later,
ceases sooner
begins later,
ceases later
begins later,
ends sooner
Decrease
Travel
begins later,
ceases earlier
begins earlier,
ceases later
begins earlier,
ceases earlier
begins earlier,
ceases later
Increase
Angular
Advance
begins earlier,
period
unchanged
begins sooner,
period
unchanged
begins earlier,
period
unchanged
begins earlier,
period
unchanged
Decrease
Angular
Advance
begins later,
period
unchanged
begins later,
period
unchanged
begins later,
period
unchanged
begins later,
period
unchanged
SHAFT-GO VERNORSANAL YSIS.
TABLE IV.
PORT AREA, PORT WIDTH, AND STEAM-PIPE DIAMETER FOR VARIOUS
PISTON SPEEDS AND STEAM VELOCITIES.
157
Velocity of Steam. Feet per Minute.
4,000.
6,000.
8,000.
10,000.
12,000.
4J
3
a
G
O
V
rt
G
a
c
3
G
o
V
rt
1
fc
1
rt
g
E
I
g
E
I
g
E
O
a
.2
8
.2
w
s
a
tt
to
1
rt
8.
<
.
Q
^
w
3
**
u
Q
u
3
"^
C .
o
G
O
S^ 1
c
o
G
O
% 5
c
I
~.t?
o
iJ^
G
c
<u >,
G
O
1
fl
.2
E
Is
.a
JJP
tn
E
tn
II
.S3
OH
H
S
53
38
Q IS
Q"" tn
i
Q8
1
co
G
8 .
a$
'II
"O4J
i.
< b
m-pipe
ameter
jgS
rt
* '
m-pipe
ameter
if
cB
QJ
fl
II
d .
m-pipe
ameter
i!
2
?
15
CO
!'
| S
So
c/5
| S
1 S
co
l]
jjjo
So
c/5
I 3
1 S
2 a
V
IOO
.025
.158
.022
.017
.129
.015
.013
.112
.011
.010
. IOO
.009
.008
.091
.007
I2 5
.031
037
.177
.194
O27
032
.025
^58
.022
.019
.125
137
.014
.017
.OI3
OI.S
.123
.013
.013
.112
.009
.Oil
1 75
.044
.209
.038
.029
.171
025
.022
.I 4 8
.019
.Ol8
.132
015
.015
. 121
.013
200
.050
.224
.043
033
183
.029
.025
.158
.022
.020
.141
.017
.017
.129
.015
225
.056
237
.049
.038
.194
33
.028
.168
.024
.023
.150
.O2O
.019
I 37
.016
250
.063
.250
055
.042
.204
37
.031
.177
.027
.025
.158
.022
.021
.144
.018
275
.069
.262
.060
.046
.214
.040
034
.I8 5
.030
.028
.166
.024
.023
.020
300
075
274
.065
.050
.224
.044
.038
'93
.033! .030
173
.026
.025
157
.022
325
.081
.285
.070
054
233
.047
.041
.201
.036
033
.180
.028
.027
.164
.024
350
.088
.296
.076
.058
.242
.051
.044
.209
.038
035
.187
.031
.029
.171
.025
375
.094
.306
.O8l
.063
.250
055
.047
.2I 7
.041
.038
.194
033
.031
.177
.027
400
. IOO
3*3
.086
.067
.258
.050
.224
.044
.040
.200
035
033
.183
.029
425
.106
.326
.092
.071
.266
.062
053
.231
.0461 .043
.206
037
035
.188
031
45
"3
335
.098
075
274
.065
.056
.2 3 8
.049
045
.212
39
.038
193
.032
475
.119
344
103
.079
.281
.069
059
.244
.052
.048
.218
.041
.040
.199
33
500
125
353
.108
.083
.288
073
.250
.055; .050
.224
.044
.042
.204
036
525
362
"3
.088
295
.077
!o66
.256
058) .053
.220
.046
044
.209
.038
550
!ifj
.119
.092
.302
.080
.069
.262
.060: .055
235
.048
.046
.214
.040
575
.i 44
^380
.124
.096
309
.084
.072
.268
.063! .058
.240
.050
.048
.219
.042
600
.15
.388
.130
.100
.316
.087
075
274
.o6s! .060
245
.052
.050
.224
.044
625
.156
395
*35
.104
323
.091
.078
279
.068 .063
.250
055
.052
.228
45
650
.163
403
.141
.108
329
.094
.081
285
.071
06 5
255
057
054
232
.047
675
.169
.411
.146
"3
335
.098
.084
.290
.074
.068
.260
059
056
237
.049
700
175
.418
.150
.117
341
.102
.088
.296
.077
.070
-265
.061
.058
.241
.051
725
.181
.426
155
.121
347
.106
.091
.301
.079
073
.269
.063
.060
.246
53
750
.188
433
.161
125
353
.109
.094
.306
.082 .075
-274
06 5
.063
.250
55
775
.194
.440
.166
.129
359
.097
3 TI
.085 .078
.278
.068
.065
254
056
800
.200
447
.172
133
365
!n6
.100
.316
.087 .080
.283
.070
.067
259
.058
825
.206
..177
137
37 1
.120
.103
.321
.090
.083
.287
.072
.0-59
.262
.060
850
213
.461
.183
.141
376
.123
.106
326
.093
.085
.292
.074
.071
.266
.062
875
.219
.468
.188
145
.382
.127
.109
.095
.088
.296
.076
.073
.270
.064
900
225
474
193
.150
-388
.113
^336
.098
.090
. 3 00
.079
075
.274
065
925
.231
.481
.198
154
393
.134
.116
340
.101
093
34
.O8l
.077
277
.067
950
.238
-.487
.204
.158
.398
.138
.119
344
.104] .095
308
08 3
.079
.281
.069
975
.244
492
.209
.162
43
.141
122 .349
.106
.098
312
08 5
.081
.285
.071
1000
.250
.500
.214
.166
.408
MS
125 .353
.109
.100
.316
.087
.083
.289
073
1025
.256
.506
.220
.170
413
.149
.128
357
.112
.103
.320
.08 9
085
.292
1050
263
.512
.225
175
.418
153
1 3 I
.114
.105
324
.092
.088
295
.076
I0 75
.269
.518
231
.179
423
.156
.134
'365
.117
.108
328
.094
.090
.299
.078
IIOO
.275
524
236
.183
.428
.160
.138 .370
.120
.110
332
.096
.092
303
.oSo
1125
.281
53
.241
.187
433
.163
J 4i -375
.123
"3
335
.098 '.094
.306
.082
1150
.288
.536
.246
.191
438
.167
144 -379
.i26| .115
339
.100
.096
.310
.084
"75
.294
542
251
195
443
.170
147 -384
.128: .Il8
343
.103
.098
3'3
085
1 200
.300
.548
256
.200
447
1 75
.150 .388
I3t
.120
346
.105
.IOO
316
.087
i 5 8
SLIDE-VALVES.
TABLE V.
AREAS OF CIRCLES.
Advancing by Eighths.
Diam.
Area.
Diam.
Area.
Diam.
Area.
Diam.
Area.
1/64
.00019
3
7.0686
u
51.849
207-39
1/32
.00077
1/16
7-3662
8
53-456
2
210.60
3/64
00173
X
7.6699
%
55-088
x^
213.82
1/16
.00307
3/i6
7.9798
%
56.745
%
217.08
.00690
u
8.2958
%
58.426
94
220.35
X
.01227
5/i6
8.6179
%
60.132
7%
223.65
5/32
.01917
%
8.9462
7%
61.862
17
226.98
.02761
7/16
9 . 2806
9
63.617
l
230-33
7/32
03758
X<
9.6211
L
65-397
X4
233-7 1
M
.04909
9/16
9.9678
8
67.201
%
237.10
9/32
.06213
%
10.321
?8
69.029
X^
240.53
.07670
11/16
10.680
X
70.882
%
243.98
11/32
.09281
94
11.045
%
72.760
94
247-45
%
.11045
13/16
11.416
94
74-662
%
250.95
13/32
.12962
%
"793
%
76.589
18
7/16
15033
15/16
12.177
10
78.540
258.02
15/32
17257
4
12.566
X
80.516
i^
261.59
\/-
19635
1/16
12.962
X^
82.516
%
265.18
17/32
.22166
X
13-364
%
84.541
x^
268.80
9/16
.24850
3/16
I3-772
1/2
86.590
%
272.45
.27688
'H
14.186
%
88.664
94
276. 12
%
.30680
5/16
14.607
%
90.763
%
279.81
21/32
.33824
15-033
/&
92.886
19
283.53
11/16
.37122
7/16
15-466
11
95-033
287.27
23/32
40574
/^
15.904
X
97.205
/4
291.04
94
.44179
9/1 6
'6.349
/4
99-402
%
294.83
25/32
47937
%
16.800
n
101 .62
X<j
298.65
13/16
.51849
11/16
17.257
n
103.87
%
302.49
27/32
559'4
M
17.728
rm
106.14
94
%
.60132
13/16
18.190
%:
108.43
?2
310.24
29/32
.64504
%
18.665
%
110.75
20
T
15/16
.69029
15/16
19.147
12
113. 10
x^
318.10
3V3 2
73708
5
19-635
x^
"5-47
X4
322.06
1
7854
1/16
20. 129
x^
117.86
%
326.05
1/16
.8866
y
20.629
%
120.28
u
330.06
X^
9940
3A6
2i.i35
x^
122.72 .
5^
334-io
3/i6
.1075
N
21.648
%
125.19
94
338.i6
Y \6
.2272
3530
5/i6
22.166
22.691
?8
127.68
130.19
21
346.36
%
.4849
7/16
23.221
13
x^
350.50
7/16
.6230
n
23-758
x^
135.30
H
354-66
Ni
.7671
9/16
24.301
X4
137-89
%
358.84
9/16
9175
%
24.850
Z
140-50
LA
363-05
%
739
11/16
25.406
x^
I 43- I 4
%
367 28
11/16
2365
N
25.967
%
94
371-54
94
453
13/16
26.535
%
148.49
%
375 83
13/16
%
.5802
.7612
'SA6
27.109
27.688
14
151.20
153-94
22
380.13
384.46
15/16
.9483
6
28.274
x^
156.70
H
388.82
2
1/16
3.1416
H
29-465
30 . 680
S
159.48
162.30
H
393 20
397-6i
X
3*5466
%
31 -919
x^
165.13
%
402.04
3/'6
/^
33-183
%
167.99
94
406 49
X^
3 .9761
!h6
34-472
94
170.87
/&
410.97
5/16
4.2000
%
35.785
%
I 73>7 8
23
415.48
%
4-4301
%
37.122
15
176.71
x^
420.00
7/16
4.6664
7
38.485
x^
179.67
X4
424-56
L2
4.9087
/4
39-87I
x|
182.65
B
429-13
31
5- I 572
%
41.282
42.718
s
185.66
188.69
H
433-74
438.36
5-6727
^
44.179
%
I9L75
94
443-or
94
5-9396
%
45.66 4
94
194.83
%
447.69
13/16
6.2126
4
47- I 73
%
^97 '93
24
452.39
7/3
6.4918
fe2
48.707
16
201.06
IX
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6-777 1 ,
8 '
50.265
^
204.22
.. ..J4
461.86
SHAFT-GOVERNORSANAL YSlS.
159
AREAS OF CIRCLES.
Diam.
Area.
Diam.
Area.
Diam.
Area.
Diam.
Area.
%
466.64
33
855-3
%
1360.8
H
1983.2
%
471.44
j
861.79
%
1369.0
%
%
476.26
/4
868.31
%
1377-2
t<
2003.0
4
481.11
%
874-85
42
1385-4
%
2012.9
H
485.98
12
881.41
/4
1393-7
%
2022.8
25
490.87
%
888.00
/4
1402 .0
%
2032.8
*A
495-79
%
894.62
%
1410.3
51
2042.8
%
500.74
%
901.26
Jl8
1418.6
i^
2052.8
%
505 -7 1
34
907.92
%
1427.0
/4
2062 . 9
m
510.71
i^j
914.61
%
M35 4
78
2073.0
%
5 I 5-7 2
J4
921.32
%
M43-8
1^
2083.1
%
520.77
%
928.06
43
1452.2
98
2093.2
%
525-84
L<J
934.82
ly
1460.7
%
2103.3
26
330-93
%
941.61
}4
1469 i
%
/^J
536-05
%
948.42
%
1477.6
52
2123.7
/4
54 I - 1 9
%
955-25
12
1486.2
2133-9
%
546.35
35
962 . i i
%
M94-7
/4
2144.2
x^
551-55
1 s
969.00
%
%
2154-5
%
556.76
^4
975-91
%
1511.9
H
2164.8
%
562.00
%
982.84
44
1520.5
Kg
2175.1
%
567.27
C2
989.80
^&
1529.2
94
2185.4
27
572.56
%
996-78
/4
1537-9
%
2195-8
^
577.87
%
1003.8
%
1546.6
53
2206 . 2
J4
583-21
%
1010.8
i .,
1555.3
22IO.6
%
588.57
36
1017.9
%
1564-0
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2227.0
zl3
593 . 96
i^
1025.0
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1572.8
&
2237.5
%
599-37
M
1032. i
%
1581.6
IX
2248.0
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604.81
fjj
1039.2
45
1590.4
%
2258.5
/^
610.27
/^s
1046.3
^x
J 599-3
%
2269.1
28
6i5-75
%
JOSS-S
14
1608.2
%
2279.6
^
621.26
^4
1060.7
N
1617.0
54
2290.2
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626.80
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1068. o
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1626.0
2300.8
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632.36
37
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1634.9
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1082.5
94
1643.9
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2322.1
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643-55
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1089.8
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1652.9
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2332.8
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649-18
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1097.1
46
1661.9
2
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654.84
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1670.9
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2354.3
29
660.52
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mi .8
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1680.0
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23^5.0
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666.23
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1119.2
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1689.1
55
2375-8
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671. g6
%
1126.7
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1698.2
2386.6
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677.71
38
1134.1
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1707.4
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2397-5
lit
683.49
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1141.6
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1716.5
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2408.3
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689.30
695-13
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1149.1
1156.6
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1725-7
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2419.2
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700.98
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1164.2
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1744.2
34
2441.1
30
706.86
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1171.7
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J 753-5
%
2452.0
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712.76
718.69
8
"79-3
1186.9
ft
1762.7
1772.1
56
2463.0
2474.0
94
724-64
39
1194.6
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1781.4
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2485.0
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730.62
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1202.3
4
1790.8
%
2496.1
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736.62
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121O.O
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1800.1
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2507.2
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742.64
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1217.7
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1809.6
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748.69
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1225.4
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31
754-77
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1233.2
14
1828 5
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2540.6
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760.87
^4
I24I.O
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1837.9
57
255L8
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766.99
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1248.8
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1847-5
2563.0
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773-M
40
1256.6
%
1857.0
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2574-2
J^
779 3i
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1264.5
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1866.5
1
2585.4
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785-51
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1272.4
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1876.1
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2596.7
$4
791.73
12
1280.3
49.
1885.7
%
2608 . o
32 ^
804.25
H
1288.2
1296.2
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1895.4
1905.0
%
2619.4
2630.7
J^J
810.54
^4
1304.2
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1914.7
58
2642.1
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816.86
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1924.4
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g
823.21
829.58
41
1320.3
1328.3
1
1934.2
1943-9
N
2664.9
2676.4
%
835-97
^4
1336.4
%
1953-7
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2687.8 .
%
842.39
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1344.5
50
^
2699.3
*
848.83
JJ
1352.7
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1973-3
N
2710.9
i6o
SLIDE-VALVES,
AREAS OF CIRCLES.
Diam.
Area.
Diam.
Area.
Diam.
Area.
Diam.
Area.
y
2722.4
34
3578.5
H
4556.4
M
5641.2
59
2734-0
%
359 x -7
34
4565-4
%
5657-8
2745-6
%
3605.0
%
4581-3
85
5674-5
n
2757.2
2768.8
68
3618.3
363 1 '7
H
4596.3
4611.4
^
5691-2
5707-9
34
2780.5
3645-0
M
4626.4
%
5724-7
%
2792.2
34
3658.4
%
4641.5
34
574 T -5
H
2803.9
%
3671.8
77
4656.6
%
5758.3
%
2815.7
34
3685-3
34
4671.8
M
5775-1
60
2827.4
%
3698.7
34
4686.9
/&
579 I -9
2839.2
%
3712.2
%
4702.1
86
5808.8
34
2851.0
%
3725-7
34
47 T 7-3
34
5825.7
a2
2862.9
69
3739-3
%
4732-5
34
5842.6
iz
2874.8
34
3752-8
M
4747-8
%
5859-6
%
2886.6
H
3766.4
%
4763 - 1
34
5876.5
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2898.6
%
3780.0
78
4778.4
%
5893.5
T^j
2910.5
34
3793-7
34
4793-7
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5910.6
61
2922.5
%
34
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%
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3821.0
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4824.4
87
5944-7
34
2946-5
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3334.7
34
4839-8
34
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2
2958-5
70
3848-5
%
4855-2
34
5978.9
34
2970.6
34
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4870.7
%
5996.0
%
2982.7
H
3876.0
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4886.2
34
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%
29 H- 8
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3889.8
79
4901-7
%
6030.4
%
3006.9
34
3903.6
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6047.6
62
3019-1
%
39J7.5
34
4932-7
%
6064.9
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4
3931-4
%
4948.3
88
6082.1
34
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3945-3
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3055.7
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4979-5
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3080.3
34
3987.1
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%
4001.1
80
5026.5
%
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3104. 9 f
34
4015 2
34
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%
6186.2
63
3117.2*
%
4029.2
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5058.0
%
6203.7
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3129.6
4
4043.3
%
5073-8
89
6221.1
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3142.0
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4057.4
34
5089.6
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3I54-5
72
407L5
%
5105.4
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6256. i
34
3166.9
34
4085.7
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5121.2
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6273.7
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3 I 79-4
34
4099.8
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4114.0
81
S^.o
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6308 . 8
%
3204-4
34
4128.2
34
5168.9
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6326.4
64
3217.0
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4142.5
34
5184.9
%
6344.1
3229.6
%
4156.8
Q
5200.8
90
6361.7
34
3242-2
%
4171.1
34
5216.8
34
6379-4
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73
4185.4
%
5232.8
34
6397 - 1
34
3267-5
34
4I99.7
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5248.9
%
6414.9
%
3280.1
3^
4214.1
%
5264.9
34
6432.6
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3292.8
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4228.5
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5281 .0
%
6450.4
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3305.6
34
4242.9
34
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$4
6468.2
65
3318.3
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34
53 J 3-3
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6486.0-
AX
4271.8
3z
91
6503.9
34
3343-9
%
4286.3
34
5345-6
34
6521.8
a2
3356-7
74
4300.8
%
53 6t.8
34
6539-7
H
3369.6
34
43 I 5-4
4
5378.1
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6557-6
M
3382-4
H
4329-9
%
5394-3
34
6575-5
a
3395-3
%
4344-5
83
5410.6
%
6593-5
%
3408.2
34
4359- 2
34
5426.9
%
6611.5
66
3421.2
3434-2
3447 - 2
s
4373-8
4388.5
4403.1
1
5443-3
5459-6
5476.0
92 ^
6629 . 6
6647.6
6665.7
%
3460.2
75
4417.9
%
5492.4
34
6683.8
34
3473 2
34
4432-6
%
5508.8
%
6701 .9
M
3486.3
34
4447-4
/&
5525-3
V^
6720.1
94
3499-4
n
4462.2
84
5541-8
%
6738.2
%
3512.5
34
4477-0
34
5558.3
8
6756.4
67
3525-7
%
4491.8
M
5574-8
?
6774-7
l^C
3538.8
4
4506.7
Z
559^4
93
6792.9
\A
3552.0
7Z
i^
5607.9
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6811.2
%
3565-2
76
4536.5
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5624.5
34
6829.5
SHA FT- G O VERNORSA NA L YSIS.
161
AREAS OF CIRCLES.
Diatn.
Area.
94
95
68 47 -8
6866.1
6884.5
6902.9
6921.3
6939.8
6958 . 2
6976.7
6995-3
7013.8
7032.4
7051.0
7069.6
7088.2
Diam.
96
Area.
7106.9
7125.6
7M4-3
7163.0
7181.8
7200.6
7219.4
7238.2
7257 - 1
7276.0
7313.8
7332 8
7351-8
Diam.
98
Area.
737 -8
7389.8
7408.9
7428.0
7447 - 1
7466.2
7504-5
7523-7
7543-
7562.2
758i.5
7600.8
7620.1
Diam.
99
100
Area.
7639.5
7658.9
7678.3
7697.7
7717.1
7736.6
7756.1
7775-6
7795-2
7814.8
7854.0
INDEX.
Admission 32
Advance, Angle of . 12
Advancing Eccentric, Effect of 41
Allen Valve 82
Angle of Advance 85.
Inside Lap 83
Laying out False Seat ; 88
Maximum Port-opening 85
Outside Lap 85
Port Area 84
Port Width.. 84
Valve-travel 85
Analysis of Shaft-governors 124
Angle of Advance 13, 14
Determination of 56, 58, 59
Effect of Changing 41
A ngle between Crank and Eccentric 13
Angle of Follow 14
Angular Advance. See Angle of Advance.
Angularity of Connecting-rod 19
Eccentric-rod 22
Balanced Valves 130;
Barr Diagram 62
Bridge 12
Width of 5I
Buckeye Governor b 123
Center, Dead 6
Placing Engine on 100
Centrifugal Force 126
Changing Angular Advance, Effect of 41
Dimensions of Valve, Effect of 41
Inside Lap, Effect of 42
163
164 INDEX.
PAGE
Changing Outside Lap, Effect of 41
Valve-travel, Effect of 41
Compression 8, 34
and Exhaust, Equalization of 67
Connecting-rod 17
Angularity of 19
Infinite 6
Construction of Eccentric 1 8
Crank.. 17
Crank Angle 10
Crank and Connecting-rod 17
Crank-end 1 1
Crank-pin 6, 17
Crank and Piston Positions, Relative 23
Crank-shaft 6, 17
Crank, Throw of 8
Cross-head 17
Cushioning 8
Cut-off 8, 32
and Lead, Equalization of 65
Dash-pot 144
Dead-center 6
Placing Engine on 100
Decreasing Speed of Engine. , 152
Variation 152
Deprez Method for Relative Crank and Piston Positions 23
Design of an Allen Valve. .... 82
a D-valve 68
a Double-ported Valve 90
a Trick-valve 82
General Problems of 54
Diagram, Barr 130
Valve 27
Diameter of Exhaust-pipes 48
Steam-pipes > 48
Dimensions of Exhaust-port 48
Steam-port 47
Valve, Effect of Changing 41
Displacement of Valve 10, 14, 27
Drag of Valve. 138
D-valve Design 68
Angular Advance 69
Bridge 74
Exhaust-port.. 74
INDEX. 165
PAGE
D-valve Design, Inside Lap 72
Laying out Valve 76
Maximum Port-opening 68
Outside Lap 69
Port Area 68
Port Width . . , 68
Valve-travel 74
Double-ported Valve Design 9 o
Bridge 95
Exhaust-port 95
Laying out the Valve 97
Maximum Port- opening ; 94
Port Area . . . 94
Port Width 94
Eccentric, Construction of 18
Following Crank 14
Leading Crank 14
Eccentric-pin 7
Eccentric-rod 7
Angularity of 22
Length of 80
Eccentric, Shifting 113, us
Swinging 113,114
Throw of 8
Eccentricity 8
Determination of 58, 60
Effect of Changing Angle of Advance 41
Dimensions of Valve 41
Inside Lap 42
Outside Lap 41
Valve-travel 41
Effect of Inside Lap 12
Outside Lap 12
Equal Cut-off, Valve-setting for . . 104
Lead, Valve-setting for 102
Equalizing Cut-off and Lead 65
Exhaust and Compression 67
Exhaust-lap 12
Exhaust-port 3
Dimensions of 48
Maximum Opening 49
Exhaust-pipes, Diameter of 48
Expansion of Steam, Economy of 8
1 66 INDEX.
PAGE
Governors, Shaft 108
Analysis 124
Buckeye 123
Diagram 130
Isochronous 133
Stable... 131
Straight-Line 122
Unstable 134
Weight, Inertia of 144
Westinghouse 122
Guides 17
Head-end 1 1
Increasing Speed of Engine 152
Inertia of Governor-weight 144
Valve 143
Infinite Connecting-rod 6
Initial Tension 148
Inside Lap 1 1
with Infinite Rod 1 1
Effect of Changing 42
Isochronous Governor. 133, 152
Lap, Determination of 54> 5$> 59
Effect of 12
Inside n
Outside 9
Lead 4
Lead-angle 40
Lead and Cut-off, Equalization of 65
Determination of. 54> 5$
Length of Eccentric-rod 80
Port 47
Valve-chest 7&
Valve-stem 79
Maximum Port-opening 49
for Exhaust 49
Outside Lap. 9
Effect of 12
with Infinite Rod 10
Pipes, Diameter of 48
INDEX. 167
PAGE.
Piston and Crank Positions 23
Piston-rod 3
Piston-speed 68, 84, 93
Piston-valve 139
Pitman ''" . . . 17
Ports 2
Area of . . . . . 44, 68
Length of 47. 48
Width 47,68
Port-opening 27
Exhaust 33
Maximum 49
Problems of Design 54
Reducing Speed of Engine 152
Variation 152
Release 34
Reversing Engine 14, 108
Rockers 7
and Bell Cranks 62
Running " Over " 15
"Under" 16
Setting Valves 98
with Chest-cover on 106
for Equal Cut-off 104
Leads 102
Shaft-governors. See Governors.
Shifting Eccentrics 113, 118
Slide-valve, Design of a D 68
No Laps 2
Slotted Cross-head , 6
Speed, Decreasing 152
Increasing 152
Variation, Decreasing 152
Spring Extension 125
Strength 1 24
Stable Governor 131
Steam-chest 2, 78
Steam-pipes ; 48
Steam-ports 2
Dimensions 47
Maximum Opening 49
Steam, Velocity of 44
Straight- Line Governor 122
1 68 INDEX.
t
PAGE
Stuffing-box 3
Swinging Eccentrics 113, 114
Trick- valve. See Allen Valve.
Throw of Crank and Eccentric 8
Unstable Governor 134
Valve, Allen. See Allen Valve.
Balanced 139
D-. See D-valve.
Double-ported. See Double-ported Valve.
Trick-. See Allen Valve.
Valve-chest, Length of 78
Valve-circle 27
'Valve Design, General Problems 54
Examples 68, 82, 90
Valve-diagrams 27
Valve-displacement 10, 14, 27
Valve, Drag of 138
Inertia of 143
Piston- 139
and Piston, Relation between 5
Valve-rod 3
Valve-seat 3
Valve-setting 99
with Chest-cover on 106
for Equal Cut-offs. . ^ 104
Leads 102
Valve-stem, Length of 79
Valve-travel , 8
Determination of 56, 59
Effect of Changing 41
Variation of Speed, Reducing 152
Velocity of Steam. 44
Westinghouse Governor 51
Wrist-pin 17
Zeuner Diagram 27
I
'NIVERSITY OF CALIFORNIA LIBRARY
BERKELEY
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
iration of loan period.
DEC 20 1916
17 1917 O)
31 191 A V
001271988
50?n-7,'16
YC !2896
U.C. BERKELEY LIBRARIES
UNIVERSITY OF CALIFORNIA LIBRARY