Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004646935 TJ 515.K6r e " UniVerS,,y Ubrary °lSEmuSS&SBSS l s,eam <*ng'ne.Note 3 1924 004 646 935 The D. Van Nostrand Company intend this book to be sold to the Public at the advertised price, and supply it to the Trade on terms which will not allow of discount. DESIGN OF A High-Speed Steam Engine. Notes, Diagrams, Formulas and Tables. BY J. F. KLEIN, Professor of Mechanical Engineering in Lehigh University. SECOND EDITION, REVISED AND ENLARGED. New York : D. VAN NOSTRAND COMPANY, 23 Murray and 27 Warren Strekts. 1903. Copyright 1903, by J. F. Klejn. PREFACE. When the first edition of this book was published the intention was to make it Part I of a treatise on the Dynamics and Design of a High-Speed Steam Engine. Part II was to treat of the Dynamics and Design of Shaft Governors. This second part is now nearly finished, but proves to be so difficult and volumi- nous as to make it unsuitable for most undergraduate work in technical schools; it will, therefore, be issued as a separate and independent work. The present volume drops its old designation as Part I, corrects the typographical errors of the first edition, and adds much new material in the form of appendices. J. F. Klein. Bethlehem, Pa., February, ipoj. NOTES, DIAGRAMS, FORMULAS AND TABLES FOR THE DESIGN OF A HIGH-SPEED STEAM ENGINE Data. Determine the principal dimensions of an engine fulfilling the following requirements : The work ordinarily required, = ioo H. P. net = no I.H. P., but the load may vary from 50 I. H. P. to 1 70 /. H. P. The engine is to run steadily under all these loads, the total variation (N T — N 2 ) of the number of revolutions of engine shaft being less than ^ of the normal number of revolutions (N) i. e., N t — N, = — when N = Nl + N * . (1) 50 2 The engine is to be of the non-condensing, high (rotative) speed type, durable, of compact and rigid form, and the steam consumption of the 100 H.P. per hour is to be a minimum. Moreover, this high-speed engine must be provided with gen- erous wearing surfaces which are continuously lubricated, the reciprocating parts must be balanced and their weights so chosen that the tangential pressures on the crank pin will be as nearly constant as possible. ORDER OF CALCULATION. It is assumed that the workmanship will be excellent, the par- allelism of crank pin and shaft, as well as the alignment of crank- shaft bearings, perfect, and that they will be maintained so by good foundations. Order of Calculation. I. Determination of diagram of effective steam pressures on piston of required engine. a. Determination of boiler pressure. b. Determination of clearance, real cut-off and apparent cut-off. c. Determination of back pressure and amount of com- pression. d. Method of constructing equilateral hyperbola and of determining points on the curves pv n = constant by tabulated coordinates. e. Methods of determinining the mean effective pressures of indicator diagrams. f. Determination of the cut-off corresponding to a given horse-power H x when the mean effective pressure p m and the total pressure p" m is known for another horse-power H of the same engine running at the same speed, with same initial pressure /, and same counterpressures. g. Diagrams of effective-steam-pressure-on-piston for for- ward and return strokes. II. Determination of ratio of length of connecting rod to that of crank. III. Determination of that mean accelerating force (per □" of piston) necessary to start reciprocrating parts and cor- responding to the most uniform tangential pressure on crank pin. a. Determination of acceleration and construction of dia- grams of accelerating forces. ORDER OF CALCULATION. 3 b. Construction of horizontal or resultant pressures on cross-head pin (or on crank pin when friction is neg- lected) by combining diagram of effective steam pres- sures on piston with diagram of acceleration. c. Discussion of effects on this diagram of valve setting, high speed and long and short cut-offs. d. Conversion of horizontal pressures on crank pin into tangential pressures. e. Construction of diagrams of tangential pressures on crank pin. IV. Determination of diameter of cylinder, length of stroke, revolutions per minute and weight of reciprocating parts. V. Preliminary estimate of dimensions of reciprocating parts, to see whether minimum weight of latter does not exceed that necessary to produce steadiness of running under the conditions assumed. VI. Determination of weight of fly-wheel rim from tangen- tial pressure diagram. VII. Determination of dimensions of crank disk and distri- bution of its material so that both crank arm and recip- rocating parts will be balanced. VIII. Method of determining the influence, on the foregoing results, of the frictional resistances, of the weight of the rod and of the exact values of the forces of inertia. IX. Calculation of width and diameter of belt pulley. X. Graphical determination of diameter of crank shaft. XI. Calculation of the length of crank shaft journal and graphical determination of the plane of division of its brasses. XII. Calculation of dimensions of steam ports and pipes, also thickness of cylinder walls. XIII. Valve diagrams and dimensions of valve and gear. XIV. Drawing of details of engine. XV. Drawing of plan and elevation of complete engine. 4 determination of boiler pressure. Plates to be Drawn. I. Six Indicator Diagrams. II. Six Diagrams of effective steam pressures on piston. III. Diagrams of accelerating and retarding pressures by approximate and exact methods. IV. Six Diagrams of horizontal pressures on cross-head pin. V. Diagram for converting horizontal pressures on cross- head pin into tangential pressures on crank pin. VI. Six Diagrams of tangential pressures on crank pin. VII. Four Diagrams of the forces shaking engine bed, for as many different degrees of counter-weighting. VIII. Distribution of material on crank disk, and determina- tion of its center of gravity. IX. Determination of diameter of engine shaft by the methods of Graphical Statics. X. Graphical determination of the direction of the resultant pressure on crank-shaft bearing. XI. Valve Diagrams. Plates containing details, plan and elevations of com- plete engine. I A. Determination of Boiler Pressure. If we consider the cylinders as perfect non-conductors of heat, and consider economy of steam only, we must, in accordance with the deductions of thermodynamics, make our initial steam pressures as high as possible, for by so doing we increase the range of temperature of the steam in the engine, and thereby increase the efficiency of the engine. This is clearly shown by the expression for the thermal efficiency tj of the engine T T V = — T — (2) 7", being the absolute temperature of the steam at the initial absolute pressure p lt and T 2 the lowest absolute temperature DETERMINATION OF BOILER PRESSURE. 5 practicably possible for the engine, being in non-condensing engines T 2 = 21 2° + 459-°4 = 671. "4 Fahr. T x increases very slowly with the higher pressures of steam, see Roentgen's Ther- modynamics, pp. 672 and 674. Even for ordinary (cast iron) conducting cylinders it is true (up to not very well defined limits) that economy in the use of steam increases slowly with the initial pressure. Thus according to Prof. Thurston the weight of steam per hour per H. P. should in good engines be not more than that given by the empirical formula 200 y£ Pounds (3) and for the best practice with large engines, dry steam, high piston speed and good design, construction and management the consumption of steam should be not more than 150 In (3) and (4) p x is the absolute pressure in pounds per sq. inch. Sometimes the highest pressure which can be employed in an engine is prescribed by the strength of the boilers which are to furnish it with steam ; at other times (with engines running under conditions in which the power occasionally sinks considerably below the average) a lower pressure (accompanied by a greater cut-off) than economy of fuel and steam would prescribe, must be employed in order that the engine may run at high speed with great steadiness and without reversing the direction of the hori- zontal pressures on crank pin. See limits to high speed, III, c. The employment of very high pressures is also limited by the greater first cost of the boilers owing to the greater strength required, and the greater care which must be exercised in their construction. The engine also becomes somewhat more costly for the same reason. A practical limit is thus set to the general use of extremely high pressures. O CLEARANCE. In the present case we will start the calculation by assuming the initial boiler pressure employed by many makers of high- speed engines when designing and rating their engines, namely, 85 pounds gauge pressure. Ib. Determination of Clearance, Real Cut-off and Apparent Cut-off. Clearance is the volume or space included between the piston (when the latter is at the end of its stroke) and the cylinder head, and also includes the steam passages. It is usually expressed as a fraction of the volume swept through by the piston, i.e., Hi repre- sents the clearance, A the area of piston and 6" the stroke we have clearance volume /.\ * _ AS The clearance should be as small as possible, for when it is at all large it involves considerable loss by presenting a large space to be filled by fresh steam, which steam is not engaged in push- ing back the piston when entering the cylinder, but in giving whirling motions to its own particles, though during expansion a portion of the energy of the steam which has filled the clear- ance spaces is utilized. On account of the loss occasioned by it, the clearance is sometimes called the hurtful space. Its hurtful influence can, however, be entirely removed by allowing the ex- haust steam to be compressed till it reaches a pressure equal to the initial pressure. When this is done no fresh steam is needed to fill the clearance spaces at the beginning of the stroke ; but as it is not always practicable (as will appear later) to compress the exhaust steam to so great an extent, it behooves the designer to diminish the clearance as much as possible. The total amount of engine clearance employed in practice varies greatly with the type and size of engine employed. For example, in Corliss Engines the four separate valves employed REAL CUT-OFF. 7 are placed close to the cylinder and the distance between end of piston and cylinder head left for variation of length of con- necting rod (when the wear is taken up) is frequently not more than %" , so that in these engines the clearance often does not exceed yh of one per cent. In high-speed engines, however, the steam passages are usually very large, as it is very important to admit the steam promptly, and at full pressure. In these engines the clearance sometimes amounts to 14 per cent., but this is very large. We will for the present assume that the clearance in our high-speed engine is as great as 7 per cent. Cut-off. See Fig. 3, p. 16. The volume of steam in the cylinder (on the live steam side of the piston) when the admission valve closes, divided by the vol- ume of this steam (after it has expanded) when the exhaust valve opens, is called the real or virtual cut-off =- = e. It is always a proper fraction. The reciprocal of this fraction is the expan- sion, and is of course a whole or mixed number. The distance of the piston from the end of its stroke when the admission valve closes, divided by the whole stroke, is the appar- ent cut-off = — = e 1 . The relation between the real and apparent cut-off is represented by the following equation : $A + iSA I -f i real cut-off = *- = <^ + iSA = f^ = <■ (6) The following tables, pp. 8 and 9, give this relation, and were taken from Du Bois' translation of Weisbach, Vol. II. If our cyl- inders were non-conductors, and economy of steam were the only element of economy to be considered, we know from thermo- dynamics that it would be wise to employ both high pressures and great expansions. Unfortunately our cylinders are con- ductors of heat, and as a consequence there is a very marked REAL CUT-OFF. TABLE I. Real Cut-off Corresponding to Apparent Cut-off for Different Fractions of Clearance. *J +J So Clearance. Cu," 1- O 2.3 .05 2.15 0.3 .05 .01 .02 ■°3 .04 ■°5 .06 .07 .08 .120 09 .128 .10 .059 .069 .078 .087 .095 .104 .112 .136 .06 .069 .078 .087 .096 .105 "3 .122 .130 .138 .145 .06 .07 .079 .088 .097 .106 .114 .123 •131 •139 .147 -155 .07 .08 .089 .098 .107 .115 .124 .132 .140 .148 .156 .164 .08 .09 .099 .108 .117 .125 •133 .142 .150 •157 .165 • 173 .09 .10 .109 .118 .126 •135 •'43 .151 •159 .167 ■174 .182 .10 .11 .119 .127 .136 .144 .152 .160 .168 .176 .183 .191 .11 .12 .129 •137 .146 .154 .162 .170 .178 .185 •193 .200 .12 •13 •!39 .147 •155 .163 .171 .179 .187 .194 .202 .209 ■13 •14 .149 •157 .165 •173 .181 .189 .196 .204 .211 .218 .14 ■15 .158 .167 • 175 .183 .190 .198 .206 .214 .220 .227 ■ l S .16 .168 .176 .184 .192 .200 .208 .215 .222 .229 .236 .16 • 17 .178 .186 .194 .202 .210 .217 .224 ■ 231 •239 .245 •17 .18 .188 .196 .204 .212 .219 .226 •234 .241 .248 •255 .18 •19 .198 .206 .214 .221 .229 .236 •243 .250 .257 .264 .19 .20 .208 .216 .223 •231 .238 •245 .252 •259 .266 •273 .20 .21 .218 .225 •233 .240 .248 •255 .262 .269 .275 .282 .21 .22 .228 ■235 • 243 .250 ■257 .264 .271 .278 .284 .291 .22 •23 .238 •245 .252 .260 .267 .274 .280 .287 •294 .300 •23 .24 .248 •255 .262 .269 .276 .283 .290 .296 •3°3 •3°9 •24 ■25 ■257 .265 .272 ■279 .286 .292 .299 .306 .312 .318 •25 .26 .267 .275 .282 .288 •295 .302 .308 ■3'5 .321 •327 .26 •27 ■277 .284 .291 .298 •3°5 •3" .318 •324 ■33° ■336 .27 .28 .287 .294 .301 .308 ■3H .321 ■327 •333 •339 •345 .28 .29 •297 •3°4 ■3H ■317 ■324 ■33° •336 •343 •349 •355 •29 ■30 •307 •3H .320 ■327 •333 •340 •346 •352 •358 •364 •30 ■31 •317 •324 •33° ■337 •343 •349 •355 .361 •367 •373 ■31 •32 .327 •334 •34o •346 ■352 ■358 .364 •37° •376 .382 ■32 •33 •337 •343 •35° •356 .362 .368 ■374 .380 .385 ■391 ■33 •34 •347 •353 ■359 .365 • 371 •377 .383 •389 •395 .400 ■34 •35 •356 ■363 •369 ■375 .381 ■387 •393 ■398 .404 .409 ■35 •36 .366 •373 •379 •385 •390 •396 .402 .407 •413 .418 ■36 •37 •376 .382 .388 •394 .400 .406 .411 ■417 .422 •427 ■37 •38 .386 •392 ■398 .404 .410 .415 .421 .426 •431 •436 .38 •39 •396 .402 .408 ■413 .419 ■425 •43° •435 .440 ■445 •39 .40 .406 .412 .417 •423 .429 •434 •439 •444 .450 •455 •40 REAL CUT-OFF. TABLE I (Continued). Real Cut-off Corresponding to Apparent Cut-off for Different Fractions of Clearance. gsri i- o .41 Clearance. OJ 1 $3 .01 .02 ■03 .04 .05 .06 .07 .08 .09 .10 .416 .422 •427 ■433 •438 •443 •449 ■454 •459 .463 •41 .42 .426 •43i •437 .442 .448 ■453 •458 •463 .468 •473 .42 ■43 •436 ■44' •447 .452 •457 .462 .467 .472 •477 .482 •43 ■44 .446 .451 .456 .462 .467 •472 •477 .481 .486 .491 •44 ■45 •455 .461 .466 •471 .476 .481 .486 .491 •495 .500 ■45 .46 .465 .471 •475 .481 .486 .491 •495 .500 •5°5 .509 .46 •47 •475 .480 ■485 .490 •495 .500 .505 .509 •5H .518 ■47 .48 ■485 .490 •495 .500 .505 .509 •5H •5i9 ■523 .527 .48 •49 •495 .500 .505 .510 .514 •5'9 •523 .528 •532 •536 •49 .50 •5°5 .510 •515 ■519 ■ 524 .528 •533 •537 .541 •545 .50 •Si ■5i5 .520 •524 •529 •533 .538 •542 .546 .550 •554 •51 .52 .525 •529 ■534 •538 • 543 ■544 ■551 .556 .560 .564 •52 •53 •535 •539 • 544 •548 ■552 •557 .561 .565 .569 •573 •53 •54 ■545 ■549 •553 •558 .562 .566 .570 •574 •578 .582 •54 •55 ■554 •559 •563 .567 •57i • 575 •579 .583 .587 •591 •55 .56 .564 .569 •573 • 577 .581 .585 •589 •593 .596 .600 .56 •57 •574 • 578 .583 .587 .590 ■594 •598 .602 .606 .609 •57 .58 .584 .588 •592 .596 .600 .604 .607 .611 .615 .618 .58 •59 •594 ■598 .602 .606 .610 .613 .617 .620 .624 .627 ■59 .60 .604 .608 .612 .615 .619 .623 .626 .630 •633 .636 .60 .61 .614 .618 .621 .625 .629 .632 .636 •639 .642 .645 .61 .62 .624 .627 .631 •635 .638 .642 .645 .648 .651 .655 .62 •63 .634 .637 .641 .644 .648 .651 .654 .657 .661 .664 •63 .64 .644 .647 .650 .654 .657 .660 .664 .667 .670 •673 .64 .65 •653 .657 .660 .663 .667 .669 •673 .676 .679 .682 .65 .66 .663 .667 .670 .673 .676 .679 .682 .685 .688 .691 .66 .67 •673 .676 ,€80 .683 .686 .689 .692 .694 .697 .700 .67 .68 .683 .685 .C89 .692 .695 .698 .701 .704 .706 .709 .68 .69 ■693 .695 ,t99 .702 .705 .708 .710 •713 .716 .718 .69 .70 •703 .706 •/09 .712 •7H .717 .720 .722 .725 .727 .70 ■7i •713 .716 .718 .721 •724 .726 .728 •731 •734 •736 •7i •72 •723 .725 .728 •731 •733 •73 6 .738 ■741 •743 •745 •72 •73 •733 ■735 •738 .740 •743 •745 .748 .750 •752 •755 •73 •74 •743 •745 .748 .750 ■752 •755 .756 •759 .761 .764 •74 •75 .752 •755 •757 •76° .762 .764 .766 .769 .771 •773 •75 IO REAL CUT-OFF. loss arising from the condensation of the live steam as it enters the cylinder and comes in contact with the surfaces of the latter which have just been cooled by contact with, and giving up their heat to, the exhaust steam. The complexity of this interchange of heat is such that thus far it has not been satisfactorily followed by calculation, and we must depend upon experiment for our knowledge of what constitutes the most economical point of cut-off with respect to the consumption of steam or fuel. This cut-off may be determined by means of an empirical formula representing the result of extensive and carefully conducted experiments by Mr. Charles Emery, namely, 22 I l _ r~ ! + A ~" 22 + P- 22 (7) e = real cut-off. r = ratio of expansion. p z = absolute initial pressure. Mr. Emery considers that the cut-offs given by this formula are " nearly correct for single engines of ordinary construction, and too large for the better class of compound engine." The first two columns of Table IV on page 18 contain the initial absolute pressure p T for various values of FOR/ IN LINEAR MEASURE MULTIPLY 7- BY fa EXPRESSED IN LINEAR A MEASURE. "o » > E- •O _- Exponent n ir formula fl z v z » = pv n . to a rt rr .2 -4-> J?3 d is rt 1 e < a a txOe '■3 g 1 0) O-rt « ™ w -~ *i ^ H be A . . 0J ■ 3 1 Ratio of assum to initial v> Equilateral hyperb curve for air expani ith temperature con n = 1. 1 = i .0646 = || ne urve for steam, dry rated during expans II 5 1 m- 3^. ° .2 ^ n= 1.135. diabatic curve for s dry saturated at be ning of expansioi ft = 1.250 = f. impression curve foi peed engine with St. jacketed cylinder » = 1-333 = i- diabatic curve for s heated steam. u **H it 1- 3 O II .a a 3 a ■5 < Vl £ O O* a ^ II £ i 1 s TO !_, 13 8g II s d 1 3 u ~3 Pi . "3 C u a i> II g II | frko 3 d ■ "(3 II fc S > 1 3 8g II I HH 4-1 1 3 "c3 4J XI a >- 11 s a" TO )_ S3 c 3 II 3 353 ^•3 5> rt •^•3 •S,w .01 W .26 w •51 W .76 W .0561 .0500 .6102 •5959 •8534 .8463 .9686 .9668 .02 .0982 .0884 .27 .6235 .6594 .52 .8600 .8532 ■77 •9713 .9696 •03 ■ 1352 .1245 .28 .6364 .6226 •53 .8665 .8599 •78 •9738 ■9723 .04 .1688 .1566 •29 .6490 •6355 •54 •8727 .8664 ■79 .9762 •9749 •OS .1998 .1866 •30 .6612 .6479 •55 .8788 .8727 .80 ■9785 ■9773 .06 .2288 .2148 •31 •6731 .6601 .56 .8847 .8789 .81 .9807 .9796 .07 .2562 .2415 •32 .6846 .6719 •57 .8904 .8848 .82 .9827 .9817 .08 .2821 .2669 •33 .6959 •6835 .58 •8959 .8906 •83 •9847 .9838 .09 .3067 .2912 •34 .7068 .6947 • 59 .9013 .8962 .84 .9865 ■9857 .10 .3303 ■3H5 •35 ■7174 .7056 .60 .9065 .9017 •85 .9881 .9874 .11 .3528 ■3368 .36 .7278 ■7163 .61 .9115 .9069 .86 .9897 .9891 .12 •3744 •3583 •37 •7379 .7267 .62 .9164 .9120 .87 .9912 .9906 .13 •3952 •379° •38 ■7477 .7368 •63 .9211 .9169 .88 .9925 .9920 .H •4IS3 •399° •39 .7572 .7466 .64 .9256 .9217 .89 •9937 •9933 .15 •4346 .4183 .40 .7665 .7562 .65 .9300 .9263 .90 .9948 •9945 .16 •4532 •437° .41 .7756 .7656 .66 •9342 .9307 •9i •9958 .9956 .17 .4712 •4552 •42 .7844 •7747 .67 •9383 •935° .92 .9967 .9965 .18 .4887 .4727 •43 .7929 •783S .68 •9423 •939i •93 •9975 •9973 .19 •5°5S .4897 •44 .8012 .7921 .69 .9460 ■943i •94 .9982 .9981 .20 .5219 .5062 •45 ■8093 .8005 .70 •9497 .9469 •95 •9987 .9987 .21 • 5377 .5223 .46 .8172 .8087 •71 •9532 .9506 .96 .9992 .9991 .22 •5531 •5378 •47 .8249 .8166 •72 ■9565 •9541 •97 .9996 •9995 .23 .5680 ■ 5S30 .48 •8323 .8244 •73 •9597 •9575 .98 .9998 •9998 .24 .5825 •5677 •49 .8395 .8319 •74 .9628 .9607 •99 •9999 •9999 •25 .5966 .5820 .50 .8466 ' .8392 •75 .9658 .9638 i8 MEAN EFFECTIVE PRESSURE. TABLE IV. Values of the Mean Effective Pressures p m Corresponding to Economical Point of Cut-off, 22 22 4- p z Compression takes place up to a pressure = _A_+_i6 _.ft =A- 2 A Here all Steam Curves follow the Law pfjil = constant. See reference to this Table on page lb. Real Cut-off Absolute Clearance. i pressures r A .01 •03 .05 .07 .09 .IO 198. 44-27 34-31 26.43 18.53 10.65 .ii 178. 40.62 33-94 27.27 20.61 13.96 .12 161.3 38.93 33-18 26.56 21.67 15-99 •13 147.2 37-3 32.16 27-37 22.39 17-4 .14 I3S-I 35-8 31-4 27.1 22.7 18.3 • 15 124.7 34-3 3°-4 26.6 22.8 19.0 .16 115.5 32.8 29.4 26.1 22.7 19.4 •17 107.4 31-4 28.4 25.5 22.5 19-5 .18 100.2 30.2 27.4 24.8 22.1 19-5 • 19 93-8 28.8 26.4 24.0 21.7 19-3 .20 88.0 27.5 25.4 23-3 21.2 19.1 .21 82.8 26.3 24.4 22.5 20.6 18.8 .22 78.0 25.1 23.1 21.7 20.0 18.4 •23 737 24.0 22.5 20.9 19.4 17.9 .24 69.7 22.9 21.5 20.0 18.8 17.6 .25 66.0 21.8 20.6 19-3 18.1 16.9 .26 62.6 20.8 19.7 18.5 17.4 16.3 .27 59-5 19.8 18.7 17.7 16.7 15-7 .28 56.6 18.8 17.9 17.0 16.1 15.1 .29 53-9 17.8 17.0 16.2 15.4 14.6 ■30 5i-3 16.9 16.1 15.4 15.2 13-9 CUT-OFF FROM MEAN EFFECTIVE PRESSURE. 19 I F. To determine the cut-off corresponding to a given horse-power H x , when the mean effective pressure p m and the mean total pres- sure p" m (see Table III and Fig. 3) is known for another horse- power H of the same engine running at the same speed, with the same initial pressure p x and the same back pressure and compression. In engines with cut-off under the control of the governor, the speed (i.e. R.p.m.) remains constant. Hence we have pm ■ pmi '■'■ H : H x andp ml = -£p„ (IO) Fig- 4- By means of Fig. 4, if we suppose the line of counter-pressures DEF to remain the same for all variations of power, we can deduce the following relations : area BCDEFB stroke HI area ACDIGA length GI = Pm — 6" — y fi AB area BC X LD Z MEFB stroke HI area ACJLDJGA _ length GI = Pmx (11) /"«.. (12) In the figure, —- = clearance i and AG = initial pressure^. HI Table III on page 17 contains the values of for various 20 CUT-OFF FROM MEAN EFFECTIVE PRESSURE. real cut-offs and for the two steam curves pv = constant and p x v r rh = constant. A = area ABFEMDIGA GI or area A CDIGA — area B C DMEFB _ p m , . ^ = GH+HI ~ P m i + i Kl) P" mi x G/ = p mi X HI + p.X GI ,, _ Pmz , .,, Pm }> mi , ■ V F m I+Z I+Z hence ^=-' = £=7=^ + ^ = ^= + ( ^—^ ) ' " («4) A (i + 0a a a \ i + 1 • '■ In a like manner it may be shown that P"m, _ P\n + // \pm Pi ~~ A \ X + * / A where p" m2 is the corresponding mean total pressure for the horse-power H 2 . The quantities in the second members of these equations are either given directly or may be found from the diagrams already constructed or from the tables. This enables us to calculate pit pi i J L^1 an( j -c_^? an( j from these, by means of Table III on page A A 17, we can get the corresponding cut-offs — and — • ' 1 '2 Calculate the cut-offs corresponding to 50 and 170 I.H.P., the extremes of power for the proposed engine. Then draw two indicator diagrams for each of these powers. DEFINITION OF FORWARD AND RETURN STROKES. 21 I G. Diagrams of Effective Steam Pressures. Draw diagram of effective steam pressures on piston both for the forward and return strokes. We will here state what is to be understood by the forward and the return stroke. In all cases suppose the engine to lie horizontally, with the crank on the right hand and the cylinder on the left. The forward stroke will then be when the piston moves to the right or when the crank moves in its upper or lower semi-circle so that it first passes through the quadrant nearest the cylinder and then through the quadrant farthest from the cylinder. This does not agree with locomotive practice where the forward end of the cylinder is its front end, and the stroke of the piston toward that end the for- ward stroke, but it agrees with the practice of many engineers when treating of stationary engines. To avoid all ambiguity as ,Jad £" I 1 — cos to ) for point T ) for point 5. 2-5 / 2-5 cos a) (16) 26 HIGH-SPEED STEAM ENGINE. TABLE V. Distance of Piston from Beginning of Stroke when Stroke is towards Crank-shaft. To find Actual Distance of Piston from end of Stroke, multiply tabular quantity by Stroke. * ft 2 Connecting-rod -h Crank = ^ D Connecting-rod -^ Crank : 4 .0004 4.5 5 5.5 6 7 8 CO 4 4.5 5 5.5 6 7 8 00 .0004 .0004 .0004 .0004 .00031.0003 .0003 92 .5809 .5737 .5679 .5632 .5594 .5533 .5487 5174 4 .0015 .0015 .0015 .0014 .0014 .0014 .0014 .0012 94 .5981 .5909 .5851 .5805 .5766 .6706 .5661 .5349 6 .0034 .0033 .0033 .0032 .0032 .0031 .0030 .0027 96 .6151 .6079 .6022 .5977 .5938 .5878 .5833 .5523 8 .0061 .0060 .0058 .0057 .0057 .0056 .0055 .0049 98 .6318 .6247 .6191 .6145 .6107 .6048 .6004 .5696 10 .0095 .0092 .0091 .0089 .0089 .0087 .0085 .0076 100 .6484 .6414 .6358 .6313 .6275 .6216 .6172 .5868 12 .0136 .0133 .0131 .0129 .0127 .0126 .0124 .0109 102 .6647 .6578 .6523 .6478 .6441 .6384 .6340 .6040 14 .0185 .0181 .0178 .0175 .0173 .0170 .0167 ■0149 104 .6807 .6739 .6685 .6641 .6605 .6548 .6505 .6210 16 .0241 .0236 .0232 .0228 .0225 .0221 .0218 .0194 106 .6964 .6897 .6845 .6802 .6766 .6710 .6668 .6378 18 .0304 .0298 .0293 .0290 .0285 .0279 .0275 .0245 108 .7119 .7053 .7002 .6959 .6924 .6870 .6829 .6545 20 .0375 .0367 .0360 .0355 .0350 .0344 .0339 .0302 110 .7270 .7206 .7115 .7115 .7080 .7027 .6987 .6710 22 .0452 .0442 .0434 .0428 .0423 .0414 .0408 .0364 112 .7418 .7356 .7307 .7267 .7233 .7181 .7143 .6873 24 .0536 .0524 .0515 .0507 .0501 .0491 .0484 .0432 114 .7562 .7502 .7455 .7416 .7383 .7334 .7296 .7034 26 .0627 .0613 .0602 .0594 .0586 .0575 .0566 .0506 116 .7703 .7645 .7599 .7561 .7530 .7482 .7445 .7192 28 .0724 .0708 .0696 .0686 .0677 .0664 .0654 .0585 118 .7841 .7785 .7740 .7704 .7674 .7626 .7591 .7347 30 .0827 .0809 .0796 .0784 .0774 .0759 .0748 .0670 120 .7974 .7921 .7878 .7843 .7814 .7769 .7735 .7500 32 .0936 .0916 .0901 .0888 .0877 .0860 .0848 .0760 122 .8104 .8053 .8012 .7979 .7951 .7908 .7875 .7650 34 .1051 .1029 .1012 .0997 .0985 .0967 .0953 .0855 124 .8230 .8181 .8142 .8105 .8084 .8042 .8011 .7796 36 .1172 .1147 .1128 .1112 .1099 .1079 .1063 .0955 126 .8352 .8305 .8268 .8238 .8213 .8173 .8144 .7939 38 .1298 .1271 .1250 .1233 .1218 .1196 .1179 .1060 128 .8470 .8426 .8361 .8362 .8338 .8300 .8273 .8078 40 .1430 .1400 .1317 .1358 .1342 .1318 .1297 .1170 130 .8584 .8542 .8509 .8452 .8459 .8424 .8398 .8214 42 .1566 .1534 .1509 .1489 .1471 .1444 .1424 .1284 132 .8694 .8655 .8623 .8598 .8577 .8544 .8519 .8346 44 .1707 .1673 .1646 .1623 .1605 .1576 .1557 .1403 134 .8799 .8763 .8733 .8710 .8690 .8658 .8635 .8473 46 .1853 .1816 .1787 .1763 .1743 .1712 .1689 .1527 136 .8901 .8866 .8839 .8817 .8798 .8770 .8751 .8597 48 .2003 .1963 .1922 .1906 .1885 .1852 .1827 .1654 138 .8998 .8966 .8941 .8920 .8903 .8876 .8856 .8716 50 .2156 .2115 .2081 .2054 .2032 .1996 .2970 .1786 140 .9090 .9061 .9038 9019 .9003 .8978 .8959 .8830 52 .2314 .2270 .2234 .2206 .2182 .2144 .2117 .1922 142 .9178 .9152 .9130 .9113 .9098 .9076 .9059 .8940 54 .2474 .2428 .2391 .2360 .2335 .2293 .2266 .2061 144 .9262 .9238 .9218 .9203 .9189 .9169 .9153 .9045 56 .2638 .2589 .2550 .2518 .2492 .2450 .2419 .2204 146 .9342 .9320 .9302 .9288 .9276 .9257 .9243 .9145 58 .2805 .2754 .2713 .2679 .2652 .2608 .2575 .2350 148 .9417 .9397 .9381 .9368 .9358 .9340 .9328 .9240 60 .2974 .2920 .2878 .2843 .2814 .2769 .2735 .2500 150 .9487 .9469 .9455 .9444 .9435 .9419 .9408 .9330 62 .3146 .3090 .3046 .3009 .2979 .2932 .2897 .2653 152 .9553 .9538 .9825 .9515 .9507 .9494 .9484 .9415 64 .3320 .3262 .3215 .3178 .3147 .3098 .3061 .2808 154 .9617 .9601 .9590 .9581 .9574 .9563 .9554 9494 66 .3495 .3435 .3387 .3348 .3316 .3266 .3228 .2966 156 .9671 .9660 .9651 .9643 .9637 .9627 .9620 .9568 68 .3672 .3610 .3561 .3520 .3487 .3435 .3397 .3127 158 .9724 .9713 .9706 .9700 .9694 .9686 .9678 .9634 70 .3850 .3786 .3735 .3694 .3660 .3607 .3567 .3290 160 .9772 .9763 .9757 .9751 .9747 .9740 .9735 .9698 72 .4028 .3963 .3911 .3869 .3834 .3780 .3739 .3455 162 .9815 .9808 .9803 .9801 .9795 .9789 .9785 .9755 74 .4208 .4141 .4088 .4045 .4009 .3954' .3912 .3622 164 .9854 .9848 .9844 .9841 .9838 .9833 .9830 .9806 76 .4388 .4319 .4266 .4222 .4185 .41281.4085 .3790 166 .9888 .9884 .9881 .9878 .9878 .9872 .9869 .9851 78 .4568 .4498 .4444 .4399 .43621. 4304! .4260 .3960 168 .9918 .9913 .9912 .9910 .9909 .9907 .9905 .9891 80 .4747 .4677 .4622 .4576 .4539 .4480 1 . 4436 1 | .4132 170 .9943 .9941 .9939 .9938 .9937 .9935 .9933 .9924 82 .4927 .4857 .4799 .4754 .4716 .46561.4612 .4304 172 .9964 .9962 .9961 .9960 .9959 .9958 .9957 .9951 84 .510; .5034 .4977 .4930 .4892 .48321.4787 .4477 174 .9979 .9979 .9978 .9978 .9977 .9977 .9976 .9973 86 .5283 .5211 .5154 .5107 .5069 .5008 .4963 .4651 176 .9990 .9990 .9990 .9990 .9990 .9990 .9990 .9988 88 .546C .5388 .533C .5283 .5245 .5185 .5139 .4826 178 .9998 .9997 .9997 .9997 .9997 .9997 .9997 .9997 90 .563: .5563 .550: .5453 .5420 .5359 .5314 .5000 180 1.000C 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 HIGH-SPEED STEAM ENGINE. 27 TABLE VI. Distance of Piston from Beginning of Stroke when Stroke is away from Crank-shaft. To find Actual Distance of Piston from end of Stroke, multiply tabular quantity by Stroke. S-3, 2 Connecting-rod -=- Crank = r \ C Connecting-rod -5- Crank 4 4.5 5 5.5 6 7 8 00 4 4.5 5 5.5 6 7 8 OS .0002 .0003 .0003 .0003 .0003 .0003 .0003 .0003 92 .4540 .4612 .4670 .4717 .4755 .4715 .4861 .5174 4 .0010 .0010 .0010 .0010 .0010 .0010 .0010 .0012 94 .4717 .4789 .4846 .4893 .4931 .4992 .5037 .5349 6 .0021 .0021 .0022 .0022 .0023 .0023 .0024 .0027 96 .4895 .4966 .5023 .5070 .5108 .5168 .5213 .5523 8 .0036 .0038 .0039 .0040 .0041 .0042 .0043 .0049 98 .5073 .5143 .5201 .5246 .5284 .5344 .5388 .5696 10 .0057 .0059 .0061 .0062 .0063 .0065 .0067 .0076 100 .5253 .5323 .5378 .5424 .5461 .5520 .5564 .5868 12 .0082 .0087 .0088 .0090 .0091 .0093 .0095 .0109 102 .5432 .5502 .5556 .5601 .5638 .5696 .5740 .6140 14 .0112 .0116 .0119 .0122 .0122 .0128 .0131 .0149 104 .5612 .5681 .5734 .5778 .5815 .5872 .5915 .6210 16 .0146 .0152 .0156 .0159 .0162 .0167 .0170 .0194 106 .5792 .5859 .5912 .5955 .5991 .6046 .6088 .6378 18 .0185 .0192 .0197 .0199 .0205 .0211 .0215 .0245 108 .5972 .6037 .6089 .6131 .6166 .6220 .6261 .6545 20 .0228 .0237 .0243 .0248 .0253 .0260 .0265 .0302 110 .6150 .6214 .6265 .6306 .6340 .6393 .6533 .6710 22 .0276 .0286 .0294 .0300 .0306 .0314 .0322 .0366 112 .6328 .6390 .6439 .6480 .6513 .6565 .6603 .6873 24 .0329 .0340 .0349 .0367 .0363 .0373 .0380 .0432 114 .6505 .6565 .6613 .6652 .6684 .6734 .6772 .7034 26 .0385 .0399 .0410 .0419 .0426 .0437 .0446 .0506 116 .6680 .6738 .6785 .6822 .6853 .6902 .6939 .7192 28 .0447 .0462 .0475 .0485 .0493 .0506 .0516 .0585 118 .6844 .6910 .6954 .6991 .7021 .7068 .7103 .7347 30 .0513 .0531 .0545 .0556 .0565 .0581 .0592 .0670 120 .7026 .7080 .7122 .7157 .7186 .7231 .7265 .7500 32 .0583 .0603 .0619 .0632 .0642 .0660 .0671 .0760 122 .7195 .7245 .7287 .7321 .7348 .7392 .7425 .7650 34 .0658 .0680 .0698 .0712 .0724 .0748 .0757 .0855 124 .7362 .7411 .7450 .7482 .7508 .7550 .7581 .7796 36 .0738 .0762 .0782 .0797 .0811 .0831 .0847 .0955 126 .7526 .7572 .7609 .7640 .7665 .7705 .7734 .7939 38 .0822 .0848 .0870 .0887 .0902 .0924 .0941 .1060 128 .7686 ,7730 .7766 .7794 .7818 .7856 .7883 .8078 40 .0910 .0939 .0962 .0981 .0997 .1022 .1041 .1170 130 .7844 .7885 .7919 .7919 .7969 .8004 .8030 .8214 42 .1002 .1034 .1059 .1080 .1097 .1124 .1144 .1284 132 .7997 .8037 .8068 .8094 .8115 .8148 .8173 .8346 44 .1099 .1134 .1161 .1183 .1202 .1230 .1249 .1403 134 .8147 .8184 .8213 .8237 .8257 .8288 .8311 .8473 46 .1201 .1237 .1267 .1290 .1310 .1342 .1365 .1527 136 .8293 .8327 .8354 .8377 .8395 .8424 .8443 .8597 48 .1306 .1348 .1377 .1402 .1423 .1456 .1481 .1654 138 .8434 .8469 .8491 .8511 .8529 .8556 .8576 .8716 50 .1416 .1458 .1491 .1518 .1541 .1576 .1607 .1786 140 .8570 .8600 .8623 .8642 .8658 .8682 .8703 .8830 52 .1530 .1574 .1609 .1638 .1662 .1700 .1727 .1922 142 .8702 .8729 .8750 .8767 .8782 .8804 .8821 .8940 54 .1648 .1695 .1732 .1762 .1787 .1827 .1856 .2061 144 .8828 .8853 .8882 .8888 8901 .8921 .8937 .9045 56 .1769 .1819 .1858 .1804 .1916 .1958 .1989 .2204 146 .8949 .8971 .8988 .9003 .9015 .9033 .9047 .9145 58 .1896 .1947 .1988 .2021 .2049 .2092 .2125 .2350 148 .9064 .9084 .9099 .9112 .9123 .9140 .9152 .9240 60 .2026 .2079 .2122 .2157 .2186 .2231 .2295 .2500 ISO .9173 .9191 .9204 .9216 .9226 .9241 .9252 .9330 62 .2159 .2215 .2260 .2296 .2326 .2324 .2409 .2653 152 .9276 .9292 .9304 .9314 .9323 .9336 .9346 .9415 64 .2297 .2355 .2401 .2439 .2470 .2518 .2555 .2808 154 .9373 .9387 .9398 .9406 .9414 .9425 .9434 .9494 66 .2438 .2498 .2545 .2584 .2617 .2666 .1704 .2966 156 .9464 .9476 .9485 .9493 .9499 .9509 .9516 .9568 68 .2582 .2644 .2693 .2733 .2767 .2819 .2857 .3127 158 .9548 .9558 .9566 .9572 .9577 .9586 .9592 .9636 70 .2730 .2794 .2844 .2885 .2929 .2973 .3013 .3290 160 .9625 .9633 .9640 .9645 .9650 .9656 .9661 .9698 72 .2881 .2947 .2998 .3041 .3076 .3130 .3171 .3455 162 .9696 .9752 .9707 .9710 .9715 .9721 .9725 .9755 74 .3036 .3103 .3155 .3198 .3234 .3290 .3332 .3622 164 .9759 .9764 .9767 .9772 .9775 .9779 .9782 .9806 76 .3193 .3261 .3315 .3359 .3395 .3452 .3495 .3790 166 .9815 .9819 .9822 .9825 .9827 .9830 .9883 .9851 78 .3353 .3422 .3477 .3522 .3559 .3616 .3660 .3960 168 .9864 .9867 .9869 .9871 .9873 .9874 .9876 .9861 80 .3516 .3586 .3642 .3687 .3725 .3784 .3828 .4132 170 .9905 .9908 .9909 .9911 .9912 .9913 .9915 ,9924 82 .3682 .3753 .3809 .3855 .3893 .3952 .3996 .4304 172 .9939 .9940 .9942 .9943 .9943 .9944 .9945 .9951 84 .3849 .3921 .3978 .4024 .4062 .4122 .4167 .4477 174 .9966 .9967 .9967 .9968 .9968 .9969 .9970 .9973 86 .4019 .4091 .4149 .4195 .4234 .4294 .4349 .4651 176 .9985 .9985 .9985 .9986 .9986 .9986 .9986 .9988 88 .4191 .4263 .4321 .4368 .4406 .4467 .4013 .4826 178 .9996 .9996 .9996 .9996 .9996 .9997 .9997 .9997 90 .4365 .4437 .4495 .4547 .4580 .4641 .4686 .5000 180 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 28 INFLUENCE OF ROD ON PISTON TRAVEL. With respect to distance of piston from end of stroke it is evi- dent from the figure that the distances mB and nF, correspond- ing to the crank positions T and S, are unequal. If the connect- ing rod were infinite m*B z and n*F z would respectively represent the distances of piston from nearest end of stroke for the crank positions 7~and S. The following figure, drawn to scale, shows graphically the inequalities of the paths traversed by piston during the first and a 8 M h a. j 7 X 1 V 1 1 a 6 m: h J ■ w a. 6' m /\o lr / A 0, '/ JT ? VVV/AO lr / y////^ a. / JV \///r ////////?'. T~„ ° Fig. ii. second quadrants of crank motion for different ratios of connect- ing rod to crank, the horizontal position of stroke (= ab) included by shaded figure representing the distance Mo of the piston from its middle position M(see preceding figure and Tables V and VI). When length of crank = R = i and a> = 90 Mo. = .063 .072 .084 5 •127 3 .172 .268 Fig. 1 1 shows clearly that as the ratio of connecting rod to crank diminishes the inequality of the paths ao and ob respec- INFLUENCE OF ROD ON ROTATIVE EFFECT. 2 9 tively described by piston while crank passes through the first and second quadrant, increases. This inequality tends to heap up the work in the first and fourth quadrants, as will now be more fully shown. In order to show how the length of connecting rod influences the tangential components of the pressures on crank pin, we will assume the horizontal pressure on cross-head pin constant, and \ / \ \ //^ V 7\\ / \ \ / / ^4 \ / / V \ / J\ N. J \ • 1 \ "^ / • « / £ , ^* -J \'' 1 c 1 *l *Q0 C I A *o\ £-*"\ r -A s\ o\ • / \ \ / ' v I / ^ 1/ VI / 7 — ~"\ \ y N. / / \ \ /1 v\/ 1 \ \/i 1 \ s/ ^^v 7 -"T Fig. 12. will lay off the tangential pressures on prolongation of the radii of the crank-pin circle, Figs. 12 and 13. The curve drawn through the extremities of these radial ordinates will show the variation of the tangential pressure for the different crank positions ; the area included between this curve and the crank-pin circle measures, ap- proximately, the work done in one revolution. The two diagrams connecting rod drawn correspond respectively to crank = infinity and - = 2 ^ as above. A glance at these diagrams shows crank 3Q INFLUENCE OF ROD ON ROTATIVE EFFECT. at once that while with the infinite rod the work done in each of -^ the quadrants is equal, with the finite rod the principal part of the work is done in quadrants I and IV. The average tangential \i pressures can easily be found by remembering that if friction is neglected the work done on piston must equal the work done on crank pin, and that therefore the average tangential pressure p t Fig. 13. must be to the average piston pressure p„ as the path 25 (S = stroke) of piston in one revolution is to the path nS of the crank pin, or 2 p t : p m ■■■■ 2S : tcS hence p t = - p m = 0.6366 p m . (17) Attention has already been called to the fact that the pressure Vi and friction on the slide increase with the ratio of crank to con- necting rod, then the INFLUENCE OF ROD ON STEAM DISTRIBUTION. 3 1 CFcLtllc max. friction on slides = uP X : (18) connecting rod fj. = coefficient of friction, P = pressure on cross-head pin. But the ratio of crank to connecting rod affects not only the distribution of work and the magnitude of friction, it also affects the distribution of steam by increasing the duration of admis- sion and reducing that of exhaust at one end of the cylinder, while it diminishes the period of admission and increases the duration of exhaust at other end. That it thus affects the admission of steam can be roughly shown by means of Fig. 10, page 25. Since the ratio of eccen- tric radius to length of eccentric rod is small, even when the ratio is great, we may neglect the obliquity of eccentric connecting rod rod and treat the latter as if it were of infinite length, conse- quently the angular positions of the eccentric radius for the points of cut-off will be diametrically opposite, and since eccentric and crank are keyed to the same shaft, the crank positions corre- sponding to points of cut-off will also be directly opposite. Let //and T, Fig. 10, page 25, represent the two crank positions corresponding to cut-off, it will then be evident from the figure that the period of admission Bm for the end of the cylinder farthest from the crank will be greater than the admission Fn for end of cylinder nearest crank. When the ratio of crank to con- necting rod is large this distribution of steam will be injurious, but when it is equal to, or less than \ this irregularity of admis- sion and exhaust at the two ends will not only not be excessive but will be a positive advantage in high-speed engines, because it gives the greater compression {i.e. diminished period of exhaust) at the end farthest from the crank where it is most needed. This will be proved later when we come to consider the proper period of compression. Moreover, with such proportions of crank to connecting rod, the valve setting obtained from Zeuner's simple valve diagram will give very satisfactory results practically. The 32 RATIO OF ROD TO CRANK. indicator diagrams for the two ends of the cylinder obtained from such distribution of steam will of course not be identical in form, but they will be nearly equal in area, and as we have already mentioned will give the greater compression where it is most needed. In marine engines where engine space is limited the ratio is sometimes as' low as i y 2 , while in other well designed stationary engines it is sometimes as high as 8. Decide upon the proper ratio for the present case. III. Determination of that Mean Accelerating Force (per □" of Piston) Necessary to Start Reciprocating Parts, which Corresponds to the Most Uniform Tangential Pressure on Crank Pin. That the weight and speed of the reciprocating parts have an important influence on the steadiness with which an engine can be run, will be evident from the following consideration : Steadi- ness of running, other things being equal, results from the driving effort, or tangential pressure on crank pin, being as nearly constant as possible for all positions of the crank ; the magnitude of the tangential pressure (= DF'm Fig. 14) depends upon the position of the crank CD, and upon the direction and intensity of the pressure DB = EB exerted by the connecting rod upon the crank pin ; this latter pressure in turn depends upon the angular position of the connecting rod and upon the magni- tude of the force K = EA. Let us now suppose the mass M oi the reciprocating parts (piston, piston rod, cross-head and con- necting rod)* to be concentrated in a ring around the cross-head * According to Weisbach-Herrmann, (Vol. Ill, Section II, p. 867) the work stored up in a connecting rod of uniform section is the same as if one- third of its mass were concentrated at crank pin and the remaining two- thirds at cross-head pin. This result is itself an approximation and as sec- tions are generally not -uniform the accuracy of the assumption is still further diminished. Usually rod tapers from wrist to crank pin. HORIZONTAL PRESSURE ON CRANK PIN. 33 pin, and that the resistance offered by this mass to a change in its state of motion, is represented by F, we will then have F= Md where represents the acceleration of the center of the cross- head pin, M = mass of reciprocating parts and F= force neces- sary to give the mass M the acceleration 0. Fig. 14. The accelerating force F is of course part of the steam pressure P, the difference between them, P — F = K = EA (see figure), is that part of the steam pressure which is transmitted through the piston rod to the lower end of connecting rod and from these to the crank pin. When the point E(i.e., the piston) has reached its maximum velocity, acceleration ceases, and we have F = and P = K = EA = DA' , but retardation now begins (for the reciprocating parts must be brought to rest at the end of the stroke) and no part of the steam pressure being absorbed in pro- ducing acceleration, the whole of the steam pressure P\s trans- mitted to the cross-head pin. This P is however not the only pressure to which the crank pin is subjected, for the reciprocating parts having once attained their maximum velocity, tend to keep it, and tend to move faster (horizontally) than the crank pin, and will thus exert a pressure F upon the latter, over and above that received from the steam (= P) hence P + F= K=EA = DA\ 34 HORIZONTAL PRESSURE ON CRANK PIN. The general expression for the horizontal pressure exerted upon the crank pin or cross-head pin will then be DA* = EA = K= P± F the upper sign of F being employed when reciprocating parts are retarded, and the lower sign, when the reciprocating parts are accelerated, F being the accelerating or retarding force necessary to produce in the mass M the acceleration or retardation 6, that is, W , , F=Md = — (19) g so that F increases with the weight W of the reciprocating parts and their acceleration. Before determining the value of for the different piston and crank positions, we will call attention to, and emphasize the point, that the horizontal pressures on crank pin or cross-head pin are not directly obtainable from the diagram of effective steam pressures on piston but can be obtained indirectly from the latter by subtracting the accelerating pressures from the steam pressures given by the diagram for the earlier part of the stroke, and adding the retarding pressures to the steam pressures given by the dia- gram for the later part of the stroke. From this we_already_s.ee that the mass and ..motion. ..of the_recipxocating parts tend to equalize the horizontal pressures upon the crank pin by.diminish- ing the larger pressures, near the beginning. and increasing the smaller pressures near the end of expansion, the energy of the steam absorbed by the reciprocating parts near the beginning, being given out again to the crank pin toward the end ^ of the expansion. With too great a speed, however, this equalizing tendency will be more than neutralized by the transfer of the whole of the steam pressures to the end of the stroke and will tend to cause shocks and dangerous stresses on all the working pieces. This point will be more fully described further on. We will now find an expression for the acceleration 6 and then substitute in W F= W -Q. ( IQ ) PISTON TRAVEL. 35 The distance s traversed by cross-head pin E from the beginning X of its stroke (see Fig. 14) is equal to j = XE = XC — EC = (R + L) — (L cos a + R cos a>) s = R(i — cos to) + Z(i — cos a). (20) This equation holds for both strokes, provided we estimate to from o° to 360 and s from the same dead point (left hand dead point X, Fig. 14). But if we estimate w and s from the opposite center or dead point for the return stroke, and this is commonly done, we must use s = R(i — cos a>) — Z(i — cos a). The cosine of the angle made by connecting rod with axis of cylinder can be obtained as follows = L sin a = R sin w hence sin a = j sin a> (21) and cos « = -J 1 — — - sin 2 w = 1 • — ^~ sin 2 m very nearly (22) Introducing this value of cosine a in equation (20) we obtain s = Ry(i — cos (o) + %- sin 2 wj (23) For the return stroke, and estimating j and w from the right hand dead point this equation becomes j = R ( 1 — cos to) — y 2 j sin 2 to From this general formula for the travel, s, of the piston we can obtain the velocity, w, of the piston or of the point E when the crank has a uniform velocity. According to Elementary Mechanics we have for linear velocity, w = — and for acceleration, = — — dt 36 PISTON VELOCITY. Now from (23) A „^ *& = ^(sin w -f- Vz j sm 2co)dco therefore „/ . ,R . \da> , . zw = ^( sin w + i^y sin 2co)-— (24) is velocity of piston. If the crank pin has a uniform circumferential velocity z/ we will have dco v , Rdco = zv# or — = - • (25) dt R x ' Substituting this value of —^ in (24) we obtain ze> = vl sin « + }£-= sin 2a»J ; (26) For return stroke and estimation of s, co and w from the right hand dead point, we have instead w = z/f sin co — y 2 — sin 2coJ* Differentiating (26) we have dw = z/( cos codco -\- *4-* cos 2co idco), reducing and dividing both members by dt we obtain dw { , R \dco — - - = V 1 COS CO + -= COS 2fl) I — r • <# V L J dt div Now since the acceleration of a unit of mass is = — r- we can substitute in preceding equation, getting = v(< , ^ \dco cos see Table VIII, p. 60. PISTON ACCELERATION. 37 but we found from (25) that — = -, therefore, v dt R = — cos 10 + - cos 2a> .* (28) For return stroke, and estimation of s, w, o> and from the right hand dead point, this becomes o f( R \ = -= I COS 0) - COS 20) J . TJ 2 ~ At this point we should note that the factor -is identical with the .A. acceleration of a unit of mass rotating in the crank-pin circle, and when the connecting rod L is infinite the value of for u> = o and if a) =1 180 is = ± =, that is with infinite connecting rod the ac- celeration of reciprocating parts, when the crank is on the dead centers and N, is identical with that possessed by a unit of mass rotating in the crank pin circle. But when the connecting qf rod is of finite length, is no longer equal to s , either on dead R center or on N, but we have instead when at = o R and when w = 180 z/V R (■ - z) (30) '-* * An exact expression for 6 is : zt* f sin 2 o> Z> cos 2 <« — I Table VII on p. 39 was not computed from this formula, but from brack- eted part of equation 28. The table, to be perfectly exact, should have been computed from the bracketed part of the equation just given. The difference is so slight, however, that it can be neglected in the problems arising in practice. The upper signs correspond to forward stroke. 38 ACCELERATING FORCE. V 2 the numerical mean of these two values is, however, equal to -=• we call this mean 6 m , or n ^0~z) +^( I + z) _^ (31) o m _ - 2 R Returning now to the expression deduced above for 6 and substituting in W , X F=M0 = ~d (32) J" we get .F= d( cos w + 7 cos 2<0 J • (33) Dividing both members of the equation by ^4, area of piston in square inches, we get 2 = Y R A \ cos M + L C ° S 2 °V ^ 34 ' If the weight W of the reciprocating parts were rotating in the crank pin circle, the centrifugal force would be F substituting this in the above expression for — we get F f , R \F a F . s - = I cos « + — cos 2 for the return stroke, but its algebraic sign will be different in these two cases. Table VII on following page contains values of c for different ' crank angles w and different values of— • TABLE OF ACCELERATION COEFFICIENTS. 39 TABLE VII. Acceleration of Reciprocating Parts = c A Table Containing Values of c ( R COS 0) ± — COS 2ti> ) The if signs in formula relate respectively to \ > stroke. t lower ) ( return i The algebraic signs in the table relate to forward stroke only ; for return stroke, signs opposite to those there given must be used. Crank Angles. Connecting-rod - -i- Crank = Crank Angles. Connecting-rod -~ Crank = For- ward. Return. 4 J 4.5 1.250 1.222 5.0 5.5 6.0 For- ward. Return. 4.0 4.5 5.0 5.5 6.0 0° 180° 1.200 1 182 1.167 90° 90° -.250 -.222 -20o|-.182 -.167 3° 177° 1.248 1.220 1.198 1.180 1.165 93° 87° -.301 -.272 -.250 -.233 -.218 6° 174° 1.240 1.212 1.191 1.173 1.158 96° 84° -.350 -.322 -.301 -.283 -.268 9° 171° 1.226 1.199 1.178 1.161 1.141 99° 81° -.394 -.367 -.346 -.329 -315 12° 168° 1.206 1.181 1.161 1.144 1.130 102° 78° -.436 -.411 -.391 -374 -360 15° 165° 1.183 1.158 1.139 1.124 1.110 105° 75° -.476 -.449 -.430 -.415 -.401 18° 162° 1.153 1.131 1.113 1.098 1.086 108° 72° -.511 -.489 -471 -.456 -.444 21° 159° 1.120 1.099 1.083 1.069 1.058 111° 69° -.544 -.523 -.516 -491 -482 24° 156° 1.081 1.063 1.048 1.036 1.026 114° 66° -.574 -556 -.541 -.529 -.519 27° 153° 1.038 1.022 1.009 .998 .989 117° 63° -.601 -.585 -.572 - 561 - 552 30° 150° .991 .977 .966 .957 .949 120° 60° -.625 -.611 -.600 -.591 -.583 33° 147° .941 .929 .920 .913 .907 123° 57° -.647 -.635 -.626 -.619 -.616 36° 144° .886 .878 .871 .865 .861 126° 54° -.665 -.657 -.650 -.644 -.640 39° 141° .829 .823 .819 .815 .812 129° 51° -.681 -.675 -.671 -.667 -.664 42° 138° .769 .766 .764 .762 .760 132° 48° -.695 -.692 -.689 -.688 -686 45° 135° .707 .707 .707 .707 .707 135° 45° -.707 -.707 -.707 -.707 -.707 48° 132° .643 .646 .648 .650 .652 138° 42° -.717 -.720 -.722 -.724 -726 51° 129° .577 .583 .587 .591 .594 141° 39° -.725 -.731 -.735 -.739 -742 54° 126° .511 .519 .526 .532 .536 144° 36° -.732 -.740 -.747 -.753 -.758 57° 123° .443 .455 .464 .469 .477 147° 33° -.737 -.749 -.758 -.765 -.771 60° 120° .375 .389 .400 .409 .417 150° 30° -.741 -.755 -.766 -.775 -.783 63° 117° .307 .323 .336 .347 .356 153° 27° -.744 -.761 -773 -.784 -.793 66° 114° .240 .259 .273 .285 .296 156° 24° -.747 -.765 -.780 -.792 -.803 69° 111° .172 .193 .209 .223 .234 159° 21° -.748 -.769 -.785 -.798 -.810 72° 108° .107 .129 .147 .162 .174 162° 18° -.749 -.771 -.789 -.804 -.816 75° 105° .032 .067 .086 .102 .115 165° 15° -.750 -.774 -.793 -.809 -.822 78° 102° -.020 .005 .025 .032 .056 168° 12° -.750 -.775 -.795 - 810 -.826 81° 99° -.082 -.055 -.034 -.017 -.003 171° 9° -.750 -.777 -.797 -815 -.830 84° 96° -.140-. 112 -.090 -.073 -.058 174° 6° -.750 -.778 -.799 -.817 -.832 87° 93° -.1971-. 169 -.147 -.139 -.114 177° 3° -.750 -.778 -.800 -.818 -.833 90° 90° -.250 -.222 -.200 -.182 -.167 180° 0° -.750 -.778 -.800 ! -.818 -833 From the above expression for — we can deduce an approxi- A mate, graphical, method, sufficiently accurate for most purposes. If we make (o = 90° and 270 we get, respectively, ' • A/ " (37) A 4 Suppose that we now draw a circle with radius CG = -^ and RF lay off CB,C'B> and C"^" each equal to -j-^ and then through Fig. 15. the three points B',B,B" thus found pass the circle B'BB" the middle of the chord B'B will always fall on CO and the center of the circle on GBR* Then the acceleration for any crank angle FCM will be represented by MN and the retardation for the crank angle FCO by ON - • For any /I * The center F of the arc B'BB" falls on the circumference CMC" only r when T = i- EXACT CONSTRUCTION OF PISTON ACCELERATION. 41 other crank angle the horizontal line included between the two circles will measure the corresponding accelerating or retarding force. Exact Graphical Method of Finding Accelerating of Piston and of any Point of the Rod. The point Pin Fig. 10 is the instantaneous center for the rela- tive motion of connecting rod to engine bed. The velocity of crank pin B is to that of slide A as the instantaneous radius PB Fig. 16. is to radius PA. As each of the radii is also at right angles to the direction of the velocity they may respectively be taken as the representatives of the velocities of points B and A. As velocities may be combined and resolved like forces, we may re- gard APB as a triangle of velocities and either side as the resul- tant of the other two. Thus PA may be considered as the resul_ 42 EXACT CONSTRUCTION OF PISTON ACCELERATION. tant of PB and BA. That is, the velocity of point A may be considered as the resultant of the velocity of point B about P and of the velocity of point A about 5asa center (the angular velocity about B being then equal to that of B around P).* If we suppose the angular velocity of pin B about center C to be uniform and equal to unity, the crank radius CB will represent the velocity of its pin and, to the same scale, the distance CD, the velocity of the slide A at the same instant. Then the sides of triangle BCD will replace the sides of BPA as the representa- tives of the velocities. The length DB thus represents on our scale the aforesaid velocity of point A of the rod when turning about its point B. We are now ready to determine the acceleration of point A. When the direction of an acceleration and the intensity of one of its two components and the direction of both are known, we can easily find the intensity of the acceleration itself by means of the triangle of accelerations, (which triangle is analogous to that of velocities or forces. In this case we know the direction of the accel- eration of the point A to be along its path CAL. We may suppose it to be resolved into two components, one along the rod AB and the other at right angles thereto. We will determine the one along the rod AB. It was shown above that the motion of A was compounded of its own motion about B and of B's motion. Now when the crank has a uniform angular velocity equal to unity its acceleration is represented in direction and intensity by BC. If we resolve BC into components along and at right angles to BA, the one along the rod will be BC. The part of the acceleration of A which is due to its rotation about B as a center is also made up of two components, one along AB and the other at right angles to AB. We need to know only the former, * This can also be regarded as an example of the resolution of a rotation about one axis P into an equivalent rotation about another parallel axis B plus a circular translation. See Weisbach-Herrmann's Machinery of Trans- mission, Vol. Ill, Sect. II, §4 of the Introduction. EXACT CONSTRUCTION OF PISTON ACCELERATION. 43 and this is ^ r-fr^-- But it was shown above that this velocity AB BD 2 = BD, hence the component along AB is equal to ■ . This quotient can be found geometrically by describing a circle on the rod AB as a diameter and finding its intersections with an arc struck from 5asa center and BD as a radius. The common chord I J joining these intersections will cut from BA the distance BD 1 BF 1 = — — and will also cut from the horizontal through C, a distance CH exactly equal to the desired total acceleration of slide A. This can now be easily shown, for the component BF' just found has a direction from F' to B and the component BC of B along the rod has evidently the direction B to C. Therefore BC — F'B = F C = that one of the two components of the total accelaration of A which acts along rod AB. Through C draw Coequal and parallel to OF and to it at F erect a perpendicular FH; this cuts from ACB„ the direction of the total acceleration of A, a distance HC equal to the direction and intensity of the total acceleration of slide A* The point F' can also be found by drawing through D, the intersection of the rod AB with CY the perpendicular to the stroke, the parallel DE and where this meets the crank draw E& ' parallel to CY; this parallel cuts the rod or its prolongation in the desired point F'- The proof of this construction is : BF:BD = BE-.BC = BD.BA, that is, BF = ~. AB * This construction is not only the simplest, but it is also applicable to all slider-crank mechanisms, whether the stroke of the slide A does or does not pass through the crank center C. In the latter as in the former case the intercept must be taken on a line CH through C and parallel to the slide stroke. It is also applicable to variable crank-pin velocity when scale is such that latter is represented by crank AB. We have then only to draw a par- allel to stroke through the end of crank-pin acceleration (which in this case does not fall at center C of shaft). On this parallel the intercept included between the end of acceleration and chord TJ, Fig. 14, will be the exact acceleration of slide A. 44 METHOD OF FINDING CENTER OF ACCELERATION. But this construction of F' evidently fails when the crank is at either dead point. The determination of the exact acceleration of A will enable us to construct its curve of acceleration NOQ, which is drawn on the stroke of A as a base. This suffices for most of the cases arising in practice. Greater accuracy is of course obtainable if we ascertain the acceleration of each point of the rod, multiply this by the elementary mass at that point and then find the resultant of all these elementary forces of inertia. This resultant can be resolved into two components, one acting at wrist pin A and the other at crank pin B. These can then be combined with the other forces acting at A and B and the exact tangential pressure on crank pin found. We will postpone this combination of forces for the present and confine ourselves to finding an easy method of con- structing the acceleration of each point of the rod and its com- ponents parallel to, and perpendicular to, the rod. We will first show that the end of the acceleration KM of any point AT of the center line of the rod lies on the straight line joining the ends Cand L of BC and AL, the accelerations of the two points B and A. To do this we must make use of the properties of the center of acceleration G* This center is that point of the moving rod which has no acceleration. The acceleration of any point K is directly proportional to the distance GK of this point * For method of finding the center of acceleration G when the directions of the acceleration of two points are given and also their ratio, see Weis- bach-Herrmann's Machinery of Transmission, Vol. III., Section I, § 21- But for this, most common case, in which the crank has uniform rotation, we would suggest connecting crank-pin center B with point H (the inter- section of line IJ with horizontal CB % Fig. 14) and on this connecting line laying off from B a distance equal to the length of connecting rod ; then through the end of this distance draw a horizontal till it cuts the prolonga- tion of the crank. The triangle thus obtained will be similar to CBH and when this triangle is swung around B till its base coincides with BA its vertex will fall at G and give exactly the position of the center of acceler- ation. The modification necessary for cases of variable rotation of crank are evident; the triangles must be similar. ENDS OF ACCELERATION FORM AN IMAGE. 45 from the center of acceleration G, and for the instant in question the acceleration of each point makes the same angle with its own instantaneous radius of acceleration. For example, let us suppose the center of acceleration G known, and let BC, AL and KM represent respectively the accelerations of points B, A and K in direction and intensity. Then will BC: AL: KM = GB : GA : GK and angles CBG = LAG = MKG. From these properties, it follows that the triangles GBC, GKM and GAL are all similar and hence their corresponding angles BGC, KGMand AGL are equal to each other. If we turn the whole system of points AKB about the center G through the angle BGC = KGM ' = AGL, the points B, K and A will fall on B', K' and A'. Now if we can show that MC is parallel to K'B' and ML to K'A', we shall have proved that MC and ML are one and the same straight line because K'B' and K'A' con- stitute one straight line. From the similarity of the triangles GBC, GKM and GAL, we have GC: GM = GB:GK = GB' : GK' GM: GL = GK: GA = GK' : GA' and therefore MC and ML are respectively parallel to K'B' and K'A' and we have proved that the ends of all the accelerations of the points of one straight line lie on one and the same straight line. This is true of any line of the system and hence of any com- bination of lines belonging to or constituting the system. More- over it is evident that any line in the plane of rotation will be reduced in the ratio of GC to GB so that CL GC BA ~ GB The ends of the accelerations of all points in the plane figure constituting the rod will form a reduced image of the shape of the rod which will be exactly similar (in the plane of motion) to the original rod. This similarity extends to every detail. 4 46 ACCELERATION OF ANY POINT ON ROD AND ITS COMPONENTS. A corollary from this proposition is that the ends of the accelerations of the various points divide the distance CL in the same ratio as do their corresponding points the rod or distance BA = B'A'. For instance, we have CM: ML = B'K'.K'A' = BK.KA. This corollary enables us to determine the acceleration of any point of the rod AB (without the help of the center of accelera- tion G) when we know the direction and intensity of the accele- rations of any two points on that line. For example, if K is half way between A and B, the end Mof its acceleration KM will be half way between the ends C and L of the known accelerations BC and AL. If the total accelerations of each of the points of the rod be resolved into two components, one parallel and the other perpen- dicular to the rod, and these components be laid off as ordinates, (taking the rod as an axis of abscissas), one set of components on one side of the rod and the other set on the opposite side of the rod, then will the components of each set terminate on the same straight line. That is, the components of the set which are at right angles to AB will terminate on the straight line A"B", and the components of the set which coincide with AB, will (when revolved through 90 ) terminate on the straight line RST. This can be proved as follows : Drop from C, M and L the re- spective perpendiculars CC ', MM' and LL'. Because they are a series of parallels they will divide the distance C'L' in the same ratio as CL, and since CL has been shown to be divided by the acceleration in the same ratio as BA, it follows that C'L' is divided in the same ratio as BA. But the ordinates C C, M'M, L'L, etc. erected on C'L', by construction, terminate in a straight line, namely CML. As AF is equal to L'C we may suppose these self-same ordinates to be laid off from the former and then they will all terminate on the straight line A"F. Now if we suppose the base AF of this set of ordinates to be stretched till it is equal to AB in length, the ordinates remaining equidistant and chang- ACCELERATION OF ANY POINT ON ROD AND ITS COMPONENTS. 47 ing only their position and not their magnitude, then will they all terminate on the straight line A" B" . For before stretching, the ratio — = a, of the difference of any two adjacent ordinates to the difference of their abscissas, was equal to a constant a, which is the characteristic of a straight line. After stretching, the ratio of these differences is -f- = a 1 = a constant also, because -r— ax 1 ax AB = ~Tpi = a constant for the points of this rod and this posi- tion of the mechanism. If we revolve the accelerations BC, KM, AL, etc., through 90° so that they occupy positions BC", KM'", AL'", etc., these new positions will also make a constant angle with their instantaneous radii of acceleration, and we may show as before that the points L"'M"'C" lie on a straight line. If from their ends C", M'" and L'" we drop on AB the perpendiculars C'"C", M'"M", L'"L", etc., and erect these as ordinates at the corresponding points of the base AB we can show as before that these all terminate on one straight line RST. But as these perpendiculars C'"C", etc., are respectively equal to the second set of components BC, etc., the second part of our proposition has been established. The forces of inertia due to this second set of components do not however act transversely on the rod, but longitudinally. Their sum or resultant combined with the resultant of all the bending forces that are due to inertia will give a total resultant, due to inertia of rod, that can be resolved into two forces acting at pins A and B* Exactly how this resultant is found will be discussed later on ; at present the shaded area and area ABTR furnish the compo- nents for constructing the acceleration of any point belonging to the center line of rod. For this purpose however area ABTR would suffice, for if its component be laid off on the rod, say KM' = KS, a perpendicular erected at M' will cut the line CA in M giving at once KM as the acceleration of the point K. * This must not be interpreted as giving the total inertia resistance of rod for distributed mass. 4 8 ACCELERATION DIAGRAMS. Ill A. Acceleration Diagrams. Obtain by means of the preceding diagrams or tables, or both, the accelerating and retarding forces corresponding : to an as- sumed value of -^, to a given ratio of connecting rod to crank, and to certain crank angles. Then find the corresponding piston positions as in the preceding plate. At these positions erect ordinates equal to the accelerating or retarding forces the former being laid off below, and the latter above, the line of piston posi- tions. It is evident from the figure that the acceleration curve for one stroke can be obtained from that of the other by revolv- ing the latter 180 about the central line. _Cong_Rod_ Crank Crank 1 Fig. 17. Construct on the admission line of : each of the six diagrams of effective steam pressure on piston, two acceleration diagrams (in the manner shown in Fig. 17) which correspond respectively to & A Pi l6 ,, , F „ = say, 40 lbs. and ~r = 2/ 2 = say, 50 lbs.; p* is 2 A the terminal pressure at the end of the expansion curve. Assume also — = 30. A 3 PRESSURE ON CROSS-HEAD PIN. 49 III B. Determination of Pressure on Cross-Head Pin, when Weight of Rod and Friction are neglected, but Inertia taken into Account by assuming whole Mass of Rod and Reciprocating Parts concentrated on plston. To obtain this diagram we must combine the diagram of effec- tive steam pressures on piston with the diagram of accelerating and retarding pressures. Moreover the combination must be so effected that the accelerating pressures will be subtracted from the effective steam pressures, and the retarding forces added. This can be done in two ways : by subtracting and adding the ordinates of the acceleration diagram, Fig. 17, to the upper lines ABC and EFG of Figs. 8 and 9, representing the pressures on driving side of piston, and getting as a result the horizontal pres- sure diagrams VTSI'H and G'POLA'DC , Figs. 18 and 19, or by subtracting and adding the ordinates of Fig. 17 to the lower lines HI' E and A'D'C, Figs. 8 and 9, representing the counter pressures, and getting as a result the horizontal pressure diagrams A' B'C'MN and E'FG'HKE, Figs. 20 and 21. A comparison of the two sets of diagrams will show that the corresponding ordinates included by the shaded portions are equal. The shaded area in each case is bounded on the one side by the resultant line (due to the combination of expansion line or line of counter-pressure with acceleration curve) and on the other side by that line of Fig. 8 or 9 which did not help to form the resultant line. It should also be noticed that where the resultant line crosses the unused line (given by Fig. 8 or 9) the horizontal pressures on cross- head pin become equal to zero, and after passing this point, see Fig. 2i, the direction of the horizontal pressures on cross- head pin is reversed before the piston reaches the end of its stroke. In Fig. 20 the steam and retarding pressures are so related as to com- 5o PRESSURE ON CROSS-HEAD PIN. Fig. 19. pletely unload the crank-pin at the end of the stroke. This is not so favorable as it seems, for at the beginning of the return stroke there is suddenly applied a great pressure in the oppo- site direction. A little more compression I'S, Fig. 18, would have changed this. It is probable that the arrangement shown in PRESSURE ON CROSS-HEAD PIN. 51 Fig. 20. Fig. 21, is less likely to cause pounding than that in Fig. 18, and this in spite of the fact that in the case of Fig. 2 1 the reversal of pressure near end of stroke occurs while piston is in motion. A Fig. n. less lead in the case of Fig. 19 would be advantageous in cutting down the large initial pressure. Another construction used by Grashof, is well worth noting. It is to lay off these resultant pressures at cross-head pin on the rectified crank-pin circle. The straight base will then represent time, and the rate at which the pressure changes can then be 52 EFFECT OF VALVE SETTING ON RESULTANT PRESSURE. easily seen. This will be particularly useful in judging of the rapidity with which the reversals of pressure occur. We shall hereafter only make use of horizontal pressure or wrist-pin diagrams similar to those shown in Figs. 18 and 19. Ill c. Effect of Valve Setting on the Horizontal Pressure on Cross-Head Pin. We give two indicator diagrams, Figs. 22 and 23, from opposite ends ofthe same cylinder to show the influence respectively of a late A 3 Fig. 22. admission and too late an exhaust. By combining the diagram ABCUV, Fig. 2 2,with diagrams of counter-pressure HIJK, Fig.23, we get the effective steam pressure diagram ABCKJIH, Fig. 24. a f Fig. 23. EFFECT OF VALVE SETTING ON RESULTANT PRESSURE. 53 Fig. 25- In like manner by combining the counter-pressure diagram EDUV, Fig. 22, with HGFKYZ, Fig. 23, we get the effective pressure diagram KDEHG, Fig. 25. By combining these real 54 EFFECT OF VALVE SETTING ON RESULTANT PRESSURE. pressure diagrams with the acceleration diagrams we get the horizontal pressures on cross-head pin, shown in Figs. 24 and 25 by shaded areas. Fig. 24 shows that, owing to the late exhaust HI, Fig. 23, there is a drag instead of a push on the crank dur- ing the first portion of the stroke, and a reversal of pressure on the crank while the piston is in rapid motion at L. This reversal will cause a shock or knock if there is any play in the joints of the mechanism. In like manner Fig. 25 shows a drag instead of a push on crank at beginning of stroke and a reversal of pressure at D, but here the piston motion is considerably slower than at L, so there will probably be no shock. The final pressure, KS Fig. 25, on piston at beginning of return stroke is about equal to that, KN Fig. 24, at end of forward stroke, and is a desirable result. It would be better however to accomplish the same result by more compression and earlier admission. Fig. 25 also shows the very unequal distribution of the horizontal pressures which attends a large cut-off when unaccompanied by suitable compression. By having the proper amount of com- pression at 5, Fig. 18, not only may the reciprocating parts be brought to rest without shock, but all load can be taken off the crank pin while it is passing the dead center. This is not always done on account of the decided advantages gained by compres- sing nearly up to the boiler pressure, some of which advantages are the filling and heating of the hurtful space by compressed steam instead of the live, fresh, steam and the avoidance of such admission lines as K,F Fig. 23. Moreover, the reversal of pressure which accompanies considerable compression is not necessarily an evil, provided it takes place when the piston is moving slowly, i. e., near the end of stroke where reversal natur- ally takes place, The ideal case is to have the piston unloaded while moving slowly (near end of stroke) and then let the driving pressure on opposite side increase gradually till after the reverse stroke has begun. Considerable compression and even a slight negative lead may be used to effect this desirable result, desirable, because of its freedom from pounding. EFFECT OF SHORT CUT-OFF ON RESULTANT PRESSURE. 55 Fig. 26. In Fig. 26 the ordinate AH at the beginning of the accelera- F tion curve ~ is greater than the corresponding effective steam pressure AG; the result is that instead of there being a push upon the crank in the direction of its motion at the beginning of the stroke, there is a drag upon the crank, this condition continuing until the piston reaches the position /, where a reversal of pressure takes place causing shock and vibrations injurious to the durability of the machine, because at that end of the stroke there is considerable piston speed at point I. In Fig. 27 the condition of affairs is still worse, for two rever- sals of pressure take place at /' and I", causing two shocks in rapid succession. In this last case, although the effective pres- sure A'G' is greater than the initial pressure of acceleration A'H, (F F R \ — — = -j(cos co -(- — cos 2 cu) J of the acceleration A A 1*1 J 56 EFFECT OF SHORT CUT-OFF ON RESULTANT PRESSURE. Fig. 27. F A'B' curve -£ are still too large for the short cut-off-— - — and the small A Lt C initial pressure A'G'. It is evident from the figures that if the effective pressure diagrams ABCDEFGA and A'B' CD' E'FG' A> of Figs. 26 and 27 are to remain unchanged, that is, if the mean effective pressure pin is to remain unchanged, the only remedy for the evils exhibited in Figs. 26 and 27 is to diminish F the value of—?. \ When the engine varies between wide limits of cut-off, as in F the present case, the value of -^ is chosen so as to favor as much as possible the particular horse-power at which the engine ordi- narily runs ; in such a case it is desirable to know what is the minimum cut-off at which reversal of pressure will be avoided. This may be ascertained by the cut and try method, but it may REVERSALS OF PIN PRESSURES DUE TO SHORT CUT-OFF. 57 also be calculated with sufficient accuracy if we make the follow- ing suppositions, namely, that the curve of expansion is an equi- lateral hyperbola, that the best value for -^ is & A A — 16 and that the crank angle is about 6o° when the horizontal pres- sure curve EFG (Fig. 28), is tangent to the back pressure line Fig. 18. DF. As the reversals in question are more apt to occur during the return stroke we must use a formula P-Pi -30 cos co ■ R COS 210 I = K'a- P 3 (40) corresponding to that stroke and introduce the assumptions just made. * Radinger holds that —~ = 2fi x = twice the terminal pressure at end of expansion will give the steadiest running of engine. 58 CONVERSION OF HORIZONTAL INTO TANGENTIAL PRESSURES. By deriving a value for v from the piston travel J for return stroke and allowing for clearance, we find that under the average conditions of running of high-speed engines, the minimum, real, cut-off should be greater than o.io to avoid reversals of pressure when piston is running swiftly. The approximate formula for this cut-off is : A_B. ± F jR x A / , _ ix ,R x 4 ( } To find the number of horse-powers corresponding to this minimum, real, cut-off we look in Table III for value of — — and then transforming, Eq. 14, we get [,- ( . + .»(^/-»0] . H Pn All the terms of the second member being known we can calcu- late readily the minimum horse-power at which the engine can be safely run at the given speed and pressure. Find minimum cut-off and horse power for present engine. Construct on each of the six diagrams of effective steam pressures on piston two diagrams of resultant pressures on cross-head pin p j, x r 6 F F a corresponding to — | = — and — - = 2p„ i.e. —2=50 and 40 lbs. Ill D. Conversion of Horizontal Pressures of Crank Pin into Tangential Pressures. The horizontal component of the pressure of the rod against the crank-pin is evidently equal to the horizontal pressure of the cross-head pin. In this problem we are to find the rotative effect of a horizontal pressure on crank-pin. CONVERSION OF HORIZONTAL INTO TANGENTIAL PRESSURES. 59 Let P* represent horizontal pressure on cross-head pin. T represent force along connecting rod. / represent tangential pressure on crank pin. P 1 then will T — — (42) cos a and t=T sin (co + «) = /" ™& +A ( 43 ) cos a t sin (a> 4- a) , s px = — ~— (44 P cos a If we substitute in this equation sin a = -= sin a> and cos a = n! i? 2 1 y s ' n2 w ( see Eqs. 21 and 22) we shall have the means of calculating the ratios — from the crank angle m. Table VIII on page 60 contains these ratios. If we wish to determine the tangential pressures from given horizontal pressures by graphical means we can proceed as fol- lows, Fig. 29. In triangle AKC, Fig. 29, we have KC _KC _ sin (a> + a) . . AC ~ ~R ~ c^~~ (45) or t KC w , ~ P>=-R= -v (46) hence if on any scale, say 20 lbs. to the inch, we lay offDE = P 1 , join K with D and prolonging KD till it intersects at F the per- pendicular to DE erected at E, we will have in the _two_giniilar tnangles^JM^T and^ DFE, the proportion FE : P 1 : : KC : R, i. e. t = FE. 6o TABLE FOR CONVERSION OF PRESSURES. TABLE VIII. Rotative Effect of a Unit of Horizontal Pressure on Crank. To find actual tangential pressure (or rotative effect) on crank, multiply tabular quantity by resultant horizontal pressure on crank or cross-head pin. Forward stroke is towards, and return stroke away from, crank shaft, i To find wrist-pin velocity multiply tabular values by crank-pin velocity. Crank Angles. Connecting Rod -=- Crank = Forw'd. Return. 4-o 4.5 .1064 5-o 5-5 6.0 00 5 175 .1089 .1045 .1030 .1016 .0832 IO 170 .2164 .2117 .2079 .2047 .2022 ■1737 15 165 •3215 ■3H5 .3089 ■3°54 .3005 .2588 20 160 .4227 ■4136 .4065 .4019 •3957 .3420 25 '55 .5189 .5081 •4995 •4925 .4866 .4226 30 150 ./ .6091 .5968 •5870 ■5791 .5724 .5000 35 i45 .6923 .6788 .6682 .6596 .6523 ■5736 40 140 •7675 •7533 .7421 .7329 •7253 .6428 45 135 ■834i .8195 .8081 .7988 .7910 .7071 5° 130 .8914 .8771 .8657 .8564 .8488 .7660 55 125 .9392 •9253 .9144 •9055 .8982 .8192 60 120 •' -9769 .9641 .9540 .9458 ■939° .8660 65 "5 1.0046 ■9932 .9842 .9769 .9709 .9063 70 110 1.0224 1.0127 1.0052 .9990 •9939 •9397 75 105 1.0304 1.0228 1. 01 69 1.0121 1.0082 .9659 80 100 1.0289 1.0237 1.0199 1.0164 1.0137 .9848 85 95 1.0186 1. 01 60 1.0139 1. 01 27 1.0109 .9962 90 90 1. 0000 1. 0000 1 .0000 1 .0000 1. 0000 1. 0000 95 85 •9738 .9764 •9785 •9797 .9816 .9962 100 80 .9407 .9460 .9500 •9532 ■9559 .9848 105 75 .9016 .9091 .9150 .9198 ■9237 .9659 no 70 .8571 .8667 .8743 .8804 •8855 •9397 115 65 .8080 .8194 .8285 ■8357 .841 6 .9063 120 60 - .7552 .7680 .7781 .7863 •7931 .8660 125 55 .6992 .7130 •7239 .7328 .7401 .8192 130 5° .6407 .6550 .6664 .6756 •6833 .7660 135 45 .5801 .5946 .6061 .6155 .6232 .7071 140 40 .5181 ■5323 •5435 ■5527 .5603 .6428 145 35 •4549 .4683 .4790 .4876 •4949 .5736 150 3° " -39°9 .4032 •4130 .4209 .4276 .5000 155 25 .3264 •3371 .3458 •3528 .3586 .4226 160 20 .2614 .2704 .2776 .2821 .2884 .3420 165 15 .1962 .2032 .2088 .2123 .2171 .2588 170 10 .1309 •1356 •1394 .1426 .1451 • 1737 175 5 .0655 .0679 .0698 .0714 .0727 .0872 CONVERSION OF HORIZONTAL INTO TANGENTIAL PRESSURES. 6 1 The value of t will be on the same scale as P 1 . If we determine in like manner, for the same pressure P = DE, the tangential pressures corresponding to the various crank positions of Fig. 29 and lay them off on the extreme left or right hand ordinates of Return Fig. 19. the section paper employed, then drawing diagonals to the assumed zero of cross-head pin pressures we will get a diagram similar to Fig. 30 from which we can determine the tangential pressures for any value of P 1 . In Fig. 29 the ratio of length of connecting rod to length of crank was taken at 2)4. In the diagram to be drawn on section paper assume this ratio to be = 6, also DE = P 1 = 200 lbs. Lay off, in Fig. 30, the tangential pressures on a scale of 20 lbs. to the inch, and the horizontal or cross-head pressures on a scale of 20 lbs. to the inch. The tangential pressures can also be obtained in the ordinary manner by resolving the force acting along the connecting rod into two components respectively tan- 5 62 DIAGRAM FOR CONVERSION OF PRESSURES. gent and normal to the path of the crank pin. But this would involve a good deal more labor than the present method. Figs. 29 and 30 are quite accurately drawn and the method of obtain- ing 30 from 29 can therefore be easily followed. Thus EL, EM and EN are the tangential pressures corresponding respectively to the angular positions 20 , 40 and 60° of crank for return -80 T 100 60°" "120° 200 160 120 80 40 40 80 120 160 Horizontal pressure on Cross-head pin. Resultant of effective steam pressure and inertia of reciprocating parts. Fig. 30. stroke and to 160 , 140 and 120 for the forward stroke. The quickest method of obtaining a diagram like Fig. 30 would be to calculate the tangential pressures from Table VIII, assuming a constant horizontal pressure of 200 lbs. on cross-head pin. It is sufficiently evident from Fig. 30 how the tangential pressures for other and smaller resultant pressures on cross-head pin can- be obtained. CONSTRUCTION OF TANGENTIAL PRESSURES. 63 III E. Construction of Diagram of Tangential Pressures on Crank Pin. We are now ready to construct the tangential pressure dia- grams from the diagrams of resultant pressures on cross-head pin corresponding to various values of —?. We proceed as in the construction of Fig. 13. By means of the diagram, Fig. 30, we / first transform the horizontal pressures into tangential pressures, and then lay off these tangential pressures as prolongations of the radii of the crank-pin circle (or of any other convenient circle), the latter forming the base of the radial ordinates. A curve is next drawn through the extremities of these ordinates and then a circle concentric with — and outside of — the crank-pin circle, is drawn, the difference between the two circumferences is equal to the mean tangential pressure, which as before is equal to p t = 0.6366 p m , Eq. 17. The deviations of the curve (drawn through the extremities of the ordinates representing the tan- gential pressures) from the circle of tangential pressures will show the irregularity - of the driving power. The more closely v the tangential pressure curve approximates to the mean-tan- gential-pressure-circle, the steadier will the engine run. Figs. 31 and 32 show two different methods of representing the tangential pressures. In the former figure the tangential pres- sures are laid off from the crank-pin circle itself; in the latter from the rectified semi-circle. Fig. 31 has the advantage of showing the phases of resistance CMNO and A QS and the curve of tangential pressures in a continuous manner, but it is subject to the following very slight disadvantage, viz : that the area en- in- closed by AQRMCBA does not represent with great accuracy the work done in a semi-revolution, the inaccuracy being due to the divergence of the radial ordinates. The objection does not apply to Fig. 32, where the area A VECBA is exactly equal to the 6 4 CONSTRUCTION OF TANGENTIAL PRESSURE DIAGRAMS. area AF'DCA = work done in one semi-revolution; moreover by adding the shaded portions CUE (= CDE) and APG as shown in Fig. 32, the objection that this mode of representation does not represent the phases of- resistance in a continuous man- %*. V Return . Fig. 33. ner disappears. If the work has been done correctly it will be found that in Fig. 3 1 the area of the shaded portion QRM = area CMN + area QAS, nearly, and that in Fig. 32 the phase of excess GVE = AGF' + DEC exactly. Inspection of either of these figures shows that the tangential pressure line (shown by full line) represents steadier running than USE OF TANGENTIAL PRESSURE DIAGRAMS. 65 the tangential pressures corresponding to greater piston speed or heavier reciprocating parts (shown by broken line). Draw for the minimum, rated and maximum power of the pro- posed engine, separately two tangential pressure diagrams, one like Fig. 31 and the other like Fig. 32. On each diagram draw two tangential pressure curves corresponding to the resultant, or horizontal, pressure on cross-head pin which were drawn for the assumed values: -J = ^ = about 40 and--°=2/ 2 = about -fi- 2 A 50 lbs. When these have all been drawn, inspect the result and F see if a different value than those chosen for — f would not eive a A s more uniform tangential pressure — i. e. steadier motion. If so, F take the new value of ~ and draw corresponding acceleration curves for both the' forward and the return stroke, and superim- pose these upon the diagrams of effective steam pressures already drawn and combine them as before to form diagrams of resultant pressures on cross-head pin. Then convert these last diagrams into tangential pressure diagrams and again inspect the latter diagrams to see if any improvement can be effected. This pro- F cess is to be continued till that value for ~ has been found which A gives the most uniform tangential pressure for the normal, rated, horse-power of the proposed engine. This most favorable value F of —| for the rated power should however be examined with refer- ence to the minimum power required, to see whether it then causes reversals of pressure. When this is the case some smaller F value of ~ less favorable to the normal or rated power should be chosen for the engine. In this connection we may quote the following from Rigg's Steam Engine. 66 CHOICE OF SPEED AND LOAD. Choice of Speed and Load. " There are, then, three elements which can be adjusted to each other, namely, steam pressure, including the rate of expansion, weight of reciprocating parts, and number of revolutions per minute, or piston velocity ; any of these elements can be altered at the option of the designer of the engine, and the problem to be solved is not to transfer the strains from one end of the stroke to the other, not to work steadily with the highest expansion, not to produce a regular uniform horizontal pressure on the crank. But the end and object of all these calculations and changes in the old acknowledged rules for the construction of engines is, by working with that amount of expansion which practical ex- perience proves economical, to obtain as nearly as possible a uni- form tangential pressure on the crank. When this is done an engine will drive its load steadily and well, and the influence of the reciprocating parts has a most direct bearing on this most important subject. In applying the foregoing reasoning it is necessary to exercise a judicious choice in balancing the evils which arise from a too low or too high speed, and so to decide a rate at which an engine can be run to the best advantage. When the load upon the engine is regular it is comparatively easy to do this, but in a great majority of cases the load is often varying, and the speed is bound to remain constant. Thus, it is necessary to select a load on the engine which shall meet most requirements, and then the indica- tor diagram either actually taken or assumed will form the prin- cipal datum required." To these considerations might be added another, namely that to avoid pounding or shocks at crank- or wrist-pin, the driving pressures should change as gradually as possible from one side to the other of piston, the change taking place as near the dead point as possible. By using considerable compression or even a slightly_ negative lead this gradual change may be accomplished for both ends of the stroke. DETERMINATION OF DIAMETER OF CYLINDER. 67 IV. Determination of Diameter of Cylinder D, Length of Stroke S, Revolutions per Minute N or Piston Speed per Minute w 1 and Weight of Reciprocating Parts W by Means of the Tables or Formulas. F W W — =0.0000142 -jSN* = .000085 —jNw 1 F , H WN w* W H , . A = ^f m -A^- 2l8 S^-A^ (47> — = 0.0000238 w 1 !) 2 (48) Pm — = 0.00000396 NSD 2 (49) A* E = 2 3 A& = 2lS -ASB> (SO) Pm H = number of indicated H. P. S = length of stroke in inches. W — = weight of reciprocating parts per □" of piston, weight of rod included. D = diameter of cylinder in inches. w 1 = average speed of piston in feet per minute. N = revolutions per minute. p m = mean effective pressure per □" of piston. F -£■ = average force accelerating reciprocating parts when crank is on dead centers. 68 MEAN EFFECTIVE PRESSURE ON PISTON. TABLE IX.* Value of — or Indicated H. P. for each Pound of Mean Effective Pressure per □" of Piston Area. Quantities in table were obtained by neglecting area of piston rod and assuming that the same mean effective pressure (pm) existed on both sides of the piston area ; consequently the tabular quantities are about I % % and ^ fo too large for small and large cylinders respectively. To correct for (d*\* piston rod multiply tabular quantities by [i — l A \~]j) ] where D — diameter of cylinder and d z = diameter of piston rod. Z. ) (59J where W represents the weight of the reciprocating parts. The frame will be in statical equilibrium and will have no ten- dency whatever to shift on its foundations w ^ W v* f , R i CO R ui ir vv v* ( . R \ /c~\ when - -— cos co = ( cos co 4- — cos 2co 1 (00) g R g R\ L ) that is when tt) = wicos to -\- - cos 2co ) . (61) cos co It is evident from this equation that equilibrium cannot exist for all values of co unless the c ounter- weight n> also varies with the crank angle co. This it is of course not practicable to do, hence there will always be a shifting force acting on frame equal in amount to Wv*( , R \ tdv* fr I COS CO -f- — COS 2CO ) -— COS CO g R v L J gR = i^f cos co + -J cos 2co — x cos co\ = F \ (1 — jr)cos co -f- — cos 2co (62) x being equal to --— . BALANCING RECIPROCATING PARTS BY COUNTER-WEIGHT. 79 The maximum shifting effort when reciprocating parts are not balanced by counter-weight corresponds to w = o and x = o; hence maximum shifting force equals •0+f) and when the reciprocating parts are balanced the maximum shifting force corresponds again to w = o and is equal to (-*+*) When x = unity, that is when counter-weight va = W = weight of reciprocating parts, the ratio of these two maximum shifting forces is It will generally suffice to make x = .; to .8 ; the unbalanced portion being resisted by the foundations. Calculate the value of the vertical components of the mass m. Instead of employing the counter- weight as shown in the pre- ceding figure it is customary to employ a crank disk of the fol- lowing shape. The crank proper is balanced by a counter-weight of similar shape shown in dotted lines, there remain therefore only the two portions ABCDE and A'B' C'D'E having the depth w which can be utilized as counter-weight for balancing the reciprocating parts, the dotted lines A'F'G'E' and AFGE representing the counter- balance for part (if any) of the weight of connecting rod, which part is supposed to be concentrated at crank pin. It is evident from the figure that the center of gravity c of the portions ABCDEGF and A'B' C'D'E' F' does not fall upon the crank-pin circle, conse- quently this deficiency of radius must be made good by increas- ing the masses ABCDEGF and A'B'C'D'E'G'F'. Let the sum 8o BALANCING RECIPROCATING PARTS BY COUNTER-WEIGHT. of these larger masses be represented by M' , its radius by t, then we must have M'(v'Y _ m'v 1 R and since — = — - we have M' = — ni. v R r (63) The value of t can be found experimentally by suspending the figure ABCDEGF ixom two of its corners and determining the 1 1 : 1. 1 1 T* J LP 1 1 • fr- U.-5» Fig. 35. intersection of the two lines of suspension, or it may be balanced over a knife edge in two different positions. M' can then be computed. The dimensions of the simple crank can be calculated from the formula given by Reuleaux. To find room on the crank disk for the large mass M' of coun- terweight, the outside circumference of crank disk, must be taken considerably larger than crank-pin circle. By casting hollows in the disk and filling them with lead, all the counter-weight desired can be placed on crank disk. When double cranks or disks are employed the disposal of counter-weight becomes an easy matter. INFLUENCE OF NEGLECTED RESISTANCES. 8 1 VIII. Effect of Frictional Resistances, of the Weight of the Rod and Exact Values of the Forces of Inertia, on the Rotative Effort, on the Pin Pressures and on the Force Shaking the Engine Bed. The effect of these three influences on the rotative effort and energy of the fly-wheel has been very fully considered by Prof. D. S. Jacobus* in two papers, published in Vol. XI, Transac- tion of American Society of Mechanical Engineers. In these papers very accurate formulas were established which took account of these three influences. They were applied to four different engines representing a wide range of practice. The exact diagrams constructed for these engines showed that the fluctuation of energy, AE, of the fly-wheel due to the variation of the rotative effort differed by an insignificant amount from the approximate diagrams found by supposing all the reciprocating parts (inclusive of rod) concentrated on piston. A similar result was obtained for the crank-pin pressures, the nearly exact ones differing but slightly in direction and intensity from those found by assuming the mass of the rod to be all concentrated at wrist- pin. The following tables are given by Prof. Jacobus and furnish the data of the four engines and the results of the numerical computations. * Prof. Jacobus credits Prof. J. B. Webb with valuable assistance in the preparation of these two papers. 82 DIMENSIONS AND SPEED OF ENGINES. TABLE XI. Dimensions and Speed of Engines to which the Formulas have been applied. Revolutions of crank shaft per minute . Length of stroke, in inches Diameter of cylinder, in inches .... Length of connecting-rod, in inches . . Distance from the wrist-pin to the center of gravity of rod, in inches Distance from the center of the crank shaft to the line of travel of the wrist-pin, in inches Principal radius of gyration, in inches . . . Weight of piston, piston-rod, and cross-head, in pounds Weight of connecting-rod, in pounds . . . Indicated horse-power Class of Engine. M 2 o J3 II 3OO 12 IO 36 20.I5 O I5.00 90. 70. 57 .C -a > ^^ '*-> rt O o o •n o o >J rt J 250 24 92 55 o 34-i 474 3°7 345* ■a I v 1 V w m (_. U o 3 (l u o M III 60 60 26 # ISO 78 o 48 1300 1200 346 IV 3 o C 320 10 II 5* 3.22 R 0.5 R 2.07 R 50 66 For one cylinder. NUMERICAL APPLICATION TO STANDARD CLASSES OF ENGINES. 83 TABLE XII. Results of Numerical Applications to Several Standard Classes of Engines. V a So c W "o c/l m a U Conditions assumed. AE fPds Pressure acting on crank pin Exact. Approx- imate. Not in- cluding acceler- ating forces. Exact. Approx- imate. Not in- cluding acceler- ating forces. I Not including the effects of friction and weight .184 •174 .238 58.1 57.3 83.0 I Not including friction but including weight . . .180 58.0 56.8 I Including both friction and weight , . . . .171 II Not including friction and weight .l8l .172 .275 56.8 55-5 108.0 III Not including friction and weight .142 •131 .I8 S 76.1 75-5 89.4 IV Not including friction and weight .160 ■H7 .247 61.0 60.8 75.5 The term approximate in these tables refers to the assumption that the whole mass of the rod is concentrated at wrist-pin. The AE expresssion - is the ratio of the periodical excess or deficiency f Pas of energy AE, to the whole energy exerted per revolution, f Pds. The particular crank-pin pressure given in table is the maximum one and is in pounds per square inch of piston area. It is evident from these tables that for most practical work the aforesaid approximation, used in the preceding sections, is suffi- ciently accurate. The remainder of this section may therefore be omitted by those who do not care to examine the refinements of this subject. 84 ORDER OF DISCUSSION. We have not reproduced here any of Prof. Jacobus' formulas for the components of the forces acting on the pins, though they are easily applied, and, after the constants have been computed and introduced, lead quickly to the desired results. We have preferred graphical determinations because they are simple, and naturally accompany designing. We will therefore follow graphical methods in finding exact values for each of the three influences under consideration. The order in which we will take them up is : a. The friction between a pin and its bearing. b. The direction of internal stress in a rod, when friction is taken into account, but gravity and inertia neglected. c. The direction of internal stress in a rod, when friction and some other force, say, resultant of gravity and in- ertia-resistance, are considered. d. Determination of the exact accelerating force of the rod corresponding to its motion in the slider-crank chain. e. Combination of weight of rod with its force of inertia. f. The components of this total force at wrist- and crank- pin, when friction is neglected. g. The components of this force at these pins, when fric- tion is considered. h. Determination of the force shaking the engine bed. i. Diagrams of shaking forces with different degrees of counterweighting. j. Diagrams of pressures at crank- and at wrist-pin. A. The Friction Between a Pin and its Bearing. The frictional resistances of motion and rest differ by quantities that are directly proportional to the coefficients of friction for motion and rest, respectively. By using the coefficient of friction for motion, we may treat the body as if it were at rest and yet THE FRICTION BETWEEN A PIN AND ITS BEARINGS. 85 determine the direction and intensity of the forces as they exist under running conditions. Let us suppose the pin to be at rest, without tendency to move in either direction, and that the resultant pressure P between the pin and its bearing acts in the direction c a e f, Fig. 36. Then the pin can be brought to the eve of, say, left-handed rotation, in one of two ways, either by the action of an additional, single, force Q, with lever arm c I, or by the action of some couple M having left-hand rotation. Fig. 36. In the first case there will be an infinite number of solutions depending upon the location and direction of the turning force Q. The moment F'r of the friction of the bearing will balance the moment Q X cl of the new force and the point of application of the resultant pressure between pin and bearing will be shifted from a to b, sufficient play being supposed to exist between the pin and its bearing to permit this to take place. The point b is where the resultant P(=eg) of the forces Q{= eh) and P„(—ef) cuts the area of contact. Evidently the new resultant Pis dif- ferent in intensity and direction from the original pressure P„. The pin is kept stationary by + P and — P; turning of the pin is prevented because Q X cT= P X cd= Pp' = F'r = rotation shifting the force to the < . , V ; I right-hand J I " g 1 in applying this rule the plane of forces must be viewed so that the force of translation is downward. The moment of the force or couple capable of producing the desired angular acceleration — -, is, according to Mechanics, at _ dii . dii ,,, N I — - = mk 2 — - (66) dt dt v ' where I is the principal, polar, moment of inertia, m the mass of the connecting rod and k the corresponding radius of gyration. This expression is perfectly general so far as the distribution of the mass of the rod is concerned. Any arrangement of this mass which will not alter its center of gravity and which pre- serves the same total moment of inertia I unchanged will require the same total accelerating force as the rod itself. The problem may consequently be greatly simplified for graphical purposes by supposing the mass of the rod concentrated at two points that are on opposite sides of the center of gravity and in line with it. FORCE NEEDED TO ACCELERATE ROD. 95 In order that the equivalence of total accelerating force may be maintained it is necessary that the following three conditions be fulfilled, 01 = 111, + m 2 , (67) mA = mA (68) nt/& 2 = mji* + mA 2 . (69) and by combining these we get k* = kX, (70) where k z and k 2 are the distances from the center of gravity of the two points at which the mass is supposed to be concentrated and m„ m 2 the masses at these points. Equations (67) and (68) keep the total mass and location of the center of gravity the same as before and therefore keep unchanged the accelerating, compo- nent, force due to the translation of the center of gravity. The fulfillment of equation (69) or (70) makes the turning moment of the rearranged, two-point, rod the same as that of the original rod with distributed mass. As the force and energy of this two- point rod are the same as in the original one, we can confine our determination of the total accelerating force to the simplified, two-point rod. The problem is now reduced to finding the acceleration of the center of gravity of the rod and that of each of its two points of concentration. The problem is therefore now a purely kinematic one and is a special case of the general problem of finding the acceleration of any point on the connecting rod of the slider- crank mechanism. In this general case of the mechanism and its motion, the accel- eration CO, Fig. 43, of the crank-pin center C is known and the acceleration Ww" (= Ow) of the wrist-pin center or slide can be obtained as in Fig. 16 and p. 43. The accelerations of two points of the rod will then be known and the center of acceleration G of the rod can be found.* The acceleration of any point of the * See Weisbach-Hermann's Machinery of Transmission, Vol. Ill, Section I, §21. 96 FORCE NEEDED TO ACCELERATE ROD. rod is directly proportional to the distance of this point from the center of acceleration G, and, for the instant in question, the acceleration of each point makes the same angle with its own Fig. 43- instantaneous radius of acceleration.* This solves the problem but not in a convenient way because the center of acceleration generally falls beyond the limits of the drawing. A method that is free from this objection and also simpler in other respects is found as follows : In Fig. 43 revolve the triangle CWG (having the rod CW as a base and acceleration-center Gasa vertex) through an angle GCG', equal to that made by the direction of acceleration of point C with its instantaneous radius of acceleration CG. Then will the revolved radii G'C, G'<&', G'H' and G'Wbe parallel to the directions of the accelerations of points C, 5°° — '3-74 —11.32 — 25.06 3-83 160 — 14.27 — 12.07 —26.34 2.62 17c -'4-55 — 12.51 —27.06 1.33 180 — 14.64 — 12.65 — 27.29 0.00 FORCES TENDING TO SHAKE THE ENGINE BED. Ill The approximation just given can be still further simplified for all cases that may arrise in practice. In such cases the friction circles are very small and far apart, and the tangent RJ to both friction circles is scarcely distinguishable from the direction HI or EF. We may therefore treat this tangent RJ as the direction of the internal stress, draw Sb parallel and then proceed as above in the determination of Wb and location of its equal P w y. The inaccuracy involved in this last, approximate, construction is much less than that connected with the best determined coefficients of friction. H. Determination of the Forces Tending to Shake the Engine Bed. The constant forces acting on the engine bed, like the weight of fly-wheel and pull of belt, tend only to shift the bed. It is the variable forces that tend to shake it, namely, the effective steam pressure P against cylinder covers, the pressure of the cross- head against the guides and the pressure of the crank-shaft against its bearings. By constructing a polygon of forces, it can be shown that the force tending to shake the engine bed is the resultant of all the unbalanced accelerating forces acting on the different links of the machine. In other words, the shaking force is the resultant of the forces of inertia of the moving parts. The same conclusion is reached from general considerations. Suppose the forces of inertia to be replaced by their resultants acting as external forces. Then the whole system of external forces will be in equilibrium, the moving pieces may be considered at rest for the instant, the whole machine acting as one rigid piece. As the steam then acts equally in opposite directions upon the machine, its shifting and shaking influence are both nil. The internal forces at the bearings occur in pairs of equal and I 1*2 DIAGRAM OF FORCES TENDING TO SHAKE ENGINE BED. opposite forces, and their shaking influence is therefore likewise nil. The only external forces remaining are the resultants due to inertia, the action of gravity and the pull of the belt. The weights are constant forces, but their points of application change with the motion. In stationary engines we may neglect the shaking influence due to this cause. As regards the pull of the belt Mr. W. Willis has shown by his experiments that the sum of the two tensions is not a constant quantity, as is generally assumed, but increases with the load. If we suppose the load constant the belt pull will also be constant, and will only tend to shift the bed without shaking it. The only external forces re- maining are the variable forces of inertia, and these do tend to shake the engine frame. In the case of the engine, therefore, the shaking forces are, the resistances due to the inertia of the purely reciprocating pieces (piston, piston-rod and cross-head), to the inertia of the connecting rod and to the inertia of the crank disc. The first and last are easily obtained by computation, and the second can be obtained by the construction shown in Fig. 44 or Fig. 45. I. Diagram of Forces Tending to Shake Engine Bed. Fig. 50 is such a diagram prepared by Prof. D. S. Jacobus and published in the Trans. Amer. Soc. Mech'l Eng'rs, Vol. XI. It contains a series of polar curves, aaa, bbb, ccc, etc., whose vectors, drawn from the pole or center V, represent in direction and intensity the resulant forces of inertia of all moving pieces. The data for the construction of the diagram are given on the same page, and under Case I, Table XI. The table accompanying Fig. 50 contains the components or the force of inertia for this particular engine, and is in convenient shape for use. The last column contains the ordinates, like C Z C 2 , of the polar curve aaaa, and the last column but one the abscissas, like VC 3 , of this same polar curve. As this polar curve aaaa, corresponds to the case of no counterweight, its vectors represent in direction and inten DIAGRAM OF FORCES TENDING TO SHAKE ENGINE BED. 113 sity the resultant of the forces of all moving pieces, e^cept-the cranky The remaining curves bbbb, cccc, dddd, etc., can there- fore be directly deduced from curve aaaa, by laying off from each point a, and parallel to the crank position corresponding to this point a the centrifugal force of counterweight on crank disc. For instance, when crank is at 40 , the vector VC 2 will be the resultant of the inertia-forces of all the reciprocating parts. To get a point on curve bbbb, for the same crank position of 40 , we lay off C 2 b= }( X 31.25 = 7.81 parallel, but opposite to VC. To get a point on curve dddd, for this crank angle of 40 , we lay off on the same line CJbd = $4 X 31.25 = 19.53 lbs. per □", and so on for the other curves. As the vertical components of curve bbbb are very small, its small amount of counterweight would be suitable for cases in which the shake occurs more readily in a vertical than in a hori- zontal direction, as when a horizontal engine is placed on an upper floor of a building. On the other hand the horizontal components of curve eeee, are small and the corresponding, heavier, counterweight is preferable when the horizontal forces produce most shake, as in engines set on tall foundations. The diagram just described is suitable for stationary engines, and can readily be constructed for such cases as soon as the ac- celeration of the center of gravity of rod has been found graph- ically, (see page 101, Eq. 71 and Fig. 44 or 45). We have then to multiply this acceleration by the mass of the rod per □" of piston, and resolve this force into two components, one horizon- tal and the other vertical. Of these two components add that which is parallel to stroke to the accelerating force (per □" of piston) of the piston, piston-rod and cross-head. This sum will be one of the coordinates of the curve aaaa (Fig. 50). The re- maining component of the rod will be the other coordinate of curve aaaa. The influence of counterweight can then be added in the manner already described in connection with Fig. 50. The diagram represents the shaking forces acting on engine bed in the vertical and axial of the engine. DIAGRAMS OF PRESSURES AT WRIST- AND CRANK-PIN. uj g- .OCl '%' \ "t ■°' .OM \ 1 *°* ■O'li ,081 .Oil 2 11 I I* 1 " . will have the fly-wheel and main bearing to the < /% > of the axis of the cylinder, and still retaining the position at the cylin- der, if the top of the fly-wheel moves j " y , [the observer, ,, . . , . f run over 1 the engine is said to < , > . b \ run under J X. Graphical Determination of Diameter of Crank Shaft. The problem is one of compound stress, and has been treated by Reuleaux in his Konstrukteur. Both bending and twisting forces strain the shaft. The bending forces are: weight of the fly-wheel, pull of the belt, the centrifugal force of the counter- weight on crank disc, the pressure of connecting-rod against crank-pin and the reaction of the bearings. The twisting force is the moment of connecting-rod against crank-pin. As the bending forces are not parallel, and do not all act in the same plane, the first step is to combine these separate bend- ing moments into one resultant bending moment. The next step is to combine this resultant bending moment with the twist- ing moment so as to form either an equivalent bending, or an equivalent twisting moment. In the present case we will reduce all the moments to one equivalent bending moment. The first step may be simplified by certain preliminary reduc- tions and approximations. The weight of fly-wheel and the pull of the belt act in the same plane, and can be reduced to a single DETERMINATION OF DIAMETER OF CRANK SHAFT. I 1 9 force. The belt pull is the sum of the tensions on the tight and slack side of the belt, and this sum is not constant, but in- creases, while running, with the load. But this does not call for special calculation at this place, because this increase in the sums of the tensions is limited by the slip, and this was assumed at 3 feet per min. as a maximum. We will therefore assume the sum of tensions to be 1^ times the tension on the tight side of the belt when transmitting the maximum power. This, under our assumptions (p. 17) will generally be somewhat in excess of the true sum. Assuming that shop shaft and cylinder are on opposite sides of crank shaft, and that the belting makes an angle of 30 with the horizon, we can now combine the belt pull with the weight of the fly-wheel by the parallelogram of forces and note the angle made by the diagonal with the horizontal. The centrifugal force of the counterweight and the pressure of the connecting rod against crank-pin do not act in the same plane and should be given separate diagrams, if great exactness is de- sired. But such a refinement is here entirely unnecessary, and we may approximate by assuming them both to act in some convenient, intermediate, plane which is nearest to the larger of these two forces. These two may then be combined by the par- allelogram of forces, again noting the angle the diagonal makes with the horizontal. The bending forces have thus been reduced to two forces and to the reactions of the bearings. We now construct by the methods of graphical statics a diagram of bending moments for each of these forces. Suppose that in Fig. 53 the scale of distances is %,ox 3" to the foot, and the scale of forces is, say 200 lbs. to the inch. The polygon of forces due to the fly-wheel resultant found above, is ABC, the length AB representing the reaction of the out-board bearing and BC the reaction in the main bearing, the two reac- tions having the ratio MN -¥- NQ. Choosing in the polygon of forces on the horizontal through B, the closing line AD of the equilibrium polygon ADGF Will also be horizontal, and thus 120 DETERMINATION OF DIAMETER OF CRANK SHAFT. convenient for subsequent combinations. The distance OB of the pole from A C is chosen so as to give a convenient factor for multiplying the vertical chords or intercepts HI of diagram of moments ADGF. We know from the principles of graphical statics that the product of the intercept HI by the constant OB measures that part of the bending moment at cross section KL of shaft which is due to the forces on the fly-wheel. As OB is constant, the intercepts themselves may be taken to represent the bending moments of the cross-sections of the shaft directly above them. The numerical value of any bending moment is found by multiplying the proper intercept by the distance OB X 200 X 4, the last two factors respectively representing the scale of forces and the scale of distances on drawing. If we take OB equal 4 inches, the product of these three factors = 3200, which is a convenient factor with which to multiply directly the intercepts ///obtained from ADGF, the diagram of moments. In like manner for the second resultant, due to counterweight and crank-pin pressures, we find the polygon of forces A'B'C'O' and the equilibrium polygon D'E' C. This second resultant is constantly varying in direction and intensity with the rotation of the crank. It is greatest at the beginning of the forward stroke. But this does not correspond with the piston position at which the maximum twisting moment exists. As the greatest stress on the shaft probably occurs near this position, we shall suppose the diagram of moments due to this second resultant to be found for the crank position corresponding to the maximum twisting moment. The vertical intercepts of polygon D'E' C represent the bending moments, which are due to the action of this second resultant and its numerical value for any cross-section is found by multiplying 3200 the intercept of this diagram immediately below the cross-section in question. These two diagrams of bending moments are both placed, for convenient combination, in the plane of the paper, though the forces calling them forth are neither parallel nor do they act in Fie. S3. U 122 DETERMINATION OF DIAMETER OF CRANK SHAFT. the same plane. As the intensities of the bending moments are represented by linear quantities and their directions are those of the forces producing them, they may be combined like forces, by means of the parallelogram of forces. This we will now proceed to do, being careful to take into account the angle between the forces. It will probably often happen that this angle is a small one, in which case the intercepts of the two diagrams may, with little error for the present purpose, be directly < , . . , > when the directions of the bending moments developed at any ( a like 1 cross-section by the two resultants are < ,., > . When the angle I' HJ = r, included between the arrowheads of the forces is large they should be combined on the principle of the triangle of forces. In the figure this is illustrated by lay- ing off the length HT of one diagram on HJ and joining yand /. We now take this line IJ of the second diagram and lay it off below the horizontal XX making it equal to RS, which is located on the vertical SRIHH' passing through the assumed cross-section KL. Repeating this process for every other cross-section we get a third (the lowest) diagram XRXZSZ whose intercepts represent the resultants of the purely bending moments attacking each cross-section. The next step is to combine this resultant bending moment M t with the twisting moment M t , so as to get an equivalent or ideal bending moment M bi . The formula connecting these three quantities is : M bi = #M t + 5/ 8 VMf + Mf. (73) Before undertaking the graphical construction of this formula we will construct the diagram of twisting moments to the same scale as the bending moments. The twisting moment varies for the different crank-positions, but we will at once assume its maximum value P c /R. To have the intercepts of the diagram of twisting moments on the same scale we must make P C f X R = O'B' X y. Making, in one of the polygon of forces, DETERMINATION OF DIAMETER OF CRANK SHAFT. 123 C'Y=Pc/-a.nd 0'V'=R =^fg, we get, by similar triangles, the distance VW = c / = y. Now lay off la = y above the OB horizontal CD and through a draw ab horizontal. The in- clined line bee can be taken to represent the varying twisting moment of crank-shaft within the fly-wheel hub. The vertical intercepts of diagram abcC represent the twisting moment on the same scale as the other moments. We can now return to the graphical construction of the last formula. The last term containing the radical can also be written Vift M b y + (s/ 8 M t y. The first term ^ M b , under the radical is found by taking $/& of each vertical intercept of the diagram XRXZ USZ. For in- stance, Rs = $4 RS. The broken line in this diagram represents this division of RS = M b into ^ and y& segments. In like manner fyfc M t =kj= $/% la can be found. Making Rm = ^ M t =H'i, the hypothenuse ms (= sT) of the right-angled triangle mRs is the graphical equivalent of the lastj radical, term of our formula. Adding it to Sj = %i M b we get ST = M bi . Its numerical equivalent, with the scales assumed above, is M bi = length of ST X OB X % X 200 inch lbs. This value of M bi is to be substituted in the general formula for shafts subjected to bending stress d = 2.17 Jl—p- inches, (74) \ fb where f b is the permissible working stress in lbs. per □" when shaft is subjected to bending stress. The result may be checked by the following empirical formula representing high-speed en- gine practice, for diameter of main, crank-shaft, journal d = 0.44 D + y 2 ; (75) here D represents the diameter of the cylinder. Having indicated the general method of procedure it will now be easy to apply the graphical solution to cases in which the fly- wheel overhangs the bed. 124 PLANE OF DIVISION OF THE BRASSES. XL Determination of Plane of Division of the Brasses and Length of the Main Bearing. The plane of division of the brasses should be at right angles to the resultant of all those pressures which exist when the en- gine is running with its average load. These pressures are com- ponents of the weight of shaft and fly-wheel, pull of belt, centri- fugal force of counterweight and the thrust or pull of connecting- rod. For each crank position these four components may be combined by the polygon of forces so as to give the resultant force on bearing. Fig. 54. Fig 55. In Figs. 54, 55 and 56 these forces are shown and their force polygon. For a given load, speed and engine, RB and B W may be regarded as constant in direction and intensity, consequently, R W may have an invariable position on the diagram Fig. 56. As the action of the counter-weight is constant in intensity and opposite to the crank in direction, we may take W as the center of a circle described with Ww as a radius. From the points of this circumference we draw to T equal and parallel to the (reduced) PLANE OF DIVISION OF THE BRASSES. 125 values of P c y corresponding to the different crank positions, then will RT represent the resultant of the two pressures exerted by the shaft against its main bearing. Fig. 56 shows two determi- nations, RT and RT, of these resultant pressures against the bearings for the 30 and 6o° crank positions. Fig. 56. The forces in this figure (56) have been chosen at random and do not represent any special case. When a graphical determina- tion, Fig. 53, of the diameter of the crank shaft has been made, the components of these pressures at the main bearing can be found from the force polygons there drawn. A resultant should be drawn for each of a series of equidistant crank posi- tions and the plane of division of the brasses placed at right angles to the resultant of these resultants. Great accuracy is not necessary, in most cases inspection will determine best posi- tion of plane. For the determination of the length / of crank-shaft journal of a high-speed engine the following empirical formula, represent- ing good practice, may be used /= i%d+ 3' (76) 126 DETERMINATION OF DIMENSIONS OF STEAM PASSAGES. where d is the diameter of same journal determined according to another empirical formula, for same class of engines, d = 0.44 D + %", D representing the diameter of piston. XII. Determination of the Dimensions of Steam Passages. The first dimension to be settled in valve-gear problems is the requisite area of the ports, for upon this depends the maintenance of the desired steam pressure. The principal factor in its deter- mination is the velocity of the steam current and as this varies with the piston's motion in different parts of the stroke, we must see to it that the minimum cross-section in the passage is adequate for every part of the stroke. This part of the problem divides it- self into two parts: (a) the ascertainment of the minimum area needed at the given piston speed and with given cylinder area, and (b) an examination of the cross-sections of the steam passages that are really available or effective with the special type used, taking into account the contraction caused by the flow past edges and the narrowing of ports caused by the motions of the valves on each other or on their valve seats. This second part presents no theoretical difficulties and we shall therefore not examine any special case here, but this should be done in the draughting-room when the type of valve has been chosen. It is practically important in case there are supplemen- tary, or multiple, ports in the main or distribution valve. Allow- ance should also be made for the contraction of the steam cur- rent when the ports are very narrow, for this effects a very notable reduction of the port area geometrically available. We will now consider the minimum port area necessary for a given area of cylinder and a given piston speed. DETERMINATION OF DIMENSIONS OF STEAM PASSAGES. 127 Radinger's criterion of sufficient passage area was whether or not the admission line of indicator card remained horizontal up to cut-off. According to Radinger's experiments, Area of port 61 average speed of piston in feet per sec. , , Area of cylinder A ~ ioo This rule evidently corresponds to an average velocity of ioo feet per second of the current of steam in the port ; but the experiments seem to have been made on engines in which the maximum cut-off was equal or greater than 0.5, consequently if we assume that in the engines experimented upon the ratio of connecting rod to crank varied from 4 to 6 we can easily deter- mine the maximum permissible velocity of the steam in the ports. „ , connecting-rod , , , . , For when -= = 4 to 6 we have the ratio 01 crank Max. velocity of steam in ports . , . „. -x j — r- 7 — ; — = 1.62 to 1.59 respectively. (78) Ave. velocity of steam in ports J ^ r J x/ ' This gives about 160 feet per second as the maximum permissi- ble velocity of the current of steam in the ports when the steam passages are long and narrow. Radinger says that when the steam passages are short and the cut-off short (for instance, when there are separate admission valves for each end of the cylinder) somewhat smaller port areas may be employed than would result from above rule ; in other words, in such cases the maximum permissible velocity of the current of steam may be somewhat greater than 160 feet per second.* Mr. Charles T. Porter gives the rule that the velocity of the current of steam in the short ports should not exceed 200 feet per * Some American engineers prefer the rule that the average speed in short steam ports should not exceed 150 feet per second, and in the exhaust ports should < 125 feet per second. When the steam is admitted and discharged through the same passage, some engineers recommend that its cross-section be proportioned for the exhaust, preferring an increase of clearance to an increase of back pressure. 128 RATIO OF PISTON VELOCITY TO CRANK- PIN VELOCITY. TABLE XIII. Velocity of Piston for a Unit of Velocity of Crank-Pin. To obtain actual velocity of piston multiply tabular quantity by actual velocity of crank-pin. Forward stroke is towards, and return stroke, away from, crank shaft. Crank Angles. Connecting Rod -=- Crank = Forw'd. Return. 4.0 4-5 .1064 5-o 5-5 6.0 00 5 175 .1089 .1045 .1030 .1016 .0832 IO 170 .2164 .2117 .2079 .2047 .2022 • 1737 '5 165 •3215 •3H5 .3089 ■3054 .3005 .2588 20 160 4227 .4136 .4065 .4019 •3957 .3420 25 155 .5189 .5081 •4995 .4925 .4866 .4226 3° 150 .6091 .5968 .5870 •579i •5724 .5000 35 H5 .6923 .6788 .6682 .6596 .6523 •5736 40 140 .7675 •7533 ■742i ■7329 ■7253 .6428 45 135 .8341 .8195 .8081 .7988 .7910 .7071 5° 130 .8914 .8771 .8657 .8564 .8488 .7660 55 125 •9392 •9253 .9144 .9055 .8982 .8192 60 120 .9769 .9641 .9540 ■9458 •939° .8660 65 "5 1.0046 •9932 .9842 .9769 .9709 .9063 70 1 10 1.0224 1. 01 27 1.0052 .9990 ■9939 •9397 75 105 1.0304 1.0228 1.0169 1.0121 1.0082 .9659 80 100 1.0289 1.0237 1. 0199 1.0164 1-0137 .9848 85 95 1.0186 1.0160 1-0139 1.0127 1. 0109 .9962 90 90 1. 0000 1. 0000 1. 0000 i. 0000 1. 0000 1. 0000 95 85 •9738 .9764 •9785 •9797 .9816 .9962 100 80 .9407 .9460 .9500 •9532 ■9559 .9848 105 75 .9016 .9091 .9150 .9198 ■9237 .9659 no 70 .8571 .8667 .8743 .8804 •8855 •9397 115 65 .8080 .8194 .8285 •8357 .8418 .9063 120 60 •7552 .7680 .7781 .7863 •7931 .8660 125 55 .6992 .7130 •7239 .7328 .7401 .8192 130 5° .6407 .6550 .6664 .6756 •6833 .7660 135 45 .5801 ■5946 .6061 .6155 .6232 .7071 140 40 .5181 .5323 •5435 •5527 .5603 .6428 145 35 •4549 .4683 .4790 .4876 ■4949 •5736 150 3° •3909 .4032 4130 .4209 .4276 .5000 155 25 .3264 ■3371 •3458 •3528 •3586 .4226 160 20 .2614 .2704 .2776 .2821 .2884 .3420 16s 15 .1962 .2032 .2088 •2123 .2171 .2588 170 10 .1309 ■ 1356 •1394 .1426 .1451 • 1737 175 1 5 •o655 .0679 .0698 .0714 .0727 1 .0872 DETERMINATION OF DIMENSIONS OF STEAM PORTS. 129 second. This is what we shall assume as suitable for short pass- ages. Representing the actual velocity of the steam or piston by V s , and by b' the minimum permissible opening of port when the crank makes an angle (83) The full width b'" of port for piston valve may of course be found by substituting for b' its maximum value b in the preced- ing formula. Formula (83) assumes that the steam passage within the main valve has everywhere a cross-section > than the maximum, annular, port opening effected by the piston valve. As regards the steam passage itself (not the port) if it is large enough for the exhaust it will be large enough for the admission. Its area should therefore be the quotient, - Average piston speed (ft. per sec.) ,„ ■, 100 to 125 The port opening for exhaust should also be tested, taking 200 feet per second as the maximum velocity of the current that is permissible. By plotting the port-openings as ordinates on the piston stroke as a base, and connecting them by a curve, the slope of this curve where it crosses the base will be a measure of the rapidity of cut-off. 132 DETERMINATION OF DIMENSIONS OF STEAM PORTS. Where many valve-gear problems must be solved, as in a col- lege draughting-room, two diagrams like Fig. 57 may be drawn which will cover the whole range of practice. In both these general diagrams the ratio of connecting-rod to crank = 6, but the maximum permissible velocity of steam is 200 feet per second in one diagram and 160 feet per second in the other, correspond- ing to short and long steam passages respectively. For these general diagrams assume - = 10 and average speed of piston in feet per minute {w') equal to 1000. The diagrams will then be applicable to any problem, giving the minimum port-openings OG (= b') to a scale of 1 -= -that is OG on the diagram 10000/ s the true size. In this way a diagram whose will be 1 0000 / w' A scale varies with the assumed data can be made to fit all cases. To reduce the port-openings given by these diagrams to their full-size values on the Zeuner, valve-circle, diagram a reduction PORT OPENING GIVEN BY DIAGRAM. Fig. 58. VALVE DIAGRAMS AND DIMENSIONS OF VALVE AND GEAR. 1 33 arrangement, like that shown in Fig. 58, will be found very con- venient. The full sizes are taken from horizontal line 1.0. As regards the thickness of the cylinder walls, Prof. W. C. Unwin* says, "(a) that it should be strong enough to resist the internal steam pressure ; (6) rigid enough to prevent any sensible alteration of form ; (c) it must be thick enough to insure a sound casting ; (d) thick enough to permit reboring once or twice when worn. Generally other considerations than strength are of so much importance, that the empirical rule agrees better with prac- tice than a rule making the thickness t depend upon steam pressure :" hence when D is diameter of cylinder t = 0.02 D + 0.5 to 0.02 D -f- 0.75. (85) XIII. Valve Diagrams and Dimensions of Valve and Gear. We shall assume that the reader is acquainted with the prin- cipal functions and common varieties of valves, and also with the terms designating the most important dimensions and parts of ordinary valve gear. This is usually given in elementary text-books on the steam engine. There is such variety in valves in the matter of arrangement of steam passages, in the method of balancing and in the division or assignment of functions, and all this involves so much special dimensioning that it would need a separate treatise to do justice to this part of the subject. We can here only attempt to point out the principal types and the methods of ascertaining the leading dimensions. The ultimate object of all valve-gear discussion is to establish the relation existing between movement of piston and movement of valve. The piston's motion has already been fully discussed on pp. 35-39, and a table on pp. 26-27 gives its travel for dif- ferent crank positions. After the travel of the valve has been ascertained for these same crank positions it will be easy to con- struct a diagram giving directly the desired relation. * Elements of Machine Design. Eleventh Edition, Part II, p. 33. 134 KINDS OF VALVE GEARS. The valve motion that has been most extensively used in practice is that which is known to mathematicians as "harmonic motion," and to engineers as, the motion of the slotted cross- head. Almost all valve-motion deviates a little from this, owing to the angular motion of the eccentric rod. Of late years a few valve-gears have been devised in which the valve's movement differs considerably from that corresponding to the " harmonic law ;" but as yet these gears have not come into extensive use. They have usually complicated mechanisms that do not follow any simple law in their valve-movements and are therefore not readily amenable to scientific treatment. We shall confine our- selves almost wholly to the first, widely used, class. Link- motions of the common forms do however come within this class, but only the Porter-Allen (Fink) form is much used in stationary, high-speed, engine work. In this section only this latter form of link motion will receive a discussion. The Appendix will contain a general, graphical, method for obtaining the valve diagrams of the common forms of link-motions. The valve-gears subject to the "harmonic law" can be divided into two groups : a. Gears with but a single valve which slides on a stationary seat. b. Gears with two valves, of which one slides on the other. The first of these will be considered under the two heads of invariable steam distribution and variable steam distribution. In second group the main part of our problem will be to show that the complex mechanism producing the desired relative motion of the valve on its seat, is equivalent to the action of one, single, eccentric capable of giving the valve in question an iden- tical, absolute, motion, on a similar, but stationary, seat. The problem before us is therefore mainly a kinematic one. In order that we may simplify the usually difficult and complicated parts of this subject it will be necessary to review the elementary parts and present them in a new light. SINGLE-VALVE GEAR. 1 35 Single- Valve Gear.— Invariable Steam Distribution. We have already stated that the main object of valve gear dis- cussion is to find the relation between piston and valve move- ment, and that the piston's position for various crank angles has already been ascertained by suitable formulas and tables. As the eccentric is nothing but a short crank, a similar relation, of course, exists between the valve-slide positions and eccentric angles, but with this difference, that in the former case the ratio of crank to connecting rod is a comparatively large fraction, say 1, while in the latter case the ratio of eccentric radius to eccentric rod is a very small fraction, often less than ^ s . In the former case the piston's motion is unsymmetrical in the first and second halves of its stroke, while the valve's motion is practically sym- metrical on each side of its middle position and for an infinitely long eccentric rod would be perfectly so, and this is the assump- tion which is always made in this sort of discussions and which will therefore underlie all our subsequent work. Moreover, it is practically most convenient to estimate piston travel from the beginning of its stroke, while valve travel is best esti- mated from its center of motion or mid-position, because of the symmetry of its motion and of the arrangement of the ports. For these two reasons the expressions for piston and valve travel are usually different. Just at present we will confine our- selves to the construction of a diagram representing the distance of the valve or slide from its mid-position at the different crank angles and leave the graphical representation of the piston's travel to the moment when we shall combine the two movements into one diagram and thus exhibit graphically their desired relation. We shall first treat the slide-valve as a simple, rectangular, plate, moving on an unperforated seat and driven by an infinitely long eccentric rod. In Fig. 59, let be the center of the engine shaft, OC z , 0C 2 , OC 3 , 0C 4 different crank positions and 0E„ OE^ 0£ 3 , OE 4 , the corresponding eccentric positions, the eccentric 136 SLIDE TRAVEL. being set ahead of the crank by a constant angle dOE x . Let NZ be the dead center line of the eccentric OE (which may or may not coincide with that of the crank OC), then will OP be the mid-position of the eccentric. When the eccentric is in this position the valve will also be in its mid-position. When the eccentric is in position OE.,, the slide will be at the distance E^Si from its mid-position and this distance we will call the travel Fig. 59. of the slide. For the other eccentric positions 0E 2 , OE 3 , OE 4 , the slide-travel is E 2 S X , E 3 S 3 , E 4 S 4 , respectively. If now we lay off this travel ES on the corresponding crank positions while the eccentric is to the right of OP and on the prolongations of the corresponding crank positions when the eccentric is to the left of OP, and making OD, = E T S lt OD, = E 2 S„ OD 3 = E 3 S 3 and 0D 4 = E 4 S 4 , we get the curve OD 3 D r D 4 D 2 , which is called the polar diagram of the slide motion. It is evident from the method of construction that the chord cut by this curve from the VALVE DIAGRAM. 137 crank or its prolongation is the travel of the slide from its middle position. It is also evident that the slide is respectively to the ) I ft- f °f * ts m iddle position when the curve cuts { the prolonged } crank ' and that the slide is g° in § { Towards"" } I- its middle position when these chords < , r \ decrease With an infinitely long eccentric rod, the curve OD^D^D^D^ becomes a circle. To prove this let us take a general case, Fig. 60, in which dead center line OX of crank makes an angle XOZ with the dead center line OZ of the eccentric. From the right- hand half of NOZ, and in a direction opposite to the crank's rotation, measure the angle ZOE = a between this dead center OZ and the eccentric position OE that corresponds to any crank position OC. Then lay off from the crank OC in the same di- rection as the rotation, this angle a — COD. Make OD = OE and on OD as a diameter describe the circle OF'D. The right- angled triangles OEF and ODF' are evidently equal, hence OF' = 0F= ES = valve travel, that is, the circle OF'D cuts a chord OF' from any crank position OC that is equal to the N-& Fig. 60. I38 RELATION OF VALVE-CIRCLE AND CRANK. valve travel ES for that instant. As the polar co-ordinates of circle OE'D and of the polar curve OD 3 D t D 4 D 2 (Fig. 59) are alike, the latter is also a circle, and may be called the slide- or valve-circle. By reversing the steps of this construction we may evidently find the position of the eccentric from the crank position, when the valve circle is given. Our rule then is, to measure from the crank, in the direction of its rotation, the angle a included be- tween it and the diameter of the valve-circle, and then lay off this angle a from the dead center line OZ, in a direction opposite to the rotation of the crank, aud this will give the desired posi- tion of the eccentric. The angle ZOE = a is estimated from the portion of the line OZ to the right of center O, and so long as the eccentric (or slide) is to the right of its middle position the angle made by the eccen- tric with the portion OZ of the reference line will be less than 90 . When this is the case the diameter OD of the slide-circle is always less than 90 from the crank position, which ensures that this circle I 7\ C l"l 1 ri 1 I will cut the < , , > crank whenever the eccentric or slide [ prolonged j is to the < r !^, > of its middle position. This is true for all values of a, for all positions of the crank, for either right- or left-handed rotation, and whether slide-seat is to the right or left of shaft 0< provided, of course, that there is no reversing lever between eccentric and slide. The position of this valve-circle does not change while the crank rotates, its diameter making with the fixed reference line OZ a.xi angle Z>0Zthat is equal to the constant angle COE be- tween crank and eccentric. That this is true is evident from : COE = COD + DOE = ZOE + DOE = ZOD. The valve-circle's position is therefore dependent only on the relative position of eccentric to crank. Provided this eccentric setting remains exactly the same in direction and magnitude, the crank may have either right-handed or left-handed rotation, RELATION OF VALVE-CIRCLE AND ECCENTRIC SETTING. 1 39 may occupy any position in any one of its quadrants and yet give the self-same valve-circle whenever the construction de- scribed above is strictly followed. For a given eccentric setting the valve-circle will have the same position, that is, the slide's distance, position and direction of motion relatively to its center of travel will always be the same for the same crank position. But when the eccentric setting is different, the motion of the slide relatively to the center of travel will be different, even though the crank position and rota- tion are exactly the same. This is illustrated in Fig. 61 : I. Slide is to the right of its middle and is going towards it. II. Slide is to the left of its middle and is going towards it. III. Slide is to the left of its middle and is going away from it. IV. Slide is to the right of its middle and is going away from it. The corresponding valve-circles are given in the second row of Fig. 61 when the crank position is OC and the rotation right- handed, and in the third row when crank position is OC and the rotation left-handed. The angle between the crank and eccentric being the same in magnitude for both rotations, the laying off angle a, used in finding location of diameter of valve- circle, will be the same for the corresponding figures of the second and third row. In the last row of Fig. 61 we have the same four, possible, types of slide valves which are kinematically exactly like the corresponding Cases I, II, III, IV of the first row, and have the same valve-circles. The eccentric driving these slide valves is not shown because convenience of illustration requires the valve to be placed above the cylinder and because the setting of the eccentric relatively to crank is shown in the first row for both right- and left-handed rotation. Each valve is closing left port. Each of these four types has its special, valve, characteristics. A valve is said to have < P OS1 ,Y e > lap when the port is I c ose I j n ^ e valve's mid-position. We will call a ■ valve I open J r I40 RELATION OF VALVE-CIRCLE AND ECCENTRIC SETTING. Fig. 61. edge {indSct^ when it closes the left port (or passage leading to left end of cylinder) by moving to the j ^ght} ■ When all edges of the valve are \ i ^ t \ we will call the valve itself {££&}. When some of the edges are direct and others indirect (as in the ordinary D valve where the steam edges are direct and the exhaust edges indirect), we will call the valve itself direct or in- direct according as the edges regulating admission of steam are direct or indirect* These definitions hold whether the ports are * English and American engineers, so far as the writer is aware, have never used these terms, direct and indirect. Wiebe makes use of them in his "Darstellung der Verhaltnisse der Schieberbewegung." They are short, clear, terms and may profitably be employed in this extensive subject. TYPES OF VALVES. I4I in the seat or the face of the valve. In Fig. 61, fourth row, we have I. A valve with positive lap and direct cut-off. II. A valve with positive lap and indirect cut-off. III. A valve with negative lap and direct cut-off. IV. A valve with negative lap and indirect cut-off. We will designate them briefly as, I positive direct, II positive indirect, III negative direct and IV negative indirect valves. The ordinary D valve in such extensive use is positive direct on its steam side and positive indirect on its exhaust side. The Meyer expansion valve with its halves placed so close together that in the middle position they fall within the outer edges of their own steam ports is an example of a negative direct valve, and the same expansion valve when its cut-off plates are placed so' far apart that in the middle position of the valve its inner edges are outside of the inner edges of their own steam ports, is an example of the negative indirect valve. Other positions of these valves between these extremes will' give positive direct and indirect types. With the same eccentric setting therefore it is possible for a slide to represent any one of the four types of valves, according to the position of the pair of ports relatively to the middle position of the slide. (But with the same eccentric setting and same throw of valve the cut-off will be different for each type.) The terms positive and nega- tive, direct and indirect, refer only to this relative position of ports to valve and are entirely independent of the eccentric setting. (When the valve controls only one port, instead of two, the terms direct and indirect disappear entirely, only the distinctions posi- tive and negative then remaining.) The angle between crank and eccentric, measured from the former in the direction of rota- tion, for any one of these valve types may vary from o° to 360°, although commonly each type has its eccentric set in the eccen- tric-setting-quadrant shown in Fig. 61. In all of the engines of the fourth row of Fig. 61, for the same rotation, the same cut-off is taking place during the same stroke. JO 142 SELECTION OF VALVE-CIRCLES. Fig. 62. Hence four different types of valves can be arranged to accomplish the self-same object. The eccentrics which drive them will have four corresponding settings relatively to the crank. There will be four different valve-circles to represent the motion. They are grouped together in the two diagrams of Fig. 62, which differ only in the direction of crank rotation. To find the valve-circle belonging to any type we draw the crank position OC corresponding to the cut-off effected by the valve, and prolong it beyond center 0. Then lay off the lap, Oa = Oc, and draw circle ehfg with the radius of eccentric. At a and c erect the perpendiculars hae and gcf and thus determine the ends e, k,fz.ridg of the diameters Oe, Of, Og, and Ok of the four valve-circles. To find the one belonging to the type of valve under consideration, notice whether this valve, when at cut-off, is to the right or left of its middle position on its seat. If it is to the {"fa} one of the two circles -jh/ni,} wu ' De the desired one, because these cut the \ ponged \ crank. This now reduces the choice to one of two circles. Now notice whether the valve in its cut-off position is moving towards or away from its middle position. If it is moving { ^yhl m } this mid-position the chord Oa (or Oc) of the proper valve-circle will {1™™^ \ as the crank continues the rotation from its cut-off position. In Fig. 62 chord PORT OPENINGS BY DIFFERENT TYPES OF VALVES. 143 Ob is accidentally equal to chord or lap Oa, but it is not so generally. Hitherto in valve-gear discussions two diametrically opposite valve-circles were employed for they helped to show the main occurrences in the steam distribution at both ends of the cylinder. But in designing valve-gears we usually pass from the valve- circle to the actual gear and there is then danger of confusion as to the eccentric setting when two circles are used. There can be no objection to the use of two circles provided they can be distinguished by some means, say, by drawing the representative one in full lines. We shall however make use of but one valve- circle when there is but one crank. When there are two cranks, diametrically opposite, as in the Westinghouse engine, we shall use two valve-circles. We will not here describe the occurrences of the steam distri- bution, as this is given in many of the elementary treatises on the steam engine and we have supposed the reader to be familiar with the elements of this subject. Besides, the occurences for the ordinary Q valve are inscribed on the Zeuner valve diagram in Fig. 63. But the following table may be of assistance when some of the unfamiliar valve-types are under consideration. TABLE XIV. Periods During which Port Controlled by Valve is Open or Closed (Given by the Limiting and Intermediate Crank Positions). Type of Valve.* Condition of Port. I II III IV Positive Direct Positive Indirect Negative Direct. Negative Indirect. Right Port. Open 3-4 1-2 2-3-4-1 4-1-2-3 Closed 4-1-2-3 2-3-4-1 1-2 3-4 Left Port. Open 1-2 3-4 4-1-2-3 2-3-4-1 Closed 2-3-4-1 4-1-2-3 3-4 1-2 * Left port closed by positive direct valve when crank is at 2. " " " " " indirect " " " " 4. ft " " " negative direct " ,( " " 3. (i « u a it j n direct " " " " 1. 144 RELATION OF PISTON TO VALVE-TRAVEL. This table is applicable to both exhaust and admission by- considering the lap circles of the adjacent figures to represent in the former case the inside, exhaust, lap, and in the latter case the steam lap, provided account is taken of the valve type of the edge in question. Intercepts of lap and valve-circle measure openings. As the valve-circle gives us the valve travel to the right or left for any crank position and Tables, pp. 26-27, tne corresponding piston-position, we can represent graphically by an oval sort of curve the desired relation between piston movement and valve- travel. The ordinates of this curve are the chords cut from the crank by the valve-circle and the abscissas are the distances of piston from end of stroke given by Tables V and VI. Lines drawn parallel to the base line, and above and below the latter a distance equal to the outside and inside laps, give the principal occurrences in the steam distribution. The diagram is easily drawn when the valve-circle has been found. There is another diagram which requires still less work and gives readily the actual piston and valve travel for any crank position. Let OA = OB, Fig. 63, equal the crank radius to some reduced scale. For convenience we will take the length of the connecting-rod three times that of crank. Then with the length CB = 3 OB of this rod describe the arc EBF and with the same radius strike off through A an equal and parallel arc GAH. The horizontal lines included between these arcs will of course be equal to the stroke, and the horizontal intercepts be- tween these arcs EF and G77and the crank-pin circle AJLBU will represent exactly the distances of the piston from one or the other end of the stroke. For instance, when the crank-pin is at_/, the intercept //will represent the distance s of the piston from the left end of stroke. This is evident from equation (20) p. 35, where s = R (1 — cos (o) + L (1 — cos a) Here the first term of the second member is equal AZ = WJ and the second term of this member is equal to VA = IW. STEAM DISTRIBUTION Fig. 63. I46 PORT OPENINGS SHOWN BY DIAGRAM. When the crank is at the cut-off positions OL and OS, the piston travel is KL and RS for forward and return strokes re- spectively, and will be found to be unequal. The valve travels for the same instants are equal to Og ; the scale for valve travel will usually be different from that for piston travel. At an- other crank position, ON, the piston travel is given by MN and the valve travel by Od. When the crank or its prolongation traverses the ■; , , > area, one or the other of the steam ports is open for \ haust ( ' When crank is at Y, the distance of the piston from beginning of the stroke is HY, the valve travel is Os, the opening of the left steam port is us and is closing, while at the other end of the cylinder the steam is exhausting through a fully open port, pt. In Fig. 63 the four crank positions 1, 2, 3, 4, are drawn which correspond to the opening or closing of the steam ports by the admission edges of the valve; the crank position i 1 , 2 1 , 3 1 , 4 1 , correspond to the opening or closing of the steam ports by the exhaust edges of valve. This diagram is for a positive valve. It will be a profitable exercise to closely compare the steam dis- tribution given by Table XIV with that given by Fig. 63. An ordinary D valve is usually set by giving it equal leads at the dead points of the crank ; but this makes every other occurrence of the steam distribution take place unequally for the two strokes. We have already pointed out that there is no necessity for equality of lead; that the lead may be unequal provided the minimum value is sufficient to maintain the steam pressure, and cushion properly ; nor is a positive lead necessary, there are cases when a negative lead is justifiable and desirable. Engine makers lay some stress upon having the indicator cards from the two cylinder ends look alike. Equality of cut-off has more influence on this than any other factor. For this rea- son and not because equality of cut-off is necessarily conducive ECCENTRIC SETTING FOR A GIVEN, EQUALIZED, CUT-OFF. I47 to smooth running, we will give a method* for finding the un- equal laps and the angle between eccentric and crank when the cut-off is equalized for both strokes, say, made equal to 0.8. In Fig. 64 take the distance 00' equal to the sum e -\- e' of the laps at the two ends, and on stroke AB and A'B' describe crank-pin circles. Strike off the equal and parallel arcs BK and A'R with a radius equal to the connecting-rod, as in Fig. 63. Make KL = RS = 0.8 X AB. In the neighborhood of L draw a series of perpendiculars to 00' which cut the lower part of crank-pin circle BLA at points u T . . . u r Then with 5 as a center, Fig. 64. and the distances, Lu t , Lu 2 , Lu y Lu A , etc., as radii, strike off the arcs that will cut the corresponding perpendiculars through «„ u 2 , u 3 , u 4 , etc., at the points x It x 2 , x y x 1 \. Join these intersections by a curve h. It will cut at Mthe other circle on A'B'. From this point M drop another perpendicular MN to 00' cutting the latter in P and the lower part of circle ALB in N. * See Burmester's 675 and 676. 1 Lehrbuch der Kinematik," pp. 672-674, and Figs. I48 EQUALIZED CUT-OFF FOR GIVEN ECCENTRIC SETTING. Then will chord LN = SM by construction and the angle LON = angle SO'M. Each of these angles is equal to the angle between crank and eccentric and hence to the angle between diameter of valve-circle and dead point line of eccentric (see p. 137). For this last angle DOB = DOL + LOB, Fig. 63, and the first part DOL has a cosine = lap -=- eccentricity. The sec- ond part LOB is represented by LOP or SO'P in Fig. 64 and the first part by PON and POM. Hence PON 4- LOP = LON = PO'M -f S0P = S0M = angle between crank and eccen- tric. By giving the greater lap OP to the end of the valve farthest from shaft and lap O'P to the other end of the valve the two cut-offs will each equal 0.8 and the eccentric setting = LON. The other intersections of auxiliary curve h with A'MB' are of no use in this connection. Dr. Burmester also solves another problem, namely, one in which, for a given setting of eccentric, the cut-off shall be equal. Construct circles on stroke AB and A'B' with same center and as before. Then take any point / in circle ALB and lay off angle fOe equal to given setting of eccentric. Through e draw */ perpendicular to 00, intersecting circle A'MB' at e'. Lay Fig. 65. ANGLE OF ADVANCE DEFINED. 1 49 off e'O'f on this circle equal to the given setting and through /' and /draw parallels z 1 /' and zf to 00' till they cut the arcs A'R and BK. Then make z'x = zf; this will give one point x of auxiliary curve h. By similar constructions other points of this curve h can be found. The auxiliary curve h cuts the circle A' MB' in some point 5. Finally make angle SO' M = fOe, drop the perpendicular MPN on to 00' and make angle NOL = fOe. Then will cut-off ^J = ^f , and OP will be the A'B' AB outside lap of the valve at the end farthest from the shaft and O'P the lap for the other end of the valve. In the mathematical discussions of valve-gears, it is found more convenient to use the angle of advance d instead of the eccentric setting or angle between crank and eccentric. It may be expressed by a formula or it may be measured in any valve- gear as follows : Start with eccentric arm half-way between its own dead points and then move the crank to its (the crank's) nearest dead center. During this motion the angle passed through by either crank or eccentric will be the angle of advance and it will be positive if this motion or rotation is in the same direction as the engine rotation, otherwise it will be negative. The formula for this angle at any crank position is : 8 = X — a>' — 9 o°* (86) Letting x represent angle between crank and eccentric, we have x = 9 o° + d ± x = * — <»' ± X, (87) where / represents the angle between the two dead point lines and * According to Rankine : " By angular advance is to be understood the angle at which the eccentric arm stands in advance of that position, which would bring the slide-valve to mid-stroke when the crank is at its dead points." X represents the angle made by the eccentric with its own dead point, a' the angle made by the crank with its dead point line ; either of these angles may be greater than 180°. ISO LEVERS BETWEEN ECCENTRIC AND VALVE. I lower I s '£ n * s use( ^ when dead point line of eccentric I nrered I *^ at °^ ^ e cran k m tne direction of rotation. When there is a reversing lever between eccentric and valve the formula for the angle between crank and eccentric then becomes x> = 180 — x = 9 o° — d =F x- (88) Sometimes a reducing lever is placed between the valve and eccentric for the purpose of reducing the size of the latter and thus diminishing its tendency to heat under high speed. The only modification which this makes in our diagram is to make the valve-circle diameter correspondingly larger. For we as- sume that the valve is driven directly by the eccentric. Some- times, as in Corliss valve-gear, a reducing lever is used to give a sort of differential motion, and the valve direction and chord of lever-pin are made to differ purposely.* The introduction of a reversing lever between eccentric and valve has the same effect on the valve motion, as if the eccentric were shifted to a diametrically opposite position and the valve then driven directly (i. e. without reversal) by the new eccentric. Hence here also, unless the contrary is specified, it is to be understood that diagram is drawn as if valve were driven directly by the eccentric. This finishes our discussion of the simplest case of valve gearing. Under the head of single-valve gearing, with invariable steam distribution, there still remains a case which is mainly of interest because of its bearing on link motions, namely, the case in which the valve stroke does not pass through the center of the shaft, but at a certain distance c from the latter. In this mechanism the stroke of slide is more than twice the length of the eccentricity ; moreover the two dead points of the crank are not diametrically opposite as in the ordinary slider-crank. * See also levers O'RP and O'RQ Fig. 82, Porter- Allen Engine. VALVE TRAVEL IN CROSSED SLIDER-CRANK. 151 In Fig. 66 V is a valve, whose line of stroke VY passes center of shaft at distance 0Y= c. It is driven by an eccentric OE =r and rod EV of length /. In valve-gears the point from which valve-travel is usually estimated is one near the middle of the valve-stroke ; the point of valve whose travel is measured may be any point rigidly attached to valve ; in this figure the right-hand end V of eccentric-rod is the one chosen. If we find the position of V corresponding to each of the crank's dead points and then bisect the distance between these two positions of V, the point of bisection will be the point from which valve- travel is measured. Let M be this center of valve-travel and OE any position of the eccentric. MVwill then be the corresponding valve-travel or distance from middle position. At the point E draw EC equal Fig. 66. and parallel to MV, and join MC. The quadrilateral CEVMis consequently a parallelogram and CM== EV= I. In other words, the locus of the point C is a circle described from M as a center, with radius /. The valve-travel for any other eccentric position OE' can therefore be found by drawing the parallel E'F up to the arc GDff* Draw the chord GOH of this arc per- * The travel of the valve from the end of the stroke for any position OE' of eccentric is evidently either E'S or E'R, according as one or the other end of the stroke is meant. The arcs MRO and NSP are struck, from the ends A and B of valve-stroke, with radius / of rod. 152 VALVE TRAVEL IN CROSSED SLIDER-CRANK. pendicular to line OM; in this figure arc and chord meet at points G and H that seem to be on eccentric circle EE'R ; but they do not necessarily lie on this circle. Even in this sort of valve gears the rod / is long in compari- son with the radius r, the rise OD of the arc is very small and the arc may then be replaced by the chord GH passing through center 0. The distance EI, parallel to stroke, now represents the valve-travel, and for any other eccentric position OE' the dis- tance from E' to the straight line GOH measures the travel of the valve. When eccentric center E and point H are on the same horizontal, the valve-travel EH will be the same wheather mea- sured up to chord GOH or arc GDH; the same may be said when E and point G are on the same parallel to stroke. These two points of E, if constructed, would be found to be diametri- cally opposite and equidistant from arc or line of reference ; if from each of these E's as a. center and with radius / we describe an arc, cutting stroke AB, and then bisect the distance between these arcs the point of bisection will be the center of reference M assumed above. But the travel may be more easily found and this case reduced to that of the ordinary eccentric, by finding an eccentric Oe whose distance eKixom the vertical OY is always equal to EI* To find such an eccentric, we drop from point la perpendicular IL upon the eccentric OE, and erect Ee perpendicular to OE. Now from the point K, where IL cuts the vertical Y, we draw Ke parallel to ii/and join O and e. It is evident that eEIK'xs a parallelogram and eK = EI. To be an equivalent eccentric Oe must not change its length or position relatively to OE, that is, Ee = IK must be a constant for all positions of the original eccentric OE. The figure shows that triangle OKL is similar to * Up to this point the demonstration is like that given by Prof. A. Fliegner in his work " Umsteuerungen der Locomotiven." SINGLE-VALVE GEARS, STEAM DISTRIBUTION VARIABLE. I S3 each of the triangles IJR and OJE. These triangles are there- fore similar to each other, and we have -pr^c— -pr- = tangent of angle HO F= - nearly. (89) Oh. UJ I But this angle is constant, being equal to the angle MO V, hence IK = Ee is a constant, and Oe has all the prcperties of a virtual or equivalent eccentric. If we connect e by a rod / with a slide V , whose stroke passes through 0, the travel of V from its middle position will be very nearly like that of the slide V from M. Evidently stroke of V is greater than if driven by E. The distance of E from chord GOH, or of e from vertical Y, does not exactly represent the travel of the valve, for it does not take account of the angularity of the eccentric rod (though it does consider the direction of line QM, and thus takes partial account of the length of this rod when distance c is given). The exact valve-travel is only given by the distance of eccentric- center E from arc GDH. Single- Valve Gears, Steam Distribution Variable. The characteristic feature of a single-valve gear is that its valve slides on a fixed, stationary seat (p. 134). In this sense an engine may contain one or more single-valve gears which divide among themselves the functions of the steam distribution. The variable elements in these gears exist mainly to vary the ex- pansion. They may roughly be arranged into two groups, those in which the driving eccentrics themselves can be varied, and those in which the eccentrics are themselves non-adjustable, the variation in the valve motion being effected by mechanism be- tween the eccentrics and the valves. To the second group belong the link-motions, and these we will take up in the Appendix, because they are more complex and less extensively used in high-speed engine work than the first group. 154 VALVE-GEAR VARIATIONS. In the first group the eccentric is varied by simultaneously- changing both its throw and angles of advance.* The principal means by which this change is accomplished is the " swinging eccentric " device. In this the center of the eccentric is moved across the shaft in an arc LL having the point P as its center, as in Figs. 68 to 71 and Figs. 73 to 78. Ten of the " Single- Valve Automatics " represented use this device, the mechanisms for producing the swinging motion differing more or less ; (see cata- logues of engine builders). The object of this change is to alter the power of the engine by altering the amount of expansion, but other changes in the steam distribution also take place at the same time when there is but a single valve to effect them. For example, the compression begins earlier when the expansion begins earlier, great cushioning going hand in hand with great expansion. These two functions of the steam distribution have perhaps the most influence on the form of the indicator card and thus are the principal means of regulating the power per stroke and the uniformity of the driving force. These are important * The only engine known to the writer that uses a single valve and varies the cut-off by varying only the angle of advance, is the oscillating engine built by the J. T. Case Co. Strictly speaking this engine has two valves, for the ports in the rocking cylinder control release, exhaust closure and admission, the cylinder itself performing the functions of the main valve. On the other hand, the Buckeye engine has apparently a double-valve gear and yet in reality is made up of two single-valve gears the first being the one to which the main valve belongs and the second having the expan- sion valve. In this second gear, the expansion valve slides on a moving main valve, but the mechanism is such that the relative motion of expan- sion to main valve is scarcely affected by the absolute motion of main valve. The latter therefore acts as a stationary seat of the expansion valve which may be discussed and designed as if the ports it controlled were fixed and as if its rocker arm had a stationary pivot. We have however discussed the Buckeye's gear in Figs. 96-98, with the double-valve gears, which it so strongly resembles externally. When an engine has its distribution ef- fected by several valves, each sliding on a fixed seat, the functions of one of them may well be varied by varying only the angle of advance of its eccentric or only the throw or lap. EQUALIZATION OF THE DISTRIBUTION. 1 55 factors in securing steady running of the engine. To effect smooth running there must be not only absence of great fluctua- tions of power per revolution, but also absence of vibrations and shocks. Shocks occur when there is " play " (or clearance) at a bearing accompanied by such sudden and complete changes of direction in the force acting on the bearing that continuous con- tact between the pin (or slide) and its bearing ceases while the space (or " play ") between the two is traversed by one or the other of the two pieces. Tremors and vibrations of course accompany shocks, but they may also arise independently, from sudden changes in the intensity of the pressure between two pieces. Thus at the end of a piston-stroke the final cushion-pressure may be considerably less than the initial pressure on piston at begin- ning of next stroke, the sudden, additional load on piston send- ing a tremor through all the connections. Sudden reversals of pressure should therefore be avoided particularly at the dead point where they are most dangerous. As the extent to which cushioning is carried, the character of the lead (positive or neg- ative) and its extent, strongly influence these dead-point-pres- sures, the variations effected by the valve gear in these quan- tities per revolution and at the different grades of expansion become a matter of consequence. In times past engineers laid great stress on having the steam lead equal at the two cylinder ends whatever the grade of expansion, particular care being taken to " set " the valve with equal lead. This desire for equalization at the two cylinder ends extended itself to the cut-off, the release and the cushioning as well as to the lead. But it was recognized, particularly in locomotive practice, that it was impossible with existing valve gear to equalize all the functions at the same time, equalization of any one function was at the expense of the others. Generally it was the lead (a positive one) that was thus favored. At the same time most engineers also held that a lead that was constant at all grades of expansion was a desideratum. It was regarded as the special merit of the Gooch link-motion that it I56 VARIATION OF THE STEAM LEAD. possessed this virtue. But since the advent of the high-speed steam engine a difference of opinion on this matter has developed itself among engineers. In the pioneer engine of this type, the Porter- Allen, the leads were deliberately made unequal at the two ends (Figs. 86, 87 and Figs. 89, 90 of Porter-Allen valve diagrams) and the lead also varied from mid-gear to full gear, as in other link-motions. The Straight Line engine in its earliest forms had a lead that was equalized at the ends and was nearly constant at all grades of expansion ; in the later forms the lead is now variable for dif- ferent cut-offs ; it is sometimes negative and there is some in- equality at the ends. In other and excellent engines the old views on this subject are still carried out so that the matter cannot be regarded as finally settled either way, Prof. Sweet, the designer of the " Straight Line " engine, going so far as to determine ex- perimentally, with the indicator, the conditions of smooth run- ning for each engine. The collection of examples given shows the variation of American practice in this respect. We shall not attempt any comparisons of these engines, for the exact effect of the steam distribution on the steadiness and smoothness of run- ning can only be determined by a knowledge of the clearances, the pressures, the weights of the reciprocating parts and the reg- ulating capacity of the governor under variable loads. We have thus far said nothing concerning the release, because it has usually less influence on smoothness of running than the other factors, though it too effects somewhat the pressure on piston at dead point. The release is of consequence to econom- ical running because of its influence on the amount of back pressure, an early release and widely opened exhaust ports tend- ing to keep this pressure small. On the other hand release begins too early when it curtails the period of expansion to any notable degree. We have given the beginning of the release for only a few cases ; it may be easily found from the exact dia- grams, by following the directions given with Fig. 67. SHAPE AND POSITION OF LOCUS. 1 57 Returning now to the influence that the shape and location of the locus LL of the eccentric centers E has on the steam distri- bution, we note first, that as the chord of the utilized portion of the arc LL is more nearly perpendicular to the dead point line of the eccentric, the lead is more nearly constant at the different grades of expansion, provided the curvature is the same, and secondly, that with this perpendicular location an arc LL of great radius {i.e., of slight curvature) causes smaller variations in the lead than an arc described with a small radius. Radii drawn from the center to the locus LL (or E E I E :i ) of centers, give the length and position of the eccentrics corre- sponding to the different grades of expansion, the locus moving with the crank to which it is rigidly attached. The broken line D D Z £> 2 is the locus of the vertices of the diameters of the valve- circles corresponding to the eccentrics E ot E r , E 2 . The position of these diameters were found from their eccentric by laying off the angle a as in Fig. 60. In Fig. 74 the locus L'L' or E m E vl E va was derived from LL, the locus actually described by the swing- ing eccentric, by constructing a series of offsets each equal to — r n , according to the method given in connection with Fig. 66. 34 The new eccentrics thus obtained are regarded as virtual eccen- trics and their centers are indicated by E m ,E vz ,E m . The valve circles were derived from the virtual eccentrics OEv , OE VI , 0E V2 . The particular eccentrics chosen for representation are those corresponding to minimum, quarter and maximum cut-off, the first and last of these being taken from data furnished, in most cases, by the manufacturers themselves. The quarter cut-off is generally an average value of the cut-offs in the two cylinder ends when angularity of connecting-rod is taken into account. The intermediate cut-offs in the case of the Westinghouse engines are somewhat larger than 0.25 or about 0.30. Each 0* these valve-circles will give the steam distribution shown in Fig. 63, when the lap and port-width circles are drawn. The crank 158 ANGULARITY OF ECCENTRIC ROD. positions and piston travel at cut-off, preadmission and beginning of compression are only drawn for the eccentric and valve-circle corresponding to .^ cut-off; the shaded area shown in the diagram cuts, from the crank, intercepts that represent the port-openings at quarter cut-off. If there is more than one port each intercept must be multiplied by the number of ports to get the total opening. It must not be forgotten that each of these valve- circles cuts from the crank or its prolongation a chord that represents the distance of the eccentric center E from the perpen- dicular YY, through 0, to the dead center line of the eccentric. With the exception of Figs. 74, 77, this perpendicular is the vertical through 0. The valve-circles entirely neglect the angularity of the eccentric rod. Whenever the length of the latter was known, a dotted, central, arc of reference was drawn from which the valve travel from the center of this arc can be found exactly. This arc center is obtained as follows : the crank is placed at one of the dead points and the center E of the eccentric in question (usually the one for maximum cut-off) is placed in its corres- ponding position, then with Ezs, a center and length of eccen- tric-rod as a radius strike off an arc cutting the valve-stroke ; repeat the operation for the other dead center of crank, the point bisecting the distance between these arcs will be the center of the arc of reference required. In the Rice, Ball, Southwark, Straight Line and Westinghouse compound engines, the central arc of reference differs so little from the vertical through that no effort was made to draw the polar curves representing exactly the valve-travel, the valve-circle being sufficiently accurate for all practical purposes. In the Westinghouse Standard there is also only a slight difference between the perpendicular Y 1 Y' and the arc Y'Y' ; still for the case of maximum eccentricity the exact polar curves were plotted on the diagram in dotted lines, the one outside the valve-circle corresponding to the positions of the valve to the left of the center of the reference arc and the one inside to the positions of the valve to the right of this center. Comparison with the intermediate valve-circle shows that the SYMBOLS USED ON DIAGRAMS. 159 steam distribution is but slightly changed by substituting the exact polar curves for the less exact circle. The cut-offs are apparent ones and are estimated from begin- ning of stroke in the same manner as in former sections. Pre- admission, beginning of compression and release are however measured from the end of the stroke and each is then divided by the stroke itself, so that the results are always expressed as fractions of the stroke. The release is not tabulated on the diagrams. The symbols used are a = width of port. I = length of port in cylinder. d = diameter of piston valve. e = steam lap. i = exhaust lap. e = apparent cut-off. v = steam lead. r = eccentricity. R = radius of crank. L = length of connecting-rod. e = beginning of release, measured from end of stroke and divided by stroke. q = beginning of compression, measured from end of stroke and divided by stroke. r/ = beginning of admission, measured from end of stroke and divided by stroke. The period of compression would properly be represented by the difference rj — y' '. The subscripts , , and 2 affixed to these symbols signify that they belong to minimum, quarter and maximum cut-offs respec- tively, as explained above. All the engines represented "run over" (see p. 118), i. e., in these diagrams, the rotation is right-handed. Then in Fig. 67, for the distribution corresponding to quarter cut-off, KT repre- l6o SINGLE VALVE GEARS. sents admission of steam to head end during forward stroke and SR to crank end for return stroke. FH represents compression n crank end during forward stroke and GU compression in head end during return stroke because exhaust lap is negative. p t q t represents the preadmission to head end before beginning of forward stroke and pq the preadmission to crank end of cylinder for return stroke. As the inside lap at crank end is different from that at head end we must carefully distinguish between the two. Then m r n t is that last part of the forward stroke during which release takes place in the head end and mn that part of the return stroke during which release takes place in the crank end of the cylinder. By dividing these quantities by 2 X CO, we get the values tabulated in the diagrams. We have omitted the release from these tabulations but it can be easily found from these exact diagrams if desired. In following the steam distri- bution by means of the diagram, it is well to bear in mind that the valve is to the right of its middle position whenever the valve-circle cuts the crank radius. The same letters are used in the other diagrams to represent the same parts of the distribution ; the letter C always represents the crank position, E„ eccentric position corresponding to C, and D„ the diameter of the valve-circle corresponding to the eccen- tric-setting COE„. The valve diagrams, for the twelve Single-valve Automatics represented, are here divided into two groups according to the location of the pivot P (or center of curvature) relatively to the crank radius. In the first group, Fig. 67-72, the center of cur- vature of that element of the locus LL which crosses the crank lies on the crank's own radius, while in the second group, Figs. 73-78, the pivot P lies outside of the crank's radius or its pro- longation. As the pivots P are often at considerable distance from their loci LL, economy of space required that the points be detached from their proper position relatively to their loci and be placed nearer the latter ; this has been done in every case and their co-ordinates given so that they can be laid out if desired. Figs. 67 and 68. RUSSELL * CO., ENGINE. 14 x20 Positive Direct Valve 13x12 Figs. 6g and 70. PAYNE ENGINE. Positive Indirect Valve. Figs. 71 and 72. BALL ENGINE. 9 x!2 -OLDER TYPE. ' Positive Indirect Valve. Figs. 73 and 74. SOUTHWARK ENGINE, 9 x 10 1 Positive Direct Valve, ST 8x12 CRANK END HEAD END CRANK END HEAD END CRANK END HEAD END i o = 0.078 0.105 T) Q = 0.458 0.45 %' = 0.075 0.10 £,=0.218 0.28 Tj= 0.292 0.285 7^=0.025 £ 2 = 0.775 0.826 71 = 0.055 0.05 Jj' = 0.O0I6 0.0316 0.0033 CRANK END £ o =o.oo £, = 0.218 £,= 0.685 HEAD END CRANK END 0.00 77 = 0.53 77 = 0.273 0.28 0.75 0.615 0.345 "CRANK END 0.00 0.00 HEAD END 0.00 0.00 0.0916 0.12 rf= 0.0033 0.0033 CRANK END = 0.00 = 0.218 = 0.543 HEAD END 0.00 0.28 0.623 CRANK ENDL HEAD END CRANK END = 0.493 0.575 7j£=0.00 -- 0.245 0.311 73,' = 0.00 = 0.12 0.15 77' = 0.0016 HEAD END 0.00 0.00 0.0016 CRANK END HEAD END CRANK END HEADEND' CRANK END HEAD END' Y ^ = 0.0033 0.0036 77 Q = 0.493 b.576 77^= -0.005 -0.0066 £,= 0.218 0.28 T^ =0.34 0.412 77* = -0.0016 -O.O0I7 i z = 0.578 0.658 ^ = 0.143 0.186 7^' = -0.0017 -0.0018 N. Y. S. S 12 x 12 D POWER CO., ENGINE. Positive Direct Valve RICE ENGINE. Positive Indirect Valve. ARLINGTON * SIMS ENGINE. 14 x ' 5 = ! Positive Indirect Valve. ,L CRANK END HEA £ =0.0158 O.i o £, = 0.218 0.: t, =0.7113 O.- BALL I CRANK END .HEAD END i^= 0.075 0.098 £,= 0.218 0.28 L= 0.77 0.8216 CRANK END HEAD END CRANK END HEAD END 77 = 0.458 0.542 7^=0.071 0.096 77 =0.276 0.35 7^ = 0.018 0.023 71 = 0.046 0.065 7?'= 0.0016 0.0016 CRANK END HEAD END CRANK END HEAD END CRANK END HEAD END £ =0.135 0.175 77 Q = 0.453 0.54 7q=O.I3 ' 0.176 £-= 0.218 0.28 7^ = 0.351 0.433 77{=0.078, 0.11 f =0.673 0.735 7^=0.105 0.143 ^=0.016 0.0216 CRANK END HEAD END CRANK END. f = 0.00 0.00 77 = 0.458 £=0.218 0.28 77= 0.226 £ 3 = 0.516 0.596 7^=0.115 ,0.54 Tj' J 0.285 T^' 0.153 7g CRANK END ^ HEAD END 0.00 0.00 0.0016 0.0016 0.0033 0.0033 7^=0.03 J 1/ / 0.028 %= 0.463 jj 0.4-86 77*= -0.03 - -0.029 £, •= 0.218 n fL/ / 0.28 ^ = 0.358 ^•=0.701 ^Ek 0.7 L' lT 1.761 7?^, = 0.116 0.386 7fi= —0.003 - 0.003 0.14 77i=°003 0.003- CRANK END HEA £^=0.00 0.' £ ( = 0.218 O. £3= 0.585 O.' Figs. 71 and 72. BALL ENGINE. OLDER TYPE. idirect Valve Figs. 73 and 74. SOUTHWARK ENGINE. 9 x 10 1 Positive Direct Valve, Figs. 75 and 76, STURTEVANT ENGINE. Figs. 77 and 78. WESTINGHOUSE STANDARD VERTICAL Positive^ Indirect Valve. cTnSuw" tCCtHTKIC VERTICAL LINE l» S«°mra lik or a, Sif ST ^TT L" I.I ANK END HEAD END 3.00 0.00 3.00 0.00 3.0033 0.0033 idirect Valve CRANK END HEAD END CRANK ENDX HEAD END CRANK END HEAD END £ o =o.oo 0.00 r/ o = 0.493 0.575 %=o.oo o.oo t,= 0.218 0.28 ^ = 0.245 0.311 7}' t ■■ 0.543 0.623 - 0.00 0.00 = 0.0016 0.0016 ARMINCTON * SIMS ENGINE. 14 x 15i Positive Indirect Valve. ,L CRANK END HEAD END CRANK END HEAD END' Y {^ = 0.0033 0.0036 TJ = 0.493 0.576 <]' a ^~ 0.005 -O.0066 4^=0.218 0.28 77 =0.34 0.412 ^=—0.0016—0.0017 i 2 = 0.578 0.658 y ? = 0.143 0.186 %' = — 0.0017 -0.0018 , STRAIGHT LJNE ENGINE./ II x 14 \ Y 1 1 Positive Direct Valve CRANK END HEAD END . CRANK END HEAD END t =0.068 082 n =0.723 0.786 (?' = 0.026 0.035 o jo to i t = 0.218 0.28 ^ = 0.512 0.595 Tj' y = 0.0025 0.005 £,=0.718 0.78 Tj 2 = 0.153 0.196 ^=0.0025 0.005 BALL ENGINE-RECENT TYPE. RIGHT CYLINDER LEET CYLINDER! RIGHT CYL LEFT CYL i RIGHT CYLINDER LEFT CYU £ q = 0.023 0.023 Tj Q = 0.631 0.631 i %= 0.035 0.035 ^ = 0.283 0.283 ^ = 0.37; 0.373 ^ 0.0083 0.0083 £,= 0.75 0.75 Tj 2 = 0.165 0.16^ %= 0.0025 0.0025 y WESTINGHOUSE COMPOUND VERTICAL iSj LINE o 1 ■J- 1 o c± -, ANK END HEADEND 3.13 ' 0.176 D.078; 0.11 3.016 0.0216 HEAD END CRANK END 'HEAD END I 0.00 Tj = 0.458 |0.54 £=0.218 0.28 Tj = 0.226 (o.285 £^= 0.516 0.596 /?_= 0.115 GIDDINGS VALVE. 161 Russell & Co. Engine, 14x20. Fig. 6j. In this valve gear the center of the eccentric is moved in a perfectly straight line |_L across the shaft, at right angles to the crank. The mechanism for accomplishing this (it is not the " swinging eccentric " device) is not given here but can be found in catalogue of the builder. Inspection of the valve-circles shows at once that the steam lead v is constant at all grades of expansion. The lead on the exhaust side is how- ever unequal at the two cylinder ends, for the inside lap i = o and — yi at the crank and head end respectively. But this is not a matter of any consequence. The inequality in the inside lap equalizes the beginning of release and of compression for the two cylinder ends as the following table shows : Crank End. Head End. Crank End. Head End. So' e,' O.54 O.36 O.075 O.54 0-37 O.075 O.458 O.292 O.065 O.450 O.284 O.050 The value of yj includes the preadmission of steam and so does not exactly represent compression proper. But it may be taken as representing the cushioning, which is practically equalized for the two ends by this arrangement. The table connected with the diagram gives the variation of cut-off and preadmission for minimum, average and maximum throw; the range of cut-off varies from ^ to &. This engine uses the Giddings' Balanced Valve, which is divided into two parts, rigidly connected, one part for each end of the cylinder. Each part admits steam through two passages that run through the valve and somewhat resemble the supplementary passage of the well-known Allen valve. Like the latter this valve belongs to the positive direct type and increases the clearance volume during the compression period by the volume of the supplementary passage. The clear- ance volume is smaller at the beginning than at the end of the 1 62 N. Y. S. S. POWER CO., AND PAYNE ENGINES. compression. As there are two ports or passages for each end of the cylinder, the admission and cut-off are prompt, and the port opening given by the shaded area of the diagram must be multiplied by 2. The valve is of the flat and balanced variety. The speed of this size of engine is 210 revolutions per minute. N. Y. S. S. Power Co., Engine, 12x12. Fig. 68. The locus LL of the center of eccentric is concave to the shaft 0. The center of this arc is on crank radius, at pivot P, and 14" from center of shaft. The lead diminishes as the cut-off increases. The cut-off ranges from ^ to |. The beginning of release is given by this table, Crank End. Head End. 0.54 0.35 O.065 0.458 O.276 O.046 Payne Engine, 13x12. Fig. 69. The location of the pivot of the pendulum arm of the swinging eccentric in the smallest engines is different from that here given. In a 5x7 for instance, the pivot P and crank- pin C are on opposite sides of center O ; in this case pivot P and pin C are on the same side of 0. The pivot is 1 1 }4" from shaft center and lies on the same radial line as crank. The radius of arc LLis io^s"; this makes the arc convex to the shaft and the lead increases with the cut-off, the latter ranging from to ^. The fraction of piston stroke occupied by preadmission is very small. The valve takes steam on the inside and is of the flat, balanced variety. This size of engine makes about 250 revolu- tions per minute. Rice Engine, 7x10. Fig. 70. This engine presents a unique appearance in that no governor is visible. The regulator is within the crank disc and RICE AND BALL ENGINES. 1 63 turns an arbor that passes through the center of the crank-pin C. This arbor is the pivot P of the pendulum arm that swings the eccentric-pin across the shaft when the expansion is to be varied. The eccentric-pin and the pendulum arm constitute a sort of return crank. In the figure, both C and P represent the center of pivot and of crank-pin. The locus LL is again convex to the shaft and has a nearly constant lead ; its radius is 4^. The fraction of the stroke occupied by the preadmission is larger in this engine than in any other of the dozen represented. The cut-off ranges from 7 to ru. The length of the eccentric-rod is only 1 5", but the influence of its angularity on the valve-travel is but slight, as the arc VY' shows. This arc was struck from the center of motion corresponding to maximum throw and with a radius of 15". (The center of motion of valve is half way between the two valve positions corresponding to the crank's two dead points.) The vectors- of the valve-circle represent the distance of point E„ from vertical YY and are very nearly equal to the valve-travel from the center of motion. The exact valve- travel at any instant is given by the distance of point E„ from the arc Y'Y'. These exact travels could be laid off on the crank positions or their prolongations and thus two exact, polar, diagrams be found (for each eccentric throw) which would cut from the crank the exact distance of the valve from the middle position. But in this case they would differ so little from the intermediate valve-circle that this nicety of construction has been omitted. The valve is of the flat, balanced variety. Ball Engine — Older Type, 9x12. Fig. 71. The diagram of this engine is given mainly for com- parison with the diagram of a more recent style of valve-gear for the same engine whose diagram is given in Fig. 76. In this earlier form the point P of the swinging eccentric was 1 3 j4" dis- tant from the center of the shaft and was on the same radial line as the crank-pin C. The locus LL of the center of the eccentric was convex to shaft and gave a lead that increased 164 ARMINGTON & SIMS ENGINE. as the cut-off increased. The range of cut-off was from o to %. The valve was flat and balanced. The length of the eccentric- rod was about 41^ inches and with this as a radius an arc YY' is struck off from the center of motion, in the manner already explained when discussing the Rice Engine ; the distance of the center E H of eccentric from dotted arc YY gives the exact travel of the valve, while the valve-circle gives the distance of E„ from vertical YOY; here again the difference is so slight that we may rest well content with the results furnished by the valve- circles. Armington & Sims Engine, 14^x15. Fig. 72. In this engine a peculiar linkage is employed to vary the position of the center E of the eccentric ; for the details of this linkage we must refer the reader to the catalogue of the engine builder. The locus LL of this diagram is not a circular arc, though it closely resembles some of the loci that have already been given. It was plotted from an accurately measured linkage belonging to a 14% x 15 engine. Here also, the lead increases as the cut-off increases, the table on the diagram show- ing the extent to which this takes place. The range of cut-off is from o to i. As in all the preceding cases the compression or cushioning period increases with the period of expansion. The eccentric-rod which connects the driving eccentric with the \oy^" rocker arm is itself 44" long. We may regard the rocker arc as a straight line and repeat the construction of the arc Y' Y already given with the Rice & Ball engines. Here also the dif- ference between vertical reference line YY and the reference arc Y Y is so slight that it may be neglected. The valve is of the hollow piston variety, is 6^" in diameter, is of the positive indi- rect type and takes steam simultaneously at two places. One of its steam passages leads through the center of the valve and in principle is like the supplementary passage of the Allen valve that is still used on some locomotives. This size of engine is run at about 280 revolutions per minute. SOUTHWARK AND STRAIGHT LINE ENGINES. 165 South wark Engine, 9x10. Fig- 73- It is the first of the second group of Single-valve Automatics. In this group the locus LL of the driving-pin or eccentric £ of the valve has its center P outside of the crank radius OC. In most of these cases the pivot P and crank-pin C are still both on same side of shaft center 0. In the present case the co-ordinates of .Pare 5 }(" and — ffi' which brings it opposite the middle of the utilized portion of the arc LL- The arc is concave to O and its chord is perpendicular to crank radius so that it gives a nearly constant lead, for all grades of expansion. The cut-off ranges from o to f$ and the cushioning from A to v. This valve-gear and its governor were designed by Prof. C. B. Richards. The substitution of a pin E in place of the eccentric sheave is an excellent feature as it removes the danger of over- heating to which the eccentrics of high-speed engines are so liable. The rod covering the driving-pin E with the 3%" rocker is 34" long; treating the rocker arc as approximately a straight line, we can repeat the constructions of the arc Y' Y' already given with the three preceding engines. Here too the deviation of the more exact reference arc Y' Y' from the vertical reference line YY is so light that there is no need of substituting exact polar curves for the valve-circles. The valve used in this engine is flat, balanced, and of the positive direct type. The speed of this 9x10 engine is 300 revolutions per minute. The Straight Line Engine, 11x14. Fig. 74. The locus LL is concave to and has the point P outside of the crank radius OC, but P and C are both on the same side of 0, the abscissa of P being 12.25" an d i ts ordinate — 2}4". The stroke of the valve does not pass through the center of the shaft, but passes it at a distance of 1 % inches. The dead point line of eccentric will therefore be different for the two ends of 1 66 STRAIGHT LINE ENGINE. the valve strokes and for the different throws of the valve. An average position has been drawn on the diagram. It is therefore like the case represented in Fig. 66, p. 151. The offset Ee is there shown to be nearly equal to -r; here c = 1.5", /== 34" and r = eccentricity corresponding to grade of expansion con- sidered. If r= 2.5", then -^ X 2,5 = 0.11 is the offset, which laid off from E x at right-angles to 0£ 2 gives E v 2 which is the center of the virtual eccentric 0E V x . In like manner other offsets may be found for other throws of the valve and thus a curve L'L', passing through the ends of these offsets, can be found which will be the locus of the centers E v of the virtual eccentrics 0E„. The valve-circle belonging to any eccentric can be found as in Fig. 60 and in either of two ways : it can be obtained from the actual eccentric (drawn from to locus LL) by laying off the angle cylinder. The I70 WESTINGHOUSE STANDARD ENGINE. center line of each cylinder passes the center of shaft at a distance equal to half the length of crank OC" In the diagram this distance and crank are drawn to a reduced scale. The pivot P of swinging eccentric is A" from the radial line of the left crank and has a negative abscissa of 8JH3". The radius of locus LL is PE 2 = gj\"- At maximum cut-off the eccentricity is 1 ^" and the radius PME 2 is tangent to the arc M described with OM — iA" as a radius; at minimum cut-off the eccentricity is uV = 0E o and the radial line PmE is tangent to the arc m, which is struck off with radius Om = w". The valve- circles are found as in Fig. 60, p. 137. The lead v, for all grades of expansion, was measured on that dead-center line of crank which corresponds to uppermost position of piston and increases with the cut-off. The engine is composed of two single- acting, vertical cylinders which receive the live steam in the top end of the cylinder. As each cylinder has its own crank and connecting-rod this arrangement will cause the angularity of the rod to have precisely the same influence on the steam distri- bution of each cylinder. This is shown by the values tabulated on the diagram. The cut-offs evidently range from o to %. The table inscribed on the diagram and valve-circles (in full lines) do not take into account the influence of the length of the eccentric rod. The vectors of the valve-circles drawn from O give the hor. distance of the eccentric center E n from the line Y'OY' which is perpendicular to the dead center line of eccentric. To take account of this angularity of the eccentric-rod, we make the following construction. Place the crank upon dead point for top of stroke and the eccentric in a corresponding position. The center E 2 , will then be 4^° in advance of position shown in the diagram. Taking this new position of E 2 as a center and the 20" length of the eccentric as a radius, strike an arc that will cut the dead-point line of eccentric ; then revolve this eccentric through 180 and use the new position of E% as a center to repeat the former operation. Now bisect the distance between WESTINGHOUSE STANDARD ENGINE. 171 these two arcs and the point of bisection will be the center of motion to which the valve travel is referred and from which as a center the arc of reference V Y' can be described with the 20" radius. The hor. distance of the eccentric center E n from this arc of reference will be the exact travel of valve. If this exact travel is laid off on the corresponding crank positions, laying off valve travel to f right 1 I left I of center of motion on { pr *£ n g ed [crank repeating this process for second crank also, we will get two pairs of polar curves instead of the two valve-circles, and the intersection of these curves with the proper valve-circles will give the exact steam distribution. One pair of these polar curves, the one for left cylinder crank has been roughly drawn on the diagram for maximum throw. They lie on opposite sides of valve-circle 0D 2 which may be regarded as their average value. The s j nner f one of tne polar curves represents travel to the I j-jg-ht r °f the center of motion. The exact steam dis- tribution given by carefully drawn polar curves is contained in the following tabulation : Half Throw Right Cylinder. £1 £ 3 = 0.34 I e/=o.2i = 0.68 1 sj= 0.08 ^=0.38 iy 2 = 0.16 5j' I =O.OI j/ 2 = O.OO Half Throw Left Cylinder. r 2 = 1% s 2 = 0.28 = 0.67 e/=0.2l s/= O.06 ?x = 0.35 -^ = 0.17 ^'=0.00 ^l 2 '= 0.00 The slight inequalities that exist for the two cylinders is wholly due to the influence of the length of the eccentric-rod. The diameter of the piston valve is 4^" and the speed is 300 revolutions per minute. 172 VALVE GEAR. Westinghouse Compound Vertical Engine, 14 and 24x14. Fig. 78. In this case the stroke of the pistons passes directly through the center of shaft. The single, piston, valve controls two ports M and 7V (see little figure to right of diagram 78). The passage P leading to high-pressure cylinder is always in communication with the hollow of the valve. The port M therefore admits live steam from pipe 5 to the high-pressure cylinder, and port N, at one edge receives steam from high- pressure cylinder and admits it to the low-pressure cylinder, and at the other edge it delivers steam from low-pressure cylin- der to the exhaust pipe E. The steam laps e at the inner edges are alike for both ports .Af and N. The exhaust lap is however equal to O. The value of e' given in table was obtained by multiplying actual lap e by ^ (= ^s, the ratio of the arms of the bent lever driving the valve) to reduce it and the stroke of the valve to the end of the eccentric-rod. This is a departure from our usual practice of assuming that the actual valve is driven directly by the eccentric, but in this case it was simpler to change only one dimension than to change several. The pivot P does not lie on either of the two cranks, but it is near to the low-pressure crank and on the same side of center 0. The abscissa and ordinate of Pare 5.42" and 1.5" respectively, its total, radial distance from O being $$£". The locus LL struck from this center has a radius of 6^" and is concave to 0. The table and data inscribed on diagram state distribution. In this Compound, as in the Standard, the steam distribution is the same for both single-acting cylinders when the eccentric-rod is taken as infinitely long. In this case the rod is finite and 49" long and its influence on the distribution can be ascertained in the now well-known manner by drawing the arc of reference Y'Y'. It is evident from the figure and from what has been said about other diagrams, that the valve-circles give the valve-travel with sufficient accuracy. Fig. 79- Fig. 80. EACH STEAM VALVE DRIVEN INDEPENDENTLY. C B - I" C A - IO" B E - 6" EF = 8" c x - bH" E L 95 C H - 3" O' R - 21" C O' 45 1" RP-3"-RQ' H 1 15V C R - 52" 1 J = I2§" OUTSIDE LAP - 0.78 JK = 6" INSIDE LAP 0.04 L M - 38l" LENCTH OF CONN. ROD - 60" M N 111?," RADIUS OF CURVED SLOT - 48 if RELATIVE POSITION OF VALVES WITH ENGINE ON FORWARD CENTER. LINK BLOCK. (FULL CEAR) 6" FROM TRUNNION. NECATIVE LE AD - ${_ Y//////////^ WtW < l/**?* AM VALVES - FORWARD CENT BACK W///yfa EXHAUST VALVI Fig 81. porter-allen valve gear. 1 73 Link-Motions. In the second group of single-valve gears (see p. 153), the eccentrics themselves are invariable, the variation in the valve motion being affected by the mechanism between the eccentrics and the valves. All the commom link motions belong to this group ; we will here describe and examine but one of them, the Porter-Allen, (or ' Fink '), the only one which in this country has been extensively used on stationary, high-speed engines.* The Porter-Allen is that special case of the Gooch link motion which is obtained by supposing the centers of the two eccentrics to coincide, thus forming only one eccentric, the two eccentric rods and link then constituting one rigid piece. The best pro- portion for this linkage are shown in Figs. 79-82, and were taken from blue-prints furnished by the South wark Foundry & Machine Co., of Philadelphia. Porter-Allen Valve-Gear, i i ^x20. Figs. 79, 80 and 81 give a side view of this engine, a section through its rocker arm and a skeleton showing the relative posi- tion of the three valves when crank is on its forward center. This view and the skeleton of the mechanism given in Fig. 82 are sufficiently full and complete to render any detailed description unnecessary. We will only ask the reader to note that eccentric strap and link constitute one rigid piece, that for a particular grade of expansion we assume that the governor occupies an invariable position vertically, thus making point of suspension / * " Prof. Zeuner's views of the unsuitableness of the ' Fink ' link as a reverstng-gear are correct, and for the reasons he gives. But the use of the half-link in the Porter-Allen engines (it is in connection with these engines that the Fink motion was first brought into extensive use) shows that when compression is judiciously applied, then the cut-off points may be made practically 'symmetrical ' up to the half-stroke at the expense of a variable lead, the variation of which is in opposite directions for the two strokes. Before cushioning was used in these engines the variability of the lead made it almost impossible to obtain quiet running under varying conditions." 174 PORTER-ALLEN VALVE GEAR. of the hanger IH a fixed stationary point and finally that even then the center G of block does not maintain an invariable dis- tance from the trunnion E, but slips a little in its slot. The view of the relative position of the valves is only a skele- ton and does not represent the multiple port-openings ; each of the three valves admits steam at four different places thus offering a generous passageway for both the live and exhaust steam. Before taking leave of these general views we desire to call special attention to the rocker arms ORPQ which communicate motion to the rods and stems of the two steam valves. From the arrangement we see that each driving arm, RP and RQ', has a period of almost complete rest and two periods of rapid motion. The rapid motion period effects a rapid opening of the port to its full width and then a rapid closing, the period of comparative rest occuring when the port is closed by valve. This arrange- ment of bent levers does excellent service but modifies greatly the " harmonic motion" of the driving point of the link and re- quires other diagrams than the valve-circle for the exact represen- tation of the valve -travel. We will return to this matter later on. We will first show that the travel of a point on the link EGL is approximately like the " harmonic motion " communicated by an eccentric that has an infinitely long eccentric-rod. Let A represent the distance from center of eccentric B to center of trunnion E, the latter joining sustaining arm FE and EGL on center line of slot of link ; let p represent the eccentricity CB, u the distance of point G above dead center line of eccentric, to the crank or (or eccentric) angle and $ the distance of eccentric center B from perpendicular through C to eccentric's dead-point line (which is approximately along CEO' , the crank and eccentric differing in position by a small angle of 2° 12' that is equal to angle included between their respective dead-point lines). Then, for the case in which — > 4, we will show that approximately : f = p cos to -f -rp sin to (90) PORTER-ALLEN VALVE GEAR. 1 75 To prove this, we consider the total movement of link to be composed of two parts, of a horizontal motion in which the link moves parallel to itself, the character and extent of this motion depending on the horizontal throw of the eccentric, and of a rocking motion about the trunnion E (or vibrating fulcrum) which is due to the vertical throw of the eccentric. Here the term p cos to is the horizontal throw common to all points of the link and the term -$ sin w is the vertical throw p sin m X the ratio u 2 of the arms of the bent lever constituting the link and rotating about the fulcrum or trunnion E. The formula for f thus estab- lished is the polar equation of the valve-circle and the coordi- u nates p and -,p of the vertex of its diameter show that this ver- tex lies on a rectilinear locus perpendicular to the dead-point line of the eccentric and at a distance p from the center of the shaft. Such a valve-circle is drawn in Fig. 83 for full gear (i.e., for block G about 6" from trunnion), and the corresponding equivalent, (virtual or resultant) eccentric can be found in the usual way, Fig. 60, from the diameter of this valve-circle.* To show that the valve-circle thus found represents with considerable accuracy the travel of a point on the link, we * As the horizontal component motion due to the rocking, is motion rela- tive to trunnion E, we can also combine the parallel and rocking compo- nents by means of the parallelogram of velocities. We need only regard the rocking component of the motion as directly and independently pro- duced by a separate eccentric, one that would follow the actual eccentric by u 90° and have a length —p. As these two eccentrics are both at right angles to, and proportional to, the velocities they impart we can construct the parallelogram of velocities on the eccentrics as sides and the diagonal will give the length and location of a resultant or virtual eccentric capable of imparting directly the whole horizontal motion to the point of the link. The parallelogram and triangle of eccentricities constructed in Fig. 83 and Fig. 84, respectively, are illustrations of this method. The valve-circle can then be found from the virtual eccentric as in Fig. 60. 1(76 PORTER-ALLEN VALVE GEAR. have in Fig. 83 compared such a valve-circle with a polar curve OMN or OM'N' representing very exactly the travel of the rocker-pin 0'. (Figs. 79 and 82.) To simplify matters we have assumed pin 0' to travel on a straight line that bisected the rise of its arc-path and passed through center C (Fig. 82) of shaft. In Fig. 83 we have called this straight line CO' (of Fig. 82) the dead center line of eccentric and con- structed the valve-circle according to formula 90 found above. The dotted, polar, curves OMN and OM'N' (Fig. 83) were found by laying off on the crank positions the exact distance of pin 0' from its center of motion, supposing O to travel on aforesaid dead point line CO 1 . We see that these curves agree very well with the valve-circles and that the latter may unhesitatingly be used to represent the travel of lower rocker-pin C parallel to dead point line C. Inspection of the setting of the upper rocker-pins Q' and P shows at once that the horizontal compo- nents of their motion could not be represented by valve-circles, and that to represent the valve travel accurately we must con- struct exact polar curves for both the front and the back valve. A fairer comparison of the actual travel of a point on the link with that given by the formula or valve-circle is furnished by Fig. 84, which represents by the dotted, polar curves the actual travel, from its center of motion, of the point ^(Fig. 82) driving the exhaust valve. It is evident that the average direction of the point M of the exhaust rocker NM is parallel to the average direction of the path of the driving point L on link. We may therefore assume that the actual travel of M is almost axactly like that of L. Now the travel of T is 0.39 of that of M; it is rep- resented by the dotted, polar curves of Fig. 84, which were obtained by directly measuring the distances of point T from its center of motion and then laying off the distances on the corre- sponding positions of the crank. The valve-circle representing, approximately, the travel of point L is drawn on diagram on diameter ODJ and reducing it 0.39 we get the valve-circles O-Dj' representing, approximately, the travel of driving point T. Fig. 8a. SKELETON OF PORTER-ALLEN VALVE GEAR. 1lg'x20 41" RQ' S==8:: D -i v -°— > , L FRONT VALVE. BACK VALVE. -* — L- — •*-'< SKELETON OF PORTER-ALLEN VALVE GEAR. 1 77 Though the deviation of the two sets of curves is now marked, nevertheless, the deviations occur where they will but slightly influence the beginning of compression or release, which are found as of old at the intersections of polar curves (or circles) with the lap circles. We have already called attention to the fact that the setting of the driving arms PR and Q'R causes them to communicate to the valves a horizontal motion that is decidedly different from the " harmonic motion " assumed for eccentrics. If we wish to represent the valve's motion accurately by polar curves whose vectors are the crank positions, we must determine accurately the distance of the valve from its center of motion for a complete series of crank positions. Inspection of the skeleton in Fig. 82 shows that it is composed of two quadric chains CBEF and HIRO' connected by the straight link G V (which in the actual mechanism is represented by the link-block G). For the assumed point of suspension / the point G of second quadric chain describes the path G, 1, 3, 6, while the point 0' of lower rocker arm describes the correspond- ing path 0' , 1, 3, 6. In the first quadric chain CBEF the point L occupies points i, 8, 13, 23, etc., of its path, while eccentric center B occupies points 1, 8, 13, 23, etc., of its circular path. The center V oi the slot GE is the point on which rod GFturns and this point is on the line BE of the first quadric chain ; the piece BE in the skeleton is supposed to be extended so as to sup- port this point V. The path V is given and the numbers cor- respond to the crank or eccentric positions. We can easily find the position of either valve for any crank position if we know the corresponding position of 0'. To get O for any eccentric (or crank) position, we suppose V placed at that point of its path which has the same number as the crank position under consid- eration ; with this point Fas a center strike off an arc cutting the path of G : then with this intersection of the arc and the path as a center and the steam rod GO' as a radius strike off another arc ; it will cut the circular path of 0' at the point 0' desired. 178 CENTER OF MOTION OF VALVE. The position of each valve can be represented by its point Q or 5 and is estimated from a center of motion that is invariable in position for all grades of expansion, i.e., for all positions of the block G in its slot. To find this center of motion we must first find two peculiar eccentric positions near the dead-point line of the eccentric. To get these positions we erect at C (Fig. 82) the perpendicular CF Z = EF to the dead-point line of eccentric. Then with F x as a center and F t C as a radius describe an arc tangent to this dead-point line; the chords Ci and C13 of this arc will be the peculiar eccentric positions needed. Now find two positions of both Q and £ corresponding to these two peculiar eccentric locations and then find a point half way be- tween them ; this point will be the desired center of motion from which the travel is to be estimated, and will be found to be the same whatever position is assumed for the point G in its slot (that is, it will be the same whatever fixed position is chosen for the point of suspension I of the hanger IH). The center of motion can also be found, and more simply still, in the follow- ing way. The steam rod is -h" shorter than radius of link slot ; near the two intersections of path of center F(Fig. 82) with arc of 0' lay off A" from arc 0' towards path of V, in the direction of middle radius of the link slot, till a place is found (on each half of V) where arc 0' and path of Fare just A" apart. When center of steam rod occupies either of these two positions of 0' the block may be moved from one end to the other of link slot without moving rocker pin or either of the two steam valves. Hence these two points on arc 0' will give exactly the two valve positions that are equidistant from the center of motion ; in like manner the two points on path Fthus found will give the corresponding eccentric positions and these will be found to be near the peculiar ones described above. The distance of each valve from its center of motion is now laid off on correspond- ing crank positions and the polar curves shown in Figs. 86 and 87 for full gear (G is then 6" from trunnion E) obtained, also the polar curves shown in Figs. 89 and 90 for half-gear (block G is 3" from trunnion E). To ascertain the lap from the data given in Figs. 83, 84, 85. Diagrams of Rocker-Pin O' — Figs. 86, 87, 88. Fi e s - 8 9. 9°. 9'- — Exact Polar and Oval Diagrams fob Both Steam Valves- Half Gear FVLL, QEAR FRONT VALVE UKiroF CRANK MOTION CURVES. 1 79 Figs. 79-82, we- must find positions of point < t~ > when engine is on < Dac k f center and block C at trunnion E; as the edge of the J back > steam port is then tV' < ri *L t J- of edge of { bS } steam valve - we la y off A " to the { right } of { S } to get its position when valve edge is exactly at port edge. Now place \ c \ at its own center of motion and measure its distance from its former position when valve and port edges coin- cided ; the distance will be the steam lap desired. Laying off the same steam lap = II for both front and back valve, and measuring in each case the distance RS (Figs. 86-90) and dividing it by the stroke = (2X CO), we get the cut-off effected at each end and at each grade of expansion. The shaded areas give in all cases the linear port openings and should be multiplied by 4 for the total openings. The numerical values of the distribution are inscribed on the diagrams and show that it is a practically perfect valve gear in this respect. The actual valve travel found above can also be laid off as ordinates, on the piston stroke as a base, and then we get another series of valve diagrams of an oval shape, that are known as "motion" curves. The exact polar and oval diagrams give of course identical results. Figs. 85, 88 and 91 respectively repre- sent the travel of the exhaust valve, that of the front and back steam valves for full gear and that of the front and back steam valves for half gear. In Fig. 85 cushioning in crank end is measured by aX -4- OX, release in head end by bX -4- OX; similarly cushioning to head end is given by dn -4- OX and re- lease in crank end by cm -4- OX. In Figs. 88 and 91, ab -4- OX measures the cut-off in crank end and Nc -4- OX the cut-off in l8o DOUBLE- VALVE GEARS. head end. The polar diagrams for the steam valve best show the variation of lead. The speed of this size of Porter-Allen engine is 230 revolutions per minute. Double- Valve Gears* The characteristic of a double-valve gear is that while the first of its two valves slides on a fixed seat, the second slides on the first, the motion of the second valve relatively to the first being affected by the absolute motion possessed by each of the two valves. In single-valve gears with variable elements, and in link- motions, we simplify our problem by finding the equivalent or virtual eccentric and then proceed as if the valve were directly moved, to and fro upon its fixed seat, by this virtual eccentric. In double-valve gears a like simplification is introduced, in fact it is the finding of this virtual eccentric which usually constitutes * This term, double-valve, has occasionally been applied to a valve con- sisting of two rigidly connecting halves and also to two separate and independent valves each sliding on a fixed seat. The gear here called double-valve gear has sometimes been defined as gear with plain slide valve and independent cut-off ; it has also been described as composed of a plain slide valve and a riding valve. But these are long expressions and are also not free from ambiguity. We wish to distinguish between gears requiring considering of only absolute motion and those in which relative motion must be considered. Gears possessing but a single valve evidently belong to the former class and the term, single valve, is sufficiently suggestive of absolute motion. As relative motion implies at least two pieces we feel justified in using the term double-valve to suggest this motion. A further justification is sup- plied by the fact that the principal functions, of opening or closing a steam passage, must be doubly performed (though not simultaneously) when one valve slides on the back of another. We must confess, however, that we are not satisfied with the terms we have chosen and hope that something else as simple but more exact may be found. PARALLELOGRAM OF ECCENTRICITIES. IS I the main part of the problem. Such an eccentric drives the ex- pansion valve, over a perfectly stationary main valve, with an absolute motion that is the same as the motion of the expansion valve relatively to the main valve when both these valves have their actual motion. In this way the new and complex problem of the relative motion of two valves, is reduced to the old and simple problem of a single-valve sliding on a fixed seat and driven by one eccentric. Now as to the method of procedure. When the main and expansion eccentrics are both given, the virtual eccentric can easily be found by means of a proposition for which Dr. Zeuner suggests the name of " Parallelogram of Eccentricities." CRANK. Fig. g». It may be stated as follows : The main, virtual and expansion eccentrics constitute, respectively, the two adjacent sides and diag- onal of a parallelogram. In Fig. 92, 0E m , 0E V and 0E e represent these eccentrics and OE m E e E v the parallelogram which they form. To avoid confu- sion hereafter it should be specially noted that the expansion eccentric is always the diagonal of this parallelogram, provided the virtual eccentric 0E V effects the motion of the expansion valve relatively to main valve. But when it is a question of 1 82 DESIGNING DOUBLE- VALVE GEAR. relative motion of main valve to expansion valve the virtual eccentric must be OEv' , just opposite to 0E V , and 0E m must then be the diagonal of parallelogram OE„ 0E m OEJ. In all our work we shall take the expansion eccentric as the diagonal of the parallelogram. In existing valve gears the main and expansion eccentrics are completely known and the examination of the steam distribution effected by their valves is easily made by means of the main and virtual valve-circles, which are deduced in the usual way from the known main and virtual eccentrics. In designing valve gears, however, the expansion eccentric is the unknown quantity whose variations must satisfy the pre- scribed conditions as to range of expansion, time of admission, rapidity of cut-off, etc. To find it we must first find the main and virtual eccentrics and with these two as adjacent sides con- struct a parallelogram whose diagonal will be the desired expan- sion eccentric. The main eccentric can be found in the usual way from its valve-circle and this in turn from the given conditions as to lead, port-opening, maximum cut-off, release or compression. The virtual eccentric is found in a similar way, its circle being first obtained from such given conditions as, type of expansion valve, cut-off and time of reopening. Having thus indicated the principal steps in designing a double-valve gear, we will return to the beginning and show that, when there are two valves and both are in motion, one rid- ing on the other, then a fixed circle can be found which will cut from the crank, or its prolongation, chords whose lengths repre- sent the distances between a point on the expansion valve and a corresponding point on the main valve, the correspondence being simply that the two points are together when the expansion valve occupies its middle position on its seat on the main vaive. The distance between two such points at any crank position is evidently equal to the (algebraic) difference of their distances from any common, fixed, reference line. Neglecting the angu- VIRTUAL VALVE-CIRCLE. 1 83 larity of the eccentric-rods, this difference is equal to the differ- ence in the distances E e S e and E m S m , Fig. 93, of the correspond- ing eccentric centers E e and E m from their common reference line OP, which is perpendicular to the dead-point line NOZ of the eccentrics. Our problem is now reduced to finding a *Fig. 93. circle that will cut from any crank position OC a chord equal to the corresponding difference E e S e — E m S m = E e S of the distances of the centers E e and E m from the middle position OP of the eccentrics. Finding the main valve-circle OD m from its eccentric OE m as in Fig. 60, p. 137, we get the chord OF J = 0F m = E m S m as the distance of E m from OP. The distance E e S e of E e from OP can be found in the same way by constructing the expansion valve-circle OD e . To do this we measure angle a e , from the right portion the eccentric's dead-point line OZ, in direction * The inner angle o v extends from OZ around to OE v . extends from OC around to OD v . The outer angle a v 184 VIRTUAL VALVE-CIRCLE. opposite to the rotation, and lay off this angle a, from the crank in the direction of rotation ; 0D e = 0E e will be the diameter of the expansion valve-circle and OFJ = 0F e = E e S e . Then be- cause D m S" is parallel to crank position OC, we have D m S" = FJFJ = F e F m = E e S. As D m FJ is perpendicular to OC, the angle D m S"D e is a right angle. But the hypothenuse D e D m is constant in position and magnitude for all crank positions ; the lccus of S" is therefore a circle. If through D m we draw a par- allel to the crank position OC at any instant, this circle will cut from this parallel a distance D m S" which is equal to the value of E e S at this instant. By moving this circle parallel to itself so that the point D m of this circle D m S'D e shall shift its position to shaft center 0, the circle will take up the new position OD v FJ and cut from the crank itself (or its prolongation) the distance OFJ = D m S" = E e S = E e S e — E m S m . In its new position it may be called the virtual valve-circle for it possesses all the properties of a valve-circle relatively to its eccentric. For in- stance, the virtual eccentric 0E V is found from the virtual valve- circle 0D V by laying of the angle a„ in the same way as the angles a m and a e . Moreover, when the \ D rolonp-ed I cran ^ cuts the virtual valve-circle, the virtual eccentric will be to the ) left i °^ ' ts m iddle position OP. As regards the periods during which the riding-valve keeps the ports in the main valve opened or closed, the virtual valve-circle, like the two others, is subject to the conditions of valve-type, etc., laid down in Table XIV. The valve driven by the virtual eccentric is of course of the same type as the expansion valve, the relative motion of the latter corresponding exactly to the absolute motion of the former. Before applying these results to the common high-speed engines, there is still another proposition enunciated by Dr. Zeuner which is of the greatest service in the solution of valve- VARIATION OF EXPANSION. 185 gear problems. But before stating what it is, we will point out that with double-valves the expansion can be varied : I. By varying the angle of advance of the expansion eccentric. II. By varying the throw of the expansion eccentric. III. By varying both angle and throw of the expansion eccentric. IV. By varying the lap of the expansion eccentric. In the last case neither throw or angle of advance of expan- sion eccentric is varied. The most general case is therefore III ; Fig. 94. it is represented in Fig. 94, where the center E, traverses any arbitrary locus ac. Its reference line is taken to be the crank at left dead point, but any other crank position might have been chosen and the locus ac would occupy a correspondingly different position in the plane of the paper. It is simplest to consider the locus ac as rigidly attached to the crank, moving with it, and preserving always the same relative position towards it. Let OX and OY be coordinate axes rigidly attached to the crank and moving with it, then will OF and FE, represent the 1 86 LOCI OF CENTERS AND VERTICES. coordinates of any point E e of the locus ac. Prolong main eccentric E m to 0', making 00' = 0E m , and draw a parallel set of coordinate axes O'X', O'Y', also rigidly attached to crank. Then the proposition of Dr. Zeuner, last referred to, consists in the statement that, The locus a'c' of the center E v of the virtual eccentric OE v is a curve exactly equal and parallel to the locus ac and the position of a'c' relatively to the origin O' is exactly like that of ac relatively to origin O. To prove this it is only necessary to show that the coordinates O'F', F'E V are respectively equal and parallel to coordinates OF and FE e . By construction 00' is equal and parallel to EcE?, which makes 00'E v E e a parallelogram, hence 0E e is equal and parallel to OE v and this makes the two sets of coordinates respectively equal and parallel. Since in designing we start with the valve-circles rather than with the eccentrics it will be conve- nient to represent this prososition as applied to the diameters of the valve-circles. Y F'g- 95- In Fig. 95 the curve ac is the locus of the vertex D t of the valve-circle diameter 0D e . This curve ac is equal to the locus of LOCUS OF CENTER OF ECCENTRIC. 1 87 E e and is similarly located. The curve dd is the locus of the vertex D v of the diameter OD„ of the virtual valve-circle, and is equal and similarly situated to the locus of E v . The locus dc' has the same position relatively to the origin (J that ac has rela- tively to the origin 0. These valve-circle-loci, unlike those in Fig. 94, have the great advantage of preserving their places in the plane of the paper during the rotation of the crank. The subscripts e, m, v employed in these diagrams relate, re- spectively, to expansion, main and virtual eccentrics and their valve-circles. The symbols e , r and d a respectively refer to the lap of the expansion valve, the eccentricity and angle of advance of the expansion eccentric. When the subscripts i and 2 are annexed to the subscript o the values at minimum and maximum grade of expansion are meant. The other symbols have the meaning already assigned them in the discussion of single-valve gears. The loci of the center of expansion eccentric vary greatly in form according as the valve-gear elements, angle of advance, throw and lap are varied. In Figs. 100 and 101, representing the expansion valve diagrams of the Cummer engine, the locus of E t is circle concentric to shaft center and has r as a radius. In Fig. 106, the expansion valve diagrams for a Meyer valve gear, this locus reduces to a point, for the setting and throw of the expansion eccentric does not vary with the grades of expan- sion but only with the lap. The inscription, " locus of D v " on this diagram is to be understood only as a geometrical construc- tion that is helpful in solving the problem. In Fig. 109, the ex- pansion valve diagram of the Rigg engine, the locus E e is OEnE^, a straight line passing through 0. In Fig. 1 12, the locus of E e is a circle that is not concentric to 0. In Figs. 115, 116 the locus E e is a straight line that does not pass through and in Figs. 118, 119 locus of E e is approximately a circle whose circumfer- ence passes through shaft-center 0. In designing a double-valve gear, the steps are : (a) to find the main eccentric suitable for release, compression, and beginning 1 88 EXACT VALUES OF VALVE TRAVEL. of admission, (b) to find the virtual valve-circle that will give the desired cut-off, reopening, etc., for the assumed type of expansion valve, (c) to find the corresponding virtual eccentric, and (d) to find the desired expansion eccentric by combining the main and virtual eccentrics. When very great accuracy in the determination of valve travel is desired, we must substitute for the virtual valve circle a pair of exactly drawn curves whose vectors represent at once the crank's position and the valve's travel from its center of motion (see definition of this center, p. 151). To construct these polar curves we must therefore find this valve travel. Accordingly we should, in Fig. 93, replace the middle position or reference line P OP of the main eccentric by a reference arc struck from the center of motion with eccentric rod as radius. The distance of 0E m from this arc will be E m S m ' . In like manner we must replace the reference line POP of the expansion eccentric by the reference arc (of the valve) struck from the corresponding center of motion and the eccentric rod as radius. The distance of E, from this arc will be E e SJ and the true relative position or travel of the expansion valve will be E,SJ — E m S m ' and this difference laid off on the corresponding crank position will give one point on the desired polar curve. The polar curve obtained by laying off on the crank the dis- tance E V S' V (of center E v of the virtual eccentric from the common arc of reference) is in general not a closer approxima- tion than the valve-circle itself. When there is a reversing lever between one valve and its eccentric, the actual distance of each from the arc of reference must still be taken, remembering however to take the] [ of E.SJ and E m S m ' when the valves are on the \ . , \ • (opposite sides] of their center of motion. Figs. 96, 97, t Figs. 99, ioo, 101. Figs. 105, 106, 107. 1 BUCKEYE ENGINE. CUMMER ENGINE. Expansion Varied by Angle of Advance. Main Valve Diagram. Positive Indirect Valve. Expansion Varied Main Valve Diagram. MEYER VALVE GEAR. Expansion Varied by Lap. Figs. 108, iog, ilo. RIGG ENGINE. by Angle of Advance. Positive Direct Valve. Main Valve Diagram. Positive Direct Valve. Main Valve Diagram Expansion Varied by Throw of Expansion Eccentric. Positive Direct Valve. Pigs. Ill, m, 113. ST URTEVANT ENGIN E. Expansion Varied by. Simultaneous Variation of Throw and Angle of Advance of Expansion Eccentric Main Valve Diagram. Positive Direct Valve. Expansion Valve Diagram. Negative Direct Valve. Minimum Cutoff =0 Expansion Valve Diagram. Negative Direct Valve. Minimum Cutoff= 0.033 Expansion Valve Diagram. Negative Direct Valve. Expansion Valve Diagram. Negative Direct Valve Minimum Cut-off = 0.1 Expansion Valve: Diagram. Negative Direct Valve. Maximum Cutoff = That of Main Valve. M aximum Cut off =0.5 Minim um Cut-off 0.05 k M aximum Cut off = 0.5 Expansion Valve Diagram. Negative Direct Valve. Maximum Cut-off = 0.78 0.30 FOR .1 CUT-OFF. ■L6 7 " .5 » •' „--" W*D Reversing Arms between Eccentric and Valve. Development of Valve and Seat in The Rider Valve Gear. Eou of A MainV r=Z.5 c e=o'7s h=Z4° jl_ e. EXPAN: r o = 1.625 6 o =90° rt e=-i.o; EXPAN 6*1.6*5 & o=90° e=-i.o? Determination of the Rod's Inertia-Resistances Og, he Total Forces O'g Exerted by Rod (i. e. Including Both Inertia and Weight). g-S' represent component of total force parallel to the latter and acting at crank-pin. letvveen S' and 0' (= also to gg) represents corresponding wrist-pin component. Heavy line 00' = weight of rod per Q" of piston area. 12 Lbs per a" "" oj Piston. Fig. U5. Determination of the Pin Pressures. Wb = Wrist-pin pressure ; Mb = Crank-pin pressure. WV= Difference of effective steam pressure and force accelerating the reciprocating parts. Vb = Action of guide on slide. IVM= Total force exerted by rod by virtue of its inertia and weight. Sb = Line of internal stress when WS and SM respectively represent the wrist-pin and crank-pin components of total force of rod. The internal stress lines Sb are drawn parallel to the correspond- ing lines in the central diagram, the latter lines being tangent to the friction circles (of the bearings in the connecting-rod). Scale or Forces LBS PER Q"or PISTON IN THIS ricURe.CVLINOER IS TO THE RIGHT OT ENCIWE SHArT-O. EXAMPLES OF DOUBLE- VALVE DIAGRAMS. 1 89 In the great majority of cases the valve-circles will be suf- ficiently accurate ; in the few remaining cases it will rarely be necessary to construct the whole polar curve ; a few points of the curve (for the crank positions of the most practical interest) will usually suffice. Examples of Double- Valve Diagrams, Figs. 99-119. In these mechanisms the expansion valve moves on the main valve and controls the ports or steam passages in the latter. Since both valves move their relative motion depends upon the separate, absolute, motions and is one of some complexity, but it can be simplified by considering the main valve stationary and supposing the expansion valve moved by a new special eccentric that im- parts to it an absolute motion equal to the relative motion which it possesses when both valves move under the influence of their own, actual, eccentrics. This eccentric has been called the vir- tual eccentric and for a particular grade of expansion has all the simplicity and properties of the ordinary fixed eccentric. Its valve-circle is found from the virtual eccentric in the usual way and gives the same sort of information concerning the crank positions at which the port in the main valve opens and closes. In this set of diagrams the port openings have not been indi- cated by sectional areas. With negative valves the port open- ing is measured inward; from negative lap-circle to valve-circle, this intercept lying wholly or partly outside of the valve-circle. When there is a reversing arm between an eccentric and its valve, we assume in diagramming the same type of valve but an eccentric capable of giving the valve motion directly, without such an intermediate lever. At the end of the procedure this is allowed for by placing the resultant expansion eccentric 180 from that given by diagram. 13 190 BUCKEYE ENGINE. As this series of figures (99-119) is intended to illustrate the double-valve system of varying the expansion, we have neg- lected the exhaust side of the distribution in our main valve diagrams. But this may easily be supplied from what has gone before. Buckeye Engine, 13x24, Figs. 96, 97, 98. We have already touched upon the kinematic character of these valves in the foot-note, p. 1 54. Generally when there are two valves there are two eccentrics and these usually give a resultant or virtual eccentric, but not always. There is one exception in the Buckeye Engine. In this case the mechanism between expansion eccentric and expansion valve is such that the rod of this expansion valve receives a motion nearly equal to that of the main valve thus making the relative motion of the expansion valve on the main valve dependent simply on the motion of the expansion eccentric. The latter therefore is at once expansion and virtual eccentric. In this case of double valves and double eccentrics we need speak only of the expansion eccentric and valve-circle. Fig. 96. The main valve in this case covers the port in its middle position and cuts off with its inner edges which brings it into the Positive Indirect class of valves. The location of the corresponding valve-circle OFD m , Fig. 96, can easily be chosen by the help of Fig. 62 and the remarks made on pp. 142, 143. The eccentric can now be found as in Fig. 60. It is evident from Table XIV that left cylinder port is open for steam admission between crank position 3 and 4 and closed during crank's motion through positions 4 — 1 — 2 — 3, closing at 4 and reopening at 3. Fig. 97. For the sake of economy of construction we will take r = r and describe a circle with as center and r as a radius. To find the valve-circle for minimum cut-off, when neg- ative lap e„ is given, we erect at H a perpendicular HD tl to dead BUCKEYE ENGINE. I9I center line OH of crank. Then will 0D ei be the diameter of the expansion valve-circle for minimum cut-off. Inspection of Table XIV and Figure 97 shows that this negative direct valve keeps its left port in the main valve closed from crank position 3 to 4 and opening during 4 — 1 — 2 — 3, the reopening taking place at 4 after main valve has closed the left cylinder port. In getting the position of the expansion eccentric it should be noticed that in this particular problem the dead center lines of eccentric and crank do not coincide differing by about 8}4°. After the manner of Fig. 60, we find 0E, x , to be the position of expan- sion eccentric at minimum cut-off, when crank is at left dead point. As there is a reversing lever between eccentric and valve, we must reverse 0E, X to get position 0E' ex of expansion eccen- tric in the actual case. Fig. 98. In order that the expansion may have the widest possible range the expansion valve at maximum cut-off should close its port at the same instant that the main valve closes the corresponding cylinder port. To realize this place the crank at position belonging to main-valve's cut-off and then find, accord- ing to Fig. 62, the location of the proper valve-circle. As the lap e does not change the perpendicular /Z?„ 2 to the crank position CO 1 will give 0Dt 2 as the diameter of the valve-circle desired. Laying off the proper angles as in Fig. 60, we get 0E t2 for the position of expansion eccentric at maximum cut-off, when crank is at cut-off position and no reversing lever exists between eccen- tric and valve. As there is such a reversing lever, we must change by 1 8o° the position of 0E ti and thus get 0E' e2 , the position of the actual expansion eccentric when crank is at position of maxi- mum cut-off. For this same position CO of the crank the main eccentric is at 0E m and the expansion eccentric's position cor- responding to minimum cut-off is at 0E' tl . The angles of ad- vance |■ crank cuts the present valve-circle when exhaust [ prolonged J r valve is to the < . , > of its middle position. The lap on the exhaust side may easily be ascertained by assuming that cushion- ing takes place during the last fifth of the stroke ; the beginning of the release can be determined from the inside lap thus found. 194 CUMMER ENGINE. Fig. ioo. Given e and r . As the eccentricity r a of the expansion eccentric does not change, the locus of the center of the latter will be a circle FLM described about with radius r . The vertex of the diameter of the expansion valve-circle has the self-same locus FLM about as a center. Consequently accord- ing to principle enunciated in connection with Fig. 95 the locus of the vertex D v of the diameter of the virtual valve-circle must be a circumference NQP struck with r as a radius and 0' as a center. As the negative lap is given, we have only to erect the perpendicular HD V1 to the crank position C'OH at minimum cut-off to get a second locus of D VI and thus get the diameter 0D m of the desired valve-circle. We see that the reopening of the expansion valve takes place just a little before the crank reaches the dead point. By laying off the angles as in Fig. 60, we find OE V i as position of virtual eccentric for tV cut-off and then, by means of the parallelogram, 0E ex as the position of the cor- responding expansion eccentric when crank is at dead point. Fig. 101. Given = LON and KL = MN = PT = E = e, ference of outside laps. Then SP = R ib and SR = SP cos Z-. 2 £1 - E given dif- sin ib 2 sin 198 MEYER VALVE GEAR. The remarks and constructions given on pp. 194, 195 in con- nection with the Cummer engine, are also applicable here. As equality of cut-offs is conducive to smooth running, Dr. Burmester has given a solution of the problem of equalization for the Meyer valve, which consists in making the laps unequal for the two ends of the valve, the difference between the laps being a constant quantity. Fig. 103. The procedure is as follows : In Fig. 103 the stroke F F° is divided into eight equal parts and through these divisions, with the connecting rod as radius, arcs are struck intersecting the crank circle at F It F 3 , F 3 , etc., then Z x F z , Z 2 F 2 , Z 3 F 3 , etc., represent )/%, %, y&, etc., of stroke, or cut-offs, and 0F Ir 0F 2 , 0F 3 , etc., the corresponding crank positions. These cut-offs y&, %, etc., MEYER VALVE GEAR. I 99 are laid off as abscissas in Fig. 104. Now let the valve-circle for this case be represented by 0Q o Q 5 ; it will cut from the crank the chords 0Q o , OQ lt etc.; these represent the travel of any point of the valve and are laid off as ordinates T m , T^m It in Fig. 104. Fig. 104. Each of these chords also represents the lap of the valve effecting the cut-off at the crank position on which the chord lies. Hence the ordinates of curve m m^m 2 . . . m z also repre- sent the laps effecting the cut-offs represented by their corres- ponding abscissas, the ordinates above the base representing positive, and those below negative, laps. In like manner the ordinates of curve m°n?m? . . . . m z represent the laps effect- ing the cut-offs during the return stroke, It is evident from the figure that for the same cut off the laps for the return stroke are greater than those for the forward stroke. In this diagram these differences can be made to (nearly) disappear by moving the lower curve upward through a distance m* m A ; it will then occupy the position shown by the dotted curve m'tfC. In the 200 MEYER VALVE GEAR. mechanism itself, the corresponding step is to start with a sym- metrical valve having the same lap on its two parts, then set the right-hand half of the expansion valve, nearer to its left-hand half by just this amount m*m r Approximately equal cut-offs will then be effected on each stroke. It is evident from the diagram, Fig. 104, that a straight line ee will approximately represent the two-lap curves between the cut- offs yi and y%. To equalize the cut-off it is therefore unnecessary to move the two halves by two screws of unlike pitch. Both screws may be of the same pitch provided the two halves of the expansion valve are placed on their seat on the main valve with the unequal laps given above. The method of finding the unknown elements in this (Meyer's) valve gear of course varies with the character of the data. We shall assume in all the special cases discussed, that the range of cut-off of the expansion valve is given and that the laps, setting and throw of the main valve are known. We therefore know the crank position at which main valve cuts off and care must be taken that the expansion valve does not permit a second admis- sion of steam during the same stroke. Although the locus of E, (and locus of Z> v ) is reduced to a point for this gear, it still remains true that the point D v must lie at the distance r a from the point 0'; this point, we know from Fig. 95, is found by prolonging 0D m and making 00' = 0D m . The problems and their solu- tions are arranged in the following order : (1.) Given r and 8 ; required e m and e oz . (2.) Given r and * M ; required d a and e al , (3.) Given r and e 01 — e OI ; required d ol e m and e OI . (4.) Given d ■ required r , e 01 and e ol . (5.) Given 8 Q and e m — e 0I ; required r a , e^ and e ot . (].) Fig. 106. If the radius r and the angle of advance d of the expansion eccentric are both given, the problem is readily solved, for then a parallelogram constructed on the main and expansion eccentrics (Fig. 92) will determine the virtual eccentric MEYER VALVE GEAR. 201 and this (Fig. 60) the virtual valve-circle. The intersections of this circle with the crank positions corresponding to the limits of cut-off will determine the laps desired, provided the expansion valve does not open prematurely. In this gear, it is only at maxi- mum cut-off that there is any danger of an opening of this valve before the main valve has closed the port at the cylinder end in question. (2.) Fig. 106. Given r a and e ov With 0' as a center and r as a radius, describe a circumference ; the vertex D v will lie some- where on it. At the cut-off point 1' erect a perpendicular i'D v to Oi', where it cuts the circular locus will be point D v desired, provided the second intersection 2', of valve-circle on 0D V with lap circle e m , is such that the expansion valve's opening does not occur till after the closure of the port by the main valve, other- wise the data must be regarded as incompatible with good run- ning. The lap e 01 can be found from the intersection of this virtual valve-circle with the crank position at minimum cut-off. By means of the parallelogram, the position of 0D e may be found from 0D m and 0D V and then by method of Fig. 60 the position of 0E e . This determines at once the setting of the eccentric and by applying the rule given on p. 149 the angle of advance d can be found if desired. (3) Fig. 106. Given r and the difference e^ — e m = £. We know from the remarks on p. 200 that D v must lie somewhere on the circumference described with C as a center and r as a radius, and from the proposition given with Fig. 102 we know that D v must also lie on a line SV, parallel to bisector OQ (Fig. 102) of angle

„ 2 is obtained in the perpendicular erected at T to the crank position KOI for maximum cut-off. We have 0'D Vi = r 0l and 0'D V ^ = f"o 2 ; the remaining unknown element d can be found by means of the parallelogram and Fig. 60, as in the preceding valve gears, or more directly as indicated in the figure. The crank is supposed to be at OX when the expansion eccentric is on radius OE v JL Vi . (2.) Given / / \ / // // // // / / / / / / / / / / A CRANK ANGLE AND MAXIMUM VELOCIXY OF SLIDE. 2\"J is perpendicular to rod's direction. (It has occasionally been erroneously assumed that this configuration existed when line 13—23 is perpendicular to rod.) Starting with this proposition we will now find a cubic equation for the relation between this particular crank-angle Q and the y ratio j of crank to rod. Figure 120 readily furnishes the relation. OW tan Q = OF (tan B + cot 8) = . ° a W . x sin a cos p or cot 6 = sin B cos B and since sin /9 : sin Q = r : I we get, after reduction, sin 2 Q -f- - sin 4 Q r sin 6 Q = 1 . / 2 A In terms of the rod angle /? the cubic equation takes an even simpler form, sin 2 /9 -f sin* B — sin 6 /? = -. In either of these equations we have an exact solution. Grashof in his Machienenlehre, Bd. II, p. 133, found approxi- mately, cos °=7-*G)' ithin 0) 5 - which is correct to within This result is of course sufficiently exact for engineering purposes. For - = \, cos = 0.1597 and 6 = %°° 49'- 2l8 CRANK ANGLE AND MAXIMUM VELOCITY OF SPEED. Fig '21 On pages 36 and 37 it is shown that the slide acceleration is approximately, 6 = — cos ± — cos 2(3 . This means that ordinary finite rods have same velocity as an infinite rod when crank angle is about 45 °. Thus, for — = \, this angle is = 45° 12' or i34°48'- This gives another convenient method of constructing approx- imately the curve of slide accelerations abcde, Fig. 121. For we have for the infinite rod the straight line fbmdg. For the finite rod we have exactly af = \hf and eg = k ek, and approximately, there is equality in the acceleration of the two rods at bl and nd. Point c we can find closely from Grashof s formula given above. This enables us to get easily five points abcde on the curve of slide accelerations and leads to a ready and sufficiently accurate construction of this important curve in Steam Engine Dynamics. INERTIA-RESISTANCE OF THE ROD. 2IQ APPENDIX B. Diagram for Determining Inertia-Resistance of the Rod, the Horizontal and Vertical Components of this Resistance and the total Force Exerted by Rod. The method of finding the resultant of the weight of the rod and of its resistance to change of velocity has been given on page 103, Figure 47, and the method of finding the pin pressures is illustrated on page 108, Figure 49. In both the figures, the main train of engine (crank, rod and slide) is supposed to be drawn to some small scale to bring it within the limits of the paper. Even then, many of the construction lines are long, and separate, enlarged, diagrams of the polygon of forces WVbMs, etc., are found necessary. These disadvantages can in large measure be removed by omitting the connecting-rod from the drawing, then using in its place the image Cw (see figures 43 to 46) of the rod, and making the constructions employed with the image, similar to those used with the rod. The following figure, 122, illustrates how this may be done. Cgwz is the image of the portion C(& WZ of the rod. Triangle jgfz is the image of triangle J®ZF and is the triangle/'^yV; these three triangles are all similar to each other by construction. The point,/ through which passes the resultant of rod's inertia resistance is obtained by finding that point Z of rod which has acceleration along the axis of rod and using it as one point of mass concentration, the other point J of mass concentration being taken so as to satisfy Eqs. 68 to 70 on p. 95, that is, so that K*=J%X%Z, This will necessitate the rod's inertia-resistance passing through point /. 220 INERTIA-RESISTANCE OF THE ROD. INERTIA-RESISTANCE OF THE ROD. 221 Point Z is horizontally projected on rod from point z of image which point in turn was found by drawing Oz parallel to the center line CW of the rod. When rod is omitted from the figure, point z is projected to z 1 on crank OC and the right-angle triangle z'/'j' is used to determine/' (the horizontal projection of point,/.) Here the perpendicular The point/' can be laid off on a circle of its own, its zero line for angles being along line 0f o . All points f are easily found from the z 1 and/' points of the figure. We will assume, without again giving the construction, that the/' points of Fig. 122 have been correctly found and then determine the q points of division on Og. Then will qg and gO respectively represent the components at wrist-pin and crank-pin that make up, and are parallel to, the whole force of acceleration of rod. To find the resultant of this force and the weight we may pro- ceed as in Fig. 47, page 103, but this involves drawing the crank train. Or on Og and OL (here the weight OL is, for conve- nience, taken very large relatively to the inertia Og) we may con- struct the parallelogram OgML and OM is the resultant desired. By laying off the weight OL in an opposite direction, 00' , and completing parallelogram OO'gMwe can get the desired resultant O'g as the upper side of the parallelogram. It is the last con- struction which is most convenient and is the one used on plate. We next resolve this force O'g = OM into its actual, com- ponent, pressure at wrist-pin and crank-pin. This is done as in Figs. 48 and 49, the distance O'g (or OM) Fig. 120, taking the place of WZ, Fig. 48 or of WM, Fig. 49 ; and the point 0' (or 0) then taking the place of point W in Figs. 48 and 49. The first step is to find the point 5 on resultant which will divide it into the two parallel components SO and SM (SW and SZ, Fig. 48) acting at wrist-pin and crank-pin respectively. It is evident from equation 72, page 107, that the point 5 must lie on 15 222 INERTIA-RESISTANCE OF THE ROD. a line QS {SVt Fig. 48 or SZ Fig. 49) which is parallel to the line of centers WC and passes through the intersection Q of any pair of components of OM, for instance OQ and QM. It is evident that we shall know Q if either component OQ or QM is completely known. In the figure we have resolved the weight OL into components OT and TL parallel to itself and acting at wrist- and crank-pin respectively, the components satisfying the condition OT:TL= C%:<&W. In like manner the resistance to inertia is resolved into the two components Eg and EO, satis- fying the proportion Eg : EO = CJ :JW. Transferring qg to 02V and combining latter with OT we get OQ, one component of OM. We now draw through Q the line QS parallel to line of centers CW of rod, and then through 5 the line Sb (see also Fig. 49) parallel to line of internal stress, i. e., parallel to the proper tangent to the friction circles. In the figure extravagantly large friction circles have been placed at Cand Win order to show the construction more clearly. In the figure on plate the forces are transferred from center of shaft to point 0' above, the latter at distance 00' = OL = weight per □" of piston. Then (Figs. 122 and 124) O'g = 0M= resultant of weight and inertia-resistance of rod. Furthermore it is evident that this resultant gO' can be divided at S' into the same segments as OM at S, by laying off NQ' = TL and draw- ing Q'S' parallel to center line CW of rod. Here it will be well to tabulate the value of component O'S' in some table of forces ; it can then be conveniently used in the diagram for finding crank- and wrist-pin pressures. In the present example, the values of O'S' are : 20° 40° 6o° 80° IOO° I20° 140° 1 60° 180' i-59 1.68 1. 10 0.60 o-39 O.OO -O.9O -4.4I 5.21 200 220° 240 260 280° 300° 320° 340° 36o c 2.21 1.54 1-39 1.28 1.40 I.9I 2.6l 3-59 6.00 The minus sign indicates that point S' has fallen outside of space O'g. CONSTRUCTION OF DIAGRAM OF INERTIA-FORCES OF ROD. 223 In order to construct curve aaaa in the diagram of shaking forces, Fig. 50, we must know the horizontal and vertical com- ponents of the inertia- resistance of the rod. By projecting the point g (Figs. 122 and 124) to g', on the vertical through 0' we get in gg' the horizontal, and in Og' the vertical, component of the inertia-resistance Ogoi rod. The values of these components should be tabulated in the table of forces shown with Fig. 50. They can then be easily transferred to the diagram for finding wrist- and crank-pin pressures ; this is particularly desirable if scale of the pin-pressure diagrams differs from that of the rod's inertia-diagram. Construction of Diagram of Inertia-Forces of Rod. For ease of construction and accuracy it is well to make the crank of diagram about 12 inches long. The exact length of crank is determined by scale of forces, which should be a con- venient one. In this case 1 pound per □" of piston area to the linear inch will be a convenient scale, making the radius for in this 14x14 engine represented in the diagram, A = 153.94, W= 136 pounds and revolutions per minute 273, while weight of reciprocating parts is 321 pounds. Take the crank positions 20 apart and then find the foot w of the image by computing the distance Ow by means of table on page 39. We have Ow = 13.02 X tabular quantity c. The rod is supposed to be six cranks long. When the segments O'S' and S'g of the total force O'g of the rod are to be found for each one of a series of crank positions, the work of locating on the image the point g corresponding to the center of gravity of the rod can be greatly simplified by dividing crank-radius OC of diagram, Fig. 124, at point/ into the seg- ment Og' so that -^ equal ratio -^.m rod, (see Fig. 122), and 224 CONSTRUCTION OF DIAGRAM OF INERTIA-FORCES OF ROD. then describing about a circle with the radius Og'. In Fig. 124 this radius is 0.573 of the diagram crank. Where this circle cuts the crank positions draw horizontal lines to point g of images corresponding to these positions. The lines Og represent the total inertia-resistances offered by rod at these crank angles. When the crank is upon its dead points the method followed above fails because then the image coincides in direction with the radius. In this case the point g on image may be found by making, for the crank-angle 0°, the distance Og = R ( 1 ± (p — ) , when

two quadrants and the axis of X is made to extend to the-; ,°, > of origin W. The following tables give the values of the rectangular com- ponents O'g 1 and gg' for the example illustrated in the present diagrams. 228 DIAGRAM OF PRESSURES AT CRANK- AND WRIST-PIN. Vertical Components of Total Force of Rod per □". + and — Respectively Represent Up and Down Components of Rod. Crank Angles- -Forward Stroke. o 180 10 170 20 160 3° 150 40 140 50 130 60 120 5.62 70 no 80 100 90 90 —0.88 040 1.67 2.86 3-93 4.87 6.18 6.52 6.72 Crank Angles- —Return Stroke. 180 360 190 35° 200 34o 210 220 33° 320 230 310 240 300 -7-38 250 290 260 280 270 270 —0.88 — 2.16 —343 — 4.62 1 — 5.69 -6.63 —7-94 —8.28 —8.48 Horizontal Components of Total Force of Rod per □". + and — Represent Force Acting to the Right and Left, Respectively. Crank Angles — First and Fourth Quadrants. 360 10 35° 20 34° 3° 33° 40 320 5o 310 60 300 6.1 1 70 290 80 280 90 270 14.02 '3-77 13.00 11.80 10.21 8.25 3-79 1.40 — °-93 Crank Angles- -Second and Third Quadrants 100 260 no 250 120 240 130 230 140 220 150 210 160 200 170 190 180 180 -3.14 -5.18 -7.02 -8.58 -9.89 -10.87 -1 1. 61 -12.04 -12.17 Having laid off the total force of rod to the assumed scale we next lay off from the origin W, the smaller O'S', Fig. 124 (or WS of Fig. 49), of the two segments of the total force and through the point 5 thus found draw the line Sb (Fig. 49 and Fig. 122) parallel to the line of internal stress, which is very nearly a tangent to the friction circles at crank and wrist-pin. These tangents are drawn, in the middle of the diagram, Fig. 125, for the inclinations of connecting-rod corresponding to the different crank angles. The proper tangents for each position DIAGRAM OF PRESSURES AT CRANK- AND WRIST-PIN. 229 are decided upon according to the method indicated in Figs. 38 to 41, pages 89 and 90. Having drawn, Fig. 125, the line Sb parallel to the proper tan- gent, we next lay off on each axis of Xthat force WV, along axis of cylinder, which equals the difference of effective steam pressure and inertia of reciprocating parts. From its end Vlay off the guide's reaction Vb, inclined, from the normal, against the slide's motion, an amount equal to the angle of friction. This reaction will determine point b and thus determine the actual wrist-pin pressure bWand the crank-pin pressure bM. Of course the total labor of taking these various steps is con- siderably diminished by taking the same step for each of the crank positions before proceeding to take the next step in the series. In laying off the total force of rod it may be of some help to notice that the vertical components of this force are alike in quadrants I and II for symetrically placed crank positions and are also alike in III and IV for such crank positions. It is evident from the figures corresponding to the different quadrants that friction has very little influence on the intensities of the pin-pressures. It does influence the vertical component of the wrist-pin pressures to a notable degree because these are small any way. As it is just these components that determine whether slide shall press upward or downward and as a very slight upward component will suffice to effect this reversal of pressure and cause a knock, we cannot say that the friction's influence is everywhere practically unimportant. Moreover we must not conclude, because pin pressures are not appreciably affected by friction, that the latter's influence on engine friction is likewise inappreciable. The crank-pin's pressure for instance makes its influence felt to a notable degree by acting tangent to friction circle in such a way as to diminish its crank leverage by the radius of the friction circle. When it is only a question of finding when reversal of pressure on slides occurs, the determination of pin-pressures will not be necessary for every crank position of Fig. 124. Inspection of 23O DIAGRAM OF PRESSURES AT CRANK- AND WRIST-PIN. Fig. 124 will show where this reversal is liable to take place and the pressure determinations of Fig. 125 may then be confined to the crank positions of that vicinity. In each of the upper and lower semi-circles, the images are exactly parallel for crank positions symmetrically situated with respect to the 90°-270° line. This parallelism can be shown as follows : Let r represent the angle Cm W made by image with stroke of slide. Then will cot r wH COt T : — : — - = R cos Q — I CH r— sin 2 R sin Q L* cos 2 Q -+- — sin 6 IZ 2 A ' 2 f^ , „\* 'J V Here has the meaning given it on page 37 and its factor — has the value R in diagram, Fig. 124. From this it is evident that cot t will be the same for any two supplementary angles and angles 180° — Q. When cot r = o, the image becomes vertical ; this is always so for infinite rods and for common ratios of -= when crank-angle R Q is nearly 45 °. With the help of this expression for cot r, we can show 01 = iL (cos Q -f- sin Q cot r) exactly, and that the distance 01 is practically a constant quantity for ordinary rods. This would enable us to find still another easy method of getting the location of the inertia-resistance of the rod ; but we omit it as it is not any simpler than the method we now have. See note on page 100. % = — -, Fig. 45. „9 rJ "A ,9 M 4 W 9 V /> 7 ^ "> o 921 -3.U FORCES ACTING IN AND UPON THE CONNECTING ROD. 23 1 APPENDIX C. Exact Construction for the Forces Acting in and upon the Connecting Rod. by r. c. h. heck, m.e., Instructor in Mechanical Engineering, Lehigh University. i. When the forces developed in the connecting-rod, due to its own mass and motion, are not considered, the rod is taken as simply a connecting link, serving to transmit force from one pin to the other. The pin-pressures, equal in intensity and opposite in direction, act along the same straight line. This force line, or line of internal stress in the rod, passes through the pin centers, friction being neglected. Taking account of friction, the line is tangent to the friction circles. But in any actual engine, there are certain forces developed in the connecting-rod, due to gravity and inertia, generally acting transverse to the center line of the rod. It is therefore necessary to discuss what takes place under these conditions. 2. In Fig. 126, let WC represent any rigid body, upon which acts the force Q along the line XZ. Impose the condition that the body shall be held in equilibrium by two forces passing through the points W and C respectively. These two must intersect in the line XZ; but there is evidently an infinite choice in the location of this point of intersection. Let WX, CX, be the lines of action of one pair of possible forces, P w acting at Wand P c at C. Lay off XY = Q, and find its components, AX — P w and BX= P c . Now, through A draw AD parallel to WC, making two force triangles AXD, YAD. In AXD, P w is resolved into the component AD parallel to WC and DX along Q. P c is likewise resolved into DA and YD. 232 FORCES ACTING IN AND UPON THE CONNECTING ROD. A similar resolution of P w , at W, by the force parallelogram WD'A'B', and of P c at C by CD"A"B", gives the same com- ponents as the figure XADY; and, what is more, it gives the components on their lines of action, or locates them. For, since the forces P w , P c act upon the rigid bar WC at the points W and C, their components, by whatever resolutions, must also act at these points. Considering now the components of P w , P c we have Wjy (= DX) and CD" (= YD), forming with Q a system of three parallel forces in equilibrium ; we will call these the equilibrium components, and designate them P we , P ce . Besides these there are WB' = AD = P ws and CB" = DA = P cs , which we will call the internal stress components, and which simply balance each other as they act along the line WC. 3. Since the equilibrium components are taken parallel to Q, they must be invariable as long as they act at W and C, no matter what two coincident P forces are applied at these points. From the laws of parallel forces we would have P m + P„= Q and P„ ■ WZ = P c < ■ CZ. This latter relation may also be demonstrated from Fig. 1 26 : From triangles AXD, WXZ, XD_ XZ_ DA ~ WZ' Whence — - = — — , or P we DY WZ 4. Since every possible P w must have WD' as one of its com- ponents — the other component being along WC — it is evident that a line drawn through D parallel to WC, as LM, Fig. 127, is a locus of the end of P w when this is laid off from J^as an origin. Similarly, DM' is the locus of the end of P c . Fig. 1 27 illustrates the properties of these loci, and shows clearly how the internal stress components vary as P w and P c change. 5. The validity of the preceding discussion depends upon the right to select some particular point on the line of action of a orce as its point of application, through which point all the com- an d from triangles A YD, CXZ, AD CZ DY~ XZ Pc, :: CZ : WZ. CONSTRUCTION FOR THE PIN PRESSURES. 233 ponents of the force must pass. Now forces acting upon a rigid body, as MNFig. 128, can act and react upon each other only- through the medium of stresses developed in the body-stresses which are distributed over its cross section, but whose effect in any direction must be represented by a single resultant force. To select a particular point of application is simply to make that the point where the external force meets the resultant of the internal stresses ; and obviously all components of the external force must pass through this point. But in the general case of Fig. 128, no reason can be assigned for the selection of any particular point on, for instance, the line AE; not even that the point must fall within the body. The total effect, along the line AE, of the internal stresses will not vary as the point moves ; but their com- ponent effect, in any assigned direction, will vary with the same component of the external force, equilibrium being always pre- served. It will be well to realize that, when we make the force resolu- tion of Fig. 126, as at EGKF, Fig. 128, and call AB the line of internal stress, that only a part of the internal stress effect acts along AB ; there are also developed stresses, closely analogous to those in a beam, under the action of the parallel forces Q, P we , Pc, Fig. 128 is intended to illustrate choice of line of internal stress (for the force resolution of Section 2), which exists in the general case of any rigid body in equilibrium under the action of three forces, showing two out of an infinite number of possible cases. It is evident that we may impose some condition, as for instance P we shall act along the line WD, which will at once determine the point of application cf P w . Construction for the Pin Pressures, when Friction is Neglected. 6. In Fig. 129. the resultant of weight and inertia of rod is sup- posed to be known, and is represented by XY = Q. The pin- pressures must pass through the centers of the pins ; the force 234 CONSTRUCTION FOR THE PIN PRESSURES. diagram for determining them will be constructed at W; for P w , besides balancing its share of Q, must also be the resultant of the known effective horizontal pressure on the wrist-pin and of the guide reaction whose direction is known ; these two conditions determine it fully. Lay off WN, parallel, equal, and opposite in direction to Q. Let it be divided at L, so that WL : LN :: CZ : ZW; then WL is the equilibrium component of P w . The line LM, parallel to WC, is, in reference to the point W, a locus expressing graph- ically the first condition, as is shown in Section 4. Now measure off WV = effective horizontal pressure, and draw VM parallel to the guide reaction ; then VM embodies the second condition; and the intersection of these two loci at M fixes WM as the actual pressure of wrist-pin on connecting-rod. The line MN, completing the triangle WMN, is the crank-pin pressure, whose line of action is CX. Pin Pressures when Friction is Considered. 7. In Figs. 130 and 136, the circles around W and Care the friction circles of the respective pins, greatly exaggerated. The mechanism is simply outlined. The pin pressures act tangent to these circles ; they must, of course, hold Q in equilibrium. But the lack of a point common to all the forces possible of action at each pin, and which may be taken as their point of application, makes the graphical expression of this condition much more difficult than in the preceding case. Fig. 130 illustrates the construction of an exact polar locus of the force P w /, the same in its conditions as LM, Fig. 127. It is obtained by drawing pairs of pin-pressure lines to various points on the line of Q as X It X 2 , X 3 , and resolving Q along these lines ; then laying off from W 3S> a pole the various P me 's thus found and drawing a curve DD through their ends. This curve has some interesting properties. If we draw WD parallel to Q, it corresponds to the case of pin-pressures parallel to Q. Note how WL differs from WL in Fig. 129 (the figures CONSTRUCTION FOR THE PIN PRESSURES. 235 being alike, aside from friction). The branch DD„ to the right of D corresponds to the case of rod in compression. It reaches 00 when the pin pressures are conceived as acting along the line EF, tangent to the friction circles. For the left-hand branch DD V rod in tension, the similar limiting case is the force direction KH. If the direction of motion of the mechanism were reversed, xhe entire locus would be changed. 8. To establish an approximate construction of practical utility, we impose the condition that the equilibrium components of the pin-pressures, parallel to Q, shall pass through the pin centers ; in Fig. 131 their lines of action are WL, CU. This makes them the same as in Fig. 129. Now draw from any point X on the line of Q the pin-pressure lines XR, XU; the intersections R, U, become the points of application of pin pressures, through which all their components must pass, and RU is the line of the inter- nal stress components (see end of section 5). If we were to draw through Wn line parallel to RX and measure off the length of P W f : then a parallel to RU, through L, would pass through the end of /"^(compare WM, LM, Fig. 129), and at that point it would agree with the locus DD of Fig. 130; so that we have here stated, in a slightly different form, the method of obtaining a point on that locus. When the force Q of the rod is not taken into account, the internal stress line is ST, tangent to the friction circles, as stated in Section 1. The approximation which we now make is to consider that this line ST practically agrees in direction with the variable line RU, which in the actual rod is true within the limits of graphical accuracy. Then LM, parallel to ST, becomes a locus exactly similar to LM of Fig. 129 or of Fig. 127. The pin-pressure construction is now the same as in Fig. 129, differ- ing only in that the guide reaction is inclined from the normal through the angle of friction. If the rod be in tension the locus is LM' parallel to a tangent drawn like KH of Fig. 1 30. Note ■that in Fig. 131 the actual lines of action of P w /, P c / are not shown. 236 THE ROD FORCE AND ITS COMPONENTS. Drawing this locus M'LM'm Fig. 130, we can get a good idea of the error of the approximate method. It is evidently greater when Q is large relatively to the force transmitted by the rod. The lines LM, LM' are asymptotes to the true locus; for at 00 RU agrees with ST and the construction is exact. In Fig. 131 the condition was stated that the pin pressures must be tangent to the friction circles. To substitute ST for RU as the line of internal stress is to make 5 and T the points of application of all pin-pressures found by the approximate loci, thus making these pressures differ slightly from those found by the exact construction. The approximation to tangency may be easily estimated by inspection. The Rod Force and its Components. 9. Referring to Sections 6 and 8, it appears that, in order to construct the pin-pressures it is necessary to know the total rod force and its parallel components at the pin centers, the latter being a function of its location on the rod. In Fig. 132 is developed a method for determining these quantities. Refer to pp. 93-100, and foot note p. 44. In Fig. 132, WCO represents the mechanism. CO is taken to if represent the acceleration — of C; and WO, equal acceleration R of W, is found as in Fig. 16, page 41 or by Eq. 36. Now by producing Cw to W , making CW equal CW, and completing triangle CW A similar to triangle CwO ; then revolving it into the position CWA, we find the center of acceleration in A. ■wO CO (For ~ = ~, and angle A WO = angle A CO.) Cw, being the acceleration image of the rod, we have only to find center of gravity, g, of the image from G, and draw gO, to get the intensity and direction of the acceleration of the center of gravity G of the rod. Again, letting OC represent to a suitable scale of forces, the inertia (centrifugal) force which the mass of THE ROD FORCE AND ITS COMPONENTS. 237 the rod would have if concentrated at C; then to the same scale, Og represents the actual inertia force of the rod. 10. To locate this force we proceed as follows : The actual distributed mass of the rod may be replaced, for dynamic purposes, by two simple concentrated masses m z , m 2 , located on any straight line through the center of gravity, at the distances k It h, from the latter ; the conditions being — 1. w, + m x = M, (same mass). 2. mJi T = mji^, (same center of gravity). 3. m-th? + mjf x = Mk 2 , (same moment of inertia). k = principal polar radius of gyration, and k 2 = hji x . The points of mass concentration chosen, and their accelera- tions found from the image, then the inertia forces of m^, m^ will have the direction thus determined ; and through the intersec- tion of these inertia forces, components of the total inertia force of the rod, the latter force must pass. If we choose one mass point so that its inertia acts along the line WC; then the direction of the inertia of the other mass point is of no importance, for it will intersect the first force in its own mass point, and the latter will serve to locate the resultant, or total inertia of rod. Construction : make angle WAB = angle CWO = a, so that angle ABW= angle A WO; whence accelera- tion of B is along BC; at G erect GK perpendicular to rod, and equal to K; complete the right-angled triangle BKJ, making BG X GJ = GK 2 , then J is the second mass point, through which must pass the inertia force QE = Og. The weight and inertia being combined at Q, their resultant g^cuts the rod at P; and QF, divided in the ratio CP: PW\s what we wish to obtain, but by a more compact and convenient construction. 11. Lay off OO = weight of rod, then Og = QF '= total rod force Q. Construct on the image Cw a figure similar to CBKJ; determine b by Ob parallel to CW; lay off kg in proper ratio Cw, and find/; draw// parallel to wO, and J'e parallel to 16 238 THE ROD FORCE AND ITS COMPONENTS. Cb, then ge : eO — Cj : jw = CJ : JW. That is, the inertia force is divided into its parallel components at the pins. Now divide the weight 00' at T so that OT : TO :: CG : GW, separating the weight also into its parallel components at the pins. Starting from 0', combine OT and TR (= eg), the wrist-pin components of weight and of inertia. Their resultant O'R is a possible wrist-pin pressure P m (acting in the direction "rod on pin "); its end R must therefore lie on the P w locus of which 0' is the pole. This locus is RS parallel to JVC, and it cuts from Og the wrist-pin component O'S. (Compare Figs. 126 and 127.) This completes the solution of the problem. Cw changes with the crank position, and with it kg. To avoid making repeated computations for the latter, we can make part of this image construction on the crank OC instead of on the accel- eration image Cw. Project b to B' , lay off G'K' in proper pro- portion to CCand obtain^'. The entire construction is now in- dependent of the rod, except for the direction of the lines Ob, RS. 12. Fig. 133 shows the whole construction for one crank position, and will serve to illustrate the following instructions. Having necessary data, including weight, center of gravity, and radius of gyration of rod: F 1. Draw circle ACD, with radius = — "-- for rod, to a conve- A nient scale of forces, and space off C points. 2. From p. 37 compute and lay out Ow ; draw images Cw. 3. Strike circle HG, G being center of gravity of crank image of rod ; obtain g by projection ; lay off 0T0' , draw Og, Og. 4. Draw GK once; then strike circle LK, and from each G locate the corresponding K on this circle. Om R 5. Rod angle construction : — — - = — ; C cuts circle mpna± & OA L r p\ project p to u ; u is parallel to rod, for crank position C (and for 180 — C); keep this construction below the center line AO. THE ROD FORCE AND ITS COMPONENTS. 239 6. Having b project to B, draw BKJ, Je. 7. Lay off OE = eg, ER = OT; draw RS parallel to uO. To avoid confusion : (a) Complete each step in the above for all the crank positions before beginning the next. (b) Mark the points C, w, G, g, O, T, K, b, BJ, e, E, R, S, as found, by inking a small circle about each, the first six and last three in black, the others in red. Letter one construction fully, and number each set of points with the degrees of crank angle, from o° to 360 ; likewise, designate the rod line uO with the angles to which they belong. {c) Draw as few lines as possible; thus, on upper half of figure, CO need not come inside of circle HG ; below, of circle mu ; ub need not come to O ; projection lines, as Gg, bB, BK-KJ, Je, need not be drawn in. 13. Fig. 133 completed, the pin-pressures are found by Figs. 134 and 135. Fig. 135 is drawn first. To as large a scale as paper will allow, strike arc AB, with rod-length L as radius; draw circle OC with radius equal R; put in crank posi tion OC, project C to C; WC is rod angle, and ST is line of internal stress, with friction. The play of forces, as affecting tangency of ST to friction circles, must be found from the diagram of effective horizontal pressures on wrist-pin ; modified at critical points by the inertia of the rod itself, as affecting direction of crank-pin pressure. It is possible to have one end of the rod in tension, the other in compression. Mark each ST with its crank angle. Fig. 134, is the same as the pin-pressure diagram of Fig. 131, except that all the forces are reversed, so as to give pressure of rod on pins rather than of pins on rod. Bunch the constructions in four groups on Plate 7, one for each quadrant of crank circle. NLWis O'Sg to reduced scale. 240 THE LINK MOTIONS. APPENDIX D. The Link Motions. In this set of valve gears, the eccentrics, the throw and location of each eccentric, are invariable. These Link Motions occur principally in the locomotive, which, to be sure, is a species of high-speed engine, but has so many other special problems con- nected with it that it cannot be fully treated within the limits of this work. Nevertheless, in order to give a certain degree of completeness to the general subject of valve-gear design we will here give an outline of the Kinematics of Link Motions. The Fink Motion, as used in the Porter-Allen Fngine, has already been discussed in the body of the text. In the valve gears known as Link Motions the same sort of steam distribution occurs as in the gears having a " Swinging Eccentric," that is, the locus of the eccentric centers is of the same general character. But the mechanism employed is very different, the adjustment of the Link Motions by hand causing the invariable eccentrics to produce the desired variations of valve motion. About the only case in which the adjustments are auto- matically made by the governor occurs in the Fink or Porter- Allen Motion. Dr. Zeuner found, in his well-known treatise on Valve-Gears, this locus of eccentric centers by analytical means, but it may be found much more easily graphically and, so far as practical ap- plications are concerned, with the same degree of accuracy. The basis of the graphical treatment is Fig. 66, which contains the method of finding the virtual eccentric for a valve motion whose stroke does not pass through the center of the shaft. We will here use the figure and demonstration given by Pro- THE LINK MOTIONS. 241 fessor A. Fliegner in his work " Die Umsteuerungen der Loco- motiven.*" Fig. 1 36 represents what may be regarded as the general case of all the link motions, extensively used in practice, which impart Fig. 136. to the valve a motion like that which it experiences when directly driven by a single, simple, eccentric. This figure and Fig. 137 show how two eccentric motions may be combined so as to be equivalent to the motion of a single eccentric, called the virtual eccentric. This, indeed, is the gist of almost all the link-motion problems. A'B' is regarded as the link provided with slots at the ends to permit the eccentric ends A' and B' to move in the perfectly straight lines, g T and g v The point C on this link directly drives the valve and is itself assumed to move in a straight line g, which is parallel to the lines g z and g 2 . The rods are of unequal lengths, Aj and A 2 ; the eccentricities p z and p 2 are also unequal and have the unequal angles of advance d t and d 2 . The position shown in full lines corresponds to crank position at left dead point ; the * Dr. Burmester has also given an elegant kinematic presentation in his " Lehrbuch der Kinematik," pp. 696-702, from which some of the special cases can be more rigorously deduced. 242 THE LINK MOTIONS. broken lines of the figure correspond to the position of the crank at the right-hand dead point. Here it will be well to note the difference between link-motions with open and with crossed rods. When crank is at dead point and eccentrics are placed between engine shaft-center and link A'B' , then the mechanism is said to have crossed rods if the ec- centric rods cross in this position and open rods if the eccentric rods do not then cross. Fig 1 36 is an example of the open-rods variety. Take A and B as the centers of travel of points A' and B' and join A and B ; then will AB pass through the intersection 5 and be the reference line from which the travel CO of any point C THE LINK MOTIONS. 243 on link A'B' can be estimated for any crank position. It is evi- dent from the figure that the travel CC at any instant is CO = A A' + -J— (BB' — A A'). J +9 Let us take these eccentrics 0K\, 0K\ separately and on a larger scale construct in Fig. 137, after the method of Fig. 66, the virtual eccentrics OK, and 0K 2 , whose distances j "C 1 "? 1 I from the vertical line OY are respectively equal to the travels AA' of point A' \ „. , _ , ,,. BB' of point B ) ' see Fl & J 3 6 - To do this we mak « K\ K 3 = j 1 X 0K\ and K\K,=—^t-X OK', for the case of open rods, and K\ *, = A7, ^ = — ■£- X ~0K\ and lC.k, = K',K = + ~ X 0K\ for crossed rods. Now join isTx and K 2 and divide at AT the distance K,K 2 so that the ratio KK _ i KK q in other words, in the same ratio that the valve-stroke-line g, (Fig. 136) divides the total distance between g, and^" 2l the lines of travel of the eccentric pins. This will make the distance KS of point AT from the vertical Y equal to KS = KS t + (KS, — K t S z ) -J— = AA'+ (BB'— AA)-J—. j + q j + q When the crank turns into some other position, the distance between pins A' and B' in Fig. 136 is still divided into the same ratio — — — by the line of stroke £• of the valve, and in Fig. 137 J -f- q the triangle OK KK, goes into some other position OE, EE„ 244 THE LINK MOTIONS. where the distances of E x and it 2 from F equal the corresponding distances of pins A' and B' from their centers of motion A and B, and the distance is from OF is equal to the distance of the driv- ing point C of the valve from its center of motion C. The complicated motion with which we started has therefore been reduced to the simple motion produced by a virtual eccentric OK = OE. We will now show the applicability of this method to Stephen- son's, Goocb's and Allan's link motions and will assume the reader to be acquainted with the general kinematic features of these mechanisms. In each of these motions the distance be- tween the lines g z and g 2 (Fig. 1 36) remains unchanged. In Stephenson's motion the block or driving point O keeps its line of stroke g, but the pins A' and B' change their lines of stroke g z and g t to some other parallel position in such a way that the distance c x -f- c 2 , between them remains the same, while the ratio — is changed. As c, and r 2 change the virtual eccen- tries 0K t and 0K„ will change; the size and position of OK depend on this change and also on the ratio of j to q. The locus of K for different positions of the link we will construct later on. In Gooch's motion the link is stationary, the pins A and B' keeping their lines of stroke g x and g 2 . The locus of the center K of the virtual eccentric OK will therefore for this mechanism always be the line K X K % because c z and c t do not change. The position of K on this line will depend on the ratio — . In Allan's motion both link and block are shifted and in oppo- site directions. Each of the three lines, g It g and g a changes its place, the values of c lt c 2 ,j and q change but so that c z + c 2 = j -f- q = constant. The locus of the center K of the virtual eccentric depends on c x , c 2 ,j and q and is found as before. The Fink or Porter-Allen link motion may be regarded as that particular case of the Gooch in which the two eccentrics THE LIMK MOTIONS. 245 are equal and have an angle of advance of 90 causing them to coincide and thus reducing it to a link motion with only one eccentric. (The link itself then becomes part of the eccentric strap and may be regarded as a bent lever rocking on a vibrating fulcrum). In the actual link motion this would have the effect of causing the two eccentric rods and the link to form one rigid piece. This would prevent any angular motion between either rod and the link and, in the Fink, the link would therefore receive the whole of the rocking motion due to the throw of the single eccentric. The result is to make the position of the locus of the centers K of the virtual eccentric in Fink just what it should be if it were strictly a special case of the Gooch, but the location of K on this locus cannot be found without adding still another approximation to the series, namely, to substitute for length i. of the eccentric rod the average distance A of the trunnion of link from the center of either eccentric. This substitute A will generally be shorter than I and gives a greater throw by making ordinate ~iP^jP- The result will then agree with that found by analytical investigation.* In constructing the locus of the centers of the virtual eccentric and the diagrams of the four grades of expansion for forward motion we will use the examples of the Gooch, Stephenson and Allan gears taken by Dr. Zeuner for the same purpose in his Valve Gears. Gooch's Link Motion. Dimensions r = 2}i", d = 20 , 2c = c z -f- c x = 12", / = 48", e = 0.91" and i = 0.23". Let the * This is a doubtful derivation of the Fink throw from that of the Gooch, for which only the writer is responsible Dr. Burmester preferred to obtain it independently. Below we have given a simple analytical proof of the travel of the valve in the Porter-Allen link motion. 246 THE LINK MOTIONS. four grades of expansion correspond to the four cases in which block {i.e., line g) is at %c, y 2 c, %c and c from the line of centers Fig. 138. =6", J % and KJKi == --^, ><,^ando. CUT (Fig. 1 36). In Fig. 1 38 c T = c, 0.297 = K 2 K 2 . The value of the ratio . Make the offsets K z K r ' = iT 2J £ 2 ' = 0.3", join ^ and K x then divide iT,^ into four equal parts and draw 0E Vi , 0E V2 , 0E Vy and OK*. The link K T K 2 is the locus of the centers of the virtual eccentric for forward and backward motion and 0£ Vl , OE Vi , 0E Vj and OJC z represent in length and position the virtual eccentric themselves for the forward motion when the crank is at the dead point. The diagram thus found agrees exactly with that obtained from Dr. Zeuner's approximate formula. The diagram shows that lead nX constant. The locus for crossed rods is shown at k z k 3 . THE LINK MOTIONS. 247 • Porter-Allen Link Motion. The relation between the Fink and the Gooch motion was pointed out above, p. 245. If the angle of advance of each eccentric is made equal to 90 , then the two eccentrics OK x ' and OKJ reduce to one OKand the locus K^K 2 of the virtual eccentrics takes the position K'KK". For the reason already given, p. 246, the offset KK' = KK" for full grade should be greater than K T K\ = K 3 K' 2 = c - p. According A to Zeuner it should be -j p for the same grade, where A is the distance from center of eccentric to center of trunnion ; (this trun- nion in Porter-Allen gear joins sustaining arm and link on the center line of the slot). For any other grade of expansion Zeuner It gives offset = -= p, where u is the perpendicular distance of block above line of centers OX. When — <4 we should plot positions of block to ascertain the valve travel but when — > 4 we may for the Fink gear, as real- ized in the Porter-Allen link motion, determine the travel £ as follows : Consider the total movement of link to be composed of two parts, a horizontal motion parallel to itself which is due the horizontal throw of the eccentric and a rocking motion about the trunnion (or vibrating fulcrum) which is due to the vertical throw of the eccentric. These two motions together cause the block to move a distance 6 equal to the sum p cos co -f- -j p sin co — c. Here the first term is the horizontal throw common to all points of the link and the second term is the vertical throw p sin co of u the eccentric X the ratio —. of the arms of the bent lever consti- A tuting the link and rotating about the trunnion. 248 THE LINK MOTIONS. The formula just given is the polar equation of the valve-circle and the coordinates p and — oof the vertex of its diameter show A that this vertex lies on a rectilinear locus perpendicular to the line of centers and at a distance p from the center of the shaft, which agrees with the result so irregularly deduced above. Allan Link Motion. Dr. Zeuner gives the following dimen- sions of an existing valve gear with crossed rods. Eccentricity, p =-. 2.75". Angle of advance, d = 30 . Length of each eccentric rod / = 49.2". Half length of link, c = 6". Length'of the radius rod, X z = 60"- Distance of trunnion of hanger \ , ,, of radius rod from valve stem / ° Short lifting arm of the lever, a 1 = 2.88" 1 Long lifting arm of the lever, b 1 = 6.81" i Outside lap, e = W. Inside lap, z = &". Ratio 7- of block to link movement, — — - X - ='2.9. u z d h Ratio of distance u of block from dead point of link to distance u z that link is shifted from line of centers is y = 3.9. For the four grades of expansion, Corresp'd'g amount of link shifting * 7 0.256c 0.192c u=y 2 c 0.128c u=-%c 0.064c u=o 0.0 Since c^= c — u z and c, = c + u r approximately, jP are 0.249 0.270 0.292 0.314 0.335 the correspond- c% ing values of 1 P m ° 42 ° °^ 99 °^ S °^ 6 °^ - are o * I | i THE LINK MOTIONS. 249 Lay off K{k x = 0.25 on a line perpendicular to OKI, also K 2 'k 2 = 0.25 on a line perpendicular to OKJ; the points k t and k 2 will be points on the locus corresponding to u = c, or full gear forward and backward respectively. On each of these two perpendiculars lay off from K x ' and KJ, in the direction Klk lt and K a 'k 3 the distance 0.335, ano * join the ends of these distances; where this connecting line cuts OC will be the points p of the locus corresponding to mid-gear. This gives us three points k It p and kz on the desired locus ; a circle through these three would closely represent the curve but the arc is so very flat that its center would fall off the drawing. We will, therefore, construct the arc. The lines are so close together that only the construction for half gear forward is shown. We make the offset at K z ' equal to 0.292 and that at KJ equal to 0.378 and join the ends of these 2 SO THE LINK MOTIONS. offsets. This connecting line is divided at D v ^ in the ratio of i to 3, and therefore D v ^ is another point on the locus. The other points are found in like manner. This locus can be shown to be a parabola and for the present case of crossed rods is convex to the center 0. For open rods the locus is K Z P for forward gear and is again a parabola but is not exactly like locus k*p for forward gear and is again a parabola but is not exactly like locus k^p for crossed rods. We have however TP=Tp and K.'K, = KJK The locus for open arms is concave to center 0. To find it we will use, without proof, a simpler demonstration given by Burmester. On the diagonal K^P, joining the known points AT, and P, construct the rectangle K^QPS; draw the second diagonal QS. To find points of the locus on the parallels through y lt y 2 and y 3 we draw through these points parallel to QS to their intersection with X T Q at q z , q a and q 3 respectively. The intersections of the lines Pg It Pq^ and Pq 3 with the corresponding parallels will be points on the desired locus. The valve-circles have only been drawn for the case of crossed rods, forward gear. They show but little variation in the lead. The corresponding eccentric positions, when crank is at dead point, are shown on the other side of the figure. Stephenson's Link Motion. The dimensions taken by Dr. Zeuner for his diagram : p = 2.36", d = 30 , A = 55", c = 5.9" e = 0.94" and i = 0.27". We will consider the grades of expansion represented by U=C, U = "i/^C, U = ]/ 2 C, U z= I^V, U = O for open rods. Since c z = c — u and c? = c -\- u we have the corresponding values of jP = o 0.064 0.127 0.191 0.254 ~p = 0.508 0.444 °-3 Sl o-l 1 ? 0.254 THE LINK MOTIONS. 251 K t ' and K a ' are points of the locus of the centers of the virtual eccentrics which correspond to u = -+- c for full gear foward and to u = — c for full gear backward. To find the other points on the locus, make the offsets K\t v K' T t 2 , K^, K\ X z respectively equal to, 0.06, 0.13, 0.19, 0.25; also on the perpendicular k 2 K, make Fig. 140. the offsets K' s 3 , K'a, K' 2 s r , K^R* equal to the corresponding values 0.44, 0.38, 0.32, 0.25. Join j, and t lt s 2 and 4, s 3 and t y — , on sj, make UE„^ = — 8 4 Ki and K 2 . On s 3 t 3 make t 3 E„ 3 and on sj x make A E t>1 = }£ s^. The points E„ 3 , E V2 , E„ z and the mid-point P of ^^ are points on the desired locus. OK\ is at once actual and virtual eccentric, while 0E vy 0E Vl , 0E Vl and 252 THE LINK MOTIONS. OP are all virtual eccentrics corresponding to the given grades ol expansion for forward motion. The locus K\PK' % for open rods is concave towards center 0. The locus K'ipK'z for crossed rods is convex towards center 0. The two loci are exactly equal and symmetrical to the line K\K' 2 . That for crossed rods can be found in the same manner as the one for open rods, but we have preferred to find it by a much simpler method given by Burmester,* omitting the proof. In the lower part of Fig. 140 on the diagonal connecting the known points K\ and/ we construct the rectangle pQK'JT and then draw the second diagonal 7& 2 . To find points of the locus on any parallels y x , y^ and y 3 we draw through points y x , jc 2 and _y 3 parallels to Tk 2 letting them cut Q-K' ' 2 in q It q 2 , q 3 . Then the intersections/^, with parallel y lt of pq? with jp 2 , of/y 3 with j 3 will give three points on the desired locus. Of course the locus for open rods might have been found in the same manner. Drawing the valve-circles on the corresponding diameters OD^, OD v , 0D Vi , 0D VV OD Vo and the lap circles we get, in the usual manner, the steam distribution for the different grades of expansion. The varying lead, nm Q , nm lt nm^, nm 3 and nm^ is particularly noticeable. We will show later how this variation may be reduced for this link motion. It is however not so serious a defect as has been commonly believed. Porter-Allen Link Motion. Dimensions are taken from the 1 1 y 2 x 20 engine shown in text, Figs. 79-82. Here p = 1", A = 6", d = 90 , c=6", £=iA, 2=xV. The engine "runs over" and the expansion and exhaust valves are both of the positive direct type. Between the link and expansion valves there are reversing levers whose arms multiply the motion of the link four-thirds. The diagram should be constructed as if the valve were driven by a correspondingly enlarged link-motion. There is also a lever between link and exhaust valve, but this does not reverse * This method and its demonstration are given in the " Lehrbuch der Kine- matik " p. 722. For another easy method see Fliegner's " Umsteuerungen der Locomotiven,'' p. 81. THE LINK MOTIONS. 253 the motion, it reduces the travel in the ratio of 4^ to 1 iH. The exhaust valve does approximately follow the law of eccentric mo- tion but the steam valve does not do it as well, for the reversing is effected by a bent lever of such an angle as to effect a differ- ential motion like that of the wrist-plate of the Corliss engine. This valve motion has been satisfactorily discussed in the text, see Figs. 83-91 inclusive, and the accompanying notes. For the purpose of comparison however the graphical methods of the preceding link motion may be followed. Then for the grades of expansion : u == c u= ^c u = y^c u = y^c u = c we have c x =Cz — c and the corresponding values of ~ — : J + q o y, y # y 2 Then on a scale of - make the offset — -p = -". 1 3 A 3 Adjustments of Link Motions so as to Equalize Motion. "The requirements of a perfectly equalized link motion are: perfect quality of cut-off, of exhaust closure, lead opening and maximum port opening, together with absence of block-slip, be- tween the forward and return stroke of the piston for every sus- pension of the link from full gear forward to full gear back. Such theoretical excellence is absolutely impossible with ordinary link motion. But good practical qualities may be obtained by sacri- ficing the non-essential to the essential points of the motion." Auchincloss, from whose well-known work these remarks are quoted, then proceeds to lay out the motion by making the leads equal at mid-gear and the cut-offs equal at half and full gear for- ward and back, thus indicating that these elements constituted the essential points of the motion. At y, cut-off however the condition favorable to minimum slip is introduced. The lead at full gear and the slip of the block then existing in the designed 17 254 THE LINK MOTIONS. link-motion are examined and the directions in which modifica- tions should be made are suggested.* Zeuner, Burmester and Fliegner, on the other hand, seem to lay greater stress on equality of lead at the at ends for each grade of expansion and on the reduction of the slip. As these link motions are usually accompanied by long connecting rods the inequalities of cut-off due to the angular motion of the rods is not very great and we believe them to be comparatively unim- portant, particularly at high speeds. We shall therefore follow the presentation given by Drs. Zeuner and Burmester. There are two leading principles or conditions: — i. A regular and correct distribution of the 'steam requires that the valve should move symmetrically on each side of a point (the center of motion) that remains fixed for all grades of expan- sion. 2. A minimum slip of block requires that the trunnions of both the link and the radius rod should have as little vertical motion as possible, in other words, the chords of the arcs de- scribed by the lower ends of the hangers should be parallel to the central line of motion and the hangers themselves should be as long as possible. The first condition is satisfied in Gooch's motion by making radius of link-slot equal to length of the radius-rod; in Stephen- son's motion by making the arc of this slot equal to the length of the eccentric rod; and in Allan's motion by making the arms * To reduce the slip of the block if it should be excessive, Auchincloss recommends alterations in the following elements, (enumerating them in the order of relative efficiency), (i) Increasing the angle of advance, (2) "Reducing the travel of the valve, (3) Increasing the length of the link, (4) Shortening the eccentric rods. When the proportions are unusual, as in link-motions for small launch-engines, considerable deviation from the standard arrange- ment is profitable, the irregularity introduced curing greater irregularities in the more important functions. In such cases it will always be well to check and correct work by a model. THE LINK MOTIONS. 255 of the lever carrying the hangers (which move link and radius- rod) bear to each other the ratio - = y ( i + -J i + - ) . In all three motions the effect is to give equal leads at the two ends for each grade of expansion, but the lead is not constant for all grades of expansion, except in the case of Gooch's motion. In Allan's motion the lead is slightly variable { de^sTnl } from ful1 S ear to mi d-gear in case of { c ^d } rods. In Stephenson's motion there is still greater variation and in the same direction as in Allan's gear. Dr. Zeuner has shown however that giving the eccentrics un- equal angles of advance would tend to correct this. A simple way of doing this is to change the angle between crank and eccentrics by moving back the crank when link drives the valve directly and by moving the crank forward (i. e. in direction of rotation) when there is a reversing lever between link and valve. The amount of this change in the setting is about 5 , or more exactly, it is the angle whose tangent is equal to y 1 . When this modification is introduced the Stephenson becomes the simpler and more compact gear of the three and equally efficient. The second condition is not satisfactorily fulfilled in the case of the Allan motion as ordinarily constructed. The upper ends of the hangers should move in parabolic curves that are convex to each other. Instead of that they are guided in circular arcs that are con- cave to each other. In this case make the hangers as long as pos- sible, make total length of lever a-\-b=l 1 — l a and give fulcrum of lever an abscissa (measured from center of shaft) equal to I A- a and an ordinate equal to length of hanger. 2I 256 THE LINK MOTIONS. In Gooch's motion the second condition is met by placing trun- nions of link at center of chord of the link-arc are utilized* and upper end of the hanger is placed so as to have an abscissa X r (estimated from center of shaft) and an ordinate ^ = length of hanger. The abscissa is obtained by placing link at mid-gear with center of chord at points on central line of motion that correspond to the dead points of crank and then bisecting the distance between them. The condition is also fulfilled for the hanger of the radius-rod by finding the abscissa of point of suspension of the hanger in a similar way, namely, by placing crank at both dead points (for each grade of expansion) and the radius rod in corresponding positions (which will be nearly parallel) and then bisecting the horizontal distance between the two trunnion positions of the rod. The ordinate will be the length of the hanger and, erected perpendicular to the central line of motion at the point of bisection belonging to each grade of expansion, will give the locus of suspension for the uppper end of the hanger of the radius-rod. This locus is a circle with radius ^ struck from a center which is at the distance k 3 above the mid-position of end of valve stem. In Stephenson's motion the second condition is only partially satisfied by the method of suspension usually employed in American practice. Concerning this matter of suspension the authorities differ according to their view of what constitutes an excellent gear. Dr. Zeuner's rule, of guiding the upper end of the hanger in the arc of a parabola, (having a parameter 2A and a c 2 vertex whose coordinates are A and ^ 2 ) or in the arc of a 2/ * Dr. Burmester chooses a different point than middle of chord for the trunnion. He finds that each point of suspension for hanger, on the vertical having the abscissa 1 — — . , has a particular trunnion point on the link-chord which will make the center of that chord travel most closely to the central line of motion. We do not reproduce the construction here, referring the reader to p. 711, Fig. 704, of Burmester's "Lehrbuch der Kinematik.'' THE LINK MOTIONS. 257 circle with radius A, , is probably best for maintaining equality of lead at each end of cylinder. Dr. Burmester's way of plotting the locus of suspension is to place the driving point of the link corresponding to each grade of expansion on the central line of motion when the crank is at dead points, then join the trunnion points for these two link positions and bisect their distance apart by a perpendicular which of course passes through the point of suspension. This way of doing it makes the locus an f shaped curve which can be approximately replaced by a straight line and its effect will be to keep the slip of the block very small. On the other hand, Auchencloss's method of sus- pension keeps the cut-offs equal, the slip being minimized only for half cut-off; if in this case the slip should be regarded as excessive, the alterations mentioned in the foot note on p. 254 can be undertaken. The principal steps in the laying out of a Stephenson link motion are: I. Find position of center of rocker from mid-gear travel of link. II. Locate cut-off positions of rocker arm and find lap of valve. III. Find position of stud (or trunnion) on saddle for one-half cut-off, lor maximum cut-off and both for the forward and back- ward motion. The lines connecting the stud positions for the same cut-offs should be as nearly parallel as possible. IV. Locate the tumbling shaft for accomplishing an equalized cut-off in all positions ; (the tumbling shaft is the approximate center of the locus of suspension). V. Find and examine the lead at full gear for both strokes, also the extreme travel and the slip of the block. VI. Modification of the results by various expedients. All this work is graphical and is made with the help of a link template having on it the central arc of the link slot and notches representing the centers of the joints connecting link with eccentric rods.