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 
 
 4S7-" 
 
 15/16 
 
 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 
 
 /4 
 
 2227.0 
 
 zl3 
 
 593 . 96 
 
 i^ 
 
 1025.0 
 
 3^ 
 
 1572.8 
 
 & 
 
 2237.5 
 
 % 
 
 599-37 
 
 M 
 
 1032. i 
 
 % 
 
 1581.6 
 
 IX 
 
 2248.0 
 
 M 
 
 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 
 
 /4 
 
 626.80 
 
 ?e 
 
 1068. o 
 
 JX 
 
 1626.0 
 
 
 2300.8 
 
 % 
 
 632.36 
 
 37 
 
 1075.2 
 
 % 
 
 1634.9 
 
 H 
 
 
 /^ 
 
 637-94 
 
 /^ 
 
 1082.5 
 
 94 
 
 1643.9 
 
 ' % 
 
 2322.1 
 
 % 
 
 643-55 
 
 n 
 
 1089.8 
 
 % 
 
 1652.9 
 
 i^ 
 
 2332.8 
 
 M 
 
 649-18 
 
 % 
 
 1097.1 
 
 46 
 
 1661.9 
 
 2 
 
 2343-5 
 
 % 
 
 654.84 
 
 Mi 
 
 1104.5 
 
 /^ 
 
 1670.9 
 
 ^4 
 
 2354.3 
 
 29 
 
 660.52 
 
 % 
 
 mi .8 
 
 /4 
 
 1680.0 
 
 % 
 
 23^5.0 
 
 ^ 
 
 666.23 
 
 % 
 
 1119.2 
 
 sx 
 
 1689.1 
 
 55 
 
 2375-8 
 
 ^4 
 
 671. g6 
 
 % 
 
 1126.7 
 
 L^ 
 
 1698.2 
 
 
 2386.6 
 
 9i 
 
 677.71 
 
 38 
 
 1134.1 
 
 % 
 
 1707.4 
 
 H 
 
 2397-5 
 
 lit 
 
 683.49 
 
 ^ 
 
 1141.6 
 
 % 
 
 1716.5 
 
 ^ 
 
 2408.3 
 
 It 
 
 689.30 
 695-13 
 
 % 
 
 1149.1 
 1156.6 
 
 47 ^ 
 
 1725-7 
 1734-9 
 
 K 
 
 2419.2 
 2430.1 
 
 n. % 
 
 700.98 
 
 i^j 
 
 1164.2 
 
 /^ 
 
 1744.2 
 
 34 
 
 2441.1 
 
 30 
 
 706.86 
 
 % 
 
 1171.7 
 
 /4 
 
 J 753-5 
 
 % 
 
 2452.0 
 
 M 
 
 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 
 
 5,g 
 
 1781.4 
 
 /4 
 
 2485.0 
 
 /-ia 
 
 730.62 
 
 /^ 
 
 1202.3 
 
 4 
 
 1790.8 
 
 % 
 
 2496.1 
 
 /^& 
 
 736.62 
 
 i^ 
 
 121O.O 
 
 TA 
 
 1800.1 
 
 iz 
 
 2507.2 
 
 % 
 
 742.64 
 
 ?1 
 
 1217.7 
 
 48 j| 
 
 1809.6 
 
 ^| 
 
 2518.3 
 
 n* / " 
 
 748.69 
 
 U 
 
 1225.4 
 
 
 1819.0 
 
 3 
 
 2529.4 
 
 31 
 
 754-77 
 
 % 
 
 1233.2 
 
 14 
 
 1828 5 
 
 % 
 
 2540.6 
 
 /^ 
 
 760.87 
 
 ^4 
 
 I24I.O 
 
 % 
 
 1837.9 
 
 57 
 
 255L8 
 
 /4 
 
 766.99 
 
 ?i 
 
 1248.8 
 
 t2 
 
 1847-5 
 
 
 2563.0 
 
 % 
 
 773-M 
 
 40 
 
 1256.6 
 
 % 
 
 1857.0 
 
 J4 
 
 2574-2 
 
 J^ 
 
 779 3i 
 
 J4 
 
 1264.5 
 
 M 
 
 1866.5 
 
 1 
 
 2585.4 
 
 /o 
 
 785-51 
 
 /4 
 
 1272.4 
 
 ^ 
 
 1876.1 
 
 IX 
 
 2596.7 
 
 $4 
 
 791.73 
 
 12 
 
 1280.3 
 
 49. 
 
 1885.7 
 
 % 
 
 2608 . o 
 
 32 ^ 
 
 804.25 
 
 H 
 
 1288.2 
 1296.2 
 
 H 
 
 1895.4 
 1905.0 
 
 % 
 
 2619.4 
 2630.7 
 
 J^J 
 
 810.54 
 
 ^4 
 
 1304.2 
 
 N 
 
 1914.7 
 
 58 
 
 2642.1 
 
 ^4 
 
 816.86 
 
 7^6 
 
 I3I2.2 
 
 vz 
 
 1924.4 
 
 1^ 
 
 26" -> r 
 
 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 
 
 l^j 
 
 2687.8 . 
 
 % 
 
 842.39 
 
 9i 
 
 1344.5 
 
 50 
 
 
 ^ 
 
 2699.3 
 
 * 
 
 848.83 
 
 JJ 
 
 1352.7 
 
 J5 
 
 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 
 
 M 
 
 2898.6 
 
 % 
 
 3780.0 
 
 78 
 
 4778.4 
 
 % 
 
 5893.5 
 
 T^j 
 
 2910.5 
 
 34 
 
 3793-7 
 
 34 
 
 4793-7 
 
 3^i 
 
 5910.6 
 
 61 
 
 2922.5 
 
 % 
 
 
 34 
 
 4809.0 
 
 % 
 
 5927.6 
 
 
 2934-5 
 
 M 
 
 3821.0 
 
 N 
 
 4824.4 
 
 87 
 
 5944-7 
 
 34 
 
 2946-5 
 
 7^3 
 
 3334.7 
 
 34 
 
 4839-8 
 
 34 
 
 5961.8 
 
 2 
 
 2958-5 
 
 70 
 
 3848-5 
 
 % 
 
 4855-2 
 
 34 
 
 5978.9 
 
 34 
 
 2970.6 
 
 34 
 
 3862.2 
 
 ^4 
 
 4870.7 
 
 % 
 
 5996.0 
 
 % 
 
 2982.7 
 
 H 
 
 3876.0 
 
 /s 
 
 4886.2 
 
 34 
 
 6013.2 
 
 % 
 
 29 H- 8 
 
 ^1 
 
 3889.8 
 
 79 
 
 4901-7 
 
 % 
 
 6030.4 
 
 % 
 
 3006.9 
 
 34 
 
 3903.6 
 
 /4 
 
 4917.2 
 
 % 
 
 6047.6 
 
 62 
 
 3019-1 
 
 % 
 
 39J7.5 
 
 34 
 
 4932-7 
 
 % 
 
 6064.9 
 
 i ^ 
 
 3Q3 1 -3 
 
 4 
 
 3931-4 
 
 % 
 
 4948.3 
 
 88 
 
 6082.1 
 
 34 
 
 3043.5 
 
 /-i 
 
 3945-3 
 
 34 
 
 4963-9 
 
 34 
 
 6099.4 
 
 2 
 
 3055.7 
 
 71 
 
 3959-2 
 
 % 
 
 4979-5 
 
 34 
 
 6116.7 
 
 i< 
 
 3068.0 
 
 34 
 
 3973 - 1 
 
 M 
 
 4995-2 
 
 % 
 
 6134.1 
 
 ^8 
 
 3080.3 
 
 34 
 
 3987.1 
 
 ^8 
 
 5010.9 
 
 34 
 
 6i=;i .4 
 
 M 
 
 3092 6 
 
 % 
 
 4001.1 
 
 80 
 
 5026.5 
 
 % 
 
 6168. 8 
 
 vb 
 
 3104. 9 f 
 
 34 
 
 4015 2 
 
 34 
 
 5042-3 
 
 % 
 
 6186.2 
 
 63 
 
 3117.2* 
 
 % 
 
 4029.2 
 
 H 
 
 5058.0 
 
 % 
 
 6203.7 
 
 i^ 
 
 3129.6 
 
 4 
 
 4043.3 
 
 % 
 
 5073-8 
 
 89 
 
 6221.1 
 
 H 
 
 3142.0 
 
 /i 
 
 4057.4 
 
 34 
 
 5089.6 
 
 34 
 
 6238.6 
 
 8 
 
 3I54-5 
 
 72 
 
 407L5 
 
 % 
 
 5105.4 
 
 3^ 
 
 6256. i 
 
 34 
 
 3166.9 
 
 34 
 
 4085.7 
 
 M 
 
 5121.2 
 
 % 
 
 6273.7 
 
 ^6 
 
 3 I 79-4 
 
 34 
 
 4099.8 
 
 ?8 
 
 5I37- 1 
 
 34 
 
 6291.2 
 
 3X 
 
 
 9& 
 
 4114.0 
 
 81 
 
 S^.o 
 
 Z 
 
 6308 . 8 
 
 % 
 
 3204-4 
 
 34 
 
 4128.2 
 
 34 
 
 5168.9 
 
 M 
 
 6326.4 
 
 64 
 
 3217.0 
 
 fa 
 
 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 
 
 a2 
 
 
 73 
 
 4185.4 
 
 % 
 
 5232.8 
 
 34 
 
 6397 - 1 
 
 34 
 
 3267-5 
 
 34 
 
 4I99.7 
 
 M 
 
 5248.9 
 
 % 
 
 6414.9 
 
 % 
 
 3280.1 
 
 3^ 
 
 4214.1 
 
 % 
 
 5264.9 
 
 34 
 
 6432.6 
 
 % 
 
 3292.8 
 
 % 
 
 4228.5 
 
 82 
 
 5281 .0 
 
 % 
 
 6450.4 
 
 7^2 
 
 3305.6 
 
 34 
 
 4242.9 
 
 34 
 
 5297.1 
 
 $4 
 
 6468.2 
 
 65 
 
 3318.3 
 
 M 
 
 
 34 
 
 53 J 3-3 
 
 % 
 
 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 
 
 ?i 
 
 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 
 
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 3552.0 
 
 7Z 
 
 
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 5607.9 
 
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 6811.2 
 
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 3565-2 
 
 76 
 
 4536.5 
 
 " 
 
 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