Valve Gears and Indicators 
 
 A Manual of 
 
 PRACTICAL INSTRUCTION IN VALVE-SETTING, USE OF INDICATORS, AND 
 OTHER DETAILS OF STEAM ENGINE OPERATION ES- 
 SENTIAL TO EFFICIENCY AND ECONOMY 
 
 PART I VALVE GEARS 
 
 By WALTER S. LELAND, S.B. 
 
 Assistant Professor of Naval Architecture. Massachusetts 
 Institute of Technology 
 
 PART II STEAM ENGINE INDICATORS 
 
 By CARL S. Dow, S.B. 
 
 American Society of Mechanical Engineers. With B. F. Sturtevant Company, 
 Boston, Mass, 
 
 ILLUSTRATED 
 
 CHICAGO 
 
 AMERICAN SCHOOL OF CORRESPONDENCE 
 11 1908 
 
Wittlttl. 
 
 COPYRIGHT 1907 BY 
 AMERICAN SCHOOI, OF CORRESPONDENCE 
 
 Entered at Stationers' Hall, London 
 All Rights Reserved 
 
Foreword 
 
 recent years, such marvelous advances have been 
 made in the engineering and scientific fields, and 
 so rapid has been the evolution of mechanical and 
 constructive processes and. Irii&thodsj* that a distinct 
 need has been created for a series of practical 
 working guides, of convenient size and low cost, embodying the 
 accumulated results of experience . and the most approved modern 
 practice along a great variety of lines. To fill this acknowledged 
 need, is the special purpose of the series of handbooks to which 
 this volume belongs. 
 
 d, I 11 the preparation of this series, it has been the aim of the pub- 
 lishers to lay special stress on the practical side of each subject, 
 as distinguished from mere theoretical or academic discussion. 
 Each volume is written by a well-known expert of acknowledged 
 authority in his special line, and is based on a most careful study 
 of practical needs and up-to-date methods as developed under the 
 conditions of actual practice in the field, the shop, the mill, the 
 power house, the drafting room, the engine room, etc. 
 
 C. These volumes are especially adapted for purposes of self- 
 instruction and home study. The utmost care has been used to 
 bring the treatment of each * subject within the range of the com- 
 
 179744 
 
mori understanding, so that the work will appeal not only to the 
 technically trained expert, but also to the beginner and the self- 
 taught practical man who wishes to keep abreast of modern 
 progress. The language is simple and clear; heavy technical terms 
 and the formulae of the higher mathematics have been avoided, 
 yet without sacrificing any of -the requirements of practical 
 instruction; the arrangement of matter is such as to carry the 
 reader along by easy steps to complete mastery of each subject; 
 frequent examples for practice are given, to enable the reader to 
 test his knowledge and make it a permanent possession; and the 
 illustrations are selected with the greatest care to supplement and 
 make clear the references in the text. 
 
 C. The method adopted in the preparation of these volumes is that 
 which the American School of Correspondence has developed and 
 employed so successfully for many years. It is not an experiment, 
 but has stood the severest of all tests- that of practical use which 
 has demonstrated it to be the best method yet devised for the 
 education of the busy working man. 
 
 C. For purposes of ready reference and timely information when 
 needed, it is believed that this series of handbooks will be found to 
 meet every requirement. 
 
Table of Contents 
 
 PART I -VALVE GEARS 
 SLIDE VALVES '. ... .. '- . ..' . Page, 3 
 
 Plain Slide or D Valve Eccentric Inside and Outside Lap Angular 
 Advance Admission Cut-Off Release Compression Lead Angle 
 of Lead Inequality of Steam Distribution Displacement of Valve 
 and Piston Compensation of Cut-Off Rocker. 
 
 VALVE-DESIGNING AND VALVE-SETTING .. . ....*. . Page 16 
 
 Valve Diagrams (Zeuner's) Area of Steam Pipe Width of Steam 
 Port Width of Exhaust Port Width of Bridge Point of Cut-Off 
 Lead Putting Engine on Center Setting Valve with Equal Lead 
 Setting Valve for Equal Cut-Off Modifications of the Slide Valve 
 (Piston Valve, Double-Ported Valve, Trick Valve) Balanced Valves 
 Link Motion (Stephenson, Gooch). 
 
 RADIAL, DOUBLE, AND DROP CUT-OFF VALVE GEARS . . . . Page 49 
 
 Hackworth Gear Marshall Gear Joy Gear Walschaert Gear Ad- 
 justable Eccentrics Meyer Double Valve Reynolds-Corliss Drop Cut- 
 Off Gear Safety Cams Brown Releasing Gear Greene Gear Sulzer 
 Gear Corliss Valve Setting. 
 
 PART II-STEAM ENGINE INDICATORS 
 TYPES OF INDICATORS -.;;;". . . . ; . . . . Page 3 
 
 Definition of Terms and Explanation of Principles Watt's Diagram 
 of Work Thompson Indicator Crosby Indicator Tabor Indicator 
 Reducing Motion Pantograph Brumbo Pulley. 
 
 USE OF INDICATOR DIAGRAMS . .... . - -V . . . . Page 22 
 
 Determination of Indicated Horse-Power Mechanical Efficiency 
 Steam Distribution Mean Effective Pressure Piston Speed Table of 
 Engine Constants Brake Horse-Power Prony Brake Rope Brake 
 Finding Area of Cards Planimeter Thermal Efficiency Theoretical 
 Indicator Diagram Atmospheric Line Admission Line Steam Line 
 Point of Cut-Off Expansion Curve Point of Release Exhaust Line 
 Back-Pressure Line Point of Exhaust Closure Compression Curve 
 Zero Line Clearance Line Drawing the Theoretical Card Cards for 
 Compound Engines Combined Diagrams Horse-Power of Compound 
 Engines Defects Revealed by Card (Events Too Early or Too Late, 
 Unequal Work at Cylinder Ends, etc.) Steam Consumption. 
 
 INDEX . ,., . , -. "*->' : - Page 59 
 
VALVE GEARS 
 
 Steam enters the cylinder of the engine through ports which 
 must, in some manner, be opened and closed alternately, in order 
 to admit and exhaust the steam at the proper time. To accom- 
 plish this purpose a valve is moved back and forth across the port 
 openings. A complete understanding of the valve and valve gear 
 is essential to the engineer as well as to the designer, for even 
 though a valve be properly designed, its economy may be seri- 
 ously impaired by improper setting. The design and adjust- 
 ment of these valves plays a very important part in the efficient 
 action of the steam engine. 
 
 o 
 
 The term " valve gear" includes the valve or valves that 
 admit steam to and exhaust it from the cylinder of the engine, 
 together with the mechanism from which the valves derive motion. 
 
 o 
 
 There may be a single valve to regulate admission and exhaust, or 
 there may be a double set of valves; one -set to admit the steam at 
 each end and another to release it. The valve may have a plain 
 reciprocating motion, moved 
 by a rod, or it may be opened 
 by some device that lets go at 
 the proper time, allowing the 
 valve to drop shut under the 
 influence of counter weights, 
 springs or vacuum dashpots. 
 To the first class belong the 
 plain slide valve and its modi- 
 fication of piston valve, gridiron valve, etc.; to the second belong 
 such valves as the Corliss, Brown, and others. 
 
 The simplest type of valve is the plain slide or D valve as 
 shown in Fig. 1. 
 
 In this figure Y is the valve, R the valve rod, K the exhaust 
 cavity, P and P* the steam ports, E the exhaust port, AB the valve 
 
 seat, and DM the bridges of the valve seat. The valve seat must 
 
 o 
 
 be planed perfectly smooth, so that pressure on the valve will 
 
 Fig. 1. 
 
VALVE GEARS 
 
 make a steam tight fit, and cause as little friction as possible when 
 the valve slides. Furthermore, the length of the seat AB must be 
 a little less than the distance from the extreme right-hand posi- 
 tion of the right-hand edge of the valve to the extreme left-hand 
 position of the left-hand edge of the valve. This allows the valve at 
 each stroke slightly to over travel the seat, thus, keeping it always 
 worn perfectly flat and smooth. If the valve seat were not raised 
 slightly above the rest of the casting, or if it were too short, the 
 constant motion of the valve would soon wear a hollow path in the 
 valve seat, and it would cease to be steam tight. 
 
 Eccentric. The valve usually receives its motion from an 
 eccentric which is simply a disc, keyed to the shaft in such a 
 
 ivfc i 
 
 Fig. 2. 
 
 manner that the center of the disc and the center of the shaft do 
 not coincide. li; is evident that as the shaft revolves, the center 
 of this eccentric disc moves in a circle about the shaft as a center, 
 just as if it were at the end of a crank. The action of the eccentric 
 is equivalent to the action of a crank the length of which is equal 
 to the eccentricity of the eccentric (the distance between the center 
 of the eccentric and that of the shaft). 
 
 Fig, 2 represents the essentials of an ordinary eccentric. O 
 is the center of the shaft, O the center of the eccentric disc E 9 and 
 S is a collar encircling the eccentric and attached to the valve 
 rod E. 
 
VALVE GEARS 
 
 As the eccentric turns in the strap, the point O moves in the 
 uotted circle around O', and the point A also moves in a circle. 
 When half a revolution is accomplished the point O will be at 
 O", the point A will be at A", and the eccentric strap and valve 
 rod will be in the position indicated by the dotted lines. 
 
 Since the revolving shaft transmits motion to the valve 
 through the eccentric, it- will be necessary to study the relative 
 motions of the crank and eccentric in order to get a clear idea of 
 the steam distribution. 
 
 The distance of the center of the eccentric from the center 
 of the shaft (GO' in Fig. 2) is known as the eccentricity, or throw, 
 of the eccentric. The travel of the valve is twice the eccentricity. 
 
 Valve without Lap. Fig. 3 shows a section through the 
 steam and exhaust ports of an engine, together with a plain slide 
 
 Fig. 3. 
 
 valve placed in mid-position, and so constructed that in this posi- 
 tion it just covers the steam ports and no more. A valve is in 
 mid-position when the center line of the valve coincides with the 
 center line of the exhaust port. 
 
 Fig. 1 shows the same valve drawn to a larger scale. 
 
 Suppose the valve is moved a slight distance to the right; 
 the port P (see Fig. 1) is then uncovered and opened to the live 
 steam which enters the cylinder and causes the piston to move. 
 Since the two faces of the valve are just sufficient to cover the 
 steam ports, it is evident that as the port P opens to live steam, 
 the port P' opens to the exhaust. The ports are closed only when 
 the valve is in mid-position. This allows admission and exhaust 
 
VALVE GEARS 
 
 to continue during the whole stroke. With such a valve there is 
 no expansion or compression; the indicator card would be a rec- 
 tangle, and the M. E. P. would be equal to the initial steam pressure, 
 assuming no frictional losses in the steam pipe or condensation in 
 the cylinder. 
 
 For a theoretical discussion of valve motion, it is assumed that 
 the eccentric rod moves back and forth in a line parallel to the 
 center line of the engine. This is not the case in practice, for the 
 eccentric rod always makes a small angle with the center line, just 
 as the connecting rod does, but the eccentricity is so small in com- 
 parison with the length of the eccentric rod that the angularity of 
 the eccentric rod is very much smaller than the angularity of the 
 connecting rod, and its influence may be neglected without appre- 
 ciable error. 
 
 I 
 
 J 
 
 Fig. 4. 
 
 When the valve shown in Fig. 3 is in mid -position, the crank 
 is on dead center, the eccentric is set at right angles to it, and the 
 piston is just ready to begin the stroke. 
 
 Fig. 4 shows the relative positions of crank, piston, eccentric 
 and valve when the crank has made a quarter turn or the piston 
 has moved to half stroke. The eccentric is now in its extreme 
 position to the right, the valve has its maximum displacement and 
 both the steam and exhaust ports are wide open. The valve will 
 not close again until the piston has reached the end of its stroke. 
 
 This type of valve is used only on small and unimportant 
 engines, and since it allows no expansion of the steam, is very 
 uneconomical. Furthermore, it will be seen that this valve opens 
 just after the stroke begins, which is impractical, for it means that 
 the piston has begun its stroke before the full steam pressure 
 
tl 
 S i 
 
VALVE GEARS 
 
 reaches it;, which will cause an inclined admission line on the indi- 
 cator diagram. 
 
 Valve with Lap. If the face of the valve is made longer 
 than showjjjn Fig. iJso that in mid-position it overlaps the steam 
 ports, ive shall have a valve such as shown in Fig. 5. The 
 amount that the valve overlaps the steam ports is called the lap of 
 the valve. In Fig. 5, DI is the inside lap, and OC the outside lap. 
 It will at once be seen that both the admission and exhaust ports may 
 remain closed during a part of the stroke, thus making expansion 
 
 Fig. 5. 
 
 and compression possible. It is also evident that steam cannot < 
 be admitted until the valve uncovers the port by moving a dis- 
 tance from mid-position equal toT)C. Admission continues until 
 the valve returns to such a position that the outer edge of the 
 valve again closes tho "port. Release will begin when the inner 
 edge,of_the inside lap begins to uncover the port. 
 
 (Fig. represents a valve with lap, at the point of admission. | 
 Since the valve must move a distance equal to the outside lap 
 before admission can take place, it is evident that the eccentric can 
 
 no longer be at right angles to the crank at the beginning of the 
 
 stroke ; but must be ahead of the right-angle point by an amount \ 
 equal to AGO. The angle AGO is known as the angular advance. 
 The maximum displacement of the valve is attained when 
 the eccentric is horizontal as shown in Fig. 7. In this position 
 
8 
 
 VALVE GEARS 
 
 both the steam and exhaust ports are wide open, and any further 
 motion of the piston will cause the valve to move toward its mid- 
 position. 
 
 Admission continues until the valve returns to the position 
 
 Fig. 6. 
 
 shown in Fig* 8. Here the outside lap just closes the left-hand 
 steam port, cut-off takes place, and the steam already in the cylin- 
 der begins to expand. As the valve continues to move toward the 
 left, the left-hand inside lap begins to uncover the left-hand port 
 and releases the steam at the position shown in Fig. 10 
 
 The dotted lines of Fig. 7 show the valve in its extreme 
 
 Fig. 7. 
 
 position to the left. Any further movement of the piston will 
 cause it to return toward mid-position. 
 
 The dotted position of crank and eccentric in Fig. 10 shows 
 the valve returned to the point of compression, which continues 
 until the conditions of Fig. G are again reached and the opening- 
 valve allows steam again to enter the cylinder. 
 
 This process has been traced step by step for one end only; 
 let us now consider what is happening at the other end. 
 
VALVE GEARS 
 
 9 
 
 Admission is the point at which the valve opens to admit 
 steam to the cylinder. Cut=off is the point at which the valve 
 closes to cut off the admission of steam. Release is the point at 
 which the exhaust is opened ; and Compression is the point at 
 which the exhaust is closed. 
 
 Fig. 8. 
 
 While the crank is moving from the position shown in Fig. 
 6 to that of Fig. 8, steam is being admitted to the head end and 
 being exhausted from the crank end. The inside lap being less 
 than the outside lap, causes the exhaust to continue longer than 
 the admission. 
 
 Fig. 9 shows the relative positions of crank, eccentric and 
 valve when the exhaust closes on the crank end and compression 
 
 Fig. 9. 
 
 V-t 
 
 begins. Between these two positions the steam is expanding in 
 the head end and exhausting from the crank end. 
 
 Between the positions of Fig. 9 and Fig. 10 both ports are 
 entirely closed, expansion is taking place in the head end and com- 
 pression in the crank end. Fig. 10 is head-end release. Fig. 11 
 shows admission at crank end of cylinder and marks the end of 
 crank-end compression. 
 
10 
 
 VALVE GEARS 
 
 By referring to Figs. 6-11, the effect of any change of lap 
 may at once be observed. If the outside lap is increased, the 
 valve must move farther from mid-position before admission will 
 occur, and on the return, after the maximum displacement is 
 reached, the outside lap, being wider, will close the port sooner, 
 and the cut-off shown in Fig. 8 will take place before the crank 
 
 Fig. 10. 
 
 reaches the angle there shown. A decrease of outside lap will 
 make cut-off later and admission earlier. 
 
 If the inside lap is increased, the valve must move farther be- 
 fore release occurs and the crank angle would be greater than shown 
 in Fig. 10. On the return of the valve to the dotted position shown 
 in Fig. 10, the port will close earlier and make an earlier compres- 
 sion; the crank angle will be less than is there shown. Decreasing 
 inside lap will cause earlier release and later compression. 
 
 Fig. 11. 
 
 Thus we see that it is the outside lap that influences admis- 
 sion and cut-off, and the inside lap that controls release and 
 compression. For this reason the outside lap is often called the 
 steam lap, and the inside lap the exhaust lap. 
 
VALVE GEARS 
 
 11 
 
 Fig. 12. 
 
 Lead. If a valve having lap is in mid-position, the port is 
 closed and the engine cannot start because no steam can enter the 
 cylinder. That the steam may be ready to enter the cylinder at 
 the beginning of the stroke it is necessary that the eccentric be 
 set more than 90 ahead of the crank and the eccentric radius will 
 take an angle as shown in Fig. 6, called the angular advance. In 
 order that the ports and 
 clearance may be prop- 
 erly filled with steam 
 at the beginning of the 
 stroke, it is necessary 
 that the valve be dis- 
 placed from its mid- 
 position an amount 
 slightly greater than 
 the outside lap. With 
 the piston at the end of the stroke the valve will have a position as 
 shown in Fig. 12. The port will be open the distance AB. This 
 causes the eccentric to be moved forward a slight amount in excess 
 of the angular advance. This excess is called the angle of lead. 
 
 In Fig. 13, O'R' represents 
 the crank at the beginning of the 
 stroke, LOA the angular ad- 
 vance, and ADA' the angle of 
 lead. The eccentric, to give 
 lead, must be set at the angle 
 EGA' ahead of the crank or 90 
 plus angular advance plus angle 
 of lead. In large, quick-run- 
 ning engines, a liberal lead is 
 essential, so that the ports and 
 clearance may be well filled 
 with steam before the stroke 
 begins. If there is no lead, a 
 
 Fig. 13. 
 
 portion of the steam will be used 
 in filling these places and full 
 pressure steam will not reach the piston until it is well advanced 
 on the stroke. This will give a sloping admission line as 
 
12 
 
 VALVE GEARS 
 
 in Fig, 14. Too much lead, on the other hand, will cause too 
 early an admission as shown in Fig 15. 
 
 If the angular advance is increased, the eccentric will be 
 moved farther ahead of the crank, and consequently will begin its 
 motion sooner. It will necessarily arrive at each of the events 
 
 Fig 14. Fig. 15. 
 
 sooner than before. If then, the angular advance is increased, all 
 of the events of the stroke will occur earlier. 
 
 Inequality of Steam Distribution. In the valve diagrams 
 thus far considered, the events of the stroke have been discussed 
 for each end separately, without reference to the relation of sim- 
 ilar events on the other side of the 
 piston. If the connecting rod 
 were of infinite length, so that it 
 would always remain parallel to 
 the center line of the engine, the 
 distribution would be the same 
 -Y for both ends of the cylinder. In 
 practice, the connecting rod is 
 from 4 to 8 times tLo length of 
 the crank, which causes the con- 
 necting rod always to be at an 
 angle to the center line of the 
 engine, and for a given crank 
 angle makes the piston displace- 
 ment greater at the head end than at the crank end. 
 
 To find the displacement of the valve, let us consider Fig. 16. 
 The circle represents the path of the eccentric center during a 
 complete revolution of the engine. OC represents the crank, and 
 OR the corresponding position of the eccentric. The diameter 
 XY represents the extent of the valve travel. Since the eccentric 
 rod is so long in comparison to the eccentricity, we make no 
 appreciable error by assuming it always to be parallel to the center 
 
 Fig. 16. 
 
VALVE GEARS 
 
 13 
 
 line of the engine. When the eccentric is at OL, the valve is in mid- 
 position. At OR the valve has moved from mid-position an amount 
 ON, found by dropping a perpendicular from R to the center line 
 XY. If the angularity of the connecting rod could be neglected, 
 't\\Q piston displacement could be found in the same manner. 
 
 To find the displacement of the piston, a diagram as shown 
 in Fig. 17 must be drawn. In this figure AB represents the 
 cylinder, P the piston, II the crosshead, HR the connecting rod, 
 and OR the crank. Suppose now the engine should stop in- this 
 position and then be clamped. The piston displacement would 
 be represented by AP. If the crank pin at R should now be 
 loosened so as to allow the connecting rod to fall to a horizontal posi- 
 tion, the point R would describe the arc of a circle RN, and XN 
 would represent the piston displacement and would be equal to AP 
 
 P' 
 
 Fig. 17 
 
 Suppose now that in this disconnected way the piston, crosshead 
 and connecting rod were moved forward until the end of the rod 
 came to O. P would then be at P' and the piston would be in the 
 middle of its stroke. Now swing the end of the rod lip to its 
 proper position on the crank-pin circle, the piston remaining sta- 
 tionary. It would describe an arc OZ. The crank pin would be 
 at Z, less than a quarter revolution from X, while the piston would 
 be in the middle of its stroke. 
 
 Suppose this engine were running with cut-off at half stroke 
 on the head end and that XOZ represented the corresponding crank 
 angle. On the return stroke the valve .would cut off at the same 
 crank angle YOT = XOZ, and OT would represent the crank cut- 
 off on the return or crank-end stroke. The piston, as we have just 
 seen, will not be at half stroke except when the crank is at OZ or OS. 
 Consequently OT is less than half stroke and cut-off takes place 
 earlier at the crank end than at the head end. When the crank is at 
 OZ the eccentric will be at OA (Fig 17#), and the valve displace- 
 
VALVE GEARS 
 
 merit will be OB. When the crank is at OT the eccentric will be at 
 OA', and the valve displacement will be OB', which is equal to OB, 
 the displacement of the valve at cut-off in the head end. The pis- 
 ton displacement will be OX in the head end and WY in the crank 
 end when cut-off occurs. If the connecting rod always remained 
 parallel to the center line, the cut-off would be the same at both ends. 
 Compensation of Cut=off. It has already been pointed out 
 that lengthening the outside lap makes the cut-off earlier, and short- 
 ening the lap makes it later. The cut-off in the case just cited may 
 then be equalized by altering the outside. laps. If we increase the 
 outside lap on the head end, or decrease the crank-end lap, the 
 inequality will be less. By changing either or hoth of the laps the 
 proper amount, the cut-off may be exactly equalized. 
 
 Fig. 11 a. 
 
 But altering the outside lap changes the lead as has already 
 been explained. If the lap is increased on the head end, the lead 
 will be less than on the crank end. If the lead becomes too small 
 on the head end, the angular advance may be increased but the 
 inequality of lead will still remain, for this increase of angular 
 advance will increase the lead at the crank end as well as at the 
 head end, and by< hastening all the events of the stroke may give a 
 bad steam distribution if care is not taken. 
 
 Unequal lead is of less consequence on a low-speed than on a 
 high-speed engine. On low-speed engines the cut-off may be 
 equalized at the expense of lead with beneficial results, but on high- 
 speed engines it will not do to give too little lead at one end. A 
 high-speed engine requires more lead than a low-speed, for there is 
 relatively less time in each stroke for the clearance to fill with steam. 
 
VALVE GEARS 
 
 15 
 
 If both inside laps are equal, compression will not occur 
 equally at both ends. To equalize it, the. inside laps may be 
 changed in the same manner as the outside laps are changed to 
 equalize the .cut-off. By altering these inside laps to equalize 
 compression, it may happen that the lap is reduced enough to 
 leave the exhaust port open when the valve is in mid-position. 
 
 Fig. 18. 
 
 This opening of the valve is called an inside clearance, or negative 
 lap. In Fig. 18, A is the inside clearance. , 
 
 Rocker. Sometimes it happens that the valve stem and 
 eccentric rod cannot be so placed that they will be in the same 
 straight line; or it may be that the travel of the valve must be so 
 great as to require an excessively large eccentric. In such cases 
 a rocker may be used. 
 
 Fig;. 19 shows a valve that is not in line with the eccentric. 
 
 O 
 
 This occurs in horizontal engines when the valve is set on top of 
 
 Fig. 19. 
 
 the cylinder instead of on one side. By means of the rocker AG 
 the valve may receive its proper motion. 
 
 In case it is more convenient to place the pivot of the rocker 
 arm between the connections to the valve stem and those of the 
 eccentric rod, such an arrangement as shown in Fig. 20 may be 
 used. Here it will be noticed that the valve stem and eccentric 
 rod are moving in opposite directions, and to give the valve the 
 
16 VALVE GEARS 
 
 same motion as in Fig. 19, the eccentric must be moved 180 ahead 
 of the position there shown. 
 
 If AB is less than AG, the valve travel will be greater than 
 twice the eccentricity, in proportion as AG is greater than AB. 
 In all cases the valve travel is to twice the eccentricity as AG is to 
 AB. Thus, if the valve travel is 4| inches, AB, 15 inches, and 
 
 AG, 18 inches, then X 4J = 3| inches, will equal twice the 
 lo 
 
 eccentricity. 
 
 / 
 
 Fig. 20. 
 
 A valve gear may be so laid out as to make both the cut-off 
 and the lead equal for both ends of the cylinder. This may be 
 done by a proper porportion between the rocker arms, and a careful 
 location of the pivot of the rocker. The eccentric must then be 
 set accordingly. In this manner the Straight Line engine equal- 
 izes the cut-off and lead. A discussion of this method will be 
 considered later. 
 
 VALVE DIAGRAMS. 
 
 Zeuner'S' Diagram. In order to study the movements of 
 valves, the effect of lap, lead, eccentricity, etc., diagrams of various 
 sorts have been devised. By the use of diagrams we may acquire 
 a knowledge of valve motion without the complex mathematical 
 expressions that such a discussion would entail. The most useful 
 of these various diagrams is that devised by Zeuner, and to avoid 
 complexity we shall confine ourselves to a discussion of this dia- 
 gram alone. The eccentric rod is assumed to be of infinite 
 length, and the positions of the crank are shown on the diagrams. 
 The displacement of the piston can easily be found if the ratio of 
 crank to connecting rod is known. 
 
 In Fig. 21 let OY be the eccentricity, then XOY will rep- 
 resent the valve travel, and the center of the eccentric will move 
 
VALVE GEARS 
 
 17 
 
 in the circle XWY. Let OK represent the position of the 
 crank and Or the corresponding position of the eccentric, which 
 is 90 -f- angle of advance 6 ahead of the crank. Draw OW 
 perpendicular to XY and lay off from it the angle WOM = 
 angle of advance 6 towards the crank. With OH as a diameter, 
 construct a circle. CM is equal to the eccentricity, and the circle 
 MFC is known as the valve circle. If Oil, the center line of 
 
 x s 
 
 the crank, cuts this valve circle at P, then OP is equal to the dis- 
 placement of the valve from mid-position. 
 
 To prove this, draw rS perpendicular to XY. Since Or 
 is the position of the eccentric, OS will represent the valve dis- 
 placement from mid-position. Draw MP. Then by geometry 
 OPM is a right angle because it is inscribed in a semicircle. 
 OSr is also a right angle ; the two right-angled triangles OSr 
 and OMP are equal because they are similar and have two cor- 
 responding sides equal. Or = OM, being radii of the same 
 circle. But we have seen that OS is equal to the valve displace- 
 ment, therefore OP is also equal to the valve displacement, for it 
 is equal to OS. 
 
 Now that the truth of our proposition has been proved, let 
 us see how we may study the valve motion from such a diagram. 
 See Fig. 22. As before, let XY represent the valve travel, then 
 the circle XEYF will represent the path of the center of the 
 eccentric. Let 6 be the angular advance and lay off EO toward the 
 crank, making an angle with the vertical. Produce EO to F, 
 and on OE and OF as diameters draw the valve circles as shown. 
 Let the outside lap.be an amount equal to OY, then with O as a 
 center and OY as a radius draw an arc intersecting the upper 
 valve circle at Y and K. Lay off OP equal to the inside lap and 
 
18 
 
 VALVE GEARS 
 
 with O as center and OP as a radius draw an arc intersecting the 
 valve circle at P and Q. Draw the crank line AO passing through 
 Y. Then, when the crank is in this position, the displacement of 
 the valve is equal to OY (the outside lap) and the steam is ready 
 to enter the cylinder. This is the position of the crank at admis- 
 sion, and the crank angle XOA is called the lead angle. The 
 valve has lead, therefore the admission takes place before the end 
 of the stroke. When the crank reaches the position OE, the 
 displacement of the valve is equal to OE, the eccentricity, and is 
 
 90* e 
 
 Fig. 22. 
 
 the maximum displacement. Further motion of the piston causes 
 the valve to move toward mid-position until, at the crank position 
 OC, the displacement OK is again equal to the outside lap and 
 the valve has reached the point of cut-off. When the position Oil 
 is reached, the crank line is tangent to both valve circles and there 
 is no displacement of the valve. At this point the valve is in 
 mid-position. 
 
 Further crank movement draws the inside lap toward the 
 edge of the exhaust port until, at the crank position OB, the dis- 
 placement is equal to OP (the inside lap) s-nd release begins. At 
 OF the maximum valve displacement is again reached -and the 
 valve moves in the opposite direction until at OD its displacement 
 
VALVE GEARS 
 
 19 
 
 from mid-position, is again equal to OQ = OP = the inside lap, 
 and compression takes place. At OH' the valve is again in mid- 
 position. At OX the displacement of the valve is OI, but since 
 the valve has to move the distance OJ before the port begins to 
 open IJ must represent the port opening when the crank is on 
 dead center and by definition we know that lead is the amount of 
 port opening at this position. Therefore IJ represents the lead. 
 
 At the position It, the port is open an amount equal to TG, 
 at E the opening is a maximum equal to EN. At C the opening 
 is nothing. If LW represents the total width of the steam port, 
 the exhaust port will be open wide when the displacement of the 
 valve is equal to OWand it will remain wide open while the crank 
 swings from OW to OK. 
 
 Fig. 23. 
 
 If the width of steam port in addition to the outside lap were 
 laid off on the other valve circle it would fall at E'. For the ad- 
 mission port to be wide open, the displacement of the valve would 
 have to be equal to OE',' which is more than the maximum dis- 
 placement. This shows that in this case the steam port is never 
 fully open and that the left-hand edge of the valve overlaps the 
 right-hand edge of the port by an amount equal to EE' when the 
 valve has reached its maximum displacement. 
 
20- 
 
 VALVE GEARS 
 
 Fig. 22, with its two valve circles, shows the diagram for the 
 head end of the cylinder only. The crank-end diagram would be 
 similar except that the laps might not be equal to those of the 
 head end. 
 
 We are now in a position to consider more in detail the effect 
 of changing in any w r ay either the valve or the setting. Let us con- 
 sider Fig. 23, which is in every way like Fig. 22 except that all 
 unnecessary letters and lines are omitted to avoid confusion. If 
 the outside lap is increased an amount equal to !NM, the ad- 
 mission will take place later, at crank position OA'; the lead will 
 
 Fig. 24. 
 
 be reduced to IG and cut-off will take place earlier at OC'. If 
 the outside lap is reduced a like amount the contrary effects will 
 be observed. If the inside lap is increased an amount equal to LS. 
 the release will take place later at the crank position OB' and com- 
 pression will take place earlier at OD'. The contrary effect will 
 be observed by decreasing the inside lap. 
 
 If the angular advance is increased (see Fig. 24) all the events 
 will occur earlier. This is evident from the figure; the crank 
 revolves in the direction indicated by the arrow and OA' (new posi- 
 tion of admission) is ahead of OA, the old position. 
 
VALVE GEARS 
 
 21 
 
 If the eccentricity is increased, Fig. 25, the valve travel will 
 increase and admission will take place earlier at OA'; the lead 
 will be increased an amount equal to II', and cut-off will take 
 place later at OC'. Release will be earlier at OB' and com- 
 
 C 1 
 
 Fig. 25. 
 
 pression will be later at CD'. The upper valve circle will now 
 cut the arc drawn from O as a center, with a radius equal to the 
 outside lap plus the width of steam port, in the points W and H', 
 and the admission port will be open wide while the crank is mov- 
 ing from OW to OH'. Similarly, the lower valve circle cuts the 
 arc drawn from O as a center, with a radius equal to the inside 
 lap plus the width of steam port, in the points W and H. The 
 steam port is then wide open to exhaust while the crank is moving 
 from W to H. From the above it will be seen that the periods 
 are all changed by changing the travel; thus, admission and ex- 
 haust begin sooner and last longer, while expansion and com- 
 pression begin later and cease sooner. With change in the angular 
 advance, however (see Fig. 24), the periods are neither increased 
 nor decreased. 
 
22 
 
 VALVE GEAKS 
 
 For convenience, these results are collected in the following 
 table which shows the effect of changing the laps, travel, and 
 angular advance: 
 
 
 Increasing 
 Outside Lap. 
 
 Increasing 
 Inside Lap. 
 
 Increasing 
 Travel. 
 
 Increasing 
 Angular Advance. 
 
 Admission. 
 
 Is later. 
 Ceases sooner. 
 
 Not changed. 
 
 Begins earlier. 
 Continues 
 longer. 
 
 Begins earlier 
 Same period. 
 
 Expansion. 
 
 Is earlier. 
 Continues 
 longer. 
 
 Beginning 
 unchanged. 
 Continues 
 longer. 
 
 Begins later. 
 Ceases sooner. 
 
 Begins earlier. 
 Same period. 
 
 Exhaust. 
 
 Unchanged. 
 
 Occurs later. 
 Ceases sooner. 
 
 Begins earlier. 
 Ceases later. 
 
 Begins earlier. 
 Same period. 
 
 Compression 
 
 Begins at same 
 point. 
 
 Begins sooner. 
 Continues 
 longer. 
 
 Begins later. 
 Ceases sooner. 
 
 Begins earlier. 
 Same period. 
 
 PROBLEMS. 
 
 All the problems on valve gears involve the relations between 
 certain variables which are : 
 
 The valve travel. 
 Angle of lead. 
 Outside lap. 
 Inside lap. 
 
 Points of stroke at which admission cut-off, release and compression 
 take place. 
 
 In designing a Slide Valve, a few of these variables depend 
 upon the conditions under which the engine is to run. For instance, 
 the valve travel is limited, cut-off must be at a certain point and 
 the engine must have a certain lead. Then, with the aid of a Zeun- 
 er's diagram, the remaining proportions of the valve may be deter- 
 mined. 
 
 Let us consider a few examples: 
 
 Given the valve travel = 3 inches. 
 Inside lap = % inch. 
 
 Angular advance = 35 
 
 Angle at cut-off = 115 
 
VALVE GEARS 
 
 23 
 
 To determine the laps, the lead and the crank angles at 
 admission, compression and release. 
 
 In Fig. 26, let XY represent the valve travel = 3 inches. 
 Draw OM perpendicular to XY, and on XY as a diameter draw 
 the circle XMYF representing the path of the center of the eccen- 
 tric as it revolves about the shaft. Lay off the angle MOE the 
 angular advance = 35 so that the angle XOE is equal to 90 
 
 Fig. 26. 
 
 minus the angular advance Produce EO to F. Then on OE 
 and OF as diameters draw the valve circles. The eccentricity 
 OE or OF, if no rocker is used, will be half the valve travel. Lay 
 off the crank angle XOC = angle of crank at cut-off = 115, and 
 OK will then represent the distance of the valve from mid-posi- 
 tion when cut-off takes place. This distance we know is the out- 
 side lap. Draw the arc KI, known as the lap circle, and it will 
 cut the valve circle again at V. When the valve is again the dis- 
 tance OV = the outside lap from mid-position, admission will take 
 place. Draw the line OVA and this will represent the position 
 of the crank at admission. 
 
24 VALVE GEARS 
 
 When the crank is at OX, the valve displacement is equal to 
 OJ. This is at dead center and the valve is open the amount IJ, 
 for it has moved this distance more than the outside lap. There- 
 fore IJ is the lead for this end. 
 
 Now on the other valve circle, draw the arc PQ with the 
 inside lap (|- inch) as a radius. It will cut the valve circle at 
 P and Q. When the valve displacement is equal to OQ, the 
 exhaust port has just closed and the engine is at compression. In 
 the same way OP is the valve displacement at release when the 
 port begins to open. OQD represents the crank position at com- 
 pression and OPB the crank position at release. 
 
 The results then are as follows : 
 
 Given : 
 
 Valve travel = XY = 3 inches. 
 
 Angular advance = angle MOE = 35. 
 Inside lap = OP = % inch. 
 
 Crank angle at cut-off = angle XOC 115 
 Found : 
 
 Outside lap = OK = % inch. 
 
 Angle of lead = XOA = 5. 
 
 Linear lead = IJ = 8 inch. 
 
 Max. port opening for admission = HE = % inch. 
 Crank angle at compression = XOD = 185 
 Crank angle at release = XOB = 65 
 
 Max. port opening for exhaust = FN = % inch. 
 
 Fig. 26 is drawn full size, and all of these measurements may 
 readily be verified. This figure is drawn for the head end only. 
 If the crank angle at cut-off is the same on both ends, the Zeuner's 
 diagram for the crank end will be exactly like Fig. 26. 
 
 ANOTHER PROBLEM. 
 Given : 
 
 The valve travel = 3 inches. 
 
 The lead angle = 6. 
 
 Crank angle at cut-off = 70. 
 
 Crank angle at compression = 75. 
 To Find : 
 
 Angular advance. 
 
 Laps. 
 
 Linear lead. 
 
 Crank angle at release. 
 
 As before, let XY represent the valve travel = 3 inches and 
 draw OM and the circle XMYF. See Fig. 27. Lay off the lead 
 
VALVE GEARS 
 
 angle XOA = 6. Then OA represents the crank position at 
 admission. Next lay off the crank angle XOC, the angle at cut- 
 off 70. Bisect the angle COA by the line OE and on OE draw 
 the valve circle. Angle MOE = the angular advance. The valve 
 circle will cut the crank lines OC and OA at Kand V respectively. 
 If the work has been carefully done, OK will be exactly equal to 
 OY and will represent the outside lap. The lead is IJ as before. 
 Draw OD at position of the crank at compression so that angle 
 XOD = 75. Continue OE to cut the eccentric circle at F. On 
 
 B 
 
 Fig. 27. 
 
 OF draw the second valve circle. It will cut OD at Q, and OQ 
 will represent the inside lap. Draw the lap circle OP, and the 
 crank position OPE. This will be the crank position at release. 
 
 The angular advance in this problem is large and all the 
 events of the stroke are early. Compression and release are excess- 
 ively early and the outside lap is unusually large. In the previ- 
 ous problem, with cut-off at about two-thirds stroke, the results 
 were nearly normal. Cut-off with the plain slide valve, earlier 
 than half stroke cannot be had without sacrificing the steam dis- 
 tribution on the other events. 
 
26 VALVE GEAKS 
 
 To sum up we have 
 
 liven : 
 
 Valve travel = XY 3 inches. 
 
 Lead angle = XOA = 6. 
 
 Crank angle at cut-off = XOO = 70. 
 
 Crank angle at compression = XOD = 75. 
 
 To find : 
 
 Angular advance MOE 58. 
 
 Outside lap = OV 1 & inches. 
 
 Lead = I J = & inch. 
 
 Inside lap = OQ = 3 7 a inch. 
 
 Orank angle at release = XOB = ISO 3 . 
 
 Suppose in this last problem the cut-off had been given at 
 half stroke instead of having the crank angle given, and that the 
 compression had been given in the same way. AVe should, of 
 course, need to know the ratio of length of connecting rod to 
 crank. Let this be given as 4, that is, the connecting rod is four 
 times the length of the crank. 
 
 In Fig. 28 let XY represent the valve travel. Extend XY 
 to the left to the point Z, and make OZ equal to four times OX. 
 With Z as a center and OZ as a radius, strike an arc OC that will 
 cut the eccentric circle at C; then draw OC, which will represent 
 the crank when the piston is at half-stroke, which is assumed to 
 be the point of cut-off. 
 
 To find the crank angle at compression, lay off YII equal 
 to .8 of the distance YX. From II lay off II W = OZ = four times 
 OX. From "VV as a center with a radius "WTI, draw an arc cut- 
 ting the eccentric circle at D. Draw OD, which will represent 
 the position of the crank at compression. 
 
 The student is advised to read over again pages 13 to 14 if this expla- 
 nation of finding the crank angle does not seem perfectly clear. 
 
 ANOTHER PROBLEM, 
 Given : 
 
 Cut off at .6 stroke. 
 
 Lead = fa inch. 
 
 Maximum port opening % inch, 
 
 Katio of crank to connecting rod - 4. 
 
 k To find : 
 
 The eccentricity. 
 Lead angle. 
 Angular advance. 
 Laps. 
 
VALVE GEARS 
 
 In Fig. 29 assume an eccentricity that will if possible be a 
 little too large. Let us take for trial 2J inches and draw XY 
 equal to twice the assumed eccentricity equal to 4^ inches. Lay 
 off XC' equal to .0 of XY, 
 and with a radius equal to 
 four times OX draw the arc 
 C'C as already explained. 
 
 Then draw OC, which 
 will represent the position of 
 the crank at cut-off, and 
 XOCwill be the crank-angle 
 
 o 
 
 at cut-off. Assume a lead 
 angle of about 7 and draw 
 OA, which, if this assump- 
 tion be true, will represent 
 the crank -angle at ad miss ion. 
 
 G> 
 
 Bisect the angle CO A by the 
 line OE, and on OE draw 
 the valve circle. Draw the 
 .lap circle VNK. With this 
 assumed eccentricity we find 
 a maximum port opening of 
 NE = .75 inch, which is 
 larger than the conditions of 
 the problem demand. We 
 may then form a proportion, 
 namely: 
 
 The actual port opening de- 
 sired : the port opening with the 
 assumed eccentricity : : probable 
 eccentricity: assumed eccentric- 
 ity. 
 
 Substituting the figures 
 we have .5 : .75 : : x : 2J .-. 
 x the probable eccentric- 
 ity; equals 1.42 inches. 
 
 Now draw on OE,a new 
 valve circle (dotted) with a diameter equal to the required eccen- 
 tricity of 1.42 inches. See Fig. 29a. It will cut the crank line 
 
28 
 
 VALVE GEARS 
 
 OC at K', and OK' will be the new outside lap and IT will be 
 the new lead (assuming the lead angle to be 7). This lead I'J' is 
 ^ inch, while the required lead is only T ^ inch. Now decrease the 
 angular advance enough to correct one half of this difference, by 
 drawing a new lap circle J"K" of -J fl inch greater radius. This 
 will make the valve circle cut OC at K", so that OK" will now be 
 
 the final lap, and I"J' ; the final lead, which is equal to the required 
 j\ inch. The lead angle is now XO A' instead of XOA. The port 
 opening at Nil' is -| inch (nearly) as required, but the change in 
 angular advance necessitates an increase of lap if cut-off is to 
 remain the same. This reduces the port opening by the amount 
 HI!', so that the maximum opening is only .46 inch. By increas- 
 ing the eccentricity this port opening may be increased. 
 
VALVE GEARS 
 
 29 
 
 s-i 
 t'c 
 
 .46 : .50 : : 1.36 : x. 
 
 x = 1.48, the true eccentricity. 
 
 Now draw the valve circle on OE' with a diameter of 1.48 
 inches. It will cut GO in lv'" and OX in J"'. The lap will be 
 OK'" = .97 inch, the lead will be T" J'" -^ inch, the angular 
 advance will be MOE' and the eccentricity ON'. 
 
 To sum up we have 
 
 Given : 
 
 The cut-off = .6 stroke. 
 
 The lead = -* g inch 
 
 Max. port opening = } inch. 
 
30 VALVE GEARS 
 
 Obtained all of the above conditions together with: 
 
 Lap = .97 inch. 
 
 Lead angle = XOA' 
 
 Angular advance = MOE' 
 
 Compression, release and inside laps are found as in the pre- 
 vious problems. 
 
 There are of course all sorts of combinations that would make 
 up different problems, but they can all be solved in the same gen- 
 eral way, as they are modifications of the above. 
 
 DESIGN OF THE SLIDE VALVE. 
 
 In designing a slide valve some of the variables are assumed 
 and the others are found by means of diagrams as we have already 
 seen. These diagrams show only the dimensions of the inside and 
 outside laps and travel of valve; the other dimensions of the valve 
 and seat must be calculated. 
 
 Area of Steam Pipe. Pipes that supply the steam chest 
 should be large enough to prevent an excessive loss of pressure due 
 to friction. If the pipes are long they should be of such si^e that 
 the mean velocity of steam in them does not exceed 100 feet per 
 second or 6,000 feet per minute. For this calculation it is. usual 
 to assume steam admitted to the cylinder during the whole stroke. 
 
 For example. Suppose an engine is 10" X 18", and makes 
 180 revolutions per minute. What is the diameter of the steam, 
 pipe? 
 
 The piston displacement or volume of the cylinder is : 
 
 TTvZ 2 3.1416 X 10' 
 
 - X I = - X 18 = 1413. i 2 cubic inches. 
 
 4 4 
 
 ' ^ = .818 cubic feet. 
 IT/oo 
 
 If the engine makes 180 revolutions it would use 2 X 180 X 
 .818 = 294.48 cubic feet per minute. 
 
 294.48 
 
 The area would be = ' := .04908 sq. ft. == 7.0675 sq. 
 
 in. 
 
 The diainoter corresponding to 7.0(575 square inches is 3 
 inches. 
 
 A three- inch pipe would be large enough, especially if the 
 engine cut off early in the stroke, 
 
VALVE GEARS 31 
 
 For a very large engine cutting off early, the allowable veloc- 
 ity may be taken as 8,000 feet per minute instead of 6,000 feet. 
 
 Width of Steam Port. The port opening at admission should 
 give nearly as great an area as the steam pipe in order to prevent 
 loss of pressure due to wire-drawing, but the actual width of the 
 port should be great enough for the free exhaust of steam. It 
 is well to have the steam port a little larger than the area of the 
 steam pipe, then with a port opening of .6 to .9 of the port area 
 for admission and full port opening at exhaust, satisfactory condi- 
 tions will result. 
 
 The length of the ports is usually made about .8 the diameter ol 
 the cylinder. Then in the 10" X 18" engine the steam ports would 
 be 8 inches long. If the area for admitting steam is 8.0675 squan 
 inches and the length of port is 8 inches, the width w T ill be 
 
 = _1 -= .8834 inch, or about J inch. 
 
 The width of port opening would be about ,9 X .8834 = .79506 
 inch or about -if inch. 
 
 Width of Exhaust Port. When the slide valve is at its 
 maximum displacement, the valve overlapping the exhaust port 
 as shown in Fig. 7 reduces the area more or less. In designing 
 the valve, the exhaust port should be of such a width that the 
 maximum displacement of the valve does not reduce the area of 
 the exhaust port to less than the area of the steam port.- It is not 
 advisable to make the exhaust port too large for this increases 
 the size of the valve and thus causes excessive friction. 
 
 The height of the exhaust cavity should never be less than the 
 width of the steam port, and may be made much higher to advantage. 
 
 Width of Bridge. The bridge must be of sufficient width so 
 that outside edges of the valve cannot uncover the exhaust port. 
 The width of the steam port plus the width of the outside lap plus 
 the width of the bridge must be greater than the maximum dis- 
 placement. 
 
 The width of the bridges should be not less than the thickness 
 of the cylinder wall in order to. make a good casting. 
 
 The Point of Cut=off. In the study of Indicators, it was shown 
 that if the point of cut-off is early, the other events are not good. 
 If a plain slide valve is used with an automatic cut-off, the cut-off 
 
32 VALVE GEARS 
 
 is hastened either by changing the eccentricity or by changing 
 the angular advance. Either of these methods will accomplish the 
 result at the expense of the compression which consequently will 
 be earlier and excessive. Except for locomotives and high-speed 
 engines, where compression is an advantage, the plain slide valve 
 is not arranged to cut-off earlier than -J or stroke. If an earlier 
 cut-off is desired, large outside laps are necessary. The cut-offs 
 may be equalized by giving the head end a greater lap than the 
 erank end, but this will cause an inequality of lead. 
 
 Lead. The lead of stationary engines varies from zero to | 
 inch according to the style of engine. An engine having high 
 compression that compresses the steam nearly to boiler pressure, 
 will give good results with little or no lead. If the ports are small, 
 and the clearance large, there should be considerable lead in order 
 to insure full initial pressure on the piston at the beginning of the 
 stroke. Valves that open slowly require more lead than quick- 
 acting valves. 
 
 Let us design and lay out the valve and valve seat for the fol- 
 lowing engine: 
 
 Diameter of cylinder = 10 inches. 
 
 Stroke = 18 inches. 
 
 Revolutions = 180 per minute. 
 
 Lead angle = 3. 
 
 Cut-off to be equal at both ends and to take place at .75 stroke. 
 
 Max. port opening .9 area of steam pipe. 
 
 Compression to be .85 of the stroke at both ends. 
 
 Length of connecting rod = 3 feet. 
 
 The 'piston displacement, or cylinder volume, will be 
 
 - X 18 == 1413.7 cubic inches or .818 cubic feet, If the 
 
 engine makes 180 revolutions, it will use 2 X 180 X .818 = 294.48 
 
 cubic feet of steam per minute. Steam pipe area = .. = .0491 
 
 square feet = 7.07 square inches. 
 
 This 7.07 square inches would also be the least possible area 
 of the steam ports. If the length of port is made .8 the diam- 
 
 7.07 
 eter of cylinder, the width will be -^ = .88 inches or about J 
 
 inch. The width of maximum port opening will be .9 X .88 = .792 
 or nearly j- j| inch. 
 
VALVE GEAKS 
 
 33 
 
 X 
 
 A 
 
 It will be necessary to draw a separate valve circle for eacli 
 end of the cylinder. First consider the head end. 
 
 The valve travel not being known, we shall lay off XY on an 
 assumption of 6 inches travel and draw the eccentric circle as 
 shown in Fig. 30. Lay off the lead angle XOA 3. Lay off 
 XC' = .75 of the assumed valve travel = 4J inches. Draw the 
 are CC' as previously explained and draw OC which will be the 
 crank angle at cut-off. The radius of the arc C'C will be equal to 
 
 G' 
 
 .H 
 
 V 1 
 
 V 
 
 rA' 
 
 1 
 
 H 
 
 Fig. 30. 
 
 4 times the radius of the eccentric circle, or 12 inches, because 
 the connecting rod is 4 times the length of the crank. Bisect 
 the angle AGO by the line ()E, and on OE draw the valve circle. 
 OV = OK is then the outside lap, with these assumed condi- 
 
34 
 
 VALVE GEARS 
 
 tions. Draw the lap circle; then EN will be the maximum port 
 opening. EN == 1^ inches, while i-J inch is all that is necessary. 
 The assumed eccentricity is 3 inches, therefore the probable eccen- 
 tricity = x : 3 : : i-J. : 1 T 7 6 . x = 1-J-j- inches. 
 
 Now draw a new eccentric circle with a radius of 1-J-J inches 
 and a new valve circle w T ith OE' = 1-J-J inches as a diameter. OK' 
 is now the outside lap and the maximum port opening is equal to 
 E'N', which from actual measurement is found to be -}|- inch. The 
 outside lap = OK' OY' |~J inch and the lead is IJ = -f^ inch. 
 
 Produce EG to F and draw another valve circle. We shall 
 use this valve circle to determine the outside laps and lead for the 
 crank end of the. cylinder. Since the cut-off is to be .75 of the 
 stroke, we may lay off Oil' OC', and with a radius of 12 inches 
 
 draw the arc HIT. Then, as already explained, OH will be the 
 crank angle at cut-off on the return stroke. OB will be the out- 
 
 o 
 
 side lap = -if- inch. Draw the lap circle intersecting the valve 
 circle at D. Then ODA' is the crank angle at admission on the 
 return stroke and LM | inch is the lead on the crank end of the 
 cylinder. The maximum port opening will always be greater at the 
 crank end than at the head end because the crank end lap is less 
 in order to get the equal cut-off. If the laps were equal, -of course 
 the port openings would be equal. 
 
 Now lay off YG' = .85 of XY and find the crank position 
 OG. This is the compression on the head end of the cylinder and 
 gives an inside lap on this end of ^\ inch, which is equal to OP, 
 
VALVE GEAKS 
 
 Draw the lap circle PQ, which allows us to draw through Q the 
 crank line Oil, which is the release on the forward stroke. 
 
 Lay off XS' = : YG' = .85 of XY, and construct the crank 
 line OS, which is the crank position at the crank end compression. 
 OS intersects the valve. circle at T, which gives OT = T ? 6 inch = 
 inside lap on the crank end. Draw this lap circle, which will 
 intersect the valve circle at U. This enables us to draw OUW, the 
 crank angle at release, on the return stroke. 
 
 From the data determined by means of these diagrams the 
 valve may now be laid out. For convenience let us tabulate the 
 results obtained as follows: 
 
 Data. Head End. Crank End. 
 
 Cut-off, per cent of Stroke 75 75 
 
 Outside Lap |J" -J-f" 
 
 Inside Lap ? y ^'' 
 
 Lead -g-V I" 
 
 Port Opening -L" lyV' 
 Width of Port -f " |" 
 
 Fig. 31 shows this valve in section.. Let us begin at the end 
 having the largest inside lap, or in this case at the crank end. 
 Layout the steam port | inch wide, and the crank-end outside lap 
 = -if inch. v ' The bridge will be, say, | inch wide. From the 
 inner edge of the steam port, lay off the crank-end inside lap = 
 r 7 6 inch. When the valve moves to the left, the point E' will 
 travel 1-J-J inches, a distance equal to the eccentricity, and in this 
 position of extreme displacement the exhaust port EF must be open 
 an amount at least equal to the steam port, J inch. Therefore we 
 lay off EF equal to 1 J - J-" -f |" = 2 T 9 6 ". The inside lap overlaps the 
 bridge nearly -J inch, so that we shall have to make the exhaust 
 port opening equal to 2| inches. Lay off | inch again for the 
 bridge and measure back ^V inch, equal to the head-end inside 
 lap. The port is J inch wide, and the head-end inside lap of |J 
 inch completes the outline of the valve seat. 
 
 VALVE SETTING. 
 
 The principles of valve diagrams are useful in setting valves 
 as well as in desionino- them. The valve is usually set as accu- 
 
 !~i ! > 
 
 rately as possible, and then, after indicator cards have been taken, 
 
36 
 
 VALVE GEARS 
 
 the final adjustment can be made to correct slight irregularities. 
 
 The slide valve is so designed that the laps cannot be altered 
 without considerable labor, and the -radius of the eccentric, which 
 determines the travel of the valve, is usually fixed. The adjust- 
 able parts are commonly the length of the valve spindle and the 
 angular advance of the eccentric. 
 
 By lengthening or shortening the valve spindle, the valve is 
 made to travel an equal distance each side of the mid-position. 
 Moving the eccentric on the shaft makes the action of the valve 
 earlier or later as the angular advance is increased or decreased. 
 
 To Put the Engine on the Center. It is usual to put the 
 engine on center before setting the valve. First put the engine 
 in a position where the piston has nearly completed the outward 
 
 RAM 
 
 Fig. 32. 
 
 Stroke, and make a mark M on the guide opposite the corner of 
 the crosshead .or at some convenient place. Also make a mark, 
 with a center punch, on the frame of the engine near the crank 
 disc or on the floor. With this punch mark Pas a center, describe 
 an arc C on the wheel rim, v/itli a tram. A tram is a steel rod 
 with its ends bent at right angles and sharpened. 
 
 Turn the engine past the center until the- mark on the guide 
 again corresponds with the corner of the crosshead, and make 
 another mark D on the wheel with the tram, keeping the same 
 center. AVith the center of the pulley or crank disc MS a center, 
 describe an arc CD on the rim, which intersects the two arcs drawn 
 with the tram. Bisect the arc CD on the rim, included between 
 the two short arcs, and turn the engine until the new point E is at 
 
VALVE GEARS 37 
 
 a distance from the point on the frame equal to the length of the 
 tram, in which position the engine will be on the center. 
 
 The engine should always be moved in the direction in which 
 it is to run so that the lost motion of the wrist pin and crank pin 
 will be taken up the right way. In case the engine has been 
 moved too far at any time, it should be turned back beyond the 
 desired point and brought up to that point while the engine is 
 moving the right way. 
 
 To Set the Vaive with Equal Lead. Set the engine on the 
 dead point and give the eccentric the proper angular advance. 
 Adjust the length of the valve spindle to give the proper lead for 
 that end. Now place the engine on the other dead point and 
 measure the lead at that end. If the leads are unequal, correct 
 half the error by changing the length of the valve spindle and 
 the other half by altering the angular advance. In case the valve 
 gear has a rocker, the length of the spindle should be such that 
 the rocker will move as designed. The angular advance should 
 not be changed, but the equal lead should be obtained by means 
 of the valve spindle or the eccentric rod. 
 
 Second Method, In case it is difficult to turn an engine the 
 following method may be used. First loosen the eccentric on 
 the shaft and turn it around until it gives maximum port opening 
 first at one end and then at the other. If the maximum port 
 openings are not equal, make them so by changing the length of 
 the valve spindle by half the difference. When the above adjust- 
 ment has been made, set the engine on dead center and give the 
 valve the proper lead by turning the eccentric on the shaft. The 
 angular advance is thus adjusted. 
 
 To Set the Valve for Equal Cut=off. Place the engine on the 
 dead point, give the eccentric the proper angular advance and 
 the valve the proper lead. Move the engine forward until cut-off 
 occurs, then measure the displacement of the crosshead from the 
 beginning of the stroke. Continue moving the engine forward, 
 until cut-off takes place on the return stroke and measure the dis- 
 placement of the crosshead from the beginning of this stroke to 
 this point. 
 
 In case the cut-off is earlier at the crank than at the head-end, 
 the valve spindle is too short. Adjust the length of the spindle 
 
38 VALVE GEARS 
 
 so that the inequality will be corrected. Now set the engine on 
 the dead point again and give the valve the proper lead by means 
 of the eccentric. By repeating the process, making slight changes, 
 the desired result will be obtained. 
 
 MODIFICATIONS OF THE SLIDE VALVE. 
 
 The ordinary slide valve is suitable for small engines; but for 
 large sizes some method must be employed to balance the steam 
 pressure on the back of the valve. With large valves, such for 
 instance as those of locomotives or large marine engines, a great 
 force is exerted by the steam, and the valve is forced against its 
 seat so hard that a large amount of power is necessary to move it. 
 This excessive pressure causes the valve to wear badly and is a 
 dead loss to the engine. The larger the valve, the greater this 
 loss will be. 
 
 Piston Valve. To prevent excessive pressure on the back of 
 the valve, the piston valve is commonly used, especially in marine 
 engines. This valve consists of two pistons, which cover and 
 uncover- the ports in precisely the same manner as the laps of the 
 plain slide valve. These pistons are secured to the valve stem in 
 an approved manner and are fitted with packing rings. 
 
 The valve seat consists of two short cylinders or tiibes 
 accurately bored to fit the pistons of the valve. The port open- 
 ings are not continuous as in the plain slide valve, but consist of 
 many small openings, the bars of metal between these openings 
 preventing the packing rings from springing out into the ports. 
 
 Steam may be admitted to the middle of the steam chest and 
 exhausted from the ends or vice versa. With the former method, 
 the live steam is well separated from the exhaust, and the valve- 
 rod stuffing box is exposed to exhaust steam only. This is a good 
 arrangement for the high-pressure cylinder; if used for a cylinder 
 in which there is a vacuum, air may leak into the exhaust space 
 through the valve-rod stuffing box. With this arrangement the 
 steam laps must be inside and the exhaust laps on the outside ends. 
 
 The piston valve may be laid out and designed by means of 
 the Zeuner diagram just as if it were a plain slide valve, and the 
 action is the same except that it is balanced so far as the steam 
 
VALVE GEARS 
 
 39 
 
 pressure is concerned; the power to drive it being only that neces- 
 sary to overcome the friction due to the spring rings. 
 
 Fig. 33 shows a section of the piston valve and the high- 
 pressure cylinder for one of the engines of the IT. S. S. " Massa- 
 chusetts." This valve consists of two pistons connected by a 
 sleeve through which the valve rod passes. This valve rod is pro- 
 
 O i -I 
 
 longed to a small balancing piston, placed directly over the main 
 
 33. 
 
 valve. The upper end of the balancing cylinder does not admit 
 steam, so that the steam pressure below the balancing piston will 
 practically carry the weight of the piston valve, thus relieving the 
 valve gear and making the balance more nearly complete. 
 
 DoubIe=Ported Valve. Sometimes it is' impossible to get 
 sufficient port opening for engines of large diameter and short 
 stroke, especially those having a plain slide valve with short travel. 
 
40 
 
 VALVE GEARS 
 
 This difficulty may be overcome by means of the double-ported 
 valve shown in Fig. 84. It is equivalent to two plain slide valves, 
 each having its laps. The inner valve is similar to a plain slide 
 valve except that there is communication between the exhaust 
 space and the exhaust space of the outer valve. Each passage to 
 the cylinder has two ports; a bridge separates the exhaust of the 
 outer valve from the steam space of the inner valve, and the outer 
 valve is made long enough 'to admit steam to the inner valve. 
 
 Fig. 34. 
 
 This valve may be considered as equivalent to two equal slide 
 valves of the same travel, each having one-half the total port 
 opening. To admit the same amount of steam as a plain slide 
 valve, the double-ported valve requires but half the valve travel; 
 this is advantageous in high-speed engines. 
 
 Fig. 35. 
 
 To balance the excessive steam pressure, the back of the valve 
 is sometimes provided with a projecting ring which is fitted to a 
 similar ring within the top of the valve chest. These rings are 
 planed true, and fit so that steam is prevented from acting on the 
 back of the valve. The space inside the rings is sometimes placed 
 in communication with the condenser. 
 
VALVE GEARS 
 
 -II 
 
 The Trick Valve. The defect of the plain slide valve, due to 
 the slowness in opening and closing, is largely remedied in the 
 trick valve, which is so made that a double volume of steam enters 
 during admission. Thus a quick and full opening of the port is 
 obtained w4th a small valve travel. 
 
 In Fig. 35 the valve is shown in mid-position. It is similar 
 to a plain slide valve except that there is a passage PF through it. 
 It has an outside lap () and an inside lap I. The seat is raised 
 and has steam ports SS, bridges BE, and exhaust port K. If the 
 valve moves to the right a distance equal to the outside lap plus 
 the lead, it will be in the position shown in Fig. 30. Steam will 
 be admitted at the extreme left tdge of the valve just the same as 
 though it were a plain slide valve; also, since steam surrounds the 
 valve it will be admitted through the passage as shown in Fig. 30. 
 
 Fig- 36. 
 
 Fig. 37. 
 
 If the lead is the same as for a plain slide valve, T 1 ^ inch for 
 instance, this valve would give double the port opening, that is J 
 inch, w r hen the valve was open a distance equal to the lead. 
 
 Fig. 37 shows the valve when it is in its extreme position to 
 the right and the port is full open to steam. 
 
 Piston valves are also made with a passage similar to that of 
 the trick valve for double admission. The valve used with the 
 Armington and Sims engine is perhaps the best example. 
 
 Balanced Valves. Since there is a wide difference between 
 the pressure of admission and exhaust, there must always be a 
 great pressure acting upon the valve, causing it to run hard and 
 wear excessively. The greater the steam pressure, the lower the 
 pressure at exhaust and the larger the valve, the greater this pres- 
 sure will be. 
 
42 VALVE GEARS 
 
 Piston valves are commonly used on the higli and intermedi- 
 ate cylinders of triple-expansion engines, and if well made and 
 fitted with spring rings, should not leak. Small piston valves are 
 often made without packing rings; but even if they fit accurately 
 when new, they soon become worn and cause trouble. 
 
 The double-ported valve, the trick valve, and others often have 
 some device for relieving the pressure, such as a bronze ring or 
 cylinder, fastened to the back of the valve. This ring is pressed 
 by springs against a finished surface of the valve client cover, and 
 the space thus enclosed by the ring may be connected to the 
 exhaust. There are numerous devices for balancing valves, but 
 
 O 
 
 they are usually more or less expensive and are liable to cause 
 trouble from leakage. 
 
 STEPHENSON LINK MOTION. 
 
 One of the earliest, and at present one of the most common 
 mechanisms for reversing engines, or changing the ratio of expan- 
 sion, is the Stephenson link motion, shown in Fig. 38. This illus- 
 tration is taken from the drawings of a recent battleship engine, 
 and may be considered the typical arrangement of the Stephenson 
 gear as applied to marine practice. 
 
 The two eccentrics E and E', whose centers are at C and C, 
 respectively, are shown in their relative positions when the crank 
 OA is at dead center. The eccentric rods II and II' are connected 
 by forked ends to the link pins II and G. The link consists of 
 two curved bars bolted together in such a manner that they may 
 slide by the link block N. On the link are three sets of trunions; 
 the two outer ones, or link pins, are fitted into the forked end of 
 the eccentric rods, and the middle one, known as the saddle pin, 
 is fitted into the end of the drag links FM. 
 
 The valve stem has, at its lower end, a pivoted block N, called 
 the link block, provided with slotted sides through which the links 
 can slide from right to left. The reverse shaft, or rock shaft, X, 
 here shown in full gear "forward," may be turned until F moves 
 over to B; in this position the link will be pushed across the link 
 block, and the valve will get its motion from the rod II' instead 
 of from It as before. The link in this position would be full gear 
 4 'astern ." 
 
VALVE GEARS 
 
 43 
 
 Pig. 38 
 
44 
 
 VALVE GEARS 
 
 In all large engines, such as marine, the reverse shaft is 
 turned by power, but in smaller engines, such as locomotives, the 
 engineer can turn the shaft by means of a lever. 
 
 When set full gear forward, as in Fig. 38, the valve admits 
 steam to the crank end of the cylinder, and the crank revolves as 
 
 Fig. 39. 
 
 shown by the arrow. As the crank turns, both eccentrics impart 
 motion to the link,- but the "go ahead" link pin II approximately 
 coincides with the link block, so that nearly all its up-and-down 
 motion is transmitted to the valve stem, while the " go astern " 
 
 Pig. 40. 
 
 eccentric exerts but little effect upon the link block. Moving the 
 drag links over to the extreme right reverses all these conditions 
 by bringing the other link pin under the link block. In this posi- 
 tion, steam will be admitted to the other end of the cylinder, and 
 the engine will run in the opposite direction. This will be clearly 
 seen by referring to Fig, 38. 
 
VALVE GEARS 
 
 45 
 
 When at full gear, either forward or backing, the valve moves 
 as if there were really but one eccentric, while at intermediate 
 points its motion is the result of the combined influence of both 
 eccentrics, one tending in a measure to counteract the other. The 
 effect of this is to shorten the valve travel the same as if the valve 
 were driven by a new eccentric having less throw than either of 
 the other two. 
 
 Fig. 4L 
 
 Decreasing the valve travel causes cut-off to occur earlier 
 compression is earlier, release later, and the lead is reduced some- 
 what. If every point of the link moved in the arc of a circle 
 when the drag link is shifted, the lead would not alter; but, since 
 the eccentric rods about which each end swings are centered at 
 different points, C and C', this is impossible. 
 
 Figs. 39 and 40 show the two principal ways of arranging 
 the eccentric rods of a Stephenson gear. The first is said to have 
 "open rods", the second . i4 crossed rods"; referring to whether 
 the rods are crossed or open when both the eccentrics face the link. 
 It can easily be seen that when the eccentrics shown in Fig. 39 
 have turned through 180 they will be in the position shown in 
 Fig. 41, but this is the same arrangement as before and is " open " 
 rods. The full lines show the positions in full gear forward, while 
 
VALVE GEAKS 
 
 the dotted lines indicate the positions in mid gear. Witb open 
 rods it will be seen that when at full gear the link block is at G, 
 and that if, \vithout turning the crank, the link is shifted to mid 
 gear, then the link block moves to J, Fig. 39, and the valve must 
 consequently be moved toward the right an amount equal to GJ, 
 thereby increasing the lead on the crank end of the cylinder. 
 With crossed rods, moving the link from full to mid gear moves 
 the link block from G to J, Fig. 40, thus reducing the lead. It 
 follows then that open rods give increasing lead from full toward 
 mid gear, and that crossed rods give decreasing lead. With crossed 
 rods there will be no lead when in mid gear. It will be apparent 
 that the shorter the rods the greater this increase or decrease 
 will be. 
 
 REVERSE ROD 
 
 Fig. 42. 
 
 Nearly all marine engines, and some English locomotives, 
 have their link blocks carried directly on the valve rod. Ameri- 
 can locomotives commonly use a rocker, one end of which carries 
 the link block while the other moves the valve rod. This arrange- 
 ment indicated in Fig. 42 makes it possible to place the valve and 
 steam chest above the cylinder. The position of the crank for the 
 same valve position is just opposite that shown in Fig. 39 because 
 the rocker reverses the valve motion; this gives an arrangement of 
 crank and eccentrics that is identical with that indicated in Fig. 
 41 and the rods, although apparently crossed, are in reality of the 
 open rod arrangement, giving increasing lead toward mid gear. 
 A rod from the bell -crank lever on the reverse shaft E, leads back 
 to the engineer's cab and connects with the reverse lever. This 
 ever moves over a notched arc, and may be held by a latch in any 
 
VALVE GEARS 47 
 
 one of the notches, thus setting the link in any position from mid 
 gear to full gear, either forward or back. 
 
 The Stephenson link is designed to give equal lead at both 
 ends of the cylinder; but to accomplish this, the radius of the link 
 arc (that is an imaginary line in the center of the slot) must be 
 equal to the distance from the center of this slot to the center of 
 the eccentric. In Fig. 38 the radius of the link arc is equal to 
 OH and C'G. 
 
 Exact quality of lead is not essential, and the radius of the 
 link arc is sometimes made greater or less than stated above in 
 order to aid in equalizing the cut-off; but the change should never 
 be great enough to affect the leads. 
 
 Stephenson originally intended to use the link simply as a 
 reversing gear, but soon found, however^ that at intermediate 
 points between the two positions of full gear, it would serve very 
 well as a means of varying the expansion and cut-off. Yery soon 
 the link came to be used not only on locomotives and marine 
 engines, but on stationary engines as well, in connection wJth the 
 reverse shaft which was under the control of the governor. The 
 mechanism proved to be too heavy to be easily moved by a gov- 
 ernor and it has gradually fallen into disuse on stationary engines 
 excepting as a means of reversing. 
 
 In marine practice, the variable expansion feature is of little 
 value, for marine engines run under a steady load and the link is 
 set either at full gear or at some fixed cut-off. For locomotives, 
 however, the variable expansion is nearly as important as reversing. 
 Locomotives are generally started at full gear, admitting steam for 
 nearly the entire stroke, and then exhausting it at relatively high 
 pressure. This wasteful use of stearn is necessary to furnish the 
 power needed in starting a train. After the train is under way, 
 less power is required per stroke, and the link is gradually moved 
 toward mid gear, or " notched up" by the engineer, thus hasten- 
 ing the cut-off; the expansion is increased and the power is reduced 
 in proportion to the load. 
 
 As the cut-off is changed, it is desirable to maintain a 
 approximately equal cut-off at each end of the cylinder; this can 
 be secured in the Stephenson gear by properly locating the saddle 
 pin and the reverse shaft, When used without a rocker, as in 
 
48 VALVE GEAKS 
 
 Fig. 38, the saddle pin ehculd be on the arc of the link or slightly 
 ahead of it. When used with a rocker, the saddle pin should be 
 behind the link arcs, and to give symmetrical action for forward 
 and backward running, it should be opposite the middle of the arc, 
 that is, equally distant from each link pin. 
 
 The Stephenson link cannot be designed directly from the 
 Zeuner diagram, but a systematic investigation can be made by 
 using a wocden model of the proposed link. This can be mounted 
 on a drawing board, and the effect of changing the position of 
 pins and the proportions of rods and levers can be determined 
 without difficulty. By a system of trials a combination can be 
 found best suited to obtain the desired results. Moreover, a 
 model makes it possible to measure directly the slip cf the link 
 block along the link. This slip should be kept as small as possi- 
 ble to prevent rapid wear. It can be controlled to some extent by 
 properly locating the link pins, by avoiding too short a link, and 
 by choosing a favorable position for the reverse shaft. 
 
 The Gocch Link. Another form of link motion, known as 
 the Gooch Link, is illustrated in Fig. 43. It has been extensively 
 used on European locomotives, although it is gradually being 
 replaced by a type of valve gear known as the Walschaert, which 
 will be described later. 
 
 The Gooch link has its concave side turned toward the valve 
 instead of toward the eccentric. The radius of curvature of the 
 link is equal to AB, the length of the radius rod. The link is 
 stationary and the link block slides in the link. The engine is 
 reversed by means of the bell-crank lever on the reverse shaft E 
 which shifts the link block instead of the link, as is the case with 
 the Stephenson. The link is suspended from its saddle pin JV1 S 
 which is connected by a rod to the fixed center F, so th^t the link 
 can move forward and back as the eccentricity is changed, or it 
 can pivot about its saddle pin as the eccentrics revolve. 
 
 Since the radius of the link arc is equal to AB, it is apparent 
 that the block can be moved from one end of the link to the other, 
 that is, from full gear " forward " to full gear " back " without 
 moving the point A, which is on the end of the valve rod. The 
 lead then is constant for all positions of the block, and the distri- 
 bution of steam for locomotives is slightly preferable to that 
 
VALVE GEARS 
 
 obtained by the Stephenson j but the gear is more complicated and 
 requires nearly double the distance between shaft and valve stem. 
 
 The variable lead is perhaps a slight advantage to the loco- 
 motive, which is a slow-speed engine in starting, thus requiring 
 but little lead. As the speed increases, and the link is " notched 
 up ", the lead is increased as the cut-off is shortened, and at high 
 speed we have a large lead. With the Gooch link, the lead can be 
 set for the average running speed, and although a little too great 
 for good work at slow speed, it is a matter of small consequence, 
 because the engine runs at slow speed but a very small fraction of 
 the time it is in service, and the loss due to large lead at slow 
 speed is of no consequence whatever in a day's run. 
 
 Several other link motions have been used; but at the present 
 time probably more Stephenson link motions are used than all 
 
 Fig. 43. 
 
 the other forms of reversing gear combined, and when a " link 
 motion " is mentioned, the Stephenson is usually meant unless 
 otherwise specified. 
 
 RADIAL VALVE GEARS. 
 
 In general, it would be desirable to have precisely similar 
 steam distribution at each end of the cylinder, and it would often 
 be of great advantage with an expansion gear like the Stephenson, 
 if the cut-off could be shortened without changing any other event 
 of the stroke. A Stephenson gear can be made to maintain equality 
 of lead for both ends of the cylinder as the cut-off is shortened, 
 but we have seen that in so doing, the lead of- both ends is either 
 
50 
 
 VALVE GEARS 
 
 increased or diminished according as the link is arranged with 
 u open rods " or " crossed rods ". Moreover, the compression is 
 hastened by bringing the link to mid-gear, all of which in many 
 instances is undesirable. 
 
 This disadvantage of the Stephenson link motion lead to the 
 design of the so-called u Radial Valve Gears : ', many of which are 
 so complicated as to be impracticable, but all of which obtain a 
 fairly uniform distribution of steain. 
 
 Fig. 44. 
 
 Hackworth Gear,, The essential features of the Hackworth 
 Gear are indicated in outline in Fig. 44. In this figure, S is the 
 center of the shaft, and the eccentric E is set 180 from the crank 
 SH,r At the right-hand end of the eccentric rod EA, is pivoted a 
 block which slides in a straight, slotted guide. The guide remains 
 stationary while the engiue is running, but can be turned on its' 
 
VALVE GEARS 
 
 51 
 
 axis P, to reverse the engine or change the cut-off. P is a pivot, 
 located on the horizontal through S in such a position that 
 DP = EA. If these two distances are equal, A will coincide 
 vvith P when the crank is at either dead point and the slotted 
 guide may be turned from " full gear forward ", as shown in the 
 figure, through the horizontal position to "full gear backing", as 
 
 Fig 45 
 
 shown by the line BL, without moving the valve. Therefore the 
 leads are constant for all positions of the guide. The valve rod 
 running upward from C, connects with the valve stem which it 
 moves in a straight line. The valve stem is made just long 
 enough to equalize both leads, and if the point C has been properly 
 chosen, the two cut-offs will be very nearly equal for all grades of 
 the gear. 
 
52 VALVE GEARS 
 
 A somewhat better valve action is obtained by slightly curv- 
 ing the slotted guide, with its convex side downward. This gear 
 is sometimes used on marine engines and on small stationary 
 engines. 
 
 Marshall Gear. The most objectionable feature of the Hack- 
 worth gear, is the slotted guide, for the sliding of the block causes 
 considerable friction and wear. The Marshall gear, shown in 
 outline in Fig. 45, is designed to obviate this feature. The point 
 A moves in the desired path by swinging on the rod FA about F 
 as a center. While the engine is running, the lever FP remains 
 stationary, but can be turned on its axis P to reverse the engine, 
 or change the cut-off. The pivot P, is located precisely as in the 
 Hackwwth gear, and the lever FP can be turned from u full gear 
 forward", as show r n in the figure, to " full gear backing ", as shown 
 by the line BP, intermediate positions give different cut-offs as 
 with the Hackworth gear. Since' FA is made equal to FP, the 
 point A will always swing through P, no matter where F may 
 be, and will coincide with P, when the engine is on dead center. 
 The leads therefore will remain constant, as in the preceding case. 
 
 The Marshall gear is sometimes made with C at the right of 
 A on a prolongation of the line EA. In this case if the same 
 kind of valve is to be used, the eccentric, E must move with the 
 crank instead of 180 from it. The Marshall gear is frequently 
 used on marine engines, the one eccentric being simpler than the 
 tw r o required by the Stephenson. 
 
 Joy Gear. Perhaps the most w r idely known, and certainly 
 one of the best radial gears is the Joy, outlined in Fig. 40. It 
 is frequently used on marine engines and on some English loco- 
 motives. No eccentrics are used, the valve motion being taken 
 from C, a point on the connecting rod. II is a fixed pivot sup- 
 ported on the cylinder casting. The lever ED has a block pivoted 
 at A, which slides back and forth in a curved slotted guide. The 
 guide and the lever PF are fastened to the reverse shaft P, and 
 by means of a reverse rod leading off from F, can be turned from 
 full gear forward, as shown, to full gear backing when the pin F 
 moves over to B. Motion is transmitted to the valve stem by 
 means of the radius rod EG. The proportions are such that when 
 the crank is on either dead point, the pivot of block A coincides 
 
VALVE GEARS 
 
 with P, so that the curved guide may then be set in any position 
 without moving the valve; therefore the leads are constant. This 
 gear gives a rapid motion to the valve when opening and closing 
 and a more nearly constant compression than the Stephen son gear, 
 and the cut-off can be made very nearly equal for all grades of 
 the gear. Its many joints cause wear and its position near the 
 crosshead, makes a careful inspection of the crosshead and piston 
 exceedingly- difficult while the engine is running. 
 
 Fig. 46. 
 
 Walschaert Gear. This radial valve gear, although seldom 
 seen in the United States, is the valve mechanism most commonly 
 used on locomotives built on the continent of Europe. Like all 
 other radial gears, it gives constant lead, and a distribution of 
 steam very nearly alike for each end of the cylinder. In this 
 respect it is superior to the Stephenson link, and gives without 
 doubt better economy, but its mechanical construction is compli- 
 cated, and not well adapted to the American type of locomotive. 
 Fig. 47 illustrates this type of gear. S is the center of the driver 
 axle. The crank pin K has forged on its center end an arm KE, 
 on which the pin E is fixed. This arm lies parallel to the plane 
 of the driving wheels, and being fixed to the crank pin, turns with 
 the wheel, allowing the connecting rod to pass between it and the. 
 driving wheels. In this manner the point E moves around S in a 
 circle, and moves the rod EH back and forth just as if it were an 
 eccentric. It is so made that ES is perpendicular to the crank 
 KS, and therefore the action of the pin E is equivalent to an eccen- 
 tric with no angular advance. 
 
VALVE GEARS 
 
 This arm reaching back from the outer end of the crank pin 
 is one of the most objectionable features on the construction, and 
 is sometimes replaced by the regular type of eccentric put on the 
 shaft between the driving wheels. 
 
 The eccentric rod EH causes the box link HP to oscillate on 
 fixed trunions P. This link has a groove curved to a radius equal 
 to GD, the length of the radius rod. A block pivoted at G, on 
 one end of the radius rod, is free to move up or down in this 
 
 TOP OF RAIL 
 
 Fig 47. 
 
 groove. The valve derives its motion from C, a pivot on the float- 
 ing lever CA. Point A receives motion from the crosshead; point 
 D from the eccentric and the curved link ; and a combination of 
 these two imparts motion to C (which can slide only along the 
 dotted line). A bell-crank lever pivoted above the link shifts 
 the mechanism from " full gear forward " when F is moved to B, 
 thus raising G above the link pivot or saddle pin. 
 
 ADJUSTABLE ECCENTRICS. 
 
 The position of an eccentric for a plain slide valve is 90 plus 
 the angular advance ahead of the crank, in the direction in which 
 the engine is to turn. Thus A, Fig. 48, is correctly placed, rela- 
 tive to the crank C, if the engine turns right-handed. For run- 
 ning in the opposite direction, the position of the eccentric is at 
 D. Some engines are provided with a reversing mechanism which 
 causes the eccentric to shift from A to D, either along the arc 
 
VALVE GEARS 
 
 ABD, or along the straight line AED. Sucli engines provide, 
 not only for reversing, but for changing the cut-off as well If 
 the eccentric moves on the arc to OB, the angular advance is 
 
 Fig. 10. 
 
 increased and all the events of the stroke are hastened as well as 
 the cut-off, but the travel of the valve is not changed. 
 
 Zeuner's diagram, Fig. 49, is lettered to correspond with Fig. 
 48, and shows the effect of changing the angular advance from 
 FOA to FOB. If OK rep- 
 resents the lap, the crank 
 angle at cut-off will decrease 
 from I1OK to IIOL, and the 
 lead will increase very much, 
 viz., from GF to PIF. If the 
 eccentric is shifted on the 
 straight line to E (Fig. 48), 
 a different valve motion will 
 result. The angular advance 
 is increased as before, so that 
 all the events are hastened, 
 but the eccentricity is now 
 only OE instead of OB and Fio . 49 
 
 the valve travel is conse- 
 quently reduced. Zeuner's diagram for this case, Fig. 50, shows a 
 decrease in crank angle at cut-off from IOM to ION, and no 
 
nr, 
 
 VALVE GEARS 
 
 change in the lead IF. Let 
 
 o 
 
 us consider the eccentric posi- 
 tion OA, Fig. 48. In .this 
 position OI represents the 
 displacement of the valve 
 from mid-position when the 
 engine is on center. If the 
 eccentric moves to OB, the 
 displacement will be OM, 
 which is greater, showing an 
 increase in lead equal to IM, 
 but if the position is OE in. 
 
 stead of OB, the displace- 
 Tig. 50. 
 
 ment from mid-position will 
 
 be OI as before. It is evident that the eccentric can move on the 
 straight line from A to I), without changing the lead, while the 
 decreased valve travel will result hi an earlier cut-off. If the 
 
 . 5L 
 
 shifting eccentric is to be used for an automatic cut-off, as in 
 the various types of fly-wheel governor, the curved patli is not 
 
VALVE GEAUS 
 
 desirable on account of excessive lead at short cut-off, but if it is 
 to bo used only as a means of reversing, it is preferable to the 
 straight line. 
 
 All fly-wheel governors operate by shifting the eccentric, 
 either to change the angular advance, the travel of the valve, or 
 both. Fig. 51 illustrates the principle of a governor arranged to 
 give decreasing lead, but as these mechanisms are described in 
 the Steam Engine Part I, under the head of governors, a further 
 discussion will not be given here. 
 
 The device shown in Fig. 52 is often used for reversing 
 engines of small launches. The eccentric E is loose on the shaft 
 between a fixed collar G, and a hand wheel II. A stud project- 
 ing from the eccentric passes through a curved slot in the disc of 
 the wheel, and can be clamped by a hand nut F. When running 
 forward with the crank at C, 
 the center of the eccentric is at 
 A, and the nut clamped at F. 
 To reverse, steam is shut off, 
 and when the engine stops, the 
 nut F is loosened, and then 
 moved to B and clamped; or 
 after F is loosened, the wheel, 
 shaft, crank and propeller are 
 turned over by hand until B 
 
 strikes the stud at F, where it is 
 clamped. The engine will then 
 run astern. 
 
 To study the application of 
 the Zeuner diagram to this 
 form of mechanism turn again 
 to Fig. 51. If OA is the de- 
 sired eccentricity for a normal 
 position of the governor, the Fig 52> 
 
 perpendicular distance of A 
 
 from OF is made equal to the lap OI, plus the desired lead. Pivot 
 D is then located equally distant from A and I. Zeuner's dia- 
 gram for this gear, drawn to an enlarged scale, is shown in Fig. 58. 
 The angular advance FOB is laid back toward ()( 1 . OB is the 
 
58 
 
 VALVE GEARS 
 
 maximum eccentricity; OI, the lap or the desired least eccen- 
 tricity. An arc, with proper radius, described through B and I 
 shows the path of the eccentric. If the eccentric moves in to A, 
 the crank angle at cut-off is decreased from COD to COE, and 
 the lead decreased from FI to GI. A slight decrease in lead is 
 
 Fig. 53. 
 
 not objectionable, since the speed is not allowed to increase more 
 than two or three per cent; and further, as the lead increases, com- 
 pression decreases, so that one influence helps to counteract the other. 
 The decrease in maximum port opening from BII to AK is un- 
 avoidable, but it is permissible, since it occurs only when the load 
 decreases, and when less steam should be admitted to the cylinder. 
 
 DOUBLE VALVE GEARS. 
 
 It has been shown in the preceding discussion, that a plain 
 slide valve under the control of a gear that gives a variable 
 cut-off, such as a shifting eccentric or a link motion, will not 
 give a satisfactory distribution of steam at short cut-off owing 
 to execessive compression, variable lead, or early release. These 
 difficulties are overcome in a measure by the use of the radial 
 gear ; and also by the use of two valves instead of one. The main 
 
VALVE GEARS 
 
 59 
 
 valve controls admission, re- 
 lease, and compression ; the 
 other, called the cut-off valve, 
 regulates the cut-off only, 
 which may be changed with- 
 out in any way affecting the 
 other events of the stroke. 
 This cut-off valve may be 
 placed in a separate valve 
 chest, or it may be placed on 
 the back of the main valve. 
 
 Meyer Valve. The most 
 common form of double valve 
 gear is the Meyer Yalve, Fig. 
 54. The cut-off valve is made 
 in two parts and works on the 
 back of the main valve. The 
 two parts are connected to a 
 valve spindle with a right- 
 and left-hand thread, so that 
 their positions may be altered 
 by rotating the valve spindle. 
 
 A swivel joint is usually 
 fitted in the valve-spindle be- 
 tween the steam chest and the 
 head of the valve rod, and the 
 valve spindle prolonged into 
 a tail rod which projects 
 through a stuffing box on the 
 head of the steam chest. See 
 Fig. 55. The end of this tail 
 rod is square in section and 
 reciprocates through a small 
 hand wheel by means of which 
 it can be rotated while the en- 
 gine is running, whatever the 
 position of the valve may be. 
 
60 
 
 VALVE GEARS 
 
 Each valve is under the control of a separate eccentric. The 
 eccentric moving the main valve is usually fixed, while the cut-off 
 valve eccentric may be under the control of a governor. Since 
 a slight compression is desired, the main valve is set to give late 
 cut-off. This will give late release and late compression, and allow 
 a wide range of cut-off for the cut-off valve. With this gear, lead, 
 release, and compression are entirely independent of the ratio of 
 expansion, and the cut-off is much sharper, because the cut-off 
 valve, when closing the ports, is always moving in a direction 
 
 Fig. 55. 
 
 opposite to that of the main valve. The valve may be designed 
 by means of Zeuner's diagrams. 
 
 Design of a Meyer Valve. Let us design a Meyer Yalve 
 having an eccentricity of 2 inches. Let the eccentricity of the 
 cut-off valve be 2J inches and the relative travel of the cut-off 
 valve in relation to the main valve be 3 inches. This will make 
 the relative motion of the cut-off valve equivalent to the .travel of 
 a plain slide valve with an eccentricity of 1-J inches. Let the out- 
 side lap on the main valve be | inch, the lead ^ inch, the com- 
 pression 95 per cent of the stroke, and let the ratio of the length 
 of the crank to connecting rod be six. 
 
VALVE GEARS 
 
 61 
 
 In Fig. 56 draw XOY equal to 4 inches, the main valve 
 travel. Lay off YD = 95 per cent of 4 = 3.8 inches, and with 
 a radius of 12 inches, and the center on YX produced draw the arc 
 DH.K. H.K.O is the crank position at compression. C.K.O, the 
 crank position at cut-off, is found in a similar manner. Lay off OI 
 equal to the lap plus the lead, and draw the valve circle for the 
 main valve through I and O with a diameter equal to its eccen- 
 tricity of 2 inches. To do this take a radius equal to 1 inch, and 
 draw arcs from I and O as centers that shall intersect at B. B is 
 
 C.K. 
 
 H, 
 
 Fig. 56. 
 
 tne center of the valve circle and QBE is the eccentricity, 2 inches. 
 With E as a center, and with a radius equal to half the relative 
 travel of the cut-off valve (in this case 1^ inches), draw an arc. 
 With O as a center and with a radius equal to 2^ inches, the 
 eccentricity of the cut-off valve, draw another arc intersecting the 
 first one at F. On OF as a diameter construct a valve circle. 
 This valve Vill represent the absolute motion of the cut-off valve, 
 independent of the motion of the main valve. This circle then 
 will show the displacements of the cut-off valve from the center of 
 the sterna chest. With E as a center and with a radius equal to 
 FO draw an arc, and with O as a center and with a radius equal 
 
62 
 
 VALVE GEAKS 
 
 to EF draw another arc intersecting the first at G. On OG as a 
 diameter construct a valve circle. This circle will then represent 
 a travel of the cut-off valve moving on the main valve. That is, 
 ,it will represent the displacements of the cut-off valve from the 
 center of the main valve. This circle is not, properly speaking, 
 
 MIN.CUT 
 OFF 15 
 
 H 
 
 Fig. 57. 
 
 a valve circle and OG is not an eccentricity, but simply represents 
 the relative motion of the two valves. This can be proved by 
 analytical geometry, but an inspection of the figure shows that 
 this must be true. 
 
 Draw the crank line OC at any position, cutting the valve 
 circles at a and 1> and c. O^ represents the absolute displace- 
 ment of the cut-off, valve, that is, from the center of the steam 
 chest and Oc represents the displacement of the main valve. The 
 
VALVE GEARS 68 
 
 relative displacement of the cut-off valve, that is, from the center 
 of the main valve, will be the difference between Oc and O#, since 
 both valves are moving in the same direction. By careful meas- 
 urement it will be found that O = Oc - O^, and any ara as ()/> 
 on the auxiliary circle OIG will correctly represent the displace- 
 ment of the cut-off valve from the center of the main valve at the 
 corresponding crank angle. 
 
 Fig. 57 shows the crank angle at head -end compression H.K., 
 and at crank-end compression O.K., the main valve circle, and the 
 auxiliary circle which are transferred from Fig. 56. The con- 
 struction lines and all lines not essential to the figure are omitted 
 to avoid confusion. 
 
 Lay off on Fig. 57, OI equal to the outside lap J inch and 
 draw the head-end lap circle H.E.O. It will intersect the valve 
 circle for the main valve at L and M. Through L draw the crank 
 position at admission (head-end) H.A. and the crank position at 
 cut-off through M. This gives the greatest possible cut-off. The 
 cut-off valve may be set to give a much earlier cut-off than this, 
 but of course a later setting would be of no avail for the port 
 would be closed by the main valve at this angle. The crank line 
 OMH cuts the auxiliary circle at Jtf , so that ON (1|J- inches) is 
 the clearance of the cut-off valve. That is, the edge of the cut-off 
 valve must be set 1 if inches from the edge of the main valve port 
 in order to cut-off at this crank angle. The full lines of Fig. 54 
 show the cut-off valve placed in this position. 
 
 The intersection of H.K.O with the lower valve circle, gives 
 the inside lap at the head end of the cylinder. This line comes so 
 nearly tangent to the valve circle that the intersection can be 
 determined only by dropping a perpendicular to H.K.O. frqm E 1 . 
 This cuts the circle at P and OF ^ inch equals the head-end 
 inside lap, and II.E.L represents the corresponding lap circle. 
 
 The crank-end angle at compression is O.K. which cuts the 
 upper valve circle at N', giving an inside lap for the crank end of 
 OK'=s=-JJ inch. To make this intersection more apparent the 
 perpendicular can be drawn from E as previously explained. 
 
 Suppose that it was required that the minimum cut-off should 
 be 15 per cent. Find the crank position at 15 per cent of the stroke 
 in the same manner as the crank position was found at compression, 
 
64 VALVE GEARS 
 
 Produce this line through O until it cuts the auxiliary circle at S. 
 Then OS = ||- inch = the required lap for the cut-off valve in 
 order to cut-off at 15 per cent of the stroke. The dotted lines in 
 Fig. 54. show the cut-off valve drawn in this position. 
 
 For a valve of this sort, the cylinder port should be 1J inches 
 wide and the valve port 1 inch wide. Fig. 54 shows this valve laid 
 out to scale, but as this process is in all respects similar to that 
 described for laying out a plain slide valve, it will not be described 
 in detail. 
 
 DROP CUT-OFF GEARS. 
 
 \The ordinary slide valve controls eight different events of the 
 stroke, that is, admission, cut-off, release, and compression for both 
 ends of the cylinder. A change in the setting of a plain slide 
 valve that affects any one event on the crank end, let us say, will 
 also change to a greater or less degree every other event of the 
 stroke, on the head end as well as on the crank end; so that in 
 setting a slide valve the desired position for one event must 
 usually be sacrificed in order to make the others less objectionable. 
 In order to provide a better distribution of steam than is pos- 
 sible with a single valve, some engines have four valves, two at 
 each end of the cylinder. In horizontal engines, two are placed 
 above the center line of cylinder and two below. The upper are 
 for admission and cut-off, the lower for release and compres- 
 sion. Since each valve controls but two events, a very satisfactory 
 adjustment can be made and the extra complication and cost for 
 large engines are more than overbalanced by the advantages 
 gained, viz.: A very much better distribution of steam, short 
 steam passages arid small clearances, separate ports for the admis- 
 sion of hot steam and the exhaust of the same steam after expan- 
 sion when its temperature has fallen, and finally the possibility of 
 opening and closing the ports very rapidly, thus preventing wire- 
 drawing. The small clearances, short ports and separate admis- 
 sion and exhaust materially reduce the cylinder condensation, and 
 thus effect a large saving in the steam consumption. 
 
 When four valves are used for high speeds, the motions of 
 all must be positive, that is, they must be connected directly to 
 some mechanism that either pushes or pulls them through their 
 entire motion, but for speeds up to 100 revolutions or so a disen- 
 
VALVE GEARS 
 
 65 
 
 gaging mechanism may be used, and the valves may shut of them- 
 selves, either by virtue of their weight or by means of springs or 
 dashpots. The valve is usually opened by means of links or rods, 
 moved by an eccentric, and at the proper point of cut-off the 
 rod is disengaged from the valve which drops shut, hence the term 
 " drop cut-off " gears. 
 
 Reynolds=Corliss Gear. The most widely known drop cut- 
 off gear is the Eeyn olds- Corliss, shown in Figs. 58 and 59; it is 
 often referred to as the Reynolds hook-releasing gear. An eccen- 
 tric on the main shaft gives an oscillating motion to a circular disc 
 
 Fig. 68. 
 
 called the wrist plate, pivoted at the center of the cylinder. It 
 transmits motion to each of the four valves through adjustable 
 links known as steam rods or exhaust rods, according to whether 
 they move the admission or exhaust valves. 
 
 The valves which are shown in section in Fig. 60 oscillate on 
 cylindrical seats, and the position of the rods is so determined that 
 they give a rapid motion to the valve when opening or closing, and 
 hold it nearly stationary when either opened or closed. 
 
 The Reynolds hook is shown in detail in Fig. 59. The steam 
 arm is keyed to the valve spindle which passes loosely through a 
 bracket on which the bell-crank lever turns, and the spindle is 
 packed to make a steam-tight joint where it enters the cylinder. 
 Motion of the steam rod toward the right will turn the fell-cran& 
 lever and raise the hook stud. The hook (from which the gear 
 derives its name) pivoted on thjs stud, has at one end a hard- 
 
66 
 
 VALVE GEARS 
 
VALVE GEARS 67 
 
 ened steel die- with sharp, square edges, and at the other end, 
 a small steel block with a rounded face. As the hook rises, the 
 hook die engages the stud die which is fastened to the steam arm, 
 and one end of the steam arm is thus raised. This turns the 
 valve in its seat and admits steam. As the hook, continues to 
 rise, its stud moves in an arc above the valve spindle, and the 
 round-faced block at its left-hand end strikes the knock-off cam 
 which causes the hook to turn about its stud and disengage the 
 hook die from the stud die. In raising the steam Mrm, the dash- 
 pot rod also is raised and a partial vacuum is created in the dash- 
 pot. As soon, therefore, as the dies become disengaged, the 
 dashpot rod quickly drops under the force of this vacuum, thus 
 turning the steam arm and closing the valve. The striking of the 
 left-hand end of the hook against the knock-off cam determines 
 the point of cut-off, by releasing the valve at that instant. 
 
 This cam is a part of the knock-off lever to which the governor 
 cam rod is fastened. Any action of the governor which would 
 cause the cam rod to move toward the right would cause this 
 knock-off lever to turn on its axis, the steam arm, and conse- 
 quently lower the position of the knock-off cam. This would 
 cause an earlier contact between the cam arid end of hook, and 
 consequently an earlier cut-off. By lengthening or shortening 
 the governor cam rod, the point of cut-off can be adjusted to suit 
 the engine load without changing the speed. 
 
 There is a limit to this adjustment, for it can be shown that 
 a Corliss gear operated by a single eccentric cannot be arranged to 
 cut-off later than half stroke. Suppose the eccentric is set just 90 
 ahead of the crank. Then it will reach its extreme position just 
 as the piston gets to half stroke. If by that time the hook which 
 was rising and opening the admission valve, has not yet struck 
 the knock-off cam, it cannot strike it at all, for any further motion 
 will cause the hook to descend to its original position, that is its 
 position at the beginning of the stroke; the hook will not disen- 
 gage from the steam arm stud at all and the bell crank will 
 return, closing the valve in the same manner in which it opened 
 it. Cut-off will then take place near the end of the stroke, but it 
 will not be the sharp cut-off produced by the sudden drop when 
 the dies are disengaged. 
 
VALVE GEARS 
 
 If the eccentric were set less than 90 ahead of the crank, the 
 cut-off could be arranged to occur later than half stroke, but this 
 is decidedly impracticable, for with such a position of the eccen- 
 tric the action of the valves at release and compression is spoiled. 
 When it is necessary to cut-off later than half-stroke, as some- 
 times happens on low-pressure cylinders of compound engines, it 
 may be arranged for by means of two eccentrics, one set more 
 than 90 ahead of the crank to operate the exhaust valves, and one 
 less than 90 ahead to operate the admission valves.' 2sj 
 
 The safety cam shown in Fig. 
 59 is an important part of a Cor- 
 liss gear. If for any reason the 
 engine governor should fail to 
 act, due, for instance, to the 
 breaking of its driving belt, the 
 governor would drop to its low- 
 est position, supply more steam 
 to the engine and allow it to 
 run away. The safety cam pre- 
 vents this by moving so far to 
 the right that it strikes the hook 
 when it descends to pick up the 
 steam arm. The hook is conse- 
 quently turned toward the right 
 and then lifted without engaging 
 the stud die; the valve conse- 
 quently remains closed and the engine stops. 
 
 Brown Releasing Gear. In addition to the Reynolds hook, 
 several other devices are in use for opening and releasing Corliss 
 admission valves. Among them the Brown releasing gear shown 
 in Fig. 01 may be noted. The steam rod and dashpot rod are 
 arranged much the same as in the Reynolds gear. The governor 
 cam rod operates a plate cam having a curved slot so shaped that it 
 takes the place of both the knock-off and the safety cam of Fig. 59. 
 The steam arm is keyed to the valve spindle and carries at its 
 lower end a steel die which is free to slip up and down a small 
 amount. The part of this gear corresponding to the Reynolds 
 bell crank becomes a straight rocker pivoted at its middle; 
 
 Fig. 60. 
 
VALVE GEARS 
 
 and the part corresponding to th.e Reynolds Look has at one end 
 a die which engages the die of the steam arm, and at its other 
 end a roller running in the curved cam slot. This hook is really 
 a bell-crank lever with arms that are not in the same plane. The 
 bearing on which it turns is carried on the lower end of the 
 rocker, and therefore is equivalent to a movable pivot similar to 
 the hook stud of the Reynolds gear. 
 
 VALVE 
 SP1NDL 
 
 CAN* V STEAM ROD 
 
 GOVERNOR CAM ROD 
 
 Fig. 61. 
 
 In the position shown the dies are engaged. Motion of the 
 steam rod toward the right will move the lower end of the rocker 
 toward the left, and consequently turn the valve spindle in a right- 
 handed direction. This will open the valve and at the same time 
 raise the dashpot rod. Mean while, the roller is moving toward 
 the left in a circular part of the cam slot, the center of which is 
 at the center of the valve spindle. This causes the steam arm and 
 the bell-crank lever, which has the roller at one end, to move in 
 such a way that there is no relative motion between them. As 
 soon, however, as the roller comes to the point where it is forced 
 to move out of this circular path and move farther from the valve 
 spindle, both arms of the bell-crank lever are turned downward, 
 
70 
 
 VALVE GEARS 
 
 the dies become disengaged, and the dashpot closes the valve. 
 The slight up-and-down motion of the steam-arm die allows it to 
 rise while the hook die passes underneath when returning to re- 
 engage for the next stroke. The makers claim that this gear per- 
 mits a much higher speed than is possible with other Corliss gears. 
 Greene Gear. Another well-known drop cut-off gear is the 
 Greene, shown in Fig. 62. The valves are of the gridiron type, 
 sliding on horizontal seats, the admission valves parallel to, and 
 the exhaust valves at right angles to the axis of the cylinder and 
 just below it. AA are rock shafts turning in fixed bearings. 
 
 Fig. 62. 
 
 BE are the admission valve stems. C is a slide bar, receiving a 
 reciprocating motion from an eccentric. TT are tappets connected 
 to the slide bar. They move to and fro with the slide bar and can 
 also move independently up and down. They are made fast at 
 their lower end to the gauge plate D which slides through the 
 guide E. The guide E is made fast to the governor rod F and 
 through this means can be raised or lowered, thus regulating the 
 height of the tappets. 
 
 As the slide bar moves toward the right, the right-hand 
 tappet is brought into contact with the toe of the rocker, causing 
 it to turn on its bearings and move the rock lever and the valve 
 stem B toward the right, thus opening the admission valve. Since 
 
VALVE GEARS 
 
 71 
 
 the tappet moves in a horizontal direction while the toe of the 
 rocker moves in an arc, it will be seen at once that they will soon 
 become disengaged and release the valve which is at once closed by 
 a dash pot not shown in the figure. If the governor raises the 
 tappets, cut-off will be later. A nut at the bottom of the governor 
 rod allows a proper adjustment of the guide and gauge plate. As 
 the slide plate C moves toward the right, the left-hand tappet comes 
 in contact with the heel of the left-hand rocker, both being beveled, 
 it rises in its socket allowing the tappet to pass under. It then 
 falls by its own w r eight and is ready to engage the tappet on its 
 
 Fig. 63. 
 
 return and open the valve. In this gear the disengagement of the 
 valve throws no load whatever on the governor which is a distinct 
 advantage over the Corliss gear. The action of the exhaust valves 
 is not shown in the cut. 
 
 The Sulzer Gear is a drop cut-off widely used in Europe. 
 The valves are of the poppet type, lifting straight from conical 
 seats, so that there is no friction. They are usually placed verti- 
 cally above and below the cylinder axis and are operated by eccen- 
 trics from a shaft geared to the main shaft. The admission valves 
 are lifted from their seats by suitable levers, then released by a 
 
72 
 
 VALVE GEARS 
 
 device equivalent in action to the Keynolds hook and are quickly 
 closed by the action of springs. 
 
 The exhaust valves of all drop cut-off gears are comparatively 
 simple in their operation and both in opening and closing they are 
 moved by the direct action of the exhaust rods. 
 
 A common form of vacuum dashpot for closing admission 
 valves is shown in Fig. G3. The rod coming down from the steam 
 arm makes a ball-and-socket joint with the dashpot piston. The 
 dashpot is often let down into the engine frame as shown. When 
 lifted, the piston produces a partial vacuum underneath it so that 
 it tends to drop quickly as soon as the valve gear is released. On 
 some of the largest modern engines where the valves are very 
 heavy, steam-loaded dashpots are used; that is, the dashpot piston 
 has steam pressure on one side, and an air cushion on the other 
 prevents it from striking the bottom of the dashpot. 
 
 Corliss Valve Setting. The setting of a Corliss valve gear 
 
 is a much longer process than 
 the setting of a plain slide valve, 
 but is nevertheless a compar- 
 atively simple matter, for the 
 various adjustments are nearly 
 all independent of one another. 
 In gears like that shown in Fig. 
 58 the length of both the eccen- 
 tric rod and carrier rod are' 
 unusually adjustable, and the 
 former should be of such length 
 that the carrier arm swings equai distances on each side of a verti- 
 cal line through its pivot, and the carrier rod should be adjusted 
 until the wrist plate oscillates symmetrically about a vertical line 
 through its pivot Nearly all Corliss engines have one mark on 
 the wrist plate hub and three on the wrist plate stand, as shown 
 in Fig. 64, and the wrist plate should swing so that A, the mark 
 it carries, moves from C to D, but not beyond either one. When 
 A is in line with B, the wrist plate is in mid-position. The valves 
 are then not in their exact mid-position, but it is customary to regard 
 them as being in mid-position, and to speak of the lap as the amount 
 the valve covers the port when the wrist plate is in mid-position. 
 
 WRIST PLATE 
 STAND 
 
 Fig 64. 
 
VALVE GEARS 
 
 To set the valves, remove the bonnets or covers of the valve 
 chambers on the side opposite the gear. The ends of the valves 
 are circular, but inside their cross-section is as shown in Fig. 65. 
 On the end, in line with the finished edge of the valve, and on 
 the seat in line with the edge of the steam port, are marks as 
 shown in Fig. 65. When these marks coincide, the valve is either 
 just opening or just closing, and when in any other position, the 
 amount of opening or the amount by which the port is closed is 
 shown directly by the distance between the marks. Block the 
 wrist plate in mid-position, hook up the admission valves and 
 adjust the length of the steam rods by means of the right and left 
 threads provided for the purpose, until the ports are covered by 
 the amount of lap indicated in the following table opposite the 
 given size of engine. 
 
 Dia. of Cyl. 
 in inches. 
 
 12 
 14 to 16 
 
 18 to 22 
 24 to 28 
 30 to 36 
 36 to 42 
 
 Steam Lap 
 in inches. 
 
 Exhaust Clearance - 
 in inches. 
 
 i 
 A 
 
 7V 
 TrV 
 
 TV 
 
 A 
 
 Next adjust .the exhaust rods until the exhaust ports are open 
 an amount equal to the clear- 
 ance given in the above table. 
 Set the engine on its head- 
 end dead point, hook the car- 
 rier rod onto the wrist plate 
 and in the direction in which 
 the engine is to run, turn the 
 eccentric enough to open the MARKS 
 head-end admission valve by 
 a proper amount of lead ; 
 then the eccentric will be 90 
 plus the angular advance 
 ahead of the crank. The 
 proper amount of lead will depend upon both the design of the gear 
 and the speed at which the engine is to run ; and may vary from 
 y for small engines to as much as --%" or ^" for large and higher- 
 
 Fig. 65. 
 
74 VALVE GEARS 
 
 speed engines. When the proper amount of lead has been obtained, 
 fasten the eccentric on the shaft by means of the set screw and 
 make sure by trial that the wrist plate moves to its extremes of 
 travel. The dash pot rods must be adjusted so that when the dash- 
 pot piston is at its lowest position, the hooks (see Fig. 59) descend 
 just far enough for the hook dies to snap over the stud dies with 
 about -fa" to rV" to s p are ? depending on the size of the gear. 
 
 To adjust and equalize the cut-off, lift the governor to about 
 the position that it will occupy when running at normal speed, 
 and put a block under the collar to hold it in this position. First 
 set the double lever at the right of the governor cam rods so that 
 it makes approximately equal angles with each rod, and then turn 
 the engine over by hand until the piston has moved to the desired 
 point of cut-off. Adjust the proper cam rod until the knock-off 
 cam strikes the hook and allows the valve to close, then turn 
 the engine to the point of cut-off on the other stroke and adjust the 
 other cam rod in a similar manner. Now set the governor in the 
 lowest position to which it could fall if there were no load on 
 the engine, and set the safety cams so that in this position the hook 
 cannot open the valve. A latch is provided on which the governor 
 can be supported slightly above its lowest position so that the valves 
 can be opened by the hook when starting the engine. As soon as 
 the engine speeds up this latch must be moved aside, so that if the 
 governor fails to act, it can drop to its lowest point, otherwise this 
 latch would hold it just high enough so that the safety cams 
 could not act. 
 
 When Corliss gears are set as here described, the position of 
 the eccentric may not be quite right, due to an incorrect estimate 
 of the amount of lead required. The error is likely to produce 
 faulty release and compression as well as poor admission, but it 
 cannot be very serious, and the engine will turn over with its own 
 steam, so that indicator diagrams may be taken. The final adjust- 
 ments can then be determined from an examination of the diagrams. 
 
STEAM ENGINE INDICATORS 
 
 A most important question concerning a steam engine is, 
 "What is its horse-power? " or u How much work will it do in a 
 given time ? " 
 
 . Work is defined as pressure, force, or resistance multiplied by 
 the distance through which it acts. 
 
 Power is work done in a specified time. 
 
 In the steam engine, steam is the agent by means of which 
 heat is transformed into mechanical work. It is the heat in the 
 steam that does the work, not the steam itself. 
 
 Work is obtained from the heat in steam by confining it in a 
 closed cylinder which is fitted with a piston and a piston-rod. 
 Steam is admitted at one side of the piston while the other is open 
 to the atmosphere or in communication with a condenser. The 
 pressure of steam, usually 75 to 150 pounds per square inch, forces 
 the piston to the other end of the cylinder, driving out the low- 
 pressure steam in front of it. When it arrives at the other end, 
 steam is admitted to that end and the piston is driven back. 
 
 The piston moves because the pressure on one side is greater 
 than that on the other. 
 
 In order to move the piston, work must be performed. The 
 amount of work is easily found, since work equals total pressure 
 multiplied by the distance through which the piston moves. 
 
 Suppose a piston is 2 square feet in area and steam at a 
 pressure of 64.7 pounds per square inch acts on it during the 
 entire stroke of 4 feet ; the other side of the piston being in com- 
 munication with the atmosphere. The total pressure is then 
 2 X 144 X 64.7 = 18,633.6 pounds. If this pressure acts 
 through 4 feet it is evident from the definition that the work done 
 per stroke will be, 
 
 18,633.6 X 4 = 74,534.4 foot-pounds. 
 
INDICATORS. 
 
 Another method is as follows: The pressure on the above 
 piston is 64.7 X 144 = 9316.8 pounds per square foot. The 
 volume swept by the piston during one stroke is 2 X 4 z= 8 cubic 
 feet. If we multiply the pressure per square foot by the volume 
 or 9316,8 by 8, we get 74534.4 foot pounds, the same result as, 
 before. Thus we see that work equals unit pressure multiplied 
 by volume. 
 
 Let P -- pressure ou the piston in pounds per square foot. 
 p =: pressure on the piston in pounds j^er square inch. 
 A = area of piston in square inches. 
 L = length of stroke in feet. 
 
 V volume swept by piston in one stroke in cubic feet. 
 W = work done in foot-pounds. 
 Then from the above example, 
 
 Work = unit pressure multiplied by volume, 
 or, W = P X V 
 
 It is evident that P = 144 , and V = -J^_, X L. 
 
 144 
 
 Then we have these expressions for work, 
 
 W = P X V = 144 p X V = 144 p X A x L = p L A. 
 
 Suppose steam is admitted to the cylinder during the whole 
 stroke, as in the above example, that is, one end of the cylinder is 
 in communication with the boiler. The other end is open to the 
 atmosphere. If we draw two lines at right angles to each other, 
 as O Y and O X in Fig. 1, the volume of steam for any position 
 may be represented by some distance measured on the line O X. 
 Similarly the pressure of the steam at any position of the piston 
 may be represented by the length of a vertical line parallel to the 
 line O Y. 
 
 In the above example, the area of the piston was 2 square 
 feet, the length of stroke 4 feet and the pressure by gage 50 
 pounds. Then we let O A =. the atmospheric pressure =14.7 
 pounds. At the beginning of the stroke the pressure (absolute) 
 is 14.7 _|_ 50 = 64.7 pounds, represented by the distance O B, or 
 A B =- 50 pounds pressure. When the piston has passed through 
 i of the stroke it is represented as the point 1, or B 1 is the 
 volume swept through when the piston has completed ^ of the 
 
OTDICATOBS. 
 
 stroke. At this point the pressure is also 64.7 pounds as repre- 
 sented at 1'. Similarly, when the piston is at 2, 3, and 4 the cor- 
 responding pressure is 2', 3', 4'. Since the pressure is constant 
 the line B D is parallel to O X. We see from the above that 50 
 pounds is the net pressure acting on the piston during the stroke, 
 and is represented by A B and lines parallel to it. The volumes 
 are represented by the horizontal lino A C. Then since W = P 
 X V it also equals O B X O X which is evidently the area of the 
 rectangle O B D X. The area of the rectangle O B D X is pro- 
 portional to the work done by the steam. 
 
 In Fig. 1, one inch on the line O Y = 40 pounds, then O B 
 is 1.6175 inches long 
 since it represents 64.7 
 pounds. Similarly O A 
 must be .3675 inch since 
 it represents 14.7 
 pounds. The line A C 
 is 2 inches long; then 
 referring to the preced- 
 ing example, one inch 
 in length = | = 4 cubic 
 feet. 
 
 Since the rectangle 
 O B D X is 1.6175 by 
 
 Fig. 1. 
 
 2 inches, the area is 3.235 square inches. But one inch in height 
 
 equals 40 pounds pressure and one inch in length equals 2 cubic 
 
 feet. Then p V 40 X 3.235 X 4 = 517.6 foot-pounds and, 
 
 W = 144 p V --= 517.6 X 144 = 74,534.4 foot-pounds. 
 
 In the above cylinder the pressure acting on one side of the 
 piston was 64.7 pounds per square inch. There was also a press- 
 ure of 14.7 pounds per square inch (the atmospheric pressure) act- 
 ing in the opposite direction. Then the work done against the 
 steam pressure is represented by the area O A C X arid is equal 
 to 144 p V = 144 X 14.7 X 8 = 16934.4 foot-pounds. Then 
 since O B D X represents the total work done on one side of the 
 piston and O A C X represents the work done against the piston 
 the difference A B D C represents the net work. This net work 
 is represented by the shaded area. Also if the amount of work done 
 
INDICATORS. 
 
 on the piston is 74,534.4 foot-pounds and the work done against 
 the piston is 16,934.4 foot-pounds, the net work is the difference, 
 or 57,600 foot-pounds. 
 
 In this theoretical discussion the same result may be obtained 
 by subtracting the atmospheric pressure or back pressure from the 
 absolute initial pressure and using the difference as the value of 
 p. This value of p is called the mean effective pressure. 
 
 Then 64.7 -- 14.7 = 50 and 
 
 W = 144 p V = 144 X 50 X 8 = 57,600 foot-pounds. 
 
 The area is proportional to the work done whatever the 
 shape may be ; provided the line B D represents the relation 
 between pressures and volumes on the steam side of the piston 
 and the lower line A C represents the relation between pressures 
 and volumes on the exhaust side. If the engine is of the con- 
 densing type the line A C will be nearer O X, which is the line 
 representing absolute vacuum. 
 
 Whatever the shape of the diagram, the area is equal to the 
 area of a rectangle of the same length and a height equal to the 
 mean height, or mean ordinate as it is called. The mean ordinate 
 represents the mean or average net pressure on the steam side of 
 the piston. Then we can follow these rules in finding the work 
 of the steam from the diagram. 
 
 Multiply the area in square inches by the scale of pressures, 
 by the scale of volumes and by 144, or ; 
 
 Multiply the length of the mean ordinate by the scale of press- 
 ures a by the length of stroke, and this product by the area of the 
 piston in square inches. 
 
 Example : The area of a diagram A B D C like that of Fig. 
 1 is 6.3 square inches and its length is 3 inches. The scale of 
 pressure is 30 pounds per inch and the scale of volumes is 1.99985 
 cubic feet to the inch. If the piston is 20 inches in diameter and 
 the length of stroke 2-| feet, what is the work done per stroke ? 
 
 Solution : 
 W = area of diagram X scale of pressures X scale of volumes X 144. 
 
 = 6.3 X 30 X 1.99985 X 144 = 54,428 foot-pounds. 
 W = mean ordinate X scale of pressures X area of piston X 
 
 length of stroke, 
 =* 2.1 X 30 X 314.159 X 2f =? 54,428 foot-pounds. 
 
INDICATORS. 
 
 Thus we see that we get the same result by both rules. The 
 latter is the more common method because the mean ordinate is 
 easily found and the scale of volumes seldom considered. 
 
 In our consideration of Fig. 1, steam was admitted to the 
 cylinder during the entire stroke. In modern engines this method 
 is rarely used; instead, steam is admitted during part of the 
 stroke then the communication to the boiler is cut off, and the 
 steam in the cylinder allowed to expand, as the piston moves for- 
 ward, until it fills the entire volume of the cylinder. This is rep- 
 resented graphically in Fig. 2. 
 
 Steam is admitted to the cylinder until the piston reaches 
 the point 2 which repre- 
 sents one-half the volume 
 of the cylinder. Then the 
 cylinder is half full of 
 steam, that is, it contains 
 | = 4 cubic feet. The 
 four cubic feet of steam 
 expand until they fill the 
 cylinder. Since there is 
 the same weight of steam 
 present at every point in 
 the stroke and the volume 
 continued to increase, the 
 pressure must diminish. 
 
 This is shown in Fig. 2. The line B 2' is horizontal because the 
 pressure remains constant to the point of cut off. Then the 
 pressure begins to fall as is represented by the curved line 2' E. 
 This curve is nearly an equilateral hyperbola. 
 
 From Fig, 1 we know that the area B 2' 2 A is proportional 
 to the work done while the piston moves from A to 2 or during 
 the first half of the stroke. If we use the same data as we did in 
 Fig. 1, the work done must be one-half the work done in the first 
 
 case, or ~Z^!?_ = 28800 foot-pounds. Also the area B 2' 2 A is 
 
 easily found since it is a rectangle. The area 2' E 2 is found 
 by dividing it up into small sections, by calculus or by the use of 
 a plauimeter. 
 
 Fi. 2. 
 
8 
 
 INDICATORS. 
 
 It is easily seen that the area of the second case Fig. 2, is 
 less than that of Fig. 1. Therefore the work done is less ; but 
 the amount of steam admitted is only one-half as much as in the 
 first case. 
 
 In the first case, Fig. 1, 8 cubic feet of steam at 50 pounds 
 pressure were admitted per stroke and the work done was found 
 to be 57600 foot-pounds. In the second case only half as much 
 
 fr A A 
 
 steam is admitted and the work done is 
 
 the amount 
 
 represented by the area 2 f E C 2. Thus we see that there is a 
 considerable gain by expanding the steam. 
 
 Watt's Diagram of Work. Fig. 8 illustrates the method 
 adopted by James Watt to show the action of steam in the cylin- 
 
 der. The horizontal line A C called the abscissa represents the 
 length of the stroke and is divided into ten equal parts. The ver- 
 tical line A B called the ordinate indicates the pressure of steam. 
 
 When the piston has moved to the point E steam is cut off, 
 that is, a volume of steam equal to J the volume of the cylinder 
 expands until it fills the entire cylinder. The area may be found 
 by adding the several pressures (shown by the dotted lines), divid- 
 ing by the number of divisions, and multiplying by the length. 
 
 If by some arrangement of steam tight pistons working in cyl- 
 inders and having pencils fastened to them, \ve could get a dia- 
 
INDICATORS. 
 
 9 
 
 gram like that shown in Fig. 3 it would be of great use but too 
 large for convenience. 
 
 To obtain the same diagram on a small scale an indicator is 
 used. The value of such a diagram has already been shown when 
 finding the work done in the cylinder. The indicator has enabled 
 engineers to bring the engine of today to its present state of excel- 
 lence. A correct idea of the action of steam in the cylinder can 
 be obtained only by means of an indicator. It shows whether or 
 not the valves are set properly and how the condenser is working. 
 It also shows the engineer which end of the cylinder is doing the 
 most work. By comparing the expansion iine with an equilateral 
 hyperbola, with a curve of constant steam 
 weight, or with an adiabatic curve for 
 steam, the cylinder condensation is 
 calculated. 
 
 James Watt was the first to see the 
 need of accurate knowledge of the action of 
 steam in the cylinder. He invented the in- 
 dicator. The improved form consisted of a 
 steam cylinder S, about one inch in diam- 
 eter and six inches long, in which a solid 
 piston P, is accurately fitted. A spiral 
 spring A, is attached to this piston, and 
 controls the motion of a pencil a, which 
 is also attached to the piston. This 
 pencil can operate on a sheet of paper 
 fastened to a sliding board, B. This board 
 moves back and forth by means of a weight at one end and a cord 
 at the other which is connected to some reciprocating part of the 
 engine. The indicator cylinder S, may be put in communication 
 with the engine cylinder by means of the cock C. With this 
 instrument a complete diagram can be taken. 
 
 Watt's first indicator had no lateral motion, therefore all it 
 showed was the pressure of steam in the cylinder and the perfec- 
 tion of the vacuum. 
 
 INDICATORS. 
 
 The diagram, or card as it is often called, obtained by the use 
 of an indicator is the result of two motions. The horizontal move 
 
 '. 4. 
 
10 INDICATORS. 
 
 ment of the paper corresponds exactly to the movement of the 
 piston, and the vertical movement of the pencil is an exact ratio 
 to that of the pressure of steam in the cylinder. The diagram 
 represents by its length the stroke of the engine and by its height 
 the steam pressure on the piston at the corresponding point of the 
 stroke. The diagram shows the action of steam on one side of 
 the piston only ; to obtain the same information in regard to the 
 other side it is necessary to take another diagram from the other 
 end of the cylinder. 
 
 The essential features of an indicator are found in the instru- 
 ment invented by James Watt. Since his time, however, the 
 many improvements have made the indicator light, compact, dura- 
 ble, and accurate. Watt's diagram was traced on paper stretched 
 on a sliding board but now a revolving drum is used. The height 
 also of Watt's diagram was equal to the movement of the spring, 
 and the pencil arrangement was a simple contrivance. In the 
 indicators of the present day, the spring has a slight movement, 
 the height of the card being obtained by a multiplying arrange- 
 ment of .levers. This method requires a parallel motion to obtain 
 accuracy in the vertical lines ; for if a lever is pivoted at one end 
 and power applied near the pivot the lever tends to rise and the 
 free end will describe an arc of a circle, not a straight vertical 
 line. 
 
 THE THOHPSON INDICATOR. 
 
 Two views of the American Thompson Indicator, the outside 
 and the inside, are shown in Figs. 5 and 6. The form of spring 
 is shown in Fig. 7. The indicator consists of a cylinder in which 
 a piston is fitted, a spring, multiplying lever and parallel motion 
 for the pencil and a cylinder or drum for the paper. The piston, 
 which is .798 inch in diameter = | square inch in area, is fitted 
 accurately to the cylinder and has a travel of about one-half inch. 
 When the pressure of steam forces the piston upward it com- 
 presses the spring above it ; the amount of compression varies with 
 the strength of the spring. The rise of the piston causes the 
 pencil to rise because of the system of levers. The cylinder to 
 which the paper is attached rotates by means of a cord which is 
 fctened to some part of the ongine, the crosshead for example. 
 
INDICATORS. 
 
 II 
 
 Fig. 5- 
 
 While the drum revolves and the steam pressure forces the piston 
 
 to rise, the pencil, lightly touching the paper, describes the diagram. 
 The parallel movement of this indicator is obtained by a link 
 
 attached directly to the lever. It is 
 
 so constructed that there is but little 
 
 lost motion, hardly any friction and 
 
 no appreciable error within the limited 
 
 movement of the pencil. 
 
 The paper cylinder is so con- 
 structed that the tension of the coiled 
 
 spring within the drum may be altered 
 
 for different speeds of the engine. By 
 
 this means the cord can be kept taut 
 
 with little trouble. The cord is led 
 
 through a hold and kept in contact 
 
 with the scored wheel by another 
 
 small one. By this means the cord 
 
 can be run to any angle. It is con- 
 venient to have the cards of about the same size. If we used a 
 
 spring of such a tension that the 
 pencil would move one inch for 
 every 00 pounds pressure, and there 
 was but 30 pounds pressure in the 
 engine cylinder, the diagram would 
 be but one-half inch~high. This 
 diagram would be too small for 
 accurate work. For this reason 
 indicators are provided with sets of 
 springs of varying stiffness which 
 may be used according to the steam 
 pressure. If a spring is of such 
 stiffness that the pencil moves 1 
 inch for every 50 pounds pressure 
 it is called a 40 pound spring. 
 Others are called 10 pound, 20 
 pound, 30 pound, etc., springs. 
 To Change the Springs. In selecting the spring for a given 
 
 pressure, care should be taken that it will easily stand that press- 
 
 Fig. 6. 
 
INDICATORS. 
 
 ure. A safe rule to follow : multiply the scale of the spring by 
 ^ and subtract 15 for the vacuum. 
 
 For example : The maximum pressure for a 50 pound spring 
 is 110 pounds, because (50 X 2|) 15 .= 125 -- 15 = 110. 
 Springs are made in the following scale: 8, 10, 12, 16, 20, 24, 30, 
 32, 40, 48, 50, 56, 60, 64, 80, 100. For pressures from 70 to 90 
 pounds a 40* pound spring should be used, as 80 pounds pressure 
 on a 40 pound spring will raise the pencil 2 inches, and this is a 
 good height for the diagram. 
 
 If very high pressures are to be indicated, an extra piston, 
 having an area of J square inch, is used. This 
 doubles the allowable pressure on the spring. For 
 instance, if a spring can be used for 110 pounds press- 
 ure when the piston is | inch in area, it can be used 
 for 220 pounds pressure if the ^ inch piston is used. 
 
 When the spring has been selected it is placed 
 in the indicator. First unscrew the milled nut at 
 the top of the steam cylinder and take out the piston 
 with arm and connections. The pencil lever and 
 piston are disconnected by unscrewing the small- 
 headed screw which connects them. If a spring is 
 connected to the piston it should be removed, the 
 selected one substituted and the indicator put 
 together. The spring should always be firmly 
 screwed to the shoulder or the pencil will not 
 properly indicate the pressure. 
 
 Care of the Indicator. Before attaching the indicator to the 
 engine cylinder it should be taken apart, cleaned and oiled. If 
 each part works freely and smoothly the spring may be put in and 
 the indicator put together. After connecting to the cylinder, 
 admit steam to it, but do not take cards until it is thoroughly 
 warmed and blows dry steam through the relief. It is not neces- 
 sary to use lead in connecting as it is likely to get into the indi- 
 cator. After using, take the indicator apart, clean and oil. Only 
 porpoise or fine watch oil should be used. 
 
 THE CROSBY INDICATOR. 
 
 The internal arrangement of the parts of the Crosby Indicator 
 * NOTE : In practice it is better to use a somewhat stiffer spring. 
 
 Fig. 7. 
 
INDICATORS. 
 
 is shown in Fig. 8. The piston, 8, is formed with shallow channels 
 on its outer surface to retain oil which prevents leakage and 
 lubricates the piston. 
 
 The socket in the center of the piston is supported by a cen- 
 tral web and* projects both upward and downward. The upper 
 
 Fig. 8. 
 
 portion is threaded inside to receive the lower end of the piston- 
 rod. It has a vertical slot which allows the ball bearing on the 
 end of the spring to drop into a concave bearing on the upper end 
 of the piston-screw 9 which is screwed into the lower part of the 
 socket. 
 
 The piston-rod, 10, is made hollow, with the lower end 
 threaded. When the piston-rod is connected to the socket, the 
 former should be screwed into the socket as far as it will go. 
 
 The height of the atmospheric line on the diagram depends 
 
14 INDICATORS. 
 
 upon the amount the swivel head, 11, is screwed into the top of 
 the piston-rod. 
 
 A small projection on the lower side of this cap is threaded 
 to screw into the top of the spring and hold it firmly in place. 
 The moving parts are kept in line by this cap. The pencil 
 mechanism is supported by the sleeve 3, which surrounds the 
 upper part of the cylinder ; it turns freely and is held in position 
 by the cap. 
 
 The pencil mechanism is made as light as is consistent with 
 strength and stiffness. The pencil moves exactly 
 parallel to the piston because the fulcrum of the 
 mechanism and the point of attachment to the 
 piston-rod are always in a straight line. The 
 pencil point moves six times as far as the piston 
 because of the multiplying levers. 
 
 The drum 24, is one and one-half inches in 
 diameter, and is rewound when the string is pulled, 
 by a short spiral spring 31. 
 
 The piston spring is made of a single piece 
 of steel wire wound from the middle into a 
 double coil. The ends are screwed into a brass head 
 having four radial wings. At the bottom of the 
 spring a small steel bead is firmly attached to the wire. This 
 forms a ball and socket joint with the lower end of the piston-rod. 
 This joint is light and allows the spring to yield to pressure from 
 any direction. These springs are made in the following scale : 
 8, 12, 16, 20, 24, 30, 40, 50, 60, 80, 100, 120, 150, and 180. 
 
 To Insert the Spring. First unscrew the cap 2, then lift the 
 connected parts free from the cylinder by means of the sleeve. 
 The hollow wrench should be held in an inverted position arid 
 the piston-rod inserted until the hexagonal 'part engages the 
 wrench. Then, having the spring shown in Fig. 9 inverted, 
 insert the combined wrench and piston-rod until the steel bead 
 and the end of the spring rests in the concave seat. Now invert 
 the piston and pass the transverse wire at the bottom of the spring 
 through the slot until the threads at the bottom of the piston-rod 
 engage those inside the socket of the piston. With the wrench 
 screw it in as far as it will go. 
 
INDICATORS. 15 
 
 The piston screw should be loosened slightly before the 
 piston-rod is screwed in, and afterward set up against the bead 
 lightly to prevent lost motion. Then with the sleeve and cap 
 upright, engage the threads of the swivel bead with those inside 
 the piston-rod and screw it up until the lower projection of the 
 cap engages the threads inside the spring top; continue the process 
 until the spring is screwed up firmly against the cap. Holding 
 only by the sleeve 3, turn the piston and the connections until 
 the top of the piston-rod is flush with the shoulder on the swivel 
 head. 
 
 Now that the piston and all the connections are in their 
 places, the whole may be inserted in the cylinder and the cap 
 screwed down, which will fix all parts in their proper places. 
 If there is a spring in the cylinder first detach by reversing the 
 above process. 
 
 THE TABOR INDICATOR. 
 
 The Tabor Indicator is shown in Fig. 10. It is used exten- 
 sively in the navy. The principle of action and details of con- 
 struction are similar to the indicators already described ; the chief 
 peculiarity being the means employed to obtain a straight line 
 movement for the pencil. Inside the steam cylinder is a lining in 
 which the piston moves. This lining can expand when heated. 
 In the side of the cylinder small holes allow any steam which may 
 leak by the piston to escape. The piston-roJ. is connected at one 
 end to the piston by means of a ball and socket joint ; the other 
 end is connected to the pencil mechanism. 
 
 The pencil mechanism consists of three pieces, the pencil 
 lever, the back link, and the piston-rod link. The two links are 
 parallel for every position of the pencil. Thus the lower pivots 
 of these links and the pencil point are always in the same straight 
 line. The straight line movement of the pencil is obtained by 
 means of a curved slot in a stationary plate. The pencil bar is 
 provided with a roller which is fitted in such a manner that it can 
 roll from one end of the slot to the other. The curve of the slot 
 guides the pencil bar and is of such a radius that the pencil is 
 caused to move in a straight line. The curve compensates for the 
 tendency of the pencil point to move in an arc of a circle. The 
 
Id 
 
 INDICATORS. 
 
 pressure of the pencil on the paper is regulated by a screw which 
 strikes a stop plate attached to the frame. The end of the pencil 
 bar is formed for either a pencil lead or a metallic marking point. 
 
 The drum for the paper is made similar to those of other 
 indicators. The backward movement is obtained by a flat spiral 
 spring placed under the 
 drum. The tension of 
 this spring is altered by 
 loosening a thumbscrew, 
 lifting the carriage, and 
 winding or unwinding. 
 A simple pulley guides 
 the driving cord in any 
 direction. 
 
 The indicator is at- 
 tached by a coupling 
 having a single thread. 
 
 The springs of the 
 Tabor indicator are of 
 the duplex type, that is, 
 they are made of two 
 spiral coils of wire. A 
 50 pound spring is 
 shown in Fig. 11. The 
 wire terminates in fit- 
 tings at each end. The spring is attached to the upper side of the 
 piston by means of threads cut on the inside of the fitting and on a 
 projection on the piston. The top of the spring is attached to the 
 under side of the cover in a similar manner. The springs are 
 made in the following scales, 8, 10, 12, 16, 20, 24, 30, 32, 40, 48, 
 50, 60, 64, 80, and 100 pounds. The maximum safe steam pres- 
 sures (absolute) to which these springs may be subjected are 
 respectively, 10, 15, 20, 24, 40, 48, 70, 75, 95, 112, 120, 140, 152, 
 180, and 200 pounds. 
 
 Change of Location of Atmospheric Line. Unscrew the cap 
 and lift the sleeve and connections from the cylinder. Then turn 
 the piston to the left or right according as the pencil is to be 
 raised or lowered. One revolution causes the pencil to rise J 
 inch. 
 
 Fig. 10. 
 
INDICATORS. 17 
 
 Care of the Indicator. Before attaching the indicator to the 
 engine, steam should be blown through the pipes and cocks so 
 that all particles of dust may be removed. After using, the indi- 
 cator should be carefully wiped and oiled. The cylinder cap 
 should be unscrewed and all the parts connected to the piston 
 removed. The piston spring and piston-rod should be detached, 
 carefully wiped dry, and then oiled. The inside of the cylinder 
 also should be oiled. Then the piston and piston-rod should be 
 placed in the cylinder and the spring placed in the box. If the 
 indicator has not been used for some time the oil may have become 
 gummed. It may be easily cleaned by wiping with 
 {i cloth saturated with naphtha or benzine. It must 
 be oiled again before using. A good test that the 
 indicator is in proper working order is to detach the 
 spring and after replacing the piston and piston-rod, 
 raise the pencil to the highest point. When allowed 
 to fall it should descend to the lowest point freely. 
 The pencil should always have a smooth fine point. 
 To Attach the Indicator to the Engine. Usually 
 all first-class engines are prepared for the indicator 
 before leaving the factory. Holes are drilled and 
 tapped in the cylinder and have plugs screwed in them. These 
 plugs are easily removed and the indicator connections screwed in. 
 When this is not the case, any engineer can perform the 
 work. Before drilling the holes, in the cylinder, the heads should 
 be removed so that the exact positions of the pistons and the size 
 of the ports and passages may be known. Also with the heads off 
 all chips and particles of dirt from drilling may be easily removed. 
 If it is impossible to remove the heads, a little steam admitted to 
 the cylinder just before the drilling completed will blow the 
 chips out. 
 
 Each end should be drilled and tapped for a one-half inch 
 pipe thread. The holes must be drilled into the clearance space, 
 so that the piston at the ends of the stroke will not cover them. 
 They should also be placed so that currents of steam will not reach 
 them. Before deciding just the points at which to drill the holes, 
 it is well to consider every plan of indicating the engine. The 
 type of engine, the position of the steam chest, the kind of cross- 
 
18 
 
 INDICATORS. 
 
 head, and the position of the eccentric and its connections should 
 all be considered, as well as the most convenient place in the 
 engine room. The holes . should not be drilled until the plan 
 shows the proper connections with the reducing motion, conven- 
 ient access and free passage of steam to the indicator. 
 
 When the plan has been adopted, the engine should be placed 
 on dead center, to determine the clearance. The holes should be 
 drilled into the middle of the clearance space. 
 
 In common practice for horizontal engines the holes are 
 drilled in the side of the cylinder at each end. Short half inch 
 pipes with quarter upward bends into which the indicator coils 
 may be screwed are inserted in these holes. 
 
 It may be more convenient to drill and tap into the top of 
 the cylinder and attach the indicators directly. 
 
 For vertical engines, the upper head or cover and the side of 
 the cylinder are often drilled and tapped for the upper and lower 
 indicators respectively. It is preferable to connect the indicators 
 
 Fig. 12. 
 
 to the sides because less pipe and fittings are required and better 
 results obtained. 
 
 If only one indicator is to be used for both ends of the cyl- 
 inder, it may be connected by side pipes and a three way cock. 
 By this method both diagrams are taken on the same card and 
 with the loss of but one revolution. Fig. 12 shows the section of 
 a three way cock. 
 
 Reducing notion. As we have already seen the length of 
 the card represents the travel of the piston. As the length of 
 
INDICATORS. 
 
 19 
 
 card is obtained by the rotation of the drum the motion of the 
 drum must be taken from some part of the engine which has a 
 motion coinciding with that of the piston. The crosshead is the 
 most common, reliable and convenient part. The length of the 
 card is much less than the travel of the piston since the stroke is 
 longer than ^he circumference 
 of the drum so that the move- 
 ment of the crosshead must 
 be reduced to the length of 
 the diagram. 
 
 There are several devices 
 employed to obtain this re- 
 duced motion. 
 
 A simple form of reduc- 
 ing motion, called the panto- 
 graph, is shown in Fig. 13. 
 Four links, a, 5, <?, d, are 
 joined in the form of a par- 
 allelogram. One link, , is 
 prolonged and pivoted at the 
 crosshead C. The point where 
 b and c join is pivoted at the 
 fixed point E. The cord is 
 fastened at D on the link d. 
 
 Fig. 13. 
 
 The point D must be in the straight line connecting E and C. 
 Then letting A B represent the stroke and h the length of the 
 indicator diagram, we have 
 
 A B : h = E C : E D, from which 
 
 ED = 
 
 h X EC 
 AB 
 
 Another form of pantograph is shown in Fig. 14. It is placed 
 horizontally with the pivot, B, resting on a support opposite the 
 crosshead when in mid-position. The pivot A, is attached to the 
 crosshead ; usually by having the stud A inserted in a hole drilled 
 in the crosshead. If the pivot B is adjusted to the proper height 
 and at the right distance from the crosshead, the cord from the 
 indicator may be attached to the pin E without any pulleys. The 
 
INDICATORS. 
 
 length of the diagram is varied by adjusting the movable bar C D; 
 the pin E must be in the straight line from A to B. 
 
 The pantograph is likely to become shaky and loose on 
 account of its many joints. If well made it gives perfect motion. 
 
 The reducing motion shown in Fig. 15, called the Brumbo 
 Pulley, is easily and quickly made and can be used on almost any 
 engine. The wooden rod A is usually about twice as long as the 
 stroke. It is pivoted by a bolt or screw at B, a fixed point. At the 
 lower end it is connected by the wooden link, C, to the crosshead. 
 
 Fig. H 
 
 This link C is usually about one-half the length of the stroke. The 
 sector S may be made either of wood or metal. It should have a 
 groove in the circular edge for the cord, and is made fast to the upper 
 end of the lever A. Its center should coincide with that of the 
 pivot B. The length of the radius of the sector may be found as 
 follows. Divide the length of the lever by the length of the 
 stroke, multiply the result by the length of the desired diagram 
 and the product will be the radius of the sector. For instance, if 
 the lever is 60 inches long and the stroke 30 inches and we wish 
 the diagram to be 3 inches long, the sector should be 6 inches in 
 radius for, 
 
 To avoid the use of guide pulleys the lever should be hung 
 so that it will swing in a vertical plane parallel with the guides and 
 in line with the indicator. When the crosshead is at mid-stroke, 
 
INDICATORS. 
 
 21 
 
 the lever must be vertical and the point D must be below the axis 
 of the cylinder because it comes above the line at the ends of the 
 stroke. 
 
 The reducing lever used for large quick running engines is 
 shown in Fig. 16. The rod A is made of pine wood, tapering 
 toward the lower point and about one inch in thickness. The 
 length is about one and one-half times the length of the stroke. 
 It is suspended by a bolt o: screw from some fixed point above the 
 engine, and should swing edgewise and parallel to the guides of 
 the crosshead. The steel stud at the bottom of the rod has a T 
 shaped slot in an iron plate which is attached firmly to the cross- 
 head. The slot should be long enough to retain the stud when 
 the crosshead is at the end of the stroke. To find the point at 
 which the indicator cord should be attached, divide the length of 
 
 Fig, 15. 
 
 Fig. 16. 
 
 the lever by the length of the stroke of the piston and multiply 
 the quotient by the length of the desired diagram. The product 
 is v tlie distance of the point from the pivot at the top of the lever. 
 Example. A lever is 45 inches long, the piston stroke 30 
 inches and the diagram to be 3J inches long. At what distance 
 from the pivot should the indicator cord be attached ? 
 
 Having placed the indicator in position and obtained the 
 
22 INDICATORS. 
 
 reducing motion, the length of the cord must be so adjusted that 
 the drum will not strike the stops at either end of the stroke. 
 
 For convenience an approximate length of cord is first found 
 and the cord cut in two parts, one attached to the reducing 
 motion ; the other to the indicator. A hook should be fastened to 
 the froe end of the piece attached to the indicator. A loop is 
 made in the free end of the piece from the reducing motion. 
 The hook is then attached to the irop and the extra length 
 of cord taken up by tying knots. Another method for adjust- 
 ing the length of the cord is the arrangement shown in Fig. 
 17. The hook A is attached to the indicator cord. The cord B 
 from the reducing motion passes through the holes in the plate P 
 as shown. To adjust the length of the cord it is slacked at the 
 point B and the plate slipped along the cord. 
 
 Fig. 17. 
 
 To Take Indicator Diagrams. The indicator should be in 
 good working order before attaching to the cylinder ; it should be 
 clean, well oiled, and the levers and springs should work smoothly. 
 The pencil point should be sharp and the pressure adjusted to 
 make a distinct fine line. 
 
 The spring should be selected that will give a diagram 1^ 
 to 2 inches in height. If the spring chosen is too light, the lines 
 are likely to be wavy from the vibration of the pencil levers. 
 
 When the indicator is in position a satisfactory reducing 
 motion obtained and the cord adjusted, the paper should be 
 wrapped smoothly around the drum. The edges projecting over 
 the clips should be folded back so that they will not touch the 
 pencil lever. Before taking the card, allow the steam to enter 
 the indicator and move the piston up and down until the parts 
 have become thoroughly warmed. Then pass the pencil against 
 the paper long enough to take the diagram. Some engineers 
 allow the pencil to remain in contact with the paper during but 
 one revolution of the engine ; others trace the diagram two or 
 
INDICATORS. 23 
 
 more times. After the diagram is taken the cock is shut and 
 without unhooking the cord the pencil is again pressed to the 
 paper to take the atmospheric line. The cord is then disconnected 
 and the card removed from the drum. The scale of spring, the 
 dimensions and speed of the engine, the date, and all useful partic- 
 ulars are written on the cards. 
 
 If one indicator is used for both ends the three way cock 
 shown in Fig. 12 is opened to admit steam from one end, the 
 diagram taken ; then opened for the other end and that diagram 
 taken. Then the steam is shut off from both ends and the atmos- 
 pheric line taken. 
 
 As has been said before, the indicator is of great importance 
 to the engineer. It is used to find the indicated horse-power of 
 the engine, and by comparison of the indicated horse-power with 
 the brake horse-power, the mechanical efficiency is obtained. The 
 indicator card shows several other things ; the time and manner 
 of the four events of the stroke, namely, the admission, cut-off, 
 release, and compression. These four events make up what is 
 called the steam distribution. It shows faults in the setting and 
 working of the valves. 
 
 We have seen how the indicator diagram represents the net 
 work done on the piston in one stroke. 
 
 Work is equal to pressure multiplied by the distance through 
 which it acts. The distance is the length of stroke multiplied by 
 the number of strokes per minute. The pressure is the average 
 net pressure acting on the piston during the stroke. This average 
 net pressure is called the mean effective pressure. If we know 
 from the indicator card the mean effective pressure per square 
 inch, we can find the total pressure by multiplying it by the area 
 of the piston in square inches. 
 
 The distance per minute is equal to the length of stroke 
 multiplied by the number of strokes. 
 
 Let 1* mean effective pressure in pounds per square inch. 
 A area of piston in square inches. 
 L length of stroke in feet. 
 N number of strokes per minute. 
 Then the work done per minute, 
 
 W = P L AK. 
 
24 INDICATORS. 
 
 Since one horse-power is the rate of doing work when 33,000 
 foot-pounds of work are done per minute the indicated horse-power 
 of an engine is obtained by means of the formula, 
 
 H P = PL AN 
 33,000 
 
 The length of the stroke, the area of the piston and the num- 
 ber of strokes are easily found. Then all that remains to be 
 determined before the horse-power is calculated is the value of P. 
 or the mean effective pressure. 
 
 Suppose one side of the piston is in communication with the 
 boiler during the entire stroke, the mean pressure is then the boiler 
 pressure. But if the supply of steam is cut off before the stroke 
 is completed, the mean pressure will not equal the boiler pressure. 
 In Fig. 1 we saw that the area of the shaded portion was equal to 
 the length multiplied by the height. The area of a figure of any 
 shape can be reduced to that of a rectangle having a length equal 
 to the extreme length of the figure. Then whenever we know the 
 area and length, we can find the height or mean height by dividing 
 the area by the length. 
 
 Suppose a diagram like that shown in Fig. 2 has an area of 
 5^ square inches and its extreme length is 3i inches ; then the 
 height is 5.25 divided by 3.5 = 1.5 inches. Then with any indi- 
 cator card the area of the diagram is equal to the area of a 
 rectangle, the length of which is known and the height can easily 
 be computed. 
 
 Suppose that we have taken a card and know that the mean 
 height is 1 J inches. In order to find the horse-power we must 
 reduce the 1|- inches to pounds pressure. If we multiply the 
 height by the strength of the spring we get the desired result. 
 For instance, if we had used. a 30 pound spring (that is one which 
 causes the pencil of the indicator to move one inch for every 30 
 pounds pressure in the cylinder) the height 1| inches would 
 equal 30 X 1^ = 45 pounds pressure. 
 
 Example. An indicator card has an area of 1.925 square 
 inches and is 2.2 inches long. If a 60 pound spring is used 
 what is the mean pressure ? 
 
 The mean height equals - '- = .875 inch and 
 
 2t.2t 
 
 .875 x 60 = 52.5 pounds, 
 
INDICATORS. 25 
 
 Suppose the engine from which the above card was taken had 
 a piston 14 inches in diameter and a stroke of 24 inches. If it 
 were running at 150 revolutions per minute what is its horse- 
 power ? Assume the mean effective pressure to be the same for 
 both sides of the piston. 
 
 H P = P L A N 
 33,000 
 
 _ 52.5 X 2 X 153.94 X 300 
 
 33,000 
 
 =147. (about) 
 
 In most engines more work is done at one end of the cylinder 
 that at the other; it is not safe then to assume the mean effective 
 pressure of one side the same as that of the other. Cards should 
 be taken from each end and calculated for mean effective pressure 
 separately, then averaged. Also the area of one side of the piston 
 is greater than the other on account of the piston-rod. The two 
 ends may be figured separately or the average area of the two sides 
 of the piston may be used as the value of A. 
 
 Another method- is to find the work done at each end of the 
 cylinder and then add the results. This enables the engineer to 
 know if his valves are set so that each end does about the same 
 
 amount of work. 
 
 An engine has the following dimensions. The piston is 12 
 inches in diameter, the piston-rod is 21 inches in diameter and the 
 length of stroke is 34 inches. While running at 92 revolutions 
 cards were taken. The area of the card from the head end was 
 5.36 square inches, that of the crank end was 5.30 square inches, 
 and a 40 pound spring was used. The cards were 3.72 inches 
 long. We wish to know what horse-power the engine developed 
 and which end was doing the most work. 
 
 The area of the piston is 113.097 square inches for the head 
 end and 113.097 3.5466 = 109.55 square inches for the crank 
 end. 
 
 Then for the head end, 
 H P PLAN . P X 34 X 113.097 X 92 _ 893g p 
 
 33,000 12 X 33,000 
 
 and for the crank end, 
 
 H p PLAN P X 34 X 109.55 X 92 8 . p 
 33,000 12 X 33,000 
 
26 INDICATORS. 
 
 The value of P is found from the cards. Since the cards 
 were 3.72 inches long and the area of the card from the head end 
 
 5 36 
 was 5.36 square inches the mean ordinate is __!__ _ = 1.44 inches 
 
 which equals 1.44 X 40 = 57.6 pounds. The horse-power would 
 be, 
 
 .8933 X 57.6 = 51.45 
 
 For the crank end the mean effective pressure is found as 
 before, 
 
 = 1.425 and 1.425 X 40 = 57. 
 3.72 
 
 The horse-power would be, 
 
 .8653 X 57 = 49.32 
 
 The horse-power is evidently the sum of these two quantities 
 or 51.45 + 49.32 = 100.77. 
 
 The head end is doing more work than the crank end but the 
 difference is slight being only, 
 
 51.45 - - 49.32 = 2.13 horse-power, 
 or about 2.1 per cent, of the total power. 
 
 Considerable arithmetical work is necessary when the I. H. P. 
 is found from the formula, 
 
 I H P. = PLAN 
 33,000 
 
 and the, chances for error are of course, great. To safe time and 
 reduce the chance for error a table of engine constants lias been 
 prepared. The number of strokes, or twice the number of revo- 
 lutions, multiplied by the length of stroke in feet is called the 
 
 piston speed. Then in the formula I. H. P. = _ , L N = 
 
 33,000 
 
 piston speed in feet per minute. In the following table, the I. H. P. 
 of an engine is easily computed by multiplying the constant, cor- 
 responding to the diameter of the piston, by the piston speed and 
 by the M. E. P. Or, in other words, the constants in the table 
 equal the horse-power for an engine with a given diameter of 
 piston having a piston speed of one foot per minute and a M. K. P. 
 of one pound. 
 
INDICATORS. 
 
 27 
 
 TABLE OF ENGINE CONSTANTS, 
 
 Diam- 
 
 
 
 ! 
 
 
 
 
 
 eter 
 
 Even 
 
 -H 
 
 + 1 +i 
 
 + * 
 
 + 1 
 
 M 
 
 + 1 
 
 of 
 
 
 or 
 
 or or 
 
 or 
 
 or 
 
 or 
 
 or 
 
 Cylin- 
 
 Inches. 
 
 .125. 
 
 .25. .375. 
 
 .5. 
 
 .625. 
 
 .75. 
 
 .875. 
 
 der. 
 
 
 
 
 
 
 
 
 1 
 
 0000238 
 
 .0000301 
 
 .0000372 
 
 .0000450 
 
 .0000535 
 
 .0000628 
 
 .0000729 
 
 .0000837 
 
 2 
 
 0000952 
 
 .0001074 
 
 .0001205 
 
 .0001342 
 
 .0001487 
 
 .0001640 
 
 .0001800 
 
 .0001967 
 
 3 
 
 .0002142 
 
 .0002324 
 
 .0002514 
 
 .0002711 
 
 .0002915 
 
 .0003127 
 
 .0003347 
 
 .0003574 
 
 4 
 
 .OUU308 
 
 .0004050 
 
 .0004299 
 
 .0004554 
 
 .0004819 
 
 .0005091 
 
 .0005370 
 
 .0005656 
 
 5 
 
 .0005950 
 
 .0006251 
 
 .0006560 
 
 .0006876 
 
 .0007199 
 
 .0007530 
 
 .0007869 
 
 .0008215 
 
 6 
 
 .0008568 
 
 .0008929 
 
 .0009297 
 
 .0009672 
 
 .0010055 
 
 .0010445 
 
 .0010844 
 
 .0011249 
 
 7 
 
 .0011662 
 
 .0012082 
 
 .0012510 
 
 .0012944 
 
 .0013387 
 
 .0013837 
 
 .0014295 
 
 .0014759 
 
 8 
 
 .0015232 
 
 .0015711 
 
 .0016198 
 
 .0016693 
 
 .0017195 
 
 .0017705 
 
 .0018222 
 
 .0018746 
 
 9 
 
 .0019278 
 
 .0019817 
 
 .0020363 
 
 .0020916 
 
 .0021479 
 
 .0022048 
 
 .0022625 
 
 .0023209 
 
 10 
 
 .0023800 
 
 .0024398 
 
 .0025004 
 
 .0025618 
 
 .0026239 
 
 .0026867 
 
 .0027502 
 
 .0028147 
 
 11 
 
 .0028798 
 
 .0029456 
 
 .0030121 
 
 .0030794 
 
 .0031475 
 
 .0032163 
 
 .0032859 
 
 .0033561 
 
 12 
 
 .0034272 
 
 .0034990 
 
 .0035714 
 
 .0036447 
 
 .0037187 
 
 .0037934 
 
 .0038690 
 
 .0039452 
 
 13 
 
 .0040222 
 
 .0040999 
 
 .0041783 
 
 .0042576 
 
 .0043375 
 
 .0044182 
 
 .0044997 
 
 .0045819 
 
 14 
 
 .0046648 
 
 .0047484 
 
 .0048328 
 
 .0049181 
 
 ' .0050039 
 
 .0050906 
 
 .0051780 
 
 .0052661 
 
 15 
 
 .0053550 
 
 .0054446 
 
 .0055349 
 
 .0056261 
 
 .0057179 
 
 .0058105 
 
 .0059039 
 
 .0059979 
 
 16 
 
 .0060928 
 
 .0061884 
 
 .0062847 
 
 .0063817 
 
 .0064795 
 
 .0065780 
 
 .0066774 
 
 .0067774 
 
 17 
 
 .0068782 
 
 .0069797 
 
 .0070819 
 
 .0071850 
 
 .0072887 
 
 .0073932 
 
 .0074985 
 
 .0076044 
 
 18 
 
 .0077112 
 
 .0078187 
 
 .0079268 
 
 .0080360 
 
 .0081452 
 
 .0082560 
 
 .0083672 
 
 .0084791 
 
 19 
 
 .0085918 
 
 .0087052 
 
 .0088193 
 
 .0089343 
 
 .0090499 
 
 .0091663 
 
 .0092835 
 
 .0094013 
 
 20 
 
 .0095200 
 
 .0096393 
 
 .0097594 
 
 .0098803 
 
 .0100019 
 
 .0101243 
 
 .0102474 
 
 .0103712 
 
 21 
 
 .0104958 
 
 .0106211 
 
 .0107472 
 
 .0108739 
 
 .0110015 
 
 .0111299 
 
 .0112589 
 
 .0113886 
 
 22 
 
 .0115192 
 
 .0116505 
 
 .0117825 
 
 .0119152 
 
 .0120487 
 
 .0121830 
 
 .0123179 
 
 .0124537 
 
 23 
 
 .0125902 
 
 .0127274 
 
 .0128654 
 
 .0130040 
 
 .0131435 
 
 .0132837 
 
 .0134247 
 
 .0135664 
 
 24 
 
 .0137088 
 
 .0138519 
 
 .0139959 
 
 .0141405 
 
 .0142859 
 
 .0144321 
 
 .0145789 
 
 .0147266 
 
 25 
 
 .0148750 
 
 .0150241 
 
 .0151739 
 
 .0153216 
 
 .0154759 
 
 .0156280 
 
 .0157809 
 
 .0159345 
 
 26 
 
 .0160888 
 
 .0162439 
 
 .0163997 
 
 .0165563 
 
 .0167135 
 
 .0168716 
 
 .0170304 
 
 .0171899 
 
 27 
 
 .0173502 
 
 .0175112 
 
 .0176729 
 
 .0178355 
 
 .0179988 
 
 .0181627 
 
 .0183275 
 
 .0184929 
 
 28 
 
 .0186592 
 
 .0188262 
 
 .0189939 
 
 .0191624 
 
 .0193316 
 
 .0195015 
 
 .0196722 
 
 .0198436 
 
 29 
 
 .0200158 
 
 .0201887 
 
 .0203624 
 
 .0205368 
 
 .0207119 
 
 .0208879 
 
 .0210645 
 
 .0212418 
 
 30 
 
 .0214200 
 
 .0215988 
 
 .0217785 
 
 .0219588 
 
 .0221399 
 
 .0223218 
 
 .0225044 
 
 .0226877 
 
 31 
 
 .0228718 
 
 .0230566 
 
 .0232422 
 
 .0234285 
 
 .0236155 
 
 .0238033 
 
 .0239919 
 
 .0241812 
 
 32 
 
 .0243712 
 
 .0245619 
 
 .0247535 
 
 .0249457 
 
 .0251387 
 
 .0253325 
 
 .0255269 
 
 .0257222 
 
 33 
 
 .0259182 
 
 .0261149 
 
 .0263124 
 
 .0265106 
 
 .0267095 
 
 .0269092 
 
 .0271097 
 
 .0273109 
 
 34 
 
 .0275128 
 
 .0277155 
 
 .0279189 
 
 .0281231 
 
 .0283279 
 
 .0285336 
 
 .0287399 
 
 .0289471 
 
 35 
 
 .0291550 
 
 .0293636 
 
 .0295729 
 
 .0297831 
 
 .0299939 
 
 .0302056 
 
 .0304179 
 
 .0306309 
 
 36 
 
 .0308448 
 
 .0310594 
 
 .0312747 
 
 .0314908 
 
 .0317075 
 
 .0319251 
 
 .0321434 
 
 .0323624 
 
 37 
 
 .0325822 
 
 .0328027 
 
 .0330239 
 
 .0332460 
 
 .0334687 
 
 .0336922 
 
 .0339165 
 
 .0341415 
 
 38 
 
 .0343672 
 
 .0345937 
 
 .0348209 
 
 .0350489 
 
 .0352775 
 
 .0355070 
 
 .0357372 
 
 .0359685 
 
 39 
 
 .0361998 
 
 .0364322 
 
 .0366654 
 
 .0368993 
 
 .0371339 
 
 .0373694 
 
 .0376055 
 
 .0378424 
 
 40 
 
 .0380800 
 
 .0383184 
 
 .0385575 
 
 .0387973 
 
 .0390379 
 
 .0392793 
 
 .0395214 
 
 .0397642 
 
 41 
 
 .0400078 
 
 .0402521 
 
 .0404972 
 
 .0407430 
 
 ,0409895 
 
 .0412368 
 
 .0414849 
 
 .0417337 
 
 42 
 
 .0419832 
 
 .0422335 
 
 .0424845 
 
 .0427362 
 
 .0429887 
 
 .0432420 
 
 .0434959 
 
 .0437507 
 
 43 
 
 .0440062 
 
 .0442624 
 
 .0445194 
 
 .0447771 
 
 .0450355 
 
 .0452947 
 
 .0455547 
 
 .0458154 
 
 44 
 
 .0460768 
 
 .0463389 
 
 .0466019 
 
 .0468655 
 
 .0471299 
 
 .0473951 
 
 .0476609 
 
 .0479276 
 
 45 
 
 .0481950 
 
 .0484631 
 
 .0487320 
 
 .0490016 
 
 .0492719 
 
 .0495430 
 
 .0498149 
 
 .0508875 
 
 46 
 
 .0503608 
 
 .0506349 
 
 .0509097 
 
 .0511853 
 
 .0514615 
 
 .0517386 
 
 .0520164 
 
 .0522949 
 
 47 
 
 .0525742 
 
 .0528542 
 
 .0531349 
 
 .0534165 
 
 .0536988 
 
 .0539818 
 
 .0542655 
 
 .0545499 
 
 48 
 
 .0548352 
 
 .0551212 
 
 .0554079 
 
 .0556953 
 
 .0559835 
 
 .0562725 
 
 .0565622 
 
 .0568526 
 
 49 
 
 .0571438 
 
 .0574357 
 
 .0577284 
 
 .0580218 
 
 .0583159 
 
 .0586109 
 
 .0589065 
 
 .0592029 
 
 50 
 
 .0595000 
 
 .0597979 
 
 .0600365 
 
 .0603959 
 
 .0606959 
 
 .0609969 
 
 .0612984 
 
 .0616007 
 
 51 
 
 .0619038 
 
 .0622076 
 
 .0625122 
 
 .0628175 
 
 .0632235 
 
 .0634304 
 
 .0637379 
 
 .0640462 
 
 52 
 
 .0643552 
 
 .0646649 
 
 .0649753 
 
 .OR52867 
 
 .0655987 
 
 .0659115 
 
 .0662250 
 
 .0665392 
 
 53 
 
 .0668542 
 
 .0671699 
 
 .0674864 
 
 .0678036 
 
 .0681215 
 
 .0684402 
 
 .0687597 
 
 .0690799 
 
 54 
 
 .0694008 
 
 .0697225 
 
 .0700449 
 
 .0703681 
 
 .0705293 
 
 .0710166 
 
 .0713419 
 
 .0716681 
 
 55 
 
 .0719950 
 
 .0724226 
 
 .0726510 
 
 .0729801 
 
 .0733099 
 
 .0736406 
 
 .0739719 
 
 .0743039 
 
 56 
 
 .0746368 
 
 .0749704 
 
 .0753047 
 
 .0756398 
 
 .0759755 
 
 .0763120 
 
 .0766494 
 
 .0769874 
 
 57 
 
 .0773262 
 
 .0776657 
 
 .0780060 
 
 .0783476 
 
 0786887 
 
 .0790312 
 
 .0793745 
 
 .0797185 
 
 58 
 
 .0800632 
 
 .0804087 
 
 .0807549 
 
 .0811019 
 
 .0814495 
 
 .0817980 
 
 .0821472 
 
 .0824971 
 
 59 
 
 .0828478 
 
 .0831992 
 
 .0835514 
 
 .0839043 
 
 .0842579 
 
 .0846123 
 
 .0849675 
 
 .0853234 
 
 60 
 
 .0856800 
 
 .0860374 
 
 .0863955 
 
 .0867543 
 
 .0871139 
 
 .0874743 
 
 .0878354 
 
 .0881973 
 
INDICATORS. 
 
 To Use the Table. If the diameter of the piston is an even 
 number, the constant is found in the second column ; if it contains 
 a fraction the constant is found by following the column horizon- 
 tally until the required fraction is reached. The constant multi- 
 plied by the piston speed in feet per minute ai cl by the M. E. P. 
 in pounds per square inch gives the I. H. P. 
 
 Example. An engine runs at 75 revolutions. The stroke is 
 4 feet; if the M. E. P. is 48 pounds and the piston 27f inches in 
 diameter what is the I. H. P. 
 
 From the table the constant for a piston 27 f inches in diam- 
 eter is .0178355. The piston speed is 150 X 4 = 600 feet per 
 minute. Then the I. H. P. is, 
 
 .0178355 X 600 X 48 = 513.66 
 
 The horse-power as above calculated is called the indicated 
 horse-power and is usually written I. H. P. Although the above 
 calculation shows the amount of power the engine develops it does 
 
 Fig. 18. 
 
 not show the available power since part of the indicated horse- 
 power is used to run the engine itself, that is, to overcome the 
 friction of the parts. To determine how much power can be used 
 to run machinery some form of absorption dynamometer or friction 
 brake is attached to the engine. The power thus obtained is 
 called the Brake Horse Power or B. H. P. It is more satisfactory 
 for both the owner and builder to know the B. H. P. than to know 
 the I. H. P. 
 
 The Prony Brake, Fig. 18, is one of the simplest absorption 
 dynamometers. The two wooden blocks A and C are held together 
 against the rim of the pulley P by bolts. The thumb-nuts, e, e, being 
 
INDICATORS. 
 
 used to adjust the pressure. By means of the bolts the arm L is 
 held to the upper block. From this arm is suspended the ball 
 weight, w, which by sliding along the arm counterbalances the 
 weight of the arm and pan at the other end. The pulley revolves 
 at the required speed in the direction indicated by the arrow. 
 The bolts are tightened until the lever remains stationary in a 
 horizontal position when a known weight, W, is hung at the end. 
 
 The amount of work absorbed by the brake depends upon the 
 weight W, the length, R, and the speed. It is independent of 
 the diameter of the pulley and the pressure of the block because 
 the moments of force about the center of the pulley are equal 
 when the lever L, is horizontal. Letting/ equal the co-efficient 
 of friction, p the pressure of the blocks and r the radius of the 
 pulley, 
 
 fpr = WR . 
 
 The work done at the face of the pulley equals the force mul- 
 tiplied by the distance or the pressure multiplied by the number 
 of feet passed through. 
 
 Let N = the number of revolutions per minute. Then the 
 distance passed through per minute equals 2 IT r N and the work 
 done equals 2 TT r N/J9. Then asfp r W R, the work done 
 at the rim of the pulley equals the left hand side of the equation 
 multiplied by 2 TT N, and to keep both sides equal we multiply 
 W R by 2 TT N. Hence the work done is expressed by the formula 
 
 2 TT N W R and, 
 B p . 27rNWR 
 
 33,000 
 = .0001904 N W R 
 
 A Prony brake with an arm 4 feet long was attached to the 
 pulley on the fly wheel of an engine. The weight in the scale 
 pan was 50 pounds and the speed of the engine 300 revolutions. 
 Find the brake horse power. 
 
 B. H. P. = .0001904 X 300 X 50 X 4 
 = 11.424 
 
 The rope brake shown in Fig. 19 is easily constructed of 
 material at hand and being self-adjusting needs no accurate fitting. 
 For large powers the number of ropes may be increased. It is con- 
 sidered a most convenient and reliable brake. In Fig. 19 the spring 
 
30 
 
 INDICATORS. 
 
 balance, B, is shown in a horizontal position. This is not at all 
 necessary ; if convenient the vertical position may be used. The 
 ropes are held to the pulley or fly-wheel face by blocks of wood, O. 
 The weights at W.may be replaced by a spring balance if 
 desirable. 
 
 To calculate the Brake Horse Power, subtract the pull regis- 
 tered by the spring balance, B, from the load at W. The lever 
 arm is the radius of the pulley plus i the diameter of the rope. 
 The formula is, 
 
 B H P 27rRN(W B) 
 
 33,000 
 = .0001904 R N (W B) * 
 
 Fig. 19. 
 
 Example. A rope brake is attached to a gas engine. The 
 average reading of the spring balance is 8 pounds. W = 80 
 pounds. If the radius of the brake wheel is 28 inches and the 
 rope 1 inch in diameter, what is the B. H. P. when the engine 
 makes 350 revolutions per minute? 
 
 E = 28 + J = 28i inches == ^ feet 
 B. II. P == .0001904 E N (V .B) 
 
 = .0001904 X 2 -^- X 72 X 350 
 = 1*1.4 Ans 
 If both the indicated horse-power and the brake horse-power 
 
 * NOTE: If B is greater than W, the engine is running in the opposite 
 direction. Use the formula B. H. P. = .0001904 R N (B W) . 
 
INDICATORS. 31 
 
 are known the power used in friction is found by subtracting the 
 B. H. P. from the I. H. P. 
 
 The mechanical efficiency of the engine is the ratio of the 
 B. H. P. to the I. H. P. or, 
 
 T> TT T> 
 
 I.H.P. 
 
 If an engine of 18.2 indicated horse-power develops at a trial 
 16.02 brake horse-power, what is its mechanical efficiency? 
 
 B. H. P. 
 
 E = 
 
 I. H. P. 
 
 16.02 
 
 18.2 
 = 88 % Efficiency. 
 
 Brakes should be well lubricated. For small powers the 
 heat generated by friction between the ropes or blocks and the 
 rim of the wheel, will be conducted away by radiation but for 
 large powers some additional means is necessary. In case there 
 are flanges on the wheel, water can be introduced into the wheel, 
 the flanges keeping it from flowing out and centrifugal force keep- 
 ing it in contact with the rim. The amount of water can be regu- 
 lated so that all may be evaporated, or a scoop can be arranged to 
 carry off the water. In all cases the water should flow 
 continuously. 
 
 To Find the Area of Cards. M. E. P. or the mean effective 
 pressure is equal to the area of the indicator diagram divided by 
 the length. The length is easily found by measurement but to 
 find the area is more difficult since the shape is irregular. If the 
 figure were regular its area could be found by geometry or by 
 simple formulas. 
 
 The area of the indicator card can be found in two ways. 
 By dividing the diagram into sections and by the use of a plani- 
 meter. The former is only an approximate method ; the area thus 
 found is nearly correct if the number of divisions is great. 
 
 Tangents at each end, perpendicular to the atmospheric line 
 are first drawn. The horizontal distance between these tangents 
 is then divided into 10 or more equal parts. The horizontal length 
 of each section is then divided into two equal parts and lines per- 
 pendicular to the atmospheric line drawn through these points of 
 
INDICATORS. 
 
 division. The sum of the lengths of all these lines is divided by 
 the number of lines to get the average. This average length or 
 average ordinate multiplied by the scale of spring gives the mean 
 effective pressure. 
 
 Fig. 20 is the card from the crank end of an engine. The 
 
 1 3' 5> 7 
 
 Fig. 20. 
 
 line C L is the atmospheric line and the lines A D and E F are 
 drawn perpendicular to it and tangent to the extreme ends of the 
 diagram. The line A E is divided into 10 equal parts and lines 
 
 Fig. 21. 
 
 are drawn through points marking the centers of the divisions. 
 On each of these lines the length is marked. The sum of the 
 lengths is 15.18 and 15.18 divided by 10 is 1.518. If the scale 
 of spring is 40 pounds, 1.518 multiplied by 40 is the M. E. P. or 
 60.7 =' M. E. P. 
 
INDICATORS. 
 
 The Jiorizontal length may be divided into any number of 
 equal parts but 10 or 20 makes the computation easy. The oper- 
 ation of finding the M. E. P. for the head end is exactly the same. 
 The average M. E. P. for one revolution of the engine is the 
 average of the two mean effective pressures. 
 
 In case the diagram is very irregular it should be divided 
 into 20 equal parts instead of 10. If there is a loop in the dia- 
 gram as shown in Fig. 21 the area of the loop must be subtracted 
 
 Pig. 22. 
 
 from the area of the other part as it represents work done by the 
 piston on the steam and therefore loss. 
 
 The lengths may be marked off on a piece of paper if a good 
 scale is not at hand. 
 
 A more accurate result is obtained by using an instrument 
 called the plani meter. There are several planimeters and aver- 
 aging instruments in common use for determining the mean effec- 
 tive pressure of indicator cards. The planimeter shown in Fig. 
 
34 INDICATORS. 
 
 22 is one of the most simple and is called the Amsler Polar Plain- 
 meter from its inventor Prof. Amsler. The cut is about one-half 
 the size of the instrument. It consists of two arms free to move 
 about a pivot and a roller graduated in inches and tenths of 
 inches. A vernier is placed with the roller so the areas may be 
 read in hundredths of a square inch. The point A is kept sta- 
 tionary and the tracer B is moved once around the outline of the 
 diagram. The area in square inches of the diagram is read from 
 the roller C and the vernier E. 
 
 To Use the Planimeter. The diagram should be fastened to 
 some flat unglazed surface, such as a drawing board, by means of 
 thumb tacks, springs or pins. The point A is pressed into the 
 paper so that it will hold in place. The point B is set at any 
 point in the outline of the diagram and the roller set at zero. 
 Follow the outline of the diagram carefully in the direction of 
 the hands of a watch as indicated by the arrows in Fig. 22 until 
 the tracer has moved completely around the diagram. The result 
 is then read to hundredths of an inch from the roller. Suppose 
 after tracing over the outline we find that the largest figure that 
 has passed the zero of the vernier is 3 ; the number of graduations 
 (tenths) that have passed the zero to be 5 and the number 
 (hundredths) of the graduations in the roller that exactly coincides 
 with a graduation on the vernier to be 9. Then the area is 3.59 
 square inches. 
 
 Often at the start the roller is not adjusted so that the zeros 
 coincide but the reading is taken and subtracted from the final 
 reading. Thus if the first reading is 4.63 and the second 7.31 the 
 area is 7.31 4.63 = 2.68 square inches. In case the second 
 reading is less than the first, add 10 to the second reading then 
 subtract. 
 
 This instrument is very valuable to an engineer who takes 
 indicator cards. The results obtained are very accurate, the error 
 being small. Ten or twelve diagrams can be measured by this 
 instrument in the same time that is necessary to measure a single 
 card by the method of ordinates. 
 
 It is well to run over the area three or four times and take 
 an average as the tracing of the diagram cannot be absolutely cor 
 rect at any time. 
 
INDICATORS. 35 
 
 THERflAL EFFICIENCY. 
 
 The thermal efficiency of the steam engine is found in the 
 same manner as that of any other heat engine. The efficiency 
 depends upon the limits of temperature and not upon the nature 
 of the working medium. 
 Let Tj = absolute temperature of the heat received by the engine. 
 
 T 2 = absolute temperature of the heat rejected by the engine. 
 
 E = efficiency of engine. 
 Then, 
 
 E = TI " ' T 2 
 
 or, the efficiency equals the temperature of the heat rejected, sub- 
 tracted from the temperature of the heat received and the result 
 divided by the temperature of the heat received. 
 
 Suppose an engine is supplied with steam at 120 pounds 
 absolute pressure and the exhaust is atmospheric pressure. What 
 is the efficiency? 
 
 The absolute temperature corresponding to 120 pounds abso- 
 lute pressure is 341.05 -|- 461 and the temperature of atmos- 
 pheric pressure is 212 -f- 461. 
 
 Then, 
 
 802.05 673 
 E = - _ = .16 or 16 per cent. 
 
 802.05 
 
 If the engine had been of the condensing type and the 
 exhaust pressure one pound above the vacuum, the efficiency 
 would be as follows : 
 
 The temperature of one pound absolute pressure is 101.99 
 + 461. 
 
 -, 802.05 562.99 
 
 -8005- 
 
 In actual engines this efficiency cannot be obtained because 
 the difference between the amount of heat received and that 
 rejected is not all converted into work. Part of it is lost by 
 radiation, conduction, leakage, etc. Also cylinder condensation 
 reduces the efficiency. 
 
 The Theoretical Indicator Diagram. An indicator diagram 
 is the result of two movements ; a horizontal movement of the 
 
3(5 INDICATORS. 
 
 paper and a vertical movement of the pencil. The horizontal 
 movement exactly corresponds to the movement of the piston of 
 the engine and the vertical movement exactly corresponds to the 
 pressure of steam in the cylinder. 
 
 The shape of the indicator card depends upon the manner in 
 which steam is admitted to and released from the cylinder. Dif- 
 ferent engines give different shaped indicator cards and the cards 
 taken from an engine vary with the conditions. Figs. 1 and 2 
 show theoretical indicator cards from a non-condensing engine 
 without clearance ; the former being for the case that has admis- 
 sion during the whole stroke. The diagram of Fig. 2 shows the 
 cut off at 4- stroke. All practical engines have clearance and 
 slight compression ; so the theoretical diagram assumes the shape 
 
 shown in Fig. 23. In this card 
 the admission line H A is verti- 
 cal, the steam line A C is hori- 
 zontal, the expansion line C D 
 an hyperbolic curve, the exhaust 
 line D B vertical, the back pres- 
 sure line B F horizontal and the 
 compression curve an hyperbola. 
 The actual shape is somewhat 
 different from the theoretical 
 
 mainly because the valves do not open and close quickly, the ports 
 offer some resistance to the passage of the steam and the back 
 pressure is neither atmospheric in the non-condensing engine nor 
 absolute vacuum in the condensing engine. 
 
 The diagram shown in Fig. 24 is a practical diagram and 
 like those taken from engines. 
 
 The atmospheric line L M is the line drawn by the pencil of 
 the indicator when the connection to the engine is closed and both 
 sides of the piston of the indicator are open to the atmosphere. 
 It is the zero of the steam gage. 
 
 The admission line H A shows the rise of pressure due to the 
 admission of steam to the cylinder. If the steam is admitted 
 qnickty when the engine is nearly on dead center this line will be 
 very nearly vertical, 
 
 The steam line A C is drawn while the valve admits steam 
 
INDICATORS. 
 
 37 
 
 to the cylinder. This line is horizontal if there is no wire- 
 drawing. 
 
 The point of cut off C, indicates the point at which the 
 admission of steam is stopped by the closing of the valve. This 
 point is rounding since the valve closes slowly. Sometimes it is 
 difficult to determine the exact point where cut off takes place ; it 
 is usually where the curve changes from concave to convex. 
 
 The expansion curve C D shows the fall in pressure as the 
 bteam expands while the piston moves toward the end of the stroke. 
 
 The point of release D shows the point at which the exhaust 
 
 N 
 
 M 
 
 rig. 24. 
 
 valve opens. The rounding is due to the slow action of the valve 
 when opening. Because of this slow action of the valve, release 
 begins a little before the end of the forward stroke. 
 
 The exhaust line D E F represents the loss in pressure which 
 occurs while the valve opens to exhaust at and near the end of the 
 stroke. 
 
 The back pressure line F G shows the back pressure against 
 which the piston acts during the return stroke. For a condens- 
 ing engine this line is below the atmospheric line L M, the dis- 
 tance below being dependent upon the state of the vacuum in the 
 condenser. For cards taken from a non-condensing engine the 
 back pressure line is a little above the atmospheric line. 
 
 The point of exhaust closure G is the point where the valve 
 
38 
 
 INDICATORS. 
 
 closes to exhaust. The exact point is not clearly defined as the 
 curve shows a change of pressure due to the gradual closing of the 
 valve. 
 
 The compression curve G H shows the rise of pressure due 
 to the compression of the steam remaining in the cylinder after 
 the valve lias closed to exhaust. 
 
 The zero line of pressure or line of absolute vacuum O X is 
 drawn below and parallel to the atmospheric line. The distance 
 between the lines O X and L M represents 14. T pounds pressure. 
 
 The clearance line O Y is drawn perpendicular to the line of 
 absolute vacuum and at a distance from the end of the diagram 
 
 , ,--}-- | 4*-- 
 
 PR X 
 
 Fig. 2o. 
 
 equal to the same per cent, of the length of the diagram as the 
 clearance volume is of the piston displacement, or 
 L N clearance volume 
 
 L M volume of cylinder 
 
 It is readily seen that the area of an actual indicator diagram 
 is less than that of a theoretical card. This is because of the 
 round corners at cut off and exhaust, the back pressure and the 
 compression. Sometimes it is useful, especially in designing 
 engines, to draw the theoretical indicator card. 
 
INDICATORS. 39 
 
 To Draw the Theoretical Card. To draw an ideal diagram 
 draw P X equal to the length of stroke and O P equal to the 
 clearance shown in Fig. 25. Draw O Y and P A perpendicular 
 to O X and draw Y S parallel to O X and at a height correspond- 
 ing to the boiler pressure. 
 
 The line of initial pressure A C is then drawn parallel to Y S 
 and is usually taken as from 90 to 95 per cent, of the boiler pres- 
 sure if there is no special cause for loss. Then take A C as the 
 
 P X 
 
 portion of the stroke at which steam is admitted so that = 
 
 A O 
 
 the ratio of expansion. The expansion line is considered a hyper- 
 bolic curve with O Y and O X as asymptotes. To draw the 
 hyperbolic curve. First draw the line A C B parallel to the 
 atmospheric line and F D B and R C perpendicular to it. Then 
 make points 1, 2, 3, 4, etc., on C B and connect them with the 
 point O. At the points 1', 2', 3', 4', etc., where these lines inter- 
 sect the line R C draw parallels to C B until they meet perpen- 
 diculars from 1 * 2, 3, 4. The point of intersection of these lines 
 are points in the hyperbolic curve C D, as shown in Fig. 25. 
 Any number of points may be used ; but there must be enough to 
 determine the curve. 
 
 The area A C D F N H is the theoretical card, with a given 
 boiler pressure, ratio of expansion and an assumed back pressure. 
 The actual card for the same data would be more nearly like the 
 shaded area which lies mostly within the outline of the theoretical 
 card. In designing engines it is well to know the ratio of the 
 actual to the ideal card for all types of engines. . , 
 
 This ratio varies with the speed, type of engine, and kind of 
 valves and has the following values. 
 
 For ordinary plain slide valve engines, .8 to .9 
 
 For small engines, about, .85 
 
 For large engines, about, .90 
 
 For engines with high speed, about, .75 
 
 For compound engines, .75 to .80 
 
 For compound engines, high speed, .65 to .75 
 
 For triple expansion engines, .60 to .70 
 
 For triple expansion engines, high speed, .50 to .60 
 
 
40 
 
 INDICATORS. 
 
 CARDS FOR COMPOUND ENGINES. 
 
 In Fig. 26 the ideal cards of a tandem compound engine of 
 the Woolf type are shown. The diagram ACDEFGH is 
 from the high pressure cylinder and the diagram L M N P Q R 
 from the low pressure. S T is the atmospheric line. 
 
 The line H A is the admission line, A C the steam line, and 
 C D the expansion line. These lines are similar to those of a 
 single cylinder engine diagram. At D, the point of release, the 
 pressure drops slightly as steam is admitted to the low pressure 
 
 Fig. 26. 
 
 cylinder. The back pressure line E F, of the high pressure cylinder 
 is parallel to the steam line L M, of the low pressure cylinder. 
 They would coincide if there were no resistance to the flow of 
 steam and no heat loss. The flow of steam from the high to the 
 low pressure cylinder is cut off at F and the steam remaining in 
 the high pressure cylinder is compressed in the cylinder and in the 
 pipes. The exhaust closes at G and steam is compressed in the 
 small cylinder from G to H. 
 
INDICATORS. 
 
 41 
 
 Cut off occurs in the low pressure cylinder at M, and the 
 steam in the cylinder expands, the curve M N being an equilateral 
 hyperbola. Release takes place at N and the pressure falls to 
 that of the condenser. The back pressure line P Q and the com- 
 pression curve Q K are like those of a single engine. 
 
 The diagrams shown in Fig. 27 are from a Woolf engine. 
 The lines are similar to those of the theoretical diagram shown in 
 Fig. 26. 
 
 With compound engines of the Woolf type the steam passes 
 
 Fig. 27. 
 
 directly from the high pressure cylinder to the low pressure. In 
 the cross compound, the cranks being at 90, the piston in one 
 cylinder is at the end of the stroke when the other is at the middle 
 of the stroke. Therefore, a receiver must be used. The ideal 
 cards from a cross compound engine are shown in Fig. 28. The 
 dimensions are as follows : 
 
 Volume of high pressure cylinder 
 
 Volume of receiver 
 
 Volume of low pressure cylinder 
 
 Cut off of high pressure cylinder 
 
 Cut off of low pivssure cylinder 
 
 Initial pressure (absolute) 
 
 Back pressure (absolute) 
 
 = 5 cu. ft. 
 
 = 5 cu. ft. 
 
 = 15 cu. ft. 
 
 = | stroke. 
 
 ^- stroke. 
 =120 Ibs. 
 
 = 3 Ibs. 
 
42 
 
 INDICATORS. 
 
 Clearance and compression are not considered. The stroke 
 of the high pressure cylinder begins at A. Steam is cut off at ^ 
 stroke at C. 
 
 Since steam is cut off at ^ stroke, the ratio of expansion in 
 the high pressure cylinder is 4. Then the volume of steam at D 
 is 4 times that at C and since p v = p^ v l , the pressure at I) is 
 
 OQ vx 1 
 
 ___ 80 pounds. Steam is released at D and passes to the 
 
 receiver. The pressure at release, L, of the low pressure cylinder 
 is found from the equation p v = p l v 1 . The volume of steam 
 in the high pressure cylinder at cut off, is | cubic feet and the 
 
 C, - .A 
 
 volume at L is 15 cubic feet; then the total ratio of expansion is 
 
 120 X 1 
 15 -r- | = 12. Then the pressure atL is L = 10 pounds. 
 
 Since the low pressure cylinder cuts off at J the stroke the 
 volume at G is J that at L. 
 
 Then as p v p l v^ p X \ = 10 X 1, or p = 20 pounds 
 
 Hence the pressure at G is 20 pounds. 
 
 We can now find the pressure at E. The pressure in the 
 receiver when the low pressure cylinder cut off, was 20 pounds 
 because the low pressure cylinder was in communication with it- 
 
INDICATORS. 43 
 
 Then as the piston in the small cylinder is at the end of the stroke 
 when the steam is cut off in the large cylinder, steam at 30 
 pounds pressure rushes from the high pressure cylinder into the 
 receiver and mixes with steam at 20 pounds pressure. When 
 steam is cut off in the low pressure cylinder the high pressure 
 cylinder and the receiver are in communication with each other ; 
 the total volume being the volume of the high pressure cylinder 
 plus that of the receiver or 5 -f- 5 10 cubic feet. What we 
 wish to find is the pressure of the steam of this volume. The 
 formula is, 
 
 P V = p v + p l v^ 
 
 Since there are 10 cubic feet in the receiver and high pressure 
 cylinder, the value of V is 10. The volume of steam in the 
 high pressure cylinder is 5 cubic feet and its pressure is 30 pounds. 
 The volume of steam in the receiver is 5 cubic feet and its pressure 
 is 20 pounds, then, 
 
 10 P = 5 X 30 + 5 X 20 
 
 P = _ = 25 pounds. 
 
 Then the pressure at E = 25 pounds. 
 
 While the piston of the high pressure cylinder is on the return 
 stroke, steam in that cylinder is compressed from E to F ; the 
 volume being 10 cubic feet. When the piston has completed ^ 
 the return stroke, the volume at F is equal to J the volume of the 
 high pressure plus the volume of the receiver or f -\- 5 = 7-J cubic 
 feet. 
 
 Then with the formula p vp^ v-^ we obtain, 
 
 25 X 10 OQ oo 
 p - 7JL = 33.33 
 
 or the pressure at F is 33.33 pounds. 
 
 Then as the-'cranks are 90 apart, and the high pressure piston 
 is at the middle of the stroke, the low pressure piston must be at 
 the beginning of its stroke ; or the pressure in the low cylinder is 
 the same as that in the high and in the receiver, or 33.33 pounds* 
 Since the stroke is beginning in the low pressure cylinder, steam 
 is being admitted to it from the receiver and consequently the 
 pressure in the large cylinder falls as the volume increases, which 
 
INDICATORS. 
 
 is shown by the line F G. Cut-off occurs in the low pressure 
 cylinder at G, and the steam expands until the piston reaches L ; 
 the curve being an equilateral hyperbola. At L the release occurs 
 and the pressure drops to that in the condenser ; in this case about 
 3 pounds. The back pressure line is of course the pressure in the 
 condenser. 
 
 The cards of Fig. 29 are from a cross compound engine. The 
 rise of pressure in the middle of the back pressure line of the 
 
 Fig. 29 
 
 diagram from the small cylinder is due to the fluctuation of pres- 
 sure in the receiver. 
 
 COflBINED DIAGRAHS. 
 
 The indicator cards of multi-cylinder engines may be com- 
 bined so that the pressures and volumes are shown in their proper 
 relations. To do this, the cards are reduced to the same scale of 
 pressure and the same scale of volume. To make the combined 
 diagram of convenient size, the length of the low-pressure card is 
 
INDICATORS. 
 
 45 
 
 left as it is and the length of the high-pressure diagram is shortened 
 in the ratio of the cylinder volumes. 
 
 Perhaps the best way to show the method is by an illustra- 
 tion. The cards shown in Fig. 30 were taken from a compound 
 condensing engine. 
 
 Ratio of cylinder volumes =1:3 
 Initial pressure = 138 pounds 
 
 Spring, high-pressure card 60 pounds 
 Spring, low-pressure card = 30 pounds 
 
 Fig. 30. 
 
 First draw the line of zero pressure A B and the line of zero 
 volume E F (see Fig. 31). The line O P is drawn parallel to E F 
 and at a distance proportional to the clearance of the low-pressure 
 cylinder. Similarly the distance between M N and E F represents 
 the clearance of the high-pressure cylinder. Now draw the atmos- 
 pheric line CD. In this case it will be .49 inch above A B 
 
 1 A 7 
 
 because a 30-pound spring is used, and . 1- =.49. Reproduce the 
 
 oO 
 
 low-pressure card without change, as shown. 
 
 Divide the high-pressure card with 10 (or more) ordinates, 
 and reproduce it with volumes and pressures of the same scale as 
 the low-pressure card. Since a 60-pound spring was used, each 
 
 f\C\ 
 
 ordinate will be twice as long (because _ = 2). The distance' 
 
46 
 
 INDICATORS. 
 
 between the ordinates will be |- as great because the high-pressure 
 cylinder is 1 the volume of the low. Draw the ordinates as shown. 
 The distance between ordinates L and M will be ^ of the length 
 of the high-pressure card. The points c and e are on the fifth 
 ordinate, and are twice as far above the atmospheric line as they 
 are in the original high-pressure card. 
 
 After locating all the points draw the curve through them. 
 Now draw the theoretical expansion curve R S through the point 
 of cut-off of the high-pressure cylinder by the method explained 
 
 Fig. 31. 
 
 on page 39. The difference between the area included between 
 the theoretical curve and the lines of no pressure and no volume, 
 and the sum of the actual areas, represents approximately the 
 
 This method is not strictly accurate, because all the steam 
 used in the high-pressure cylinder does not pass to the low-pressure 
 cylinder : a small portion is left for compression. By thus com- 
 bining the cards the action of the valves may be discussed, provided 
 such data as size, type and speed are known. 
 
 The combined diagram of a multi-expansion engine drawn 
 
INDICATORS'. 
 
 47 
 
 according to the above method is shown in Fig. 32. The volumes 
 and pressures are first reduced to the scale of the low-pressure 
 card. The cards are then redrawn at the proper distances from 
 the lines of no pressure and no volume ; the clearance in each 
 cylinder being considered. 
 
 Fig. 32 
 HORSE=POWER OF COMPOUND ENGINES. 
 
 The I. H. P. of multi-cylinder engines may be found by add- 
 ing the I. H. P. of the several cylinders. Another method is to 
 reduce all the pressu^s to the area of the low-pressure cylinder. 
 This is done by dividing the M. E. P. of each cylinder by the 
 inverse ratio of ics volume to that of the low-pressure cylinder. 
 
 Suppose the M E. P. of the high-pressure cylinder of a com- 
 pound engine is 78 pounds as found from the indicator card. It 
 the volume of the high-pressure cylinder is 1 that of the low, the 
 
 78 
 M. E. P. of the high referred to the low would be = 26 pounds. 
 
 o 
 
 Then if the card from trie large cylinder shows a M. E. P. of 30 
 pounds, the total M. E. P. is 30 -f- 26 56 pounds, and the 
 I. H. P. is found by inserting 56 as the value of P and using the 
 area of the low pressure as A, in the formula for L H P. 
 
48 INDICATORS. 
 
 INDICATOR CARDS. 
 
 The indicator reveals defects in the steam distribution. That 
 is, if the valve or valves are set so that the events of the stroke 
 are too early or too late or if more work is done at one end of the 
 cylinder than at the other, the engineer finds it out by examining 
 the indicator card. 
 
 Sometimes an engine appears to run well and the owner is 
 perfectly satisfied with it ; but from the indicator diagrams it is 
 found that considerable saving might be effected by correcting the 
 defects in the valve setting. 
 
 On looking at a diagram one might say it was a faulty card 
 
 B 
 
 Fig. 33. 
 
 and yet for that type, size and speed it is perhaps the best that 
 could be obtained from the engine. The same form of diagram is 
 not possible from different types of engines. The diagram from a 
 Corliss engine, having four valves, is different from that of the 
 plain slide valve engine ; also the diagrams from high speed engines 
 differ from those of low speed. 
 
 The most common faults in the distribution of steam in the 
 cylinder can be divided into four classes. 
 
 a = Admission too early or too late. 
 /> Cut-off too-early or too late. 
 c = Release too early or too late. 
 d Compression too early or too late. 
 In the following figures, 
 
 A is the point of admission, 
 C is the point of cut off, 
 B is the point of release, 
 G is the point of compression, 
 
INDICATORS. 
 
 49 
 
 The diagram shown in Fig. 33 shows too early admission. 
 
 The admission line curves backward instead of being straight and 
 perpendicular to the atmospheric line as it is in Fig. 24. The 
 diagram also shows cut off, release and compression early. When 
 the valve is of the plain slide valve type all the events are likely 
 to be too early if one of them is. "The reason for the event being 
 
 Fig. 34. 
 
 too early is that the eccentric has too much angular advance, 
 Hence the remedy is to decrease the angular advance. 
 
 Fig. 34 shows a diagram having the opposite defects to those 
 of Fig. 33. The events are too late. The admission line curves 
 forward and the line shows that admission does not take place 
 until after the stroke is well begun. Release occurs at the end of 
 
 B 
 
 Fig. 35. 
 
 the strike. In this case the angular advance of the eccentric 
 should be increased until the admission line is perpendicular to the 
 atmospheric line. 
 
 Fig. 35 shows a card having too much back pressure. This 
 may be due to a small exhaust port or pipes, or the passage of steam 
 
50 
 
 INDICATORS. 
 
 through coils of pipe for heating. The card shows the other events 
 to be good. A diagram having too late cut off is shown in Fig. 
 36. The pressure at release is high. When the engine is running 
 under this condition much of the benefit from expansion is lost. 
 
 B 
 
 Fig. 30. 
 
 A diagram from a condensing engine is shown in Fig. 37. 
 These oscillations are caused by the vibration of the indicator 
 piston and spring. To avoid these vibrations never use a very 
 light spring for high speed. 
 
 The diagram of Fig. 21 shows too early cut off. In this case 
 the cut off is so early that the expansion line extends below the 
 
 atmospheric line making a loop. The area of the loop must be 
 subtracted from the area of the card as explained on page 33. 
 
 Fig. 38 shows a pair of diagrams from a plain slide valve 
 engine. The admission lines are good. The sloping steam lines 
 
INDICATORS. 
 
 51 
 
 show wire-drawing due to the slow action of the valve or too 
 small ports or pipes. This wire-drawing decreases the area of the 
 card which indicates loss. The greatest fault is the inequality of 
 area of the diagram for the ends. The late cut off and conse- 
 quent late compression of one end causes more area than the too 
 early cut off and too early compression of the other end. These 
 
 Fig. 38. 
 
 cards can be improved by adjusting the angular advance of the 
 eccentric and the length of the valve rod. 
 
 The diagram of Fig. 39 indicates too early compression. The 
 compression curve extends above the initial pressure line. The 
 area of the loop must be subtracted from the card area, when com- 
 
 B 
 
 Fig. 39. 
 
 puting the I. H. P. If the cut off is kept the same and the com- 
 pression made what it should be, the gain in area would be the 
 area included -between the full line and the dotted line plus the 
 urea, of the loop. The remedy for this case is to decrease 
 the inside lap. 
 
52 INDICATORS. 
 
 The amount of compression varies with the speed and type of 
 engine. Slow speed engines require less compression or cushion- 
 ing than high speed. The exhaust steam should never be com- 
 pressed higher than the boiler pressure. 
 
 For high speed engines the compression should extend to 
 about .9 the initial pressure. For medium speeds about ./> and for 
 low speed .2 to .4. 
 
 In the case of a slide valve engine it is not always possible to set 
 the valve so that the card may have all the events as they should 
 be. Sometimes the' laps of the valves should be altered. For too 
 much compression decreases the inside lap. For too early cut off 
 decrease the outside lap. 
 
 If the valve travel is increased, compression is retarded, that 
 is, decreased ; release occurs sooner. 
 
 STEAM CONSUMPTION. 
 
 The amount of steam used by an engine is called its Steam 
 consumption and for comparison, it is customary to state the 
 amount of steam consumed per indicated horse-power per hour. 
 By means of the indicator diagram the steam consumption can be 
 computed approximately. 
 
 To find the Steam Consumption from the Diagram. The 
 
 diagram shown in Fig. 40 is from a 20 X 36 engine, running at a 
 speed of 80 revolutions per minute. A 40-pound spring was used. 
 
 By measuring the card, we find the mean ordinate to be .91 
 inch and the M. E. P. = .91 X 40 ---- 36.4 pounds. 
 
 I. H. P. = Engine Constant X M. E. P. X piston speed. 
 ;= .00952 X 36.4 X 480. 
 = 166.33. 
 
 In Fig. 40 L M is the atmospheric line and O R the line 
 of zero pressure drawn so that O L -= 14.7 pounds. O N is tin 1 
 clearance volume 8 per- cent of the piston displacement. The 
 line P Q is drawn from O li to some point 011 the compression 
 
INDICATORS. 
 
 53 
 
 curve. From D, a point on the expansion curve before release, 
 the line D F is drawn perpendicular to O R. 
 
 Then from the diagram, * 
 
 O R 3.24 inches. 
 O F = 3.00 inches. 
 O P = .345 inch. 
 D F = .795 inch. 
 P Q .795 inch. 
 
 The length of stroke is 36 inches or 3 feet, and the length of 
 the diagram 3 inches. Then 1 inch of the length of the card cor- 
 responds to 1 foot of the stroke. The scale of spring used is 40. 
 Therefore we can easily reduce the above dimensions to pounds 
 pressure and to feet. 
 
 O R = 3.24 feet. 
 O F = 3.00 feet. 
 OP .345 feet. 
 D F = 31.80 pounds. 
 P Q = 31.80 pounds. 
 The area of the piston (head end) is, 
 
 Z*L = 8 ' 1416 >< 4 <> Q = 314.16 sq. in. = 2.18166 so, ft. 
 
 Fig. 40. 
 
 We can now find the volume of steam at any point of the 
 
 irrroke. 
 
 When the piston is at 1), the volume is, 
 
 2.18166 X 3 = 6.54498 cubic feet. 
 
54 INDICATORS. 
 
 When the piston is at Q, the volume is, 
 
 2.18166 x -345 .75267 cubic foot. 
 
 From the steam tables we can find the weight of a cubic foot 
 of steam at a given pressure. 
 
 The weight of 1 cubic foot at 31.8 pounds absolute pressure 
 is .07773 pound. Then the weight of steam present when the 
 piston is at D is, 
 
 G.54498 X .07773 == .50887 pound. 
 The weight of steam present when the piston is at Q is 
 
 .75267 X .07773 == .05852 pound. 
 
 The weight of steam in the cylinder is .50887 pound and the 
 weight of steam kept for compression is .05852 pound. The 
 weight exhausted per stroke is, 
 
 .50887 .05852 = .45035 pound. 
 The amount used per I. H. P. per hour is, 
 
 .45035 X 2 X 80 X 60 ^ 
 
 166.33 
 
 This may be stated in words as follows : 
 
 Measure the pressure from the vacuum line to some point in 
 the expansion curve before release and from the steam tables find 
 the weight of a cubic foot at that pressure. Multiply the volume 
 (in cubic feet) of the cylinder, including clearance to that point, 
 by the weight per cubic foot. The result is the weight of steam 
 in the cylinder. As this weight includes the steam used for com- 
 pression it must be corrected to obtain the weight used per stroke. 
 Take some point on the compression curve and measure its 
 absolute pressure. Then compute the weight of steam to this 
 point. Subtract this weight from the weight of steam to the 
 point in the expansion curve and the result is the weight of steam 
 used per stroke. 
 
 Multiply the weight of steam used per stroke by the number 
 of strokes and divide by the indicated horse-power as found from 
 the card. The final result is the number of pounds of steam con- 
 sumed per horse-power per hour. 
 
 We may also calculate the steam consumption by taking the 
 point of the expansion curve near the cut-off. 
 
 O B = 1.23 inches = 1.23 feet of the stroke. 
 B C ='1.8 inches = 72 pounds. 
 
INDICATORS. 55 
 
 The weight of 1 cubic foot of steam at 72 pounds absolute 
 pressure is .1671 pounds. Then the volume of steam at C is, 
 
 2.18166 X 1.23 = 2.68341 cubic feet. 
 The weight at C is, 
 
 2.68344 X .1671 = .44840 pound. 
 
 The weight kept for compression is the same as previously 
 found, L e., .05852 pound. 
 
 Then the weight of steam used per stroke is, 
 
 .44840 .05852 = .38988 pound. 
 The steam consumption per I. II. P. per hour is, 
 
 .38988 X 2 X 80 X 60 
 
 166.33 
 
 = 22.50 pounds. 
 
 If the valve doesn't leak, the amount of sfeam just after cut 
 off should equal the amount just before release, but our calculation 
 shows tlui 4 ; there is .45035 -- .38988 = .06047 pound more at 
 release than at cut off. This shows that at entrance .06047 pound 
 was condensed before the piston reached C and was re-evaporated 
 before release. This calculation gives an idea of the amount of 
 cylinder condensation. 
 
 If there is considerable compression as hi I^ig. 40 the above 
 method may be simplified by taking the two points D and Q 
 at the same height above the vacuum line. The pressures will 
 then be the same. 
 
 Let W = weight of steam used per I. II. P. per hour. 
 
 w weight of one cubic foot of steam at the absolute 
 
 pressure D F. 
 
 L = length of the diagram, N R. 
 / distance between Q and D. 
 P mean effective pressure. 
 
 Then, 
 
 w _ 13,750 XwXl 
 P L 
 
 Example. A card was taken from a 1 2 X 14 engine. Length 
 of card L 3.5 inches, I 2.875 inches. Absolute pressure at 
 
56 INDICATORS. 
 
 D = 30 pounds. M. E. P. = 30.75 pounds. What is the steam 
 consumption per I. H. P. per hour ? 
 
 13,750 w X I 
 
 Mr 
 
 P X L 
 13.750 X .0736 x 2.875 
 
 30.75 X 3.5 
 27.03 pounds. 
 
 This method gives only an approximate steam consunr don 
 On account of the leakage of valves and the initial condensa 
 tion of steam in the cylinder, the actual consumption is somewhat 
 greater. The excess varies considerably and makes all results so 
 obtained of little value. In our calculation we used one diagram 
 only, that for the head end; we assumed the diagram from the 
 crank end to be the same. The indicated horse-power for each end 
 should be computed, as it is subject to considerable variation. In 
 the above formula the average mean effective pressure should be 
 used. The results of calculations of steam consumption from 
 indicator cards are so unreliable that engineers place little depend- 
 ence upon them. The results may be used as a basis for estimates, 
 but for accurate knowledge, an engine test must be resorted to 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Given the following data to find the M. E. P. Area of 
 card 2.79 square inches, length of card 3.1 inches, scale of spring 
 50. Ans. 45 pounds. 
 
 2. Steam enters a cylinder at a pressure of 210 pounds by 
 gage and leaves at 3 pounds pressure (absolute). What is the 
 thermal efficiency? Ans. 29 per cent. 
 
 3. An engine has a B. H. P. of 78.90. The I. H. P. is 
 86.73. What is the mechanical efficiency. Ans. 91 per cent. 
 
 4. An engine develops 149.97 I. II. P. If the piston 
 diameter is 17| inches, the stroke 30 inches, and the M. K. P. 40 
 pounds, what is the speed ? Ans. 100 revolutions per minute. 
 
 5. The theoretical M. E. P. of a triple expansion marine 
 engine is 48 pounds. What is the probable actual M. E. P.? 
 
 0. Tlu3 cylinders of a triple expansion are 12, )>(), and 75 
 inches in diameter respectively. The stroke is 24 inches. The 
 
INDICATORS. 
 
 57 
 
 mean effective pressures from the cards were 118.4, 51.6 and 
 19.84 pounds respectively. What is the I. IT. P. when the speed 
 is 125 revolutions per minute? Ans. 2,050 I. H. P. 
 
 7. Find the steam consumption from the following card. 
 The engine was running at 75 revolutions per minute, and devel- 
 oping 230 I. H. P. when the card was taken. A GO-pound spring 
 was used, and the M. E. P. for this end was 40 pounds. Assume 
 crank-end card to be the same as the head end. 
 
 Ans. 22 pounds (about). 
 
 H 
 
 \ 
 
 Suggestion: First draw a line .3 inch above the atmospheric line, 
 then reduce this ordinate to absolute pressure, and use formula at 
 the bottom of page 55. 
 
 VlfcBiaLA 
 
 Of THE 
 
 UNIVERS 
 
 OF 
 
INDEX 
 
 Part T VALVE GEARS; Part II STEAM ENGINE INDICATORS 
 
 Part Page 
 
 Adjustable eccentrics I, 54 
 
 Admission I, 9 
 
 Admission line II, 36 
 
 Area of cards, to find II, 31 
 
 Atmospheric line II, 36 
 
 Back pressure line II, 37 
 
 Balanced valves I, 41 
 
 Brake horse-power II, 28 
 
 Bridge, width of I, 31 
 
 Brown releasing gear I, 68 
 
 Clearance line II, 38 
 
 Combined diagrams II, 44 
 
 Compensation of cut-off I, 14 
 
 Compound engines, horse-power of II, 47 
 
 Compression I, 9 
 
 Compression curve II, 38 
 
 Corliss valve setting , I, 72 
 
 Crosby indicator : . . . II, 12 
 
 inserting spring II, 14 
 
 Cut-off.... .- I, 9 
 
 compensation of I, 14 
 
 point of I, 31 
 
 Double-ported valve I, 39 
 
 Double valve gears I, 58 
 
 Meyer I, 59 
 
 Drop cut-off gears I, 64 
 
 Brown releasing. I, 68 
 
 Greene I, 70 
 
 Reynolds-Corliss I, 65 
 
 Sulzer I, 71 
 
 Eccentric I, 4 
 
 Eccentrics, adjustable I, 54 
 
 Engine, putting on center I, 36 
 
 Engine constants, table II, 27 
 
 Exhaust lap I, 10 
 
 Exhaust line -.: II, 37 
 
 Exhaust port, width of I, 31 
 
II INDEX 
 
 Part Page 
 
 Expansion curve II, 37 
 
 Gears 
 
 Hackworth. '. I, 50 
 
 Joy I, 52 
 
 Marshall I, 52 
 
 Walschaert. I, 53 
 
 Goochlink... . I, 43 
 
 Greene gear I, 70 
 
 Hackworth gear I, 50 
 
 Horse-power of compound engines II, 47 
 
 Indicator cards II, 48 
 
 Indicator diagram, theoretical II, 35 
 
 Indicator diagrams, to take II, 22 
 
 Indicators. II, 9 
 
 Crosby II, 12 
 
 Tabor II, 15 
 
 Thompson.. .'. II, 10 
 
 Inequality of steam distribution I, 12 
 
 Joy gear I, 52 
 
 Lead, angle of I, 11 
 
 Lead of engines . I, 32 
 
 Link motion, Stephenson I, 42 
 
 Marshall gear .' I, 52 
 
 Mechanical efficiency of engine II, 31 
 
 Meyer valve I, 59 
 
 design of I, 60 
 
 Negative lap I, 15 
 
 Pantograph II, 19 
 
 Pencil mechanism of Tabor indicator II, 15 
 
 Piston valve I, 33 
 
 Plain slide valve. I, 3 
 
 Planimeter II, 33 
 
 to use II, 34 
 
 Point of cut off II, 37 
 
 Point of exhaust closure II, 37 
 
 Point of release II, 37 
 
 Power of engine II, 3 
 
 Prony brake II, 28 
 
 Radial valve gears. . . . I, 49 
 
 Reducing motion '. II, 18 
 
 Release I, 9 
 
 Reynolds-Corliss gear I, 65 
 
 Rocker '.. .1, 15 
 
 Rope brake. II, 29 
 
 Safety cam I, 68 
 
 Slide valve I, 3 
 
 designing I, 22, 30 
 
 modifications of . . . I, 38 
 
INDEX III 
 
 Part Page 
 
 Steam consumption II, 52 
 
 Steam distribution, inequality of I, 12 
 
 Steam lap I, 10 
 
 Steam line.... II, 36 
 
 Steam pipe, area of I, 30 
 
 Steam port, width of I, 31 
 
 Stephenson link motion I, 42 
 
 Sulzer gear I, 71 
 
 Tables 
 
 engine constants.. II, 27 
 
 laps, travel, and angular advance, effect of changing I, 22 
 
 Tabor indicator II, 15 
 
 atmospheric line, change of location of II, 16 
 
 attaching to engine II, 17 
 
 care of II, 17 
 
 pencil mechanism II, 15 
 
 springs II, 16 
 
 Theoretical card, to draw II, 39 
 
 Theoretical indicator diagram II, 35 
 
 Thermal efficiency II, 35 
 
 Thompson indicator . II, 10 
 
 care of II, 12 
 
 change springs II, 11 
 
 Three-way cock II, 18 
 
 Trick valve I, 41 
 
 Valve diagrams I, 16 
 
 Zeuner's I, 16 
 
 Valve gears. 7 ; . . . I, 3 
 
 double I, 58 
 
 problems on , I, 22 
 
 Valve setting I, 35 
 
 Valves 
 
 balanced I, 41 
 
 double-ported I, 39 
 
 with lap I, 7 
 
 without lap f . . . . 'I, 5 
 
 piston I, 38 
 
 plain slide I, 3 
 
 setting for equal cut-off. I, 37 
 
 setting with equal lead . I, 37 
 
 trick I, 41 
 
 Walschaert gear I, 53 
 
 Watt's diagram of work II, 8 
 
 Work of engine, definition of II, 3 
 
 Zero line of pressure II, 38 
 
 Zeuner's diagram I, 16, 55 
 
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