º is º.º. tº º: rº *...* ºr º, º f * ºº:: * * * * : * * * * * * * * , ºf tº ºr gº º Aº Aº gº º * : * : , ; ºr f : * :: f ſº * * * * * * * * * * * * ; : * * * * * : * * * gº * : * : * * * * * * * * *.*.*.*.*.*.*.*.*.*, & tº - * * * * * * * * * * * [* * ; : sº. # ſº sº 2- gº g : * * : [' : ; # * * : * : * : * ~ * - * * * * : * : * * * * * : { & º * * : * : is º.º. ºf a * ... s. 3 ºr ºf ºr ; ; ; , ºf A . º. º f : P ºf ºf a p * * * * * * * * * * * * : * * * * * * * * * # * : * * ºf . º. . . . ; ; : A *. º º ºr * * * * * * * * ºf º 'ºZ. sº ſº º * : * : * : ; :: *. sº tº fºr tº º *.*.*.*.*.* . . . . . . . * * : * * * * * * * * & # * * : *, *. sº gº * . . * * º -- - 㺠º ſº. º.º. º gº ºf r ºr a ſº. 4 º- is a tº 4 is . . . . . . .º.º.º. º z.; ºr º, ſº º ſº, º ºr ſº a gº ºf . J. º f : º, ºf . º.º. * + 2 ≤ . . . . º. º. º.º. ºf . º. º. º. º. º. º. º. º sº :: * : * * * * * * * *.*, *.* * †. # 1. º. iſ ºr x * * * º ſº * * tº ºf ſº ſº. º. i. i : º **, * : * º : " : -- * c § º * * º > * * * : * : * * * : * : : - - * : * ~ * （ ; ;-- * * * : *, + ... & . . . . . . . . . . . . * : * : * : * * * sº iſ * : * * * * g. s r. º. ºf ſº, ſº 8. tº º – * : * --~~ * . . . . ; # * : * : *...* ; * * * : * : * : * : * * * * * : * * : * : * : * * * * * * ... * * * * * * * * * : * * * * * * • * * * - * * * * * : 3 & # , º f : . . ; * : * * * * * * * º §º - º sº, e - # ºf º º * gº v . . . ; ; & i. º. º 'º ºf * : * * * * . . . . . --- ... º. º. & tº. . . ; * : * ºf zºº -º-º-º: * º º º ºg ſº * . . . º gº º :* - & º f tº º - .* “º w - “. . . tº ºr ºf . - * : * * * * ºr sº ºf , ºr sº º żº º: & * : * : * * * * * : * ºr ºf , º ºf ºr ºf -ºº ºf is r ºf a wººz & a sº gº, º sº. ºr . . . . . . . . . . . ºf ºr ºf . º. 2 . . . . . . . . *- : * ~ * * * * * * *. & & *.*.* ºf i. - - -- ºf a ºr ºf ſº, gº ºf . g . * * * * * # , ºff. . . . g ... º. ºr ºf ºf ºr gº ºf 4 s : . . . . º.º. º.º. A 3 * g º a . . . ºf . . g . . .” * = F * * * * * * * * * * * * * * 8 a. # * : * ~ * * * * : *, * : tº º * * . . . . . ; ; ; ; , ; ; ; # ºr & . º. º. 6 ~ º º -- # . . . . . * * * * f * , ; ; * . . . . . . . . . . º. º.º. ºf . a 2, º f *.*, *, *, * * * - aſ a , º sº a i. f : . . . . . . . . . . . . .º.º.º. *** º, º: º - ſº tº * * : * : * * * * * * * * * * * * * * *.*.* * * * * * * *.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*, * a' s r. º. 3 º' tº sº ºr r * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ºr º' ºf & . . ; * * * : * r * * * * * * * * * * * * * * *.s.º.º.º.º.º. # ſº sº r, ſº * * * : * * * * * * * * * * *, *, *.*. : - d º º, ſº P. º.º.º.º. 3 º * : * - <--> g ~ 5 * *.*.*, *, *.*.*.*.*, * * * * º' ºr º, 6 : *, * * * : *.*.* * * *.*.*.*.* : * * * * * * * * * *, *, *, * tº & Pº ºr ºf , P. A., & & , i. 3. * * * * * * * * * * * * * * * * * * *.*.*.*.*.*.*, *, * * * * * * * * * *.*.*.* ſº C. º. º,” *: P : - * * * * : * : * * . . . . . . . . . . . . .” * . . . . . ºf h . . . .” ; : . （ - ºr- - º, º ſº. tº . g . .” º & . . . § - º ºw 2 *. º - ~ º, º f iſ º.º.º.º. r º E. w 2--> - & . . . ; - º º *, *, *, *, *, * * * * * : A & © 2: ..& sº # ſº jº # º º, & # 5 # 5 - 3. . . . ; "...º, “ , ”..." . . . . . . . . . . . .*, * * * : 3 º' # 5 º' tº tºº º § ºf . . . . ºr * : * : * * * * * * º E. L} sº a sº. # *** : * , , ; , , . . . . . . . . . . . . . . . . . - & ºr 2 : ºr e º 3: º . ; gº ſº *: , ºf § # * , 2- sº ** is º.º. * * * *.*.*.*, º sº sº . . . . . ; : - º ºś * * *.*.*.* jº.º.º.º.º.º.º.º.º. §§ *, *.*.*.*.*. º tº º º Pº º * . . . . . . As ? d * ,” “... *. .# , º, . . º.º. º, º ſº.º.º.º.º.º.º. 4, tº 3. * . . . ºf ºf . º. *...*, *; # ºf .º.º. ºr, 3. ~. Y. º-º-º- º *, *g, *.*.*.*.*.*, *.*.*.*.*.*.*.*, *,”, *º sº 3. 8 Đ sº...º.º.º. - P. º.º. - º n º '','º',',', irº g º - ſº sº a tº ºr * , ; , ; , . ***"...º.º." # . . - º sº : sº º *...*.*.*.*.*.*.* ºjº" aſ º ºf *...*, *, * * * : º: º * , º ºf º ºf ----- º . . . . º. - * * * * * * * * * ***, *, *, º º # * *.*, * ſº - *******, ºf º- º ſº fº - # , ; 'º','º','º','º','º','º','º', º, º f : . . . . . . . . . . . . . º.º. º. # * * , , , ; ; ; # *, *, *, *, *, *, * * * * * * * * *.*.*.*.* 2.8 s...} E. w sº º, º ºſ', º.º. . tº º ºsº • * * * * * * § { § tº 9 *, *, * : * * > *. I. * * * * * #: ºf ſº º * : * : * : # * * * * * * * * º - *, *, *, *s *, *, *, *, *, *, *, *, *, *, *, - * * * * * * * tº sº ºf * † tº º * ,”, º º .*, § . ..", # , º, . ; . º tº . A iº ºf: sº Sº, §: º r # , . . - º . - * : --> - - ~~ w - * - a . *.*.*.*.*.*.*.* *.*.* º : *º. º sº sº, sº a sº sº º ºf . º - ſº § º º - - - - - § * . . ; gº i. sº # , º, º iſ r. º.º.º. º. º.º.º.º.º.º.º. * Fºº F. º.º.º. º.º. º.º.º.º.º. gº p : F.3. º ..º.º. º * † tº #: *, *, *. ºº º * . . . . 3. f.º.º.º. § -> * # * *.*.*.*.*.* § º º tº º º, º.º. º.º. º * † tº ſº º > º § 3. A. º.º.º.º.º.º.º.º.º.º. º *.*,'º','º','º', w & ſº -- sº § : º, §º º º tºº.º. º.º. . . .","..?. s's sº tºº". j º º *.*.*.*.*.*.º.º.º.º.º.º. *... º. º.º. . . . . $, º, . , º ---------- is sº ºr ºf º * , , , , ; , ; , ; , ; sº gº e * * º Fº * ... Yº 3. § {} ºf ºf t = * } . . . . . . . . . * * º - º º sº, Jº {} iº º º sº s". º: . § º º º, tº º,” * . º. º.º. .*, *, *.* 3. *** º º º: § w w º º, º żº º - sº º sº sº º º f ~ §§ - * . ." § . 3 * , . . . . ; : ; ; ; # . A yºsº, tº sº. º.º.º. ***, *, *, *, *, *, *, * * * * , . . . . . º, a ºf * : V. 3 * * * * * .* # * * * * * , - º *.*.*, *, *, *.*.*.*.*. ; , ; w a r > - ; : * 'º','º','º', # , º, º a §. 3.2 º, ºf * @º º, ~ §§ t º - º *ś i. º º 3. *. v. * , s 5 # , º, º, . . . . . . . § 2 - ; : * : * * : * : º, º, ...? sº & º º i.º.º.º.º.º. º º Sºº º º Nº. §: sº : ºr tº a tº - ſº, ºº, º º, º, ...?' ºº, º.º. , sº º, . . . . . *... º.º. º.º.º.º.º.º.º.º.º.º.º. sº º, º sº: Tº º sº º jº d sº *.*.*.*.*.*. - - **.*.* sº *º ğı y É!!！!!!!!！!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!！!!！!!!!!!!!!!!!!！!!!!!!!!!!！!!!!!!!!!!!!!!!!!!!!!!！!!!!!!!!!！!!!!!！HJØRºſſºſººlºlºſſºſ (）MŘ√≠√∞EXPE›‹rrrrrrrrrrrrr!） ſſſſſſſ|[[[[[[[[[[[[[[[[[[[[[[[[[[[[[#*N3：4ĶÅ!!}}}|||||[IIIIIIIIII]]|[Ilſ[[[[[[[]]|[[[[JITHE№te {{№ſſae， syſºſ， Qae ĶĒR*;§§ E=0 ） [−] --★ º itli '1" * º V. Sº, Nº Nº.S. Sº Z1 wwn F).ER immunmarrinuºuntrilliºl º © tº e e º e º ºr e º ºr ºc was ºr º e º a º ºw ºr ºr e º ea º ºr ºilſtºniºiºniſhinº'ſ SULA*::: NY Sºº-º-º-º-º-º-º-º-ºººººººººº- Nº.2 .* N ź ºm r: „m ſł.” Q. , ' º^ œ A LEX lºº TTTTTTTTTTTTTTTTIſ: PIRO Iº. imºinhºmºiºnini, -----+---+”，|- ~ ſſſſſſſſſſ IIIIIIIIIIIIIIIIIIIIIIIIIIIII ºtnºt tº rarxxxxxxxxxxrrrrrrrrrrrrrrrr. ș#ffffffffffffffffĚĎ Cº-º-º-º-º-º-º-º-º-º-º-º-º-º-º-º-º-º- Flſifiſt East Enº, library T.J. 3, 2 C, , 7) / A#, THE BALANCING OF ENGINES t \. w” E. DALBY, M.A., B.Sc., Minst.c.e., M.I.M.E. PROFESSOR OF ENGINEERING AND APPLIED MATHEM ATICS IN THE CITY AND GUILDS OF LONDON INSTITUTE, TECHNICAL COLLEGE, FINSBURY. L O N DO N E D W A R D A R N O L D I902, 200/60||859 tº - E. E. UEST OF KGF. ALEXAf{}}ER ZIWET. &oer } 6 - / 7 J– / ? 3 2. . Y : PREF ACE. § ; ) DURING the last ten years the subject of Engine Balancing has gradually forced itself upon the attention of Marine Engineers, chiefly because the unbalanced periodic forces of the engine and the natural periods of vibration of the hull have mutually approached the sensitive region of synchronism. Electrical - Engineers have had vibration troubles at Central Stations and on Electric Railways, and many cases of undue wear and tear and hot bearings in Mills and Factories undoubtedly arise from unbalanced machinery, though the actual vibration produced may not be great. In general, the running of an unbalanced engine or machine provokes its supports to elastic oscillations, and adds a grinding pressure on the bearings, and the obvious way to prevent these undesirable effects from happening is to remove the cause of them, that is to say, balance the moving parts from which the unbalanced forces arise. The balancing of the marine engine and the peculiar problems connected therewith have been investigated by many engineers, and most of the original papers on the subject are to be found in the Transactions of the Institution of Naval Architects. The gradual introduction of the Yarrow-Schlick-Tweedy system of balancing the reciprocating parts of an engine amongst them- selves, is familiar to all who are in touch with modern marine IV AAEAE AWA CE. engine design. The Balancing of Locomotives is carried out in a traditional way, and the compromise which makes a hammer blow On the rails a necessary accompaniment to approximate uni- formity of tractive force is accepted by Railway Engineers as the best practical solution of the problem. The advent of the four-cylinder locomotive, however, brings with it practical possi- bilities of balancing the inertia forces as great as in a four-cylinder marine engine. The main object of this book is to develop a semi-graphical method which may be consistently used to attack problems connected with the balancing of the inertia forces arising from the relative motion of the parts of an engine or machine. In the case of a system of revolving masses, or a system of reciprocating masses where the motion may be assumed simple harmonic without serious error, the application of the method is simple in the extreme, as it requires nothing but a knowledge of the four rules of arith- metic and good draughtsmanship. Moreover, the work can be easily checked, and in the case of symmetrically arranged engines, like locomotives, for example, the method is self-check- ing. The application of the method to the case of a recipro- cating system, in which the motion of the Several masses is con- strained by connecting-rods which are short relatively to the cranks they turn, is considered in Chapter W. The use of the method to compute the unbalanced forces arising from the running of an engine or machine of given dimensions in which the mass of each moving part is known, is illustrated in Chapter VI. The precise effect of an engine on its supports cannot be predicted from a knowledge of the magnitudes of the unbalanced forces alone. The effect depends upon the elastic peculiarities of the support in relation to the periodic times and places of action of the unbalanced external forces acting upon it. A brief AREATA CAE. V. discussion of the principles governing the behaviour of elastic supports under the action of external forces is given in Chapter VII. I am indebted to Lord Rayleigh’s “Sound,” Vol. I., for the fundamental ideas of the first seven articles of the chapter. The motion of the connecting-rod and its action upon the frame is considered in Chapter VIII. Those who are approaching the subject for the first time are recommended to work up to Art. 30, and then to check the balanced system given there by drawing out the force and couple polygons for several different positions of the reference plane. Having done this, proceed to Art. 33, and then go straight to Chapter III. Those interested in locomotive work should begin Chapter IV. after working the examples of Arts. 48 and 49, leaving Example 50 for subsequent consideration. Progress should be tested by working the exercises at the end of the book. Exercises 1 to 42 are based upon Chapters I. to IV. A knowledge of the principles explained and illustrated through the book, will enable an engineer to apply the method to the many problems of balancing which he will find on every hand, not only with regard to engines, but in connection with machinery of all kinds. In fact, there is a balancing problem proper to every machine which has a moving part, and the con- sideration of this should form an essential part of the drawing- office work connected with the design of the machine. I must thank the Council of the Institution of Naval Architects for permission to make free use of the two papers I have had the honour to communicate to the Institution, entitled respectively, “The Balancing of Engines, with Special Reference to Marine Work” (March, 1899); “On the Balancing of the Reciprocating Parts of an Engine, including the Effect of the Connecting-rod" (March, 1901). I must also thank the Council of the Institution of Mechanical Engineers for permission to use the substance of a vi AA’Aº AºA CE. paper I had the honour to communicate to the Institution, entitled, “The Balancing of Locomotives” (November, 1901). My acknowledgments are due to Dr. W. E. Sumpner for help in connection with Chapter W.; and to Mr. C. G. Lamb for kindly reading the proofs. I am also indebted to Mr. J. A. F. Aspinall for the data of the Lancashire and Yorkshire Engines; to Mr. F. W. Webb for suggestions regarding the balancing of four-cylinder loco- motives; to Mr. J. Holden for details of locomotive connecting- rods; to Mr. C. A. Park for details of the balancing of Carriage Wheels; and to Mr. Yarrow for data supplied for the applica- tion of the method to the balancing of a Torpedo Boat Destroyer. It is too much to hope that a book involving so much numerical computation will be free from error, and I shall be grateful for any corrections. W. E. DALBY. CITY AND GUILDS OF LONDON INSTITUTE, TECHNICAL COLLEGE, FINSBURY, LONDON. December, 1901. A. 10. T C O N T E N T S. CHAPTER. I. THE ADDITION AND SUBTRACTION OF VECTOR QUANTITIES. Scalar and Vector Quanties . . Definition of a Vector Quantity . Addition of Vector Quantities - º - . Condition that the Vector Sum may be Zero . . Displacement Vectors . . Subtraction of Vectors - º . Definition of Direction to include Sense º s - . On the Two Quantities determined by closing a Polygon 11. 12. CHAPTER II. THE BALANCING OF REVOLVING MASSES. . The Force required to constrain Motion in a Circle. & © º & Methods of applying the Constraining Force: the Reaction on the Axis Dynamical Load on a Shaft . -> º e º e e e - Balancing a Single Mass by means of a Mass in the Same Plane of I3. 14. Revolution. Example . º e º g e º e © Balancing Two rigidly connected Masses by means of a Third Mass, all being in the Same Plane of Revolution . e e º e - Balancing any Number of Masses, rigidly connected to an Axis, by means of a Single Mass, all being in the Same Plane of Revolution. Example . Magnitude of the Unbalanced Force due to a Given System of Masses in the Same Plane of Revolution . . Mass Centre. Example . Experimentally Testing the Balance. Example of L.N.W.R. Carriage Wheels . Centrifugal Couples. Digression on the Properties of Couples . Definition of a Couple . . Equivalent Couples . Axis of a Couple . e º . Addition of Couples . e - . Condition for no Turning Monent e & e & º º - . Effect of a Force Acting on a Rigid Body free to turn about any Axis through a Fixed Point . e e P A G E 10 12 13 15 16 IS IS 22 22 24 25 26 27 27 viii COAVTEAVTS. ART, PAGE 25. Effect of any Number of Forces Acting simultaneously on a Rigid Body free to Turn about any Axis through a Fixed Point . e * . 27 26. Effect of a Centrifugal Force with Reference to a Fixed Point on the Axis of Rotation. Reference Plane. Example . sº 29 27. Balancing a Single Mass by means of Masses in Given, Separate Planes of Revolution. Example º * tº & * is iº , 30 28. Balancing any Number of Given Masses by means of Masses placed in Two Given Planes of Revolution. Rules for the way of drawing Force and Couple Vectors . º * . . * * * * , 33 29. Nomenclature . tº * & * º g g * . 35 30. Typical Example, illustrating the General Method t tº 36 31. To find the Unbalanced Force and the Unbalanced Couple, with respect to a Given Reference Plane due to a System of Masses rotating at a Given Speed . & e * sº {º & * * te & 32. Reduction of the Unbalanced Force and Couple to a Central Axis. Example 39 33. Reduction of the Masses to a Common Radius e e * * > . 41 34. Conditions which must be satisfied by a Given System of Masses so that they may be in Balance amongst themselves . º * tº , 42 35. On the Selection of Data . * sº s g tº t • . 43 36. Relation between the Polygons. Example . e g & * . 44 37. Geometrical Solutions of Particular Problems.—Four-crank Systems . 50 38. Experimental Apparatus . & e ve e * tº g . 52 CHAPTER, III. THE BALANCING OF RECIPROCATING MASSES.—LONG CONNECTING-RODS. 39. The Force required to change the Speed of a Mass of Matter moving in a Straight Line . & § tº e ſº g { } tº e . 54 40. Value of the Acceleration of One Set of Reciprocating Masses, when the Connecting-rod is infinitely Long . & * * > * e . 55 41, The Accelerating Force and its Action on the Frame . º * . 56 42. The Balancing of a Reciprocating System. Definition of a Reciprocating System . º gº e ſº g * tº e e ſº . 57 43. Simple Harmonic Motion and its Relation to Circular Motion . & . 57 44. Method of investigating the Balancing Conditions of a System of Recipro cating Masses whose Motion is Simple Harmonic, or may be considered so without Serious Error . te * ſº ſº º tº e . 58 45. Estimation of the Unbalanced Force and Couple due to a Given System of Reciprocating Masses assuming Simple Harmonic Motion . * . 6] 46. Elimination of the Connecting-rod t te ſº ſº ge & . 62 47. General Method of Procedure for Balancing an Engine * & * 48. Example. Given the Stroke, Cylinder Centre Lines and Three Masses . 66 49. Example. Given Three Cranks, Cylinder Centre Lines . t & , 69 50. Example. Typical of Torpedo Boats. Includes the Valve-gear . . 71 51. Balancing the Crank-shaft of the Previous Example e © * & 52. Conditions that an Engine may be balanced without the Addition of Balancing Masses either to the Reciprocating Parts or to the Crank-shaft 78 COMTEAVTS. ART. 53. 54. 55, 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. To find the Resultant Unbalanced Force and Couple due to the Revolving and Reciprocating Parts together . - º & º & - Experimental Apparatus . º o e g º e p Balancing Reciprocating Masses by the Addition of Revolving Masses CHAPTER IV. THE BALANCING OF LOCOMOTIVES. General Consideration of the Effects produced by the Unbalanced Recipro- cating Parts of a Locomotive . * - & o sº s sº Example . º º e º -> & º º Method of Balancing the Reciprocating Parts of a Locomotive A Standard Set of Reciprocating Parts. 71. 72. . Four-cylinder Locomotives . • & 74. 75. . Comparative Schedules 77. Corresponding Set of Revolving Parts Scales e e º tº Balancing an Inside Single Engine Balancing an Outside Single Engine Balancing an Inside Six-coupled Engine Variation of Rail-load ; “Bammer Blow ’’ Example e e º Speed at which a Wheel lifts Slipping o º e Example . e Distribution of the Balance Weights for the Reciprocating Parts amongst the Coupled Wheels American Practice - º e e Example—Eight-coupled Goods Engine Crank Angles for the Elimination of the Horizontal Swaying Couple Estimation of the Unbalanced Force and Couple - º Experimental Apparatus CHAPTER V. SECONDARY BALANCING. . Analytical Expression for the Acceleration of the Reciprocating Masses, to include the Secondary Effect . On the Error involved by the Approximation . - e . Graphical Interpretation of Expression, (2), Art. 78 e º te - . The Effect of the Primary and Secondary Unbalanced Forces with respect to a Plane a Feet from the Plane of Revolution of the Crank . Effect of More than One Crank on the Same Shaft . . The Conditions of Balance . - t e © . Analytical Representation of a Vector Quantity * tº e . Relation between the Quantities defining the Directions a and 26. , 125 126 I27 128 130 130 131 132 X COAVTEAVT.S. ART. PAGE 86. Application to the Balancing Problem e {e & * & . 133 87. The Eight Fundamental Equations . & * tº º * . 133 88. On the Relation between the Number of Conditional Equations and the Number of Variables tº ſº * º e e º . 134 89. On the Number of Variables * * * g ſº e * . 135 90. On the Selection of the Conditional Equations . e g * . 135 91. Application of the Method to One- and Two-crank Engines . e . 137 92. Application to Three-crank Engines 138 93. Application to a Three-crank Engine Cylinder, Centre Lines fixed . 141 94. Application to a Four-crank Engine tº tº g . 143 95. Example . it. te e g ſº & t . 146 96. Four-crank Engine satisfying Six Conditions e & e © . 147 97. Examples. tº º t tº © e & * : º . 150 98. Symmetrical Engine. On the Variation of the Different Quantities in Terms of the Pitches of the Cylinders and Two Graphical Methods . 154 99. Five-crank Engine . * q {º e & te e tº 158 100. Six-crank Engine g © ve ſº * te & & g . 161 101. Extension of the General Principles to the balancing of Engines when the Expression for the Acceleration is expanded to contain Terms of Higher Order than the Second tº * ge * g º . 163 102. General Summary . e p e g * º wº g . 168 CHAPTER VI. ESTIMATION OF THE PRIMARY AND SECONDARY UNEALANCED FORCES AND COUPLES. 103. General Nature of the Problem . e e º e & ſº . 172 104. Klein’s Construction for finding the Acceleration of the Piston ë . 172 105. Proof wº © * º © º e * ſº ſº tº . 173 106. Data of a Typical Engine. Schedules 19 and 20. & & * . 175 107. Calibration of a Klein Curve to give the Accelerating Force. . 176 108. Derivation of Curves to represent the Forces due to the Other Recipro- cating Masses in the Engine, the Ratio of Crank to Rod being Constant 177 109. Combination of the Curves for their Phase Differences to obtain the Total Unbalanced Force in Terms of one crank Angle © © ë º 110. Calibration and Combination of Klein Curves to obtain the Total Un- balanced Couple belonging to the Reciprocating Masses . tº . 180 111. Addition of Forces and Couples due to the Revolving Parts . ſe . 181 112. Acceleration Curve corresponding to the Approximate Formula (2) 178 Art. 78 . e e * g e º e e © e . 182 113. Process for finding the Primary and Secondary Components of the Resultant Unbalanced Force and Couple Curves . º g e g . 184 114. Application to the Couple Curve of the Example . e * º . 185 115. Valve-gear and Summary . te e g ſº e e ſº . 186 116. Calculation of the Maximum Ordinates of the Components of the Resultant Force and Couples Curves. Extension to any Number of Terms in the General Series e g g tº f : * e tº º . 188 117. Application to the Example g o º te tº us e . 189 118. General Formulae for Typical Cases . * & g e * . 191 119. Comparative Examples. Schedule 21 . e t iº ſº e . 195 COWTENTS. XI. CHAPTER VII. THE WIBRATION OF THE SUPPORTS. ART. PAGE 120. Preliminary e Q º tº e tº { } º º 199 121. Natural Period of Vibration of a Simple Elastic System tº ſº . 200 122. Damping . * > © g * ge ſº * * tº * . 204 123. Vibration of the System under the Action of a Periodic Force g . 207 124. Natural Vibrations of an Elastic Rod of Uniform Section e tº . 209 125. On the Point of Application of a Force and the Vibrations produced . 211 126. Longitudinal and Torsional Vibrations * * 212 127. Simultaneous Action of Several Forces and Couples of Different Periods 213 128. Possible Modes of Vibration of a Ship's Hull and the Forces present to produce them * we 214 129. Experimental Results tº e g gº gº c º tº . 216 130. Turning Moment on the Crank-shaft . e tº g e g . 219 131. Uniformity of Turning Moment . ſº * * * * © . 225 132. Example . tº te º g g & tº wº º º . 229 133. Short-framed Engines wº e e e e 9. g * . 230 CHAPTER VIII. THE MOTION OF THE CONNECTING-ROD. 134. General Character of the Problem “s g © º gº & . 232 135. Dynamical Principles on which the Investigation is based . º . 232 136. Graphical Method for finding the Acceleration of the Mass Centre of the Rod . tº © & ğ & tº & & ſº * . 234 137. Equivalent Dynamical System . e * e ge e º . 235 138. Constructions for fixing a Point in the Line of Action of R . e . 237 139. Combined Construction is {º e g º © o * . 239 140. Effect on the Frame and on the Turning Moment exerted by the Crank. 240 141. Examples . e 4. e * * e & tº tº º . 242 142. Balancing the Rod . tº & * º g º * © . 246 143. Particular Form of Balanced Engine . e © e * tº . 248 144. Analytical Method of finding R and L te e tº º te . 250 145. Values of b, p, and 2, y, at the Dead Centres e & * . 253 146. The Acceleration of the Cross-head in the Line of Stroke . & . 254 147. Example • * * te tº * tº tº te * tº . 254 EXERCISES * e º * º e * tº g g gº . 257 INDEX . * * * e ; : e º º e • * . 279 T H E BALAN CING OF ENGINES, CHAPTER I. THE ADDITION AND SUBTRACTION OF WECTOR, QUANTITIES, SIR WILLIAM HAMILTON divided quantities into two kinds. The one kind called Scalar quantities, the other Vector quantities. 1. A Scalar Quantity does not involve the idea of direction. Sums of money; the capacity of a tank; a quantity of matter, say a ton of coal; the energy stored in a moving body; all these are scalar quantities, and they are defined completely by the simple statement of their magnitudes. Quantities of this kind are added and subtracted by the Ordinary rules of arithmetic. 2. A Wector or Directed Quantity involves the idea of direction as well as magnitude. The simple statement of the magnitude of a quantity of this kind is not enough to define it. The direction in which the quantity is active must be given in such a way that there can be no ambiguity. In general, a vector quantity is said to have— Magnitude. Direction. Sense or Way of action. It will be shown immediately that direction may be defined in a way which will include the last two properties of a vector in a single statement. B 2 THE BA /A WCYWC OF EAWG/AWE.S. Force, acceleration, velocity, displacement, momentum, couples, an electric current, are all examples of vector quantities. A vector quantity may be represented by a line, whose length is proportional to the magnitude of the quantity, and whose direction is parallel to the direction of the quantity, the sense or way of action being indicated by an arrow-head placed on the line. The upper line AB (Fig. 1) represents a vector quantity whose magnitude is the length of AB to scale, whose direction is parallel to AB, and B whose way of action is from A towards B. The line is referred to as “vector AB.” If + AB represent a given vector JB A. quantity, a reversal of its sign, denoted N…' by a reversal of the arrow-head, shown on the lower line of Fig. 1, is another FIG. I. vector quantity specified by — AB or + BA, so that a change of sign, or the reversal of the order of the letters specifying a line, is equivalent to reversing the way of action of the vector. A 3. Addition of Vector Quantities.—The extension of the idea of addition to quantities of this kind is already familiar to every draughtsman through the use of the polygon of forces, to find the resultant of a number of forces acting at a point. The sum of the separate effects of the forces is equivalent to the effect of the single force, “the resultant.” Or, the resultant is the vector sum of the several vector quantities, which in this case happen to be forces. The term “vector sum ” is therefore synonymous with the term “resultant.” Both terms may be used with reference to vector quantities of all kinds, though in each particular addition the vectors must represent quantities of the same kind. The rule for addition may be stated as follows:— Starting from any point, set out the lines representing the vector quantities as if to form a polygon, the arrow heads all pointing round in the same direction: the line drawn from the starting-point, closing the polygon, represents their sum or resultant. The Order of setting out the lines is immaterial. Let A, B, C, D, E (Fig. 2) be a series of lines set out in order ADDITION & SUBTRACTION OF VECTOR QUANTITIES. 3 representing a set of vector quantities, as forces acting on a point Notice that the letter at a corner denotes the end of one line and the beginning of the next. The opera- tion of Setting them out may be conveniently indicated by the ex- pression— Vector sum (AB+ BC+CD+DE) the direction of each being specified, either numerically or by a drawing. The sum is given by the line AE, so that the first and last letter of the series are the two letters in order specifying the line representing the sum. In what follows an expression FIG. 2. of the kind— Vector sum (AB + BC + CD + DE) = AE . . (1) must be understood to mean that the lines indicated by the letters in the brackets are to be set out in order, their several directions being otherwise specified, generally by a drawing, and that the length, direction, and sense of AE, the closing line, are to be found graphically. 4. Condition that the Wector Sum may be Zero.—Bring the right- hand side of equation (1) to the left; then— Vector sum (AB + BC + CD + DE – AE) = 0 But — AE = +FA (Art. 2) therefore, Vector sum (AB + BC + CD + DE + EA) = 0 . (2) The first and last letters of the series in the brackets denote the same point. This expression represents the fact that when the lines are set out in order, they will form a closed polygon, in which case the vector sum of the quantities is zero, since no line has to be drawn to close the polygon. If the vectors represent forces, this expresses the fact that the forces are in equilibrium. It must be carefully remembered that the expressions (1) and (2) are not equations in the ordinary sense. The first is merely a convenient way of indicating an operation and its result, the Second only a convenient way of stating that the vector polygon closes, and in both the sign “ = * should be interpreted to mean “is 4 TA/F BA LANCING OF EAWGINES. equivalent to.” The greatest care must be taken when drawing a polygon to get the arrow-heads rightly placed. The accidental reversal of an arrow-head on a line means that a quantity, pre- sumably specified by the line, has been left out, and one exactly equal and opposite included. When the lines are set out, the arrow-heads must all point in the same circuit, with the exception of the one on the line representing the sum. This must point against the rest. 5. Displacement Vectors.-The properties of vectors may be illustrated by the displacement of a point from a position A to a position B. Let the vector AB (Fig. 3) represent a displacement from A to B. A further displacement from B to C is represented by the line BC. The sum of the two operations, that is— Vector sum (AB + BC) = AC has resulted in a change of position from A to C, which might have been attained by the single displacement AC. If the FIG, 3. two transferences are simultaneous, as when the point is carried in the direc- tion AB, across the deck of a ship steaming the distance BC in the same time, the result of the two transferences is a displace- ment from the position. A to the position C. The idea of the transference of a point, kept in mind, is of great assistance in thinking about vectors. Every individual unconsciously illustrates the principles of vector addition by every movement and by every walk he takes. A return to the same spot means that so far as transference is concerned the vector sum of all his displacements, reckoned from the starting-point, is zero; and if the lines repre- senting the successive straight parts of the walk are plotted, they will form a closed polygon. They need not even be plotted; the series of places passed through, joined up on a map, is a vector polygon, and is obviously closed by a return to the starting-point. Or wherever he gets to, the position attained might have been attained by walking straight to it from the start. The Sense of the vector being traced out at any instant is given by the direc- tion of walking. ADDITION & SUBTRACTION OF VECTOR QUANTITIES. 5 6. The Subtraction of two vectors is performed by setting them out from the same point; the line joining their ends represents their difference. The meaning of Vector sum (AB + BC) = AC has been defined in Art. 3, also the meaning of the minus sign in Art. 2. Remembering these, the following expressions, obtained by transposing the terms of the above expression, evidently indi- cate the operation of finding the vector difference:— Vector difference (AB – AC) = — BC = CB . . (3) Vector difference (AC – AB) = +BC . . . . (4) The corresponding triangles are shown in Figs. 4 and 5. The only difficulty in performing this graphical subtraction is, that TIG, 4. FIG. 5, having drawn the triangle, it is not at once obvious how to place the arrow-head on the line representing the difference. The rule is, that it must always be placed in circuit with the quantity being subtracted. This method is of great use in finding the value of one quantity relative to a second, both being originally expressed relatively to a third quantity. For instance, if AC (Fig. 4) represent the Velocity of a train relative to the earth and AB represent the velocity of a second train relative to the earth, CB represents the velocity of the second train relative to the first and BC (Fig. 5) represents the velocity of the first train relative to the Second. In the well-known proposition of the parallelogram of forces, which is merely a method of adding two vectors, equivalent to the rule already given, one diagonal of the parallelogram repre- sents the vector Sum, the other the vector difference. G THE BAZAAVC/AWG OF EAVG/AWE.S. The principle of taking a vector difference is the basis of the geometrical constructions used in the design of turbines and cen- trifugal pumps, for finding the angles of the vanes; it is the key to the geometrical methods given to find relative velocities and accelerations in kinematics and in mechanisms. Illustrations of its utility in this respect will be found in Art. 105. 7. Definition of Direction to include Sense.—Let OX (Fig. 6) be a line from which direction is to be measured. Suppose an arm to be hinged at O, a A. point called the origin, forming with the line OX 2’ S a kind of open compass. Q N Opening the arm out, v counter-clockwise, to an \ angle 6, the direction it W * indicates is always to be / measured from O outwards Y along the arm, or radius 2^ vector, as it is called, for 2^ a positive vector quantity. A. No ambiguity in sense can FIG. 6. arise, because the sense from O along the dotted line would be specified by the angular direction being given 180° greater. Thus OA might be inclined 60°; OA1 or AO, the direction of the opposite sense, would be inclined 240°. In the case where the vector is negative, it would be measured from O in the direction opposite to the direction in which the arm points. For instance, – OA when 0 is 30°, is equivalent to +OA, where 0 = 180 + 30°. The initial line OX, and the lines measured from it, may be all moved about together in any way, without altering the direc- tions of lines relatively to it, or to one another. If the hub of a cart-wheel, for instance, be selected for an origin, and one of the spokes be fixed upon as the line from which to measure the angular position of all the other spokes, any motion whatever may be given to the wheel without in the least affecting the inclination of the other spokes either to the initial spoke or to one another. A rotating initial line, or line of reference, is one A DD/TION & 9 S UAE TRACTION OF VECTOR QUANTITIES. 7 of the essential features of the method explained in the next chapter. 8. On the Two Quantities determined by closing a Polygon.— Measuring direction by the method of Art. 7, a vector quantity is defined by two quantities, a Magnitude and a Direction. In a closed polygon of n sides there are n magnitudes and ºv directions, 2n, quantities in all. These 2n, quantities cannot all be chosen at will, since they are subject to the condition that they form a closed polygon. This condition requires that two, and two only, of the quantities shall remain unspecified, their values being found, graphically or by calculation, by closing the polygon. If less than two of the quantities are left undetermined, the data will in general be inconsistent; if more, the closing of the polygon is indeterminate. The two unknown quantities may be— 4, . A magnitude and its direction. B, Two magnitudes. C, Two directions. D. A magnitude and the direction of another known magnitude. Case A.—Consider the case of a five-sided polygon. Ten quantities in all are concerned in the specification of its sides; of these eight must be completely specified. Let the eight quantities be specified by the following schedule, leaving a magnitude and its direction unknown:— SchEDULE 1. Magnitude. Direction. AB 2-0 Oo BC 2-5 300 CD 3-0 1500 DE 2-5 2100 EA 1-61 29.10 Setting out the four vectors, AB, BC, CD, DE, they might by 8 THE BA LA WCIAVG OF EAVG//VE.S. Some fortuitous chance form a closed polygon; in general, how- ever, an unknown term, EA, involving a magnitude and a direction, is required to close the polygon. Setting them out as shown in Fig. 7, it will be found that the closing side EA measures 1-61, inclined 29.1° to the initial direction A.B. Case B.—Let AB, BC, CD be completely given, and the directions, 6, and 6, of JDE and E.A. Set out the first three quantities (Fig. 8). At D set out the direction 01, and at A the FIG, 8, direction 180° – 02. The magnitudes of DE and EA are deter- mined by the intersection at E. Case C.—Let AB, BC, CD be given completely, and the magnitudes only of DE and E.A. Set out the completely given vectors as in Fig. 9, arriving at the point D. From centre D describe an arc with radius DE, and from centre A describe an arc with radius AE. These two arcs will— (1) cut one another in two points, in which case there are two Solutions to the problem; (2) they will touch, in which case DE and EA are in the same direction; (3) they will not intersect, showing that the data are incon- sistent, and that there is no solution possible. Case D.—Let AB, BC, CD be completely given, together with the direction of DE and the magnitude of EA, Set out the completely given vectors (Fig. 10), arriving at the point D. Erom D set out a line in the direction 6, and from A draw an arc of radius AE. This arc will— ADDITION & SUBTRACTION OF VECTOR QUANTITIES. 9 (1) cut the direction of DE in two points, giving two solutions; (2) touch DE, in which case EA is at 270° with DE; (3) not touch the line, showing that there is no solution to the problem. In general, therefore, two, and only two, quantities can be FIG. 9. FIG, 10. found from a vector polygon, and if the polygon have n sides, of the 2n, quantities concerned, (2n − 2) must be given, but no more, to make the problem determinate. Even then the data may possibly be inconsistent, as exemplified in Cases C and D above. CHAPTER II. THE BALANCING OF REVOLVING MASSES. 9. The Force required to constrain Motion in a Circle.—The natural mode of motion of a mass of matter, unacted upon by any external force, is in a straight line with uniform speed. The action of an external force is required to change either the direction of motion, or the speed. The force, in lbs. weight, required to constrain a mass of M pounds to move in a circular path, r feet radius, the mass centre moving at a uniform speed of v feet per Second, is given by the expression— Maj% grº and its direction of action is in a line through the mass centre of the body, towards the centre of the path. If w is the angular velocity of the mass in radians per second, v = wr, and the above expression may be written— Majºr g or simply, Majºr, when the force is measured in absolute units. The radius of the path, r, means the distance measured from the centre of the path to the mass centre of the circularly con- strained body. For all practical purposes the mass centre is coincident with the centre of gravity of the body, so that the usual methods for finding the centre of gravity may be employed to find the mass centre. The above expressions may be adjusted (1) 1() THE BA/A/VC//VG OF AZE VOLV/AVG MA.S.S.E.S. 11 for revolutions per minute, N, or revolutions per second, n, by the relations— 27-N 60 The different forms in which the magnitude of the constraining force may be expressed are collected together in the following schedule for convenience of reference :— w = 2irm = ScHEDULE 2. M in pounds; r in feet; g = 32°2. Angular velocity in Speed along the Constraining force F. path = 0 feet per radians per revolutions per revolutions per second. Second = 0. second = m. minute = N. tº M4tr2N2). Mºjº F in poundals ... Majºr M4trº,” ++ --- l f 3600 7° tº wº Majº 2, 20- M4TN2). My” F in lbs. weight ... Majºr M4 fºr —FFF- 2. ſ/ g g × 3600 g! * * ſº 4) - A Mr 9 •) -, a.a.s. Mº F in lbs, weight ... 0.031Moºr | 1.224Mn2r |0.000034MNºr 0 031+ The most convenient form to use in balancing problems is generally— Majºr 10. Methods of applying the Constraining Force : the Re- action on the Axis.-The constraining force F may be applied either as a push or a pull towards the centre. In either case the force has two aspects, the action on the body it is constraining in a circular path, and an equal and opposite reaction on some other body. A railway train is constrained to pass round a curve by the forces exerted between the outer rail and the flanges. The rails push the train continually away from the straight line, the train tries to push the curved constraining rail straight. These two aspects of the push constitute a pressure between the rails and the wheels. For a given speed the opposite aspect of the 12 THAE AEA/CAAWCW/WG OF EAVG/WE.S. A. () constraining force F may be supplied by tilting the sleepers, so that the resolved component of the weight of the train is equal to F, thus removing the pressure from the outer rail. The force F is frequently of necessity applied to a body by means of a radial connector. A stone whirled in a circle by means of a sling is constrained in its path by the pull exerted by the string, which is necessarily accompanied by an equal and opposite pull at the centre. The two aspects of the constraining force exerted by a radial connector, which together always form a tension, have been named respectively the centripetal and centri- fugal forces. By centripetal force is to be understood the action of the radial connector with reference to the body it is constrain- ing, and by centrifugal force the action of the radial connector on the axis. Evample.—A mass of 10 pounds is constrained to move in a circle 4 feet radius, at a speed of 5 feet per second, by means of a radial connector. Find the reaction on the axis, i.e. the centrifugal force. The tension F in the connector is— Mvº 10 × 5* gr T 32.2 × 4 Hence, if 1.94 lbs. weight is the pull on the axis, 1.94 is the pull On the mass necessary to constrain the motion. = 1.94 lbs. weight. 11. Dynamical Load on a Shaft.—A shaft supporting a rotating body whose mass centre is not on the axis of rotation becomes loaded therefore with what may be called a dynamical load—a load varying directly as the radius of the mass centre of the body, as the Square of the angular velocity, and changing continuously in direction. At a high speed, such a load tends to set up vibrations of the framework carrying the shaft, and of the floor or foundation to which the framework is attached. In some cases even moderate speeds cause trouble, and if the period of vibration of any part of the supporting framework or foundation should happen to coincide with the period of revolution of the mass, the disturbances set up may become dangerous. A mass of 1 pound at 1 foot radius attached to a shaft and turned at 12,000 revolutions per minute, imposes on the shaft a dynamical load of nearly 5000 lbs. weight. Or a mass weighing only 1 pound 6 requires a force nearly 5000, times as great as its own weight to THE BALANCING OF REVOLVING MASSES. I3 constrain the motion. Heavy foundations, strong holding-down bolts, and stiffened frames are of little avail against these dis- turbances; and moreover, even if such means reduce the extent of the vibrations to a practical minimum, the dynamical load has still to be taken by the bearings, causing unnecessary wear and tear, often heating, and always a loss of energy, thereby decreasing the efficiency of the machine. The Balancing of a Rotating System consists in the arranging of the masses forming the system, so that the centrifugal forces acting on the shaft, in consequence of the rotation, form a system in equilibrium. In developing the method of effecting such an arrangement, it will be convenient to consider first those problems where the masses are all in the same plane of revolution. Definition.—The plane in which a mass is said to revolve is the plane in which its mass centre revolves. Definition.—A plane of revolution is a plane at right angles to the axis of revolution. - An exact knowledge of the next three paragraphs is necessary for the proper comprehension of the method developed in the rest of the book. * 12, Balancing a Single Mass by Means of a Mass in the Same Plane of Revolution,-Let a mass, FIG. 12. Mi, be attached to a truly turned E-2 disc (Fig. 11), at radius T1. Let Mo, at radius ro, be the mass which will balance the effect of Mr. The condition that there may be no unbalanced centrifugal force acting on the axis, is, that the resultant or vector sum (see Arts. 3 and 4) of the forces due to M, the disturbing mass, and Mo, the balancing mass, is zero for all values of w, the angular velocity of the system. This condition is expressed by— (Miri + Mºr)0° = 0 FIG, 11. to being put outside the bracket because it is the same for every 14 THAE AERA ZAAVC/AVG OF EAVG/MAES’. point in the system. Now to is by the terms of the problem not zero; therefore, to satisfy this condition, the sum of the terms in the brackets must be zero. Each term is a vector quantity, and is called a mass moment. The magnitude of such a term is specified by the numerical value of a product of the form Mr, and its direction by the radius r, specified by a drawing, the sense being determined by the rule that the way of drawing the vector is always from the axis of the shaft towards the mass. The condition that the vector sum of the terms in the brackets is zero, means simply that when the vectors representing them are set out in order, they must form a closed polygon. The angular velocity o may have any value without affecting the result. Let it equal unity, then a term of the form Mr is the centrifugal force when to - unity. In this way the use of the term “mass moment " may be avoided. The solution is to be carried out as follows:– Set out AB (Fig. 12) parallel to OG to represent Mir, ; then BA is the vector required to close the polygon, in this case a line returning on itself, so that the sum— AB + BA = 0 BA, therefore, represents in magnitude and direction the quantity Mºro. Draw OM, (Fig. 11) parallel to BA, remembering that it is to be drawn in the direction from B to A. The balancing mass M, can be found directly the radius at which it is to be placed is given. The analytical condition is in this case expressed by the equation— Mºri = — Mºro M is always to be considered positive; the negative sign therefore refers to the radius, and since the radii r1 and ro are in the same straight line, the negative sign indicates that, measuring ri from the axis outward along a diameter, ro must be measured from the axis along the diameter in the opposite direction. Evample.—If the given mass M, is 5 pounds at a radius of 2 feet — Mºro = - 10 Fig. 13 shows graphically the way in which Mo varies with ro, so that their product may remain constant and equal to 10. The choice of the mass to be used is evidently a very wide one if the radius is not specified. Thus a mass of 5 pounds at 2 feet radius, THE BA/A MCIWG OF AF VOLVIAWG MASSES. 15 equally with a mass of 1000 pounds at Tho foot, or 1 pound at 10 feet, will effect balance. The centrifugal forces at the axis are not eliminated; they are merely balabreed. The connector is in tension, along the line connecting the masses under the action of the constraining force Fi, acting on Mi, and an equal and opposite force F., acting on Mo. The one force is the reaction to the other at any speed of rotation. 70 o & > & th Cl 2 > O 0. 2 4 3 (ſ) CJ) < > RAD I U S N F E E T. FIG. I.3. As the speed of rotation increases, F may become sufficiently great to rupture the connector. For example, a pair of 100-pound balls, attached to an iron rod 1 inch in diameter, at 10 feet centre to centre, and rotated ten times per Second about a central axis at right angles to the rod, Would require a tension in the V rod of 61,250 lbs. weight to constrain their motion—quite enough to break the rod. 13. Balancing Two rigidly connected Masses by Means of a Third Mass, all being in the Same Plane of Revolution.—Let M, and M, (Fig. 14) be the two given masses at radii r and r, respectively, and M, the balancing mass at a radius r. When the 16 THE BAZAAVC/AVG OF EAVG/AWE.S. angular velocity is w, the masses give rise to centrifugal forces Majºri, M20°ra, Motºro, a being the same for all. In order that there may be no dynamical load upon the shaft— Vector sum (Miri + Mºr, + Mºro)a,” = 0 that is, (AB + BC + CA) = 0 where Mºri, Myra, Mºro are represented in magnitude, direction, and sense by the lines AB, BC, CA, respectively. To find Mºro, therefore, set out AB (Fig. 15) to scale equal to Miri, drawing C FIG. 14. FIG. 15. in a direction from the axis to the mass M1, and BC equal to Mºre, drawing from the axis to M2. Then CA, the closing side of the triangle, represents Mºro, the balancing product. Transfer the direction CA to Fig. 14, remembering to draw from the shaft. The magnitude of Mo can be fixed as before, when the radius at which it is to be placed is given. The method of this article may be extended to any system of co-planar masses, and the next article is a general statement of the proposition. 14. Balancing any Number of Masses, rigidly connected to an Axis, by Means of a Single Mass, all being in the Same Plane of Revolution.—Let M1, M2, M3, . . . M., be the given masses, at radii ri, r2, ra, . . . r, feet respectively. The angular positions of the radii are to be specified by a drawing. Let Mo be the balancing mass at radius ro, and to the common angular velocity of the system. THE BAZAAVC/WG OF A&E VOLVIZVG MASSES. 17 When the angular velocity is w, the masses give rise to centrifugal forces M109°ri, M20°r2 . . . M,0°r, Moºro. The condition that there may be no dynamical load on the shaft is— Vector sum (Miri + Mºra -- . . . -- Mºr Jº," ſº + Moro)0° - «) =#:0 In order that this may be true for Jall ić of a) the vector Sum of the terms in the brackets must be zero. Représenting them by the lines AB, BC, . . . , these lines set out in order must form a closed polygon. The necessary closure is effected by the side corresponding to Mºro, which, being measured to scale, gives the value of the product. Its direction, transferred to the drawing specifying the angles between the radii, fixes the direction of the radius r, relative to the given radii. The operation of finding the balancing product Mºro may therefore be stated thus— Set out the magnitudes of the products of the given masses and their radii as if to form a polygon. The closing side, taken in order with the rest, represents the product Mºro. Example.—Masses of 3 pounds, 4 pounds, and 3 pounds are attached to a disc rotated by the shaft O, at the respective radii FIG. 16. FIG. 17. 2 feet, 1 foot, and 1.75 foot, at angles specified by Fig. 16. Find the balancing product Mºro. The centrifugal forces, when to = 1, or products of the given masses and their respective radii, are 6, 4, and 5.25. Starting T C 18 THE BA / AAVC/AWG OF EAVG/WE.S. from A (Fig. 17), AB, BC, CD, represent these products set out in order. The closing side measures 2-5; hence— Mºro = 2.5 Assuming Mo to be conveniently placed at 2 feet radius, its magnitude is 1:25 pounds, and its direction is completely specified by D.A. 15. Magnitude of the Unbalanced Force due to a Given System of Masses in the Same Plane of Revolution.—If Mºro is the balancing product for the system, the centrifugal force due to the rotation of Mo is— 2 Mºrto Since this balances the centrifugal forces due to the given system of masses, it is equal and opposite to their resultant. Consider- ing the previous paragraphs, the length of Aſ! to scale, Fig. 17, multiplied by 0”, gives the magnitude of the unbalanced force for the system of masses shown in black by Fig. 16; similarly, ACo)” is the unbalanced force due to the two masses M, M, of Figs. 14 and 15. In general, the unbalanced force is found by taking the vector sum of the centrifugal forces, assuming w = 1, and multiplying this sum by w”. 16. Mass Centre.—The examples considered will have shown how necessary it is to make as exact a determination as possible of the position of the mass centre of each individual mass before proceeding to find the balancing mass. Where the unbalanced masses have been machined, their form is generally simple, and there is little difficulty in finding this point near enough for ordinary work. Any of the methods generally used for finding the position of the centre of gravity may be used, since the mass centre is a point which may be looked upon as coincident with the centre of gravity. Locating it for irregular masses is more difficult and sometimes impossible, and then recourse must be made to experiment. For instance, a pulley running at a high speed will sometimes cause trouble through being out of balance, even though the rim has been turned inside and out, and obviously the only possible way of balancing it is by experiment. It may be of interest to show that the general method ex- plained for finding the balancing mass may be extended to find THE BAZAAVC/WG OF REVOLVING MASSES. 19 the mass centre of a system of masses whose individual mass Centres are known. tº The resultant centrifugal force acting on the axis due to the rotation of a given system of co-planar masses, is equal to the centrifugal force due to a single mass equal in magnitude to the arithmetical sum of the masses forming the given system, concentrated at the mass centre of the system. The combination of this principle with the methods of Art. 14 gives a simple rule for finding the mass centre. Evidently the mass balancing the concentrated mass at the mass centre is the balancing mass for the system. From Art. 12 it follows that the mass centre of the given system and of the balancing mass are on a diameter, and are on opposite sides of the axis. Hence, if a, be the distance of the mass centre of the given system from the axis of rotation— (M, + M, + Ms + g is tº M.)aw” - Mºrow” - (Miri + Mºre + tº tº a Mºr,)0° from which— _vector sum (Miri + Morº -- . . . M.T.) Scalar sum (M1 + M2 + . . . M.) If the numerator of the fraction is zero, a = 0; hence, if the vector sum of the centrifugal forces about a given axis is zero, the mass centre of the system is on the axis. Referring to Arts. 12, 13, 14, it will be seen that the result of the balancing operations is in each case to move the mass centre of the given system on to the axis of rotation by means of the balancing mass. Ea’ample.—Find the mass centre of the system of masses specified in the example of Art. 14 and shown in black in Fig. 16. The vector sum of the centrifugal forces is represented by AD, the magnitude of which is by measurement 2-5. The scalar sum of the masses is 10. Therefore— * =# = 0.25 foot measured from O in the direction from A to D. 17. Experimentally Testing the Balance.—The positions of the mass centres of the several masses forming a system, and 20 THE BAZANCING OF EAVGINES. consequently the position of the mass centre of the system, cannot be found with mathematical accuracy, even when the parts are machined, by any method of calculation, because of the small varia- tions of density throughout the material and the slight deviations from the form assumed in the calculations. There will always be a Small error in a presumably balanced system on this account. The error may be quite negligible at a relatively low speed of rotation, but may become important at a higher speed. After a system has been balanced as nearly as possible by the methods given, the most delicate test which can be applied is to run the system at a high speed, mounted, if possible, on springs. The mass which must be added to make the system run quietly may then be found by trial. The carriage-wheels on the London and North-Western Rail- way are all balanced experimentally. The system formed by a pair of wheels and their axle, each part of which is of regular form and placed so that the mass centre of the whole is presumably on the axis of rotation, is placed in bearings mounted on springs, as shown in Fig. 18. The system is driven by gear seen at the left hand of the figure, so that the peripheral speed of the wheels is one mile per minute. This corresponds to about 465 revolutions per minute for the standard 43%-inch wheel. Any want of balance is at once shown by the vibration of the bearings on their spring supports. Plates are attached to the inside of the wood centres of the wheels until the system runs steadily. When the mass and position of a balancing plate is known, the original deviation a of the mass centre of the wheel may be readily calculated. Suppose, for instance, that a mass of 1 pound attached at 1:2 foot radius effectively balanced a wheel weighing 969 pounds— 969a0 = 1.2 or a = 0.00124 foot measured from the axis on the line joining the mass centre of the balancing plate to the axis, produced. This seems a small amount to trouble about, yet at 80 miles per hour the centrifugal force due to this slight deviation of the mass centre from the axis of rotation is 15.7 lbs. weight approxi- mately. This force changes its direction of action at the horns y 618 times per minute, and tends to set up unpleasant tremors in 8 [ ':)I, HI 22 THE BALA WCING OF EAVGINES. the carriage body. However excellent the permanent Way may be, for smooth running at high speeds it is essential that every carriage-wheel in the train should be experimentally balanced. 18. Centrifugal Couple.—If the two masses of Art. 12 are placed in different planes of revolution (Fig. 19), their centrifugal forces, though always exactly equal in magnitude and of opposite sign, do not free the shaft from dynamical loading. They form a couple tending to turn the shaft in a plane containing the axis AXIS OF ON FIG. 19. of revolution and the centrifugal forces, and therefore the mass centres of the two masses. This may be conveniently referred to as an axial plane, because it always contains the axis of revolution. An axial plane is indicated by shading in Fig. 19. It will be noticed that the radii of the masses lie on its intersections with the planes of revolution. DIGRESSION ON THE PROPERTIES OF COUPLES. 19. A Couple.—A couple is the name given to a pair of equal and opposite forces acting in parallel lines. The perpendicular distance between the lines of action of the forces is called the arm of the couple. In Fig. 20 the pair of equal and opposite forces F, acting in parallel lines a feet apart, form a couple whose arm is a feet long. Proposition 1–The turning effort of a couple with respect to any axis at right angles to its plane is the same, and is measured by the product of one of the forces and the arm of the couple. THE BAZAMCING OF RE VOLV/AWG MASSES. 23 Let AB, CD (Fig. 21) be the directions of action of two equal, opposite, and parallel forces, acting upon a rigid body free to turn about the axis, O, at right angles to the plane of the couple. From O draw a common perpendicular to the forces, cutting them respectively at C and A. Let AB = CD represent the - 2 | respective magnitudes of the forces. Join BC. The turning effort of the couple is equal to the sum of the moments of the two forces with respect to O. The moment of AB about 0 is positive, and is represented by twice the area of the triangle OAB. Similarly, the moment of CD is negative, and is represented by twice the area of the triangle OCD. The resultant moment is represented by twice the difference of these areas—that is, by twice the triangle ABC—since the triangle OCB is equal to the triangle OCD, both being on the same base and of equal altitude. The turning effort or moment of the couple is therefore equal to the product of one force and the arm of the couple, and this is evidently constant for a given couple and independent of the posi- tion of the axis O. The product is usually expressed in “foot-lbs.” Corollary.—Since the moment of the couple is the same with respect to all axes at right angles to its plane, it follows that the couple may be moved to any new position in its plane, without affecting its moment with respect to a given axis. The several 24 THE BAZAAVC/WG OF EAVG/NWE.S. couples shown in Fig. 22 exert equal turning moments on the disc, though they are applied in such different positions. 20. Equivalent Couples.—The moment of a couple is represented by the product of two factors, the arm and a force. Provided that this product remains constant the turning effort of the couple remains constant, however the individual factors are varied. If F, the magnitude of the forces of a couple, be changed to F, the arm a must be changed to al, so that— Ta = Fla, Hence the arm varies inversely as the force for constant turning effort. This is exhibited graphically in Fig. 23. The arm a, of a couple whose moment is 10 foot-lbs., is plotted horizon- : FIG, 22, TIG. 23. tally against the force vertically. Taking an arm of length CA, the corresponding force is represented by the ordinate AB, to the curve marked 10. Notice how rapidly the magnitude of the force must increase to keep the turning effort constant, as the arm is shortened. Curves are also added for moments of 5 and 15 foot-lbs. These curves and the curve of Fig. 13 are the same in form, and follow precisely the same law, being in fact rectangular hyperbolas; THE BAZAAVC/WG OF RE VOA. VVMG MASSA2S. 25 but they represent different quantities, though the quantities them- selves have apparently the same name, viz. foot-pound. The pound in the case of the mass moment refers to the quantity of material in the mass, and in the case of the couple to the magni- tude of the forces. To save confusion the moment of a couple might be written “foot-lbs. weight,” but this is an awkward combination. Generally the context is enough to indicate the meaning of the term “foot-pound.” There is no ambiguity if forces are measured in absolute units, because then whilst a mass moment is measured in “foot-pounds,” the moment of a couple is measured in “foot-poundals.” Again, neither the terms foot- pound nor foot-poundal must be confused with the corresponding work units. To avoid this, it has been suggested to invert the order of the words when the combination refers to the moment of a couple—that is, to write “a moment of so many lbs.-feet,” or poundals—feet.” The usual way of writing the moment of a couple, viz. “foot-lb.,” will be followed, the abbreviation lb. distinguishing it from a mass moment. 21. Axis of a Couple.—A line drawn at right angles to the plane of a couple, whose length, measured from the plane, is proportional to the moment of the couple, and on the side of the plane such that, looking along the axis towards the plane, the couple appears to be exerting a counter-clockwise turning effort, is called the axis of the couple. The axis of the couple shown in Fig. 20 is a line Fa units long, projecting at right angles above the surface of the paper. The axis of each couple in Fig. 22 is a line AB × AC units long, projecting below the paper. To find which side of the plane of a given couple the axis is to be drawn, think of an instrument consisting of a handle attached to a cardboard disc FIG. 24. (Fig. 24). Suppose a circle to be * A drawn on the disc indicating the direction of positive rotation as shown. To find the axis, imagine the disc placed in the plane of the couple so that its director circle indicates the direction in which the couple tends to turn. The handle then shows the side 26 THE BA/AAVC/AWG OF AZAVG/AVES. of the plane on which the axis is to stand, or, as it is called, the Sense of the axis. The moment of the couple is then to be set out in a direction from the plane along this axis. 22. Addition of Couples.--A couple is a directed quantity, and it has been shown in the preceding paragraph how to represent it by a line, the axis. Couples may therefore be added by taking the vector sum of their axes. Three classes of problems present themselves. The couples may act— (1) In the same or parallel planes; (2) In planes mutually inclined to one another, but all at right angles to a given plane; (3) In planes inclined anyhow. In the first case the axes of the couples are parallel, and their Vector Sum is taken by adding their lengths algebraically, or, What is the same thing, adding the moments of the couples algebraically. This is the case of a large number of familiar |problems; for instance, questions on the equilibrium of levers, of beams and girders, roofs, spur-wheel gearing, all afford examples in which the couples involved have parallel axes. The system of planes in the second case is represented by an Open Japanese screen. The leaves are all inclined to one another, but they are all at right angles to a given plane, the floor. Suppose each leaf of the screen to be the plane of a couple. Each couple will be represented by its axis, standing out at right angles to the one or other side of its leaf. All the axes will be parallel to the floor. To find their vector sum, suppose each to be moved parallel to itself so that the whole group may be laid out in order on the floor. The single line representing their sum is the axis of the resultant couple—that is, it represents the united turning effort of the several couples on the screen, considered as a rigid system. Examples illustrating this case are to be found in questions relative to the equilibrium of three-legged derrick cranes, tripods, the gyrostat; and the application of this principle to find the vector sum of a system of centrifugal couples is one of the leading features of the sequel. In the third case, the axes of the couples form a system of lines inclined to one another in all directions. Setting them out in order from a selected origin, they form with the closing side a THE BAZAAVC/WG OF REVOLVING MASSE.S. 27 gauche polygon. The actual setting out of the lines can be done by the principles of solid geometry. 23. The Condition that there shall be no turning moment is, in each case, that the axes form a closed polygon. The closed polygon in the first case is a line returning on itself to the origin. With this slight digression on the properties of couples the course of the main argument may be resumed. 24. Effect of a Force acting on a Rigid Body free to turn about any Axis through a Fixed Point.—Let F (Fig. 25) be a given force acting at a perpen- dicular distance, a, from the EF fixed point O. The force causes a pressure on the point, equal and parallel to itself. The two aspects of this pressure, together with the original force, form a system of three forces, which split up into— (1) A force equal and parallel to F acting on the | point O; (2) A couple, whose moment is Fa, tending to FIG. 25. cause rotation about an axis through O, at right angles to its plane. Fig. 26 illustrates this in a more general way. A sphere is supported at its centre. A force, F, acts upon it at the point p. The ellipse shows the plane of the couple. OC, at right angles to this, is the line about which the sphere will tend to turn, urged by a clockwise or negative couple of magnitude Fa. The dotted force F is the action on the fixed point of support. The line AB, representing the axis of the couple, must be set out parallel to OL, drawing from O towards L. 25. Effect of any Number of Forces acting simultaneously on a Rigid Body free to turn about any Axis through a Fixed Point.— Each force is equivalent to an equal and parallel force at the fixed 28 THE ARA LA WCING OF EAWGIAWE.S. point, and a couple. The resultant force on the fixed point is the vector sum of the “equal and parallel forces” there. The resultant turning effort is specified by the axis, which is the vector sum of the axes of the different couples. The condition that there may be no force acting on the fixed point is, that the vector sum of the forces be equal to zero; and the condition that there may be no turning effort is, that the A FIG. 26. vector sum of the axes of the couples be equal to zero. These are two independent conditions, and must be separately satisfied. The extension of this way of considering the effect of a force to the centrifugal forces acting at the axis of a rotating system is the key to the solution of many balancing problems, and is the feature of the paper communicated by the author to the Institution of Naval Architects, March 24th, 1899. The rotating Z'HE BAZAAWCING OF REVOLVIZVG MASSES. 29 system is supposed to be free from the constraint of bearings, and to be held only at a fixed point selected at any conve- nient place on the axis of rotation. The effect of the weight of the system is to be entirely neglected; it acts constantly in one direction, and only loads the bearings with a constant load. In fact, the system to be balanced may be imagined at the centre of the earth, where it would be weightless, but every other condition of the problem would remain the same. 26. Effect of a Centrifugal Force with Reference to a Fixed Point on the Axis of Rotation.-Consider the effect of the mass M (Fig. 27) attached to a truly turned disc D, when rotated by the FIG. 27. shaft OX, which is held only at the fixed point O, distant a feet from M’s plane of revolution. A force, Majºr, is exerted on the shaft in the plane of the disc D. This is equivalent to— (1) an equal and parallel force, Majºr = F, acting at the fixed point O, and shown by a dotted line; (2) a couple whose moment is Majºra, tending to cause 30 THE BAZAAVC/WG OF EAVGINES. rotation about an axis through O, at right angles to the plane of the couple, in the positive direction. A plane through the fixed point O, at right angles to the axis of rotation, and revolving with it, will be called the reference plane. It contains both the force at the fixed point O, and the axis about which the system is assumed to be free to turn under the action of the centrifugal couple. The reference plane may be thought of as a sheet keyed to the shaft, or as the drawing-board on which all the vector summation which is required in the problem may be imagined carried out. Example.—The effect of a mass of 10 pounds, revolving 4 times per second at 5 feet radius, in a plane distant 5 feet from the reference, is— 2a22 (1) A force *** = 980:47 lbs. weight, acting at the fixed point O, in the plane of reference; . (2) A couple of moment 980:47 x 5 = 4902 foot-lbs., tending to turn the system about the axis shown in Fig. 27. If a balancing mass or masses be applied to the system, giving rise to an equal and opposite centrifugal couple, there will be no tendency to turn about the fixed point. If at the same time the balancing masses have a resultant centrifugal force, equal and opposite to the resultant centrifugal force at the fixed point, there will be no pressure acting on it. Under these circumstances, the constraint applied to the fixed point may be removed, and the system will continue to rotate without trying to change the direction of the main shaft. It would be a balanced rotating system, and held in bearings, would put no dynamical load upon them. 27. To balance a Single Mass by Means of Masses in Given, Separate, Planes of Revolution.—Many cases arise in practice in which it is inconvenient or impossible to apply the balancing mass in the same plane of revolution as the disturbing mass, in the way illustrated in Arts. 12, 13, and 14. Under these circumstances at least two balancing masses are required, placed in separate planes of revolution, which may be selected either with the disturbing mass between them, or outside both of them, whichever may be the most convenient arrangement. Having selected the position of these two planes, choose one of THE BAZAAWCING OF REVOLVING MASSES. 31 them for a reference plane. Then the conditions to be fulfilled by the three masses—that is, the disturbing mass and the two masses balancing it—are— (1) The sum of the forces at 0, the supposed fixed point, must be zeró; (2) The sum of the centrifugal couples must be zero. Let M (Fig. 28) be the mass to be balanced, by masses placed FIG. 29. FIG. 30. FIG. 28. I in planes Nos. I. and II. respectively. Choose No. II. for the reference plane, O being therefore the fixed point. Let plane No. I. and the given plane be distant respectively an and a feet from the reference plane. The mass in the reference plane, whatever be its magnitude or position, will have no moment about O. Let Mi be the mass in plane No. I., which will balance the couple due to M about O, acting at radius r. The condition that the sum of the moments of the couples about O vanish is expressed by— (Miria, + Mra)0° = 0 That is, the vector sum of the centrifugal couples in the brackets must be zero. The direction of the vector representing the couple Mra is at right angles to the axial plane, and it may be drawn by the rule of Art. 21. It is, however, more convenient 32 THE BAZAAWCIAWG OF EAWGZAVES. to imagine that its axis is turned through 90°, so that the direction of the axis corresponds with the direction of the crank, measuring jrom the shaft outwards. Then the side closing the couple polygon is parallel to the crank to be added with the balancing mass, and shows the sense in which the crank radius is to be drawn, viz. from the axis of the shaft, outwards, in the direction indicated by the arrow-head on the closing side, when, as in the present case, the cranks are all on one side of the reference plane. Hence set out AB (Fig. 29) to represent Mra to scale. The line closing the polygon is BA, equal in length to AB, but points in the opposite direction. This at once fixes the direction of No. 1 crank. The magnitude of the mass it carries is found from- Mar 0.1 M111 == (1) when T1 is fixed. The separate centrifugal forces due to M and M1 are each accompanied by an equal and parallel force at O in the reference plane. If M2 at radius T, is the mass in the reference plane, whose centrifugal force will balance the resultant of the transferred forces, the expression— Vector sum (Mr + MIT, -i- M37'2) = 0 states the condition of equilibrium. Setting out ab, be (Fig. 30) to respectively represent Mr and Mºri, the line ca, closing the polygon, represents the force at O in magnitude and direction required to balance the forces there. ca is the crank direction, and the magnitude of the mass it is to carry is evidently given by— Mr – MIT, . . . . . . . . (2) for the case shown. When the radius at which M2 can be con- veniently placed is given, the magnitude of M2 can be found at OL106. If the reference plane is between plane No. I. and the given plane, a and a1 must be considered of opposite sign. - Bacample.—Suppose M = 10 pounds at 2 feet radius, and a and an are 2 feet and 5 feet respectively. Mr. = 8, from equation (1) and Mgr. = 12, from equation (2) THE BAZAAWCING OF REVO/L V/AWG MASSES. 33 M, and M, in this case are to be placed in opposition to M, as shown in Fig. 28. - The effect on Miri and Mºra of varying the position of the plane No. I. relative to the others, is shown by the curves A and B respectively (Fig. 31). In any position of the plane—for instance, when it is a feet from the reference plane—the Ordinate PLAN E N9 II. 30 E. 20 i. 5 e 10 ... O - 2 - 1 => s? : CURVE A. 2 ſº º O CURVE. B. li. H. – 2 la! §: it ſº He 2 00 lil (ſ) O 3 > FIG. 31. of curve A, wa, gives the value of Miri, and the Ordinate of curve B, wb, the value of Møre. The figures refer to the case where Mr = 10 and a = 2. The curves show clearly that Miri is always negative relative to Mr, and Mrs is negative so long as plane |No. I. is to the right of the given plane, and positive when it is on the same side as plane No. II., the reference plane. 28. Balancing any Number of Given Masses by Means of Masses D 34 THE BALA NCIWG OF ENGINES. placed in Two Given Planes.—This and the next three articles state the general method of which the problems and examples already given are particular cases. The earlier and simpler propositions have, however, many practical applications, and should be carefully studied. Let a, be the distance between the two given planes in which the balancing masses are to be placed. Choose one of these planes for the reference plane, thus fixing the point O. Let M1, M2, Ms, etc., be the given masses, at radii Ti, r2, rs, etc., respectively, revolving in planes ai, az, as, etc., feet from the reference plane. - Let Ms be the balancing mass in the plane of reference at radius r, and M, the balancing mass at radius r, in the plane, which is by the terms of the problem a feet from the reference plane. When the system rotates, the centrifugal force corresponding to each mass acts upon the axis, which in turn causes an equal and parallel force to act at the fixed point O, and a couple. The condition that there may be no couple is expressed by— + Vector sum (Miria, + Myra, + . . . 4. Mºra.)0° = 0 . (1) and the condition for no force on O by— Vector sum (Mr. -- Mºr, -- . . . -- Mr. -- Mºrs)0% = 0 . (2) The artifice used to obtain a solution of the problem consists in taking the reference plane coincident with the plane of revolu- tion of one of the balancing masses, so that it has no moment about O, and consequently the balance for couples may be adjusted without it. The reference plane being at No. 5 plane, in the case under discussion, M575 does not appear in equation (1). Whatever be the value of w, the two conditions are separately fulfilled if the vector sums of the terms in the brackets are in each case Zero—that is, if when set out to scale they form a pair of closed polygons. Consider equation (1). All the terms are given but Marqaa, of which, however, the factor a, is given. Calculate their arithmetical values and set them out to scale, their relative directions being specified by a drawing. The axes of the couples THE BALA/VCIWG OF RE VOL V/AWG MA.S.S.E.S. 35 the terms represent are of course at right angles to the axial planes in which they respectively act. A little consideration will show, however, that the directions of the cranks themselves may be used in , actually drawing the couple polygon, if the following rules are observed:— Rules for the Way of drawing Couple Vectors.-If the masses are all on the same side of the reference plane, the direction of drawing is from the axis, outwards, to the mass, in a direction parallel to the respective crank directions. If the masses are some on one side of the reference plane and some on the other, the direction of drawing is from the axis, outwards, towards the mass, for all masses on one side ; and from the mass, inwards, towards the axis for all masses on the opposite side of the reference plane, drawing always parallel to the respective crank directions. The line closing the polygon represents Mariat. Scaling this off and dividing by a 4, Mºra is known. - Again, calculate the arithmetical values of the terms in equation (2), and set them out to scale, the relative directions being given by the drawing, and include of course the value of Mary just found from the couple polygon, observing the following rule :- Rule for the Way of drawing Force Vectors.-Draw always from the axis outwards towards the mass parallel to the respective crank directions. The line closing the polygon represents Mgrs. Choosing the radii, r, and rs, the magnitude of the balancing masses may be calculated at once. These added to the given system, so that their radii are placed in the relative positions specified by the closing sides of the two polygons respectively, completely balance it for all speeds of rotation. Checking the Accuracy of the Work-Having found the balancing masses, add them to the drawing in their proper positions relatively to the given masses; choose a new reference plane anywhere, and draw a new couple polygon relatively to it. If it close, it is safe to infer that no mistake has been made in the work. The force polygon is the same for all positions of the reference plane. 29, Nomenclature. It will be noticed that each term in the 36 THE BA/AAVC/WG OF EAWGIAVES. brackets of equation (2), Art. 28, is a mass moment, and that each term in the brackets of equation (1) is the moment of a mass moment with reference to a, the reference plane. The term, “pro- duct of inertia,” is also used to denote terms of this form. The con- ditions of balance may evidently be concisely stated as follows:— (1) The sum of the products of inertia about the axis of rotation must vanish ; (2) The sum of the mass moments about the axis of rotation must vanish. When these conditions are fulfilled, the axis is called a principal axis. To avoid using these somewhat unfamiliar terms, the angular velocity may be supposed equal to unity, since it may have any value, in the problem under discussion, without affecting the balance in any way. Then, as already indicated in Art. 12, a term of the form Mr may be referred to as a centrifugal force, and a term of the form Mra as a centrifugal couple. 30. Typical Example.—Three masses (shown black, Fig. 32), rigidly connected to a shaft, are specified in the following list, the distances, a, being measured from a given plane of reference :— M = 1.0 pound r, - 1.5 foot an = 7-0 feet M2 = 2-0 ,, 72 = 1°0 ,, a2 = 3.5 , Ma = 1.8 5 y 7'3 = 1.25 35 Q3 = 1-8 53 The angles between the mass radii are specified by the dotted lines (Fig. 33). Find the magnitude and position of two balancing masses, which are to be placed, one in the plane of reference at unity radius, the other in the plane of Mi, also at unity radius. It will be found convenient to arrange the data in a schedule of the following kind, in order to calculate the arithmetical values of the different terms. THE BAZAAWCING OF REVOLV/WG MASSES. 37 SCHEDULE 3. to − 1. Number Magnitude of Mass radius º: from of plane of the mass in measured in the i. §§º | The products | The products revolution. pounds (M). feet (r). Fº Mr, the Mºra, the eet (a). centrifugal centrifugal forces. couples. No. 1 1 1°5 7:0 1.5 10-5 No. 2 2 1-0 3-5 2-0 7:0 No. 3 1-8 1.25 1-8 2-25 4-05 No. 4 unknown 1-0 7-0 0-63 4'4 i (closure) No. 5 unknown 1-0 O-0 1-7 (closure) Draw the couple polygon first (Fig. 34), setting out AB, BC, CD parallel respectively to the crank directions given by Fig. 33, and representing 10-5, 7-0, and 4:05 to scale. The closure DA measures 44, and this is the value of Maraat, in which as is given equal to 7 feet ; therefore— M4r4 = '63 The angular position of the radius in the plane of M4, which in this case is given coincident with the plane of M1, is determined by drawing a line parallel to DA, from the axis. Further, if 74 = 1 foot, 63 pound placed at this radius, as shown in Fig. 32, will balance the centrifugal couples. Mara = 63 may now be written in the schedule. Again, set out the force polygon (Fig. 35), abcde, the sides being parallel to the radii given by Fig. 33, and proportional to 1:5, 20, 2:25, and 63 respectively, the forces acting at the point O. The closure ea measures 1-7, and this is the value of M5rs. Taking rs = 1, M5 is equal to 1.7 pound. The angular position of the radius in the reference plane is determined by drawing a line parallel to ca, from the axis, as indicated in Fig. 33. These two masses completely balance the given system ; in mathematical language, they convert the axis of rotation into a principal axis. 31. To find the Unbalanced Force and the Unbalanced Couple, 38 THE BAZAAVC/WG OF EAWG//VE.S. with respect to a Given Reference Plane due to a System of Masses rotating at a Given Speed,—The unbalanced force is the vector Sum º : of the forces at O, equal and parallel to the centrifugal forces. The unbalanced couple is the vector sum of the centrifugal couples. Considering the example of Art. 30, the unbalanced couple is THE AEA ZAAVC//VG OF ACE WOZVZVG MASSAE.S. 39 2 represented by AD (Fig. 34), the magnitude of which is 44; foot- lbs. The axis of the couple is at right angles to AD, and is shown by OY (Fig. 33). The magnitude of the resultant force is represented by ad (Fig. 35), which measures 1-8 feet to scale; 2 the magnitude of the force is therefore 18; lbs. weight acting at O, parallel to AD. If the shaft rotates at 10 revolutions per second, these values are 543 foot-lbs., and 221 lbs. weight respectively. The force and the couple may now be balanced in a more general way than that given in Art. 30. To balance the force, a mass, M, must be placed in the reference plane in a direction, Da, at Such a radius, r, that— Mr = 18 To balance the couple, masses Ma, Mº, at radii ra, r, respec- tively, may be placed anywhere in the axial plane, of which XX (Fig. 33) is the trace, so that if a, be the axial distance between their radii— Maraa. = 4.4 = Mºrºd, their disposition being such that they give rise to a couple opposite in sign to the unbalanced couple. This method gives in general three balancing masses, which of course may be combined into the two of Art. 30. The first method is by far the most convenient for practical use, because it gives the balancing masses without any necessity of thinking of their way of action, these being determined automatically by the closure of the polygons. 32, Reduction of the Unbalanced Force and Couple to a Central Axis.--It will readily be perceived that the magnitude of the unbalanced force is independent of the position of the reference plane, but the magnitude of the couple is different for every new position the reference plane is moved into. No fair comparison of the unbalanced couples belonging to two different systems can be made unless the respective reference planes be moved into the positions for which the magnitudes of the couples are respectively a minimum. But this is a property of Poinsot's central axis, so that the problem resolves itself into finding the central axis for 40 THE BALANCING OF ENGINES. the unbalanced force and couple found with respect to any reference plane. This may be done in the following way:- Let OR (Fig. 36) be the resultant force for a revolving system; OC the axis of the resultant couple. Resolve the axis in and at FIG. 36. O C Si N () OR 3-9. Fº Fig. 37. right angles to the direction of R. Thus OC is equivalent to a couple, CP, acting in a plane containing OR and the axis of revolu- tion, and OP in a plane at right angles to this—the axis of rotation is the common intersection of these two planes. This resolution is shown in Fig. 37 without distortion. Adjust the couple CP so that 7"HAZ AAAAAWCZWG OF AZAZ VOA. V/AVG MA.S.S.E.S. 41 its forces are each equal to R, and so that one of them, R1, acts in opposition to R. R, R1, being equal, annul one another, leaving a CP OC sin 6 R1 T R1 reference, and a couple, OP, whose axis coincides in direction with the remaining force. The disturbing forces are then reduced to a single force R1, and a couple whose axis is along R1. The line of action of R1 is the central axis for the system. It has the property that the disturbing couple about it is a minimum. The central axis may be looked upon as fixed to, and revolving with, the system. Its position relative to the cranks never alters, being determined solely by the disposition of the masses in motion. The magnitudes of the force and couple set off along it vary with the Square of the speed. In Figs. 36 and 37, if OR = 5 tons and the resultant couple OC = 20 foot-tons, the resolved couples acting in and at right angles to the plane containing the resultant force and the axis of rotation are 19:5 and 5-5 foot-tons respectively. The plane con- single force, R., at a distance from the plane of • ſº taining the central axis is therefore º = 3.9 feet from the reference plane. Q is the new origin, and marks the point in the shaft through which the central axis passes. A fair comparison of the want of balance of different systems can now be made by comparing their central axes. The reduction to the central axis is not of practical importance in a general way. 33. Reduction of the Masses to a Common Radius.-It is often convenient to make a preliminary reduction of the masses to a common radius, the crank radius in an engine problem, or unity in a general problem. The centrifugal force Majºr is proportional to the product Mr, and the individual factors, M and r, may have any value, providing that their product remains constant. This point has already been exemplified in Art. 12, Fig. 13. If R represents the common radius, the reduced masses M1, M2, M3, etc., are obtained from the several equations— MilK - Miri, M2B == M37'2, MaR = M3rg, etc. Substituting these equivalent products in equations (1) and (2) of Art. 28, they become— 42 THE BAZA MCIWG OF ENGINE.S. Vector sum (Mia! -- M302 + . . . -- Miaº)0°R = 0 . (3) Vector sum (M1 -- M3 + . . . -- M5)0%R = 0 . (4) If the vector sums in the brackets separately vanish, the two conditions of balance are fulfilled for all speeds. M is really a mass moment, and Ma a moment of a mass moment, but, to avoid using unfamiliar terms, the factor o’ſ may be supposed equal to unity; then M may be referred to as a centrifugal force, and Ma as a centrifugal couple. To calculate the value of the centrifugal force and couple in general, if the vector sum in the brackets be not zero, the quantities M and Ma representing the respective sums must be multiplied 2 by *-R, the g being introduced to obtain the result in lbs. weight 9 units of force. In what follows the ordinary capital M denotes mass at crank radius unless otherwise stated. 34. Conditions which must be satisfied by a Given System of Masses so that they may be in Balance amongst themselves.—Suppose all the masses to be first reduced to equivalent masses at a common radius so that the terms “ equivalent mass” and “equivalent mass moment” may be used instead of “mass moment” and “moment of mass moment * respectively. Then— (1) It must be possible to draw a closed polygon whose sides are proportional to the equivalent masses, and parallel in direction to the corresponding mass radii; (2) It must be possible to draw a closed polygon whose sides are proportional to the equivalent mass moments taken with respect to any reference plane. If condition (1) is satisfied and not (2), there is no unbalanced force, but there is an unbalanced couple. Condition (2) cannot be satisfied unless (1) is satisfied, for although the couple polygon may be closed for any reference plane, yet if the plane is moved into a new position, the couple polygon for the new position will close only if there is no force in the old reference plane—that is, only if condition (1) is satisfied. Balancing problems are conditioned, therefore, by the geome- trical properties of two polygons, whose sides are parallel, but of different lengths, the sides of the one being obtained from the sides of the other by multiplication, the multiplier in each case THE BAZAAWCIAVG OF REVOLVING MASSES. 43 being the distance of the equivalent mass from the reference plane. 35. On the Selection of Data.-The balancing of a given system in a specified way can only be done if there are four independent variables left to be determined by the pair of closed polygons which condition the balancing, since the closure of each polygon requires that two, and two only, quantities be left unknown (Art. 8). In fixing the preliminary data, therefore, it is necessary to know exactly how many variables there are concerned in the proposed problem. To completely specify any one mass in the System there must be given— (1) The equivalent magnitude of the mass at unity radius; (2) The direction of the radius of the mass measured from a line of reference in the reference plane; (3) The distance of the plane of revolution of the mass from the reference plane; So that, if there are n masses there will be 37 quantities altogether in the complete system. But these quantities are not all inde- pendent variables. Consider the number of independent variables as regards— (1) The Magnitudes of the Masses.—The closed force polygon only determines the ratios of the several equivalent masses repre- Sented by its sides, and not their absolute magnitudes. An infinite number of similar polygons could be drawn satisfying the same conditions. Consequently, any one side may be considered unity, and the magnitudes of the rest expressed in terms of it. The number of independent variables is, therefore, one less than the number of sides forming the closed polygon—that is, one less than the number of masses concerned in the problem—so that if there are n masses, the number of independent variables of magnitude is— m – 1 (2) With regard to the Directions of the Mass Radii or Cranks.- The specification of (n − 1) variables of direction is sufficient to determine the n angles of a closed polygon, so that the number of independent variables of direction corresponding to n masses is— m – 1 (3) With regard to the Distances apart of the Planes of Revolution.—Measuring from any arbitrarily chosen reference 44 THE BAZAAWCING OF EAWGIWE.S. plane, there are n quantities concerned in fixing the position of n planes of revolution relative to it: dividing each distance by any one of the distances, there will result— 77 – 1 variables of distance. The total number of independent variables corresponding to n masses is therefore in general— 3(n − 1) The first step in the process of balancing a system is to ascertain how many masses there are concerned in it, including, of course, any masses it may be proposed to add as balancing masses. Căll this number n. The number of independent variables is then 3(n − 1). The number of independent variables which must be left to effect closure of the two polygons is four; therefore 3(n − 1) – 4 = 3n - 7 independent quantities must be fixed, and I] O IQOI’é. In choosing these quantities, it must not be forgotten that the number specifying the magnitude of a mass is only to be con- sidered an independent variable if one of the masses is called unity; when this is not done, it is the ratio of a pair of masses which is an independent variable, so that fixing the magnitude of two masses is equal to fixing one variable quantity, fixing three masses equivalent to two, and generally fixing n masses is equi- valent to fixing n – 1 independent variables. Also, if the reference plane does not coincide with the plane of revolution of one of the masses, fixing n distances from it is equal to fixing m — 1 independent variables. Coincidence between the reference plane and a plane of revolution determines that one of the m – 1 variables of distance = 0. By Way of example, suppose a balanced system is to consist of five masses; in general, 3 × 5 — 7 = 8 of the independent quantities must be settled to start with, but no more. Suppose now that the distances of the five planes of revolution are given from any arbitrarily chosen reference plane; this is equivalent to fixing four of the eight quantities. Next, suppose the magni- tudes of three masses to be fixed; this is equal to fixing two of the variables, leaving two more to be fixed, which may be two of the crank angles. If anything else is fixed, the data becomes inconsistent, and the problem cannot be solved. THE BALANCING OF REVOLVING MASSES. 45 REF. *AN4 FIG. 41. >] FIG. 40. 46 TAIAE BAZAAVC/WG OF EAWGIZVE.S. 36, Relation between the Polygons,—Let equivalent masses, M1, M2, Mg, MA, be distant al, a2, a3, as, feet respectively from the reference plane at O (Fig. 38), and suppose the system to Satisfy the conditions of balance, and that ABCD (Fig. 39) and abcd (Fig. 40) are the force and couple polygons, the order of drawing the sides being 1, 3, 2, 4, to avoid re-entrant angles. Then— AB : ab = M1 : M1a1 = 1 : at BC : be = M3 : Mgas = 1 : as CD : că = M2 : M202 = 1 : a 2 DA : da = M4 : M40.4 = 1 : a 4 If the polygons are similar, the ratio between corrésponding sides will be constant, in which case— Q1 = 0.3 F 0.2 = 04 that is, all the masses are in the same plane of revolution. If a pair of ratios are equal, the corresponding masses are in the same plane of revolution. This principle shows that two masses cannot balance one another unless they are in the same plane of revolution, for the only pair of polygons which can be drawn are a pair of lines returning on themselves, ABA and aba, in which the ratio— AB : ab = BA. : ba Assuming three masses to be in balance, let ABC (Fig. 42) C \ C / M A Nº. N. A. JB OU b FIG, 42. FIG. 43. be the force polygon, in this case a triangle. No other triangle, as abc (Fig. 43), can be drawn with parallel sides unless it be TAZAE BAZAAVC/WG OF REVOLVING MASSES. 47 similar; therefore the three masses must be in the same plane of revolution. If, however, the force polygon closes up into a line returning on itself, as AB + BC + CA (Fig. 44), a second line may be drawn, as ab + be + ca (Fig. 45), so that the ratios of correspond- ing sides, or segments, are different. It follows that the masses may be placed in different planes of revolution, but now they must all lie in an axial plane. A pair of quadrilaterals for four masses may be drawn in which the four ratios of corresponding sides are all different, <-- ~~ => sº sº- *- A. JB CL =b |FIG. 44. FIG. 45. providing opposite pairs of sides in one of the polygons are not parallel. Let Figs. 38, 39, and 40 illustrate this case. Alter the scale of the couple polygon until ab is the same length as AB. This is equivalent to making an equal to unity. Then— AB : db = 1 : 1 CD: a = 1 . . l BC : be = 1 : * DA: da = 1 : * Cb1 (0.1 Then the couple polygon may be superposed on the force polygon so that ab coincides with AB, Da with da, and Be with be; ca. remaining parallel to CD (Fig. 39). From this it is evident that if any quadrilateral, as ABCD, is drawn, and a line de is drawn anywhere parallel to one side, cutting the other sides, produced if necessary in d and c, then— (1) The common side as AB represents unity mass at unity distance from the reference plane; (2) The sides of the quadrilateral ABCD are proportional to the equivalent masses, that is, to the centrifugal forces; (3) The directions of the sides transferred from the force polygon to an end view of the shaft give the crank angles; (4) The ratios of a pair of corresponding sides of the two quadrilaterals, ABCD, ABCd, is the distance of the corresponding mass from the reference plane, a being unity. These things are true for a pair of polygons of any number of 48 THE BA LA WC/WG OF EAWGIAWE.S. sides, Superposed so that they have a common side, the remaining sides being parallel; if the ratios between two pairs of sides are equal the corresponding masses will be in the same plane of revolution. Eacample.—Suppose the quadrilateral ABCD and the parallel cd of Fig. 39 to have been drawn at random. Measure off the different lengths concerned to any convenient scale and arrange them as in the following schedule:– SCHEDULE 4. Proportional numbers Proportional numbers | Proportional numbers | for the distances of Plane of revolution. for the masses at for mass the several planes of unit radius. moments. revolution from the reference plane. No. 1 AB = 1.5 AB = 1°5 a = 1-0 No. 2 CD – 5'85 cal - 4:55 a = 0.778 No. 3 BC = 4.9 BC – 3:5 a = 0.715 No. 4 DA = 3-3 dA = 2-2 a = 0. 667 Divide the numbers representing the mass moments by the corresponding numbers representing the masses, to obtain the numbers proportional to the distances of the planes of revolution from the reference plane. These numbers are given in the column to the extreme right of the schedule. Then if a system be arranged in which the magnitudes of the masses are in the proportion of the figures of column 2 of the schedule, the crank directions being given by the direction of the sides of the force polygon, and the spacing of the planes of revolution in the proportion of the figures of column 4, the system will be perfectly balanced, and will run at any speed without putting any dynamical load on the shaft. Mr. Macfarlane Gray * has shown how the related polygons for a four-crank system may be built up of wooden laths, leaving them freedom of distortion. Then pulling the frame into any position, the crank angles, etc., for balance can be derived from it. Obviously polygons of any number of sides may be built up to form a flexible frame, the balancing conditions being determined * Trans. I.N.A., vol. xlii., 1900. THE BAZAAVC/WG OF REVOLVING MASSES. 49 for any configuration in the way already illustrated by Schedule 4. The parallel cq (Fig. 39) may be drawn anywhere; it merely fixes the position of the reference plane (Fig. 38) relatively to the planes of revolution. Consider Fig. 46. Suppose cd to coincide with CD. d <–3 C REF. PLANE AT co D C 3 dºl A C REF, PLANE AT N24. 1 B d C REF. PLANE AT N93, dy VC - REF. PLANE AT N°2. FIG, 46. Then— Ad : AD = Bc : BC = 1 this cannot be true unless the reference plane be at an infinite distance from the plane of revolution. If cd passes through the point A– Ad : AD = 0 therefore a1 = 0, and the reference plane coincides with the plane of M4. Similarly, if it passes through B, the reference plane coincides with the plane of M3. At Q— cd : CD = 0 and the reference plane coincides with the plane of M2. Notice that the ratios as : a 1 and as : ai become negative when cd cuts either DA or CB produced. All the ratios approach infinity as the reference plane approaches the plane of M1, simply because its distance from that plane is reckoned unity. E 50 THE BA LA WCING OF EAWG/NES’. 37, Geometrical Solutions of Particular Problems. Four-Crank Systems. The problem of finding the positions of the planes of revolution for any force polygon drawn at random may be solved geometrically. Let ABCD (Fig. 47) be drawn at random. Number the sides in the way shown. Select the point of intersection of the two shorter sides; B in the figure. Draw the diagonal which this point subtends, AC. Draw Bc and Bd parallel to AD and CD respectively, intersecting AC in e and f. Then if AC represents the distance between the extreme planes of revolution, e and f give FIG. 48. N° 1. FIG, 47. the positions of the two inner planes. If the lines meeting at B are produced indefinitely, forming a pencil of four rays, any line drawn across the pencil parallel to AC is divided in the same ratio as AC, and may therefore be taken to represent to some scale the relative positions of the planes of revolution. Notice that the position of No. 2 plane is fixed by a parallel to No. 3 side, that No. 3 plane is fixed by a parallel to No. 2 side, and that Nos. 1 and 4 planes are determined respectively by Nos. 4 and 1 rays of the pencil. There is thus a reciprocal connection between the THE BAZAAVC/WG OF REVOLVING MASSES. 51 positions of the inner planes and the sides of the corresponding force polygon, and a similar connection between the outer planes and the force polygon. The proof of this construction depends upon the fact that if the plane of reference is at No. 4 plane, ABC. is the couple triangle (see Fig. 46); if at No. 1 plane, BCC is the couple triangle. Under these circumstances— CC : CD = as : a 4 and— Ad : AD = a 3 : 61 But in the triangle ACD, cf is parallel to AD, therefore— Cf: CA = Ce: CD = as : aſ Similarly— * A6 : AC = Ad : AD = as : aſ Therefore if CA represents a1, Cf represents ag, measuring from C. The reference plane is at No. 1, and therefore point C corre- sponds with No. 1 plane. Also if AC represents a1, Ae represents as, measuring from A. The reference plane is at No. 4 plane, therefore A corresponds with No. 4 plane. But as = a1, therefore AC is divided by e and f, so that the four points A, e, f, C represent to scale the relative positions of the planes of revolution along the axis corresponding to the force polygon ABCD. If the positions of the four planes of revolution are given, an inverse construction will disclose the force polygon, the first step in the process discovering the crank angles for which balance is possible. Stating the method categorically, suppose the shaft and the position of the planes 1, 2, 3, 4 (Fig. 47) to be given. Take any point B. There is no restriction in the selection of this point; it may be taken anywhere. Join the points on the axis to B, forming a pencil of four rays. The directions of these rays fix the relative positions of the crank angles, though in drawing the cranks from them the reciprocal relation already stated must not be forgotten. That is, the ray B4 (the 4 referring to the figure 4 on the shaft) gives the direction of No. 1 crank (Fig. 48), B1 No. 4 crank, B3 No. 2 crank, and B2 No. 3 crank. The way the cranks radiate from the axis is best fixed by drawing parallels to the side of the force polygon; ambiguity of sense is thereby avoided. 52 THE ARA LA MCING OF EAWGIWES. To determine the force polygon, draw any line across the pencil, as AC in the figure, parallel to the axis of revolution. This determines the two sides AB and BC. Draw CD and AD parallel to B3 and B2 respectively. Then the sides of the polygon so formed represent the masses at crank radius which will be in balance for the crank angles determined by the position of the point B and the given planes of revolution. The method of fixing the crank angles for which balance is possible, by drawing a pencil of rays to a point B, was indicated by Dr. Schubert, in Mr. Schlick’s paper on Balancing Engines, Insti- tute of Naval Architects, 1900, and earlier in a paper contributed by Dr. Schubert to the Hamburg Mathematical Society, 1898. 38. Experimental Apparatus. The principles of this chapter may easily be verified experimentally by means of the apparatus shown in Fig. 49. A wood frame carries an accurately turned and well-mounted steel shaft, on which are arranged four carefully turned and balanced discs. One disc, shown to the front in the . . . . . . . . . . . . . figure is fixed to the shaft and & carries a protractor, the others # are capable of angular and longitudinal adjustment rela- tively to the shaft. The radius defined by a disc is fixed by a small hole, drilled near the periphery. The system is driven by a motor also carried on the = | frame, so that the discs may —ºl be driven at any speed by FIG. 49. the motor, free from the action of any external driving force. The only unbalanced forces acting on the system are therefore those due to the rotation of the discs, which should be nothing. This is tested by slinging the frame on chains in the way shown in the figure. Any want of balance is at once apparent, when the System is driven, by the vibration of the apparatus. Assuming THE BAZAA’CING OF REVOLVING MASSES. 53 the system to be balanced, the four discs now serve to carry any assigned set of crank-pin masses. These are bolted to the discs at the crank-pin holes (two such masses are shown in the figure), the discs are set to the proper crank angles by means of the pro- tractor on the front one, and the proper distance apart by means of the longitudinal adjustment. Any want of balance is at once shown by the oscillations set up when the system is driven. The first apparatus of this kind was designed by Professor Ewing for the Engineering Laboratories at Cambridge. CHAPTER III. THE BALANCING OF RECIPROCATING MASSES,-LONG CONNECTING-ROIDS, 39. The Force required to change the Speed of a Mass of Matter moving in a Straight Line.—The last chapter was devoted to the consideration of the effect of forcing a system of masses to move in circular paths at uniform speeds. Turn now to the case where the natural straight path of a mass is not interfered with, but the speed in that path is changed from instant to instant by the action of a force. In general, if M is the mass in pounds of the moving body, and A the acceleration produced by the force F, acting at the maSS Centre— F = MA poundals Ol’— F = MA lbs. weight (ſ Since in balancing problems the magnitudes of the forces are generally not concerned, it is more convenient to use the first expression, avoiding thereby the introduction of g into the work. In the steam-engine mechanism, when the motion of the crank-pin is given, the corresponding acceleration of the piston can be found for any given position of the gear, though the expression giving it is a complicated one ; its consideration will be deferred for the present. On the other hand, if the connecting-rod be imagined infinitely long, the expression is a simple one. In many cases in practice the difference between the true accelerating force acting on the piston, and the force calculated on the assumption that the 54 RECIPROCATING MASSES.—LONG CONNECTING-RODS. 55 connecting-rod is infinitely long, is small enough to be negligible. The present chapter is concerned in showing how the effects of the accelerating forces, acting on - sº e Z the reciprocating masses, may Fixed be balanced, supposing these 2 * masses to be operated by in- == finitely long connecting-rods, # that is to say, supposing their rº- motion to be simple harmonic. 40. Value of the Accele- ration of One Set of Re- ciprocating Masses, when the Connecting - rod is infinitely Long.—The motion given by the crank (Fig. 50) to the slotted bar, which here repre- sents a set of reciprocating parts, is precisely the same as if the bar were connected to the crank by an infinitely “-s º long connecting - rod. The o " mechanism is, in fact, a / ractical way of expressing | ul * - fr}<- --- 71 - 5 ^-1 : T tal 0: N83. FIG. 57. IFIG. 58. which may be made by crank No. 4, since, in whatever direction it is placed, the mass it operates has no moment about the reference plane, and consequently it may be fixed in any desired position without disturbing the balance amongst the couples. Choosing a convenient scale, make Ab (Fig. 57) = to 5:1, be = 6'49 and parallel to crank No. 2, cd = 7-025 and parallel to crank No. 3. The polygon fails to close by the side dA. - Close it by means of the fourth crank. Thus, dA is the direction of crank No. 4 relatively to the others, and its length RECIPROCATING MASSES.—LONG CONNECTING-RODS. 69 represents the equivalent mass of the reciprocating parts attached to the crank—dA scales 4.95 tons. Check the work in this way— Suppose the reference plane to be at No. 1 crank. Make a new schedule for the masses with reference to this plane, including, of course, No. 4 crank. Draw the couple polygon. If it closes, the work is correct. A consideration of the couple triangle (Fig. 57) will show that the lightest mass should be placed in plane No. 1; that in plane No. 2 the crank should be arranged oppositely to crank No. 1, otherwise the mass at No. 3 would have to be relatively very great to effect balance. 49. Example.-Given the cylinder centre lines of a four-cylinder FIG. 59. N23. Nº.2) FIG. 60. engine, find the crank angles and the masses so that the recipro- Cating parts may be in balance amongst thomselves. It will be noticed at once that not enough data are given to 70 THE BAZAAVC/AWG OF EAVGIZVE.S. solve the problem. There are nine variables concerned in the balancing, and of these five must be given to get a solution. Since only three of the variables are fixed by the cylinder centre lines, two more must be fixed. Let these be two angles. Assume the angles, which are shown in Fig. 59, between cranks 1, 2, and 3. Take a reference plane at No. 4 cylinder. Set out AB (Fig. 60) in the direction of No. 1 crank, and make it = 100, to Some Con- venient scale. Draw BC, CA parallel respectively to No. 3 and No. 2 crank; measure them off, and enter them in Column II. of Schedule 7. SCIIEDULE 7. RECIPROCATING MASSES. Teference plane at No. 4 crank. Column I. | cºlumn II. Number of crank. Distance from reference Proportional equivalent ..º.º." plane. mass or centrifugal force, fugal couple, when when co”R = 1. g ºp" i. { Fect. No. 4 ... e - e. O-0 2.45 -* No. 3 ... tº º º 12-16 3.72 45'2 No. 2 ... tº e - 22.87 4.375 100.0 No. 1 ... e - tº > 33-04 3.025 100-0 The actual couples about No. 4 plane must be in the ratio— 45°2 : 100 : 100 for balance Divide each of these by the corresponding distance from the reference plane given in the schedule. The quotients are the ratios of the masses at crank-pin radius. These are entered in Column I. To find the proportional number for the mass at No. 4 crank, set out (Fig. 60)— Ab = 3.025 parallel to No. 1 crank, bc = 372 parallel to No. 3 crank, cd = 4:375 parallel to No. 2 crank RECIPROCATING MASSES.—LONG CONAVECTING-RODS. 71 dA, the closure, gives the direction of crank No. 4, and its length, 2.45, is the proportional equivalent mass number. The masses must be adjusted so that they are in the ratio of— 2.45 : 372 : 4:37 : 3.02 50. Example.—Both the preceding examples ignore the effect of the valve-gear. The next example includes it, and is worked out in somewhat greater detail to serve as a typical illustration of a way of dealing with torpedo-boat engines. The peculiarity of the problem is that the eccentric sheave angles are functions of the corresponding main crank angles. The method of dealing with the problem will be apparent in the working out. Given the centre lines of four cranks and the corresponding ahead and astern eccentric sheaves; the mass of the different parts of the valve-gears and the mass of one piston; to fix the crank angles and the masses of the pistons so that the reciprocating masses may be in balance amongst themselves, and to find the balancing masses for the crank-shaft. Fig. 61 shows the crank-shaft and centre lines. Above each crank and sheave is written the reciprocating and revolving masses at the crank radius, it being understood that the con- necting-rods and eccentric rods are included by the method of Art. 46. In calculating the reciprocating masses, it may be noted that the engine is supposed to be in forward gear, and that the equivalent reciprocating mass for the ahead eccentric in each case includes the mass of the valve, valve spindle, etc., an appropriate part of the eccentric rod, and one-half the link. The valve motion of No. 4 crank cannot be taken into con- sideration in the general method, because its crank angles are functions of crank No. 4, the last angle to be determined. Assume the mass of the reciprocating parts of No. 1 crank to be 1000 pounds. Fill in Schedule 8. 72 THE BALA/VC//VG OF AZAVGIZVE.S. SCHEDULE 8. RECIPROCATING MASSES. Reference plane at No. 4. E º at E º, * Number of crank. º, ; *: wº. º o°R = l. co"R = 1. No. 4 ast h º; * O. 4 aStel’n eCC, Shea V6 — `" º No. 4 ahead ecc. sheave – 3-0 140 }Not included No. 4 crank ... tº tº º 0-0 Unknown (1040) e=== No. 3 crank ... © º º 4-0 Unknown (1275) | Unknown (5,100) No. 3 astern ecc. sheave 7.0 60 420 No. 3 ahead ecc, sheave 7.3 150 1,095 No. 2 crank ... tº e e 10-0 Unknown (1500) | Unknown (15,000) No. 2 astern ecc. sheave 13-0 70 910 No. 2 ahead ecc. sheave | 13.3 150 1,995 No. 1 crank ... tº G & 16-0 1000 16,000 No. 1 astern ecc. sheave 20-0 70 1,400 No. 1 ahead ecc. sheave 20-3 145 2,943 Crank angles between Nos. 1 and 2, 200°; between 1 and 3, 100° (Fig. 70). Set out each crank and the centre lines of its eccentric sheaves (Figs. 62, 64, 66, and 68). The sheaves are all drawn for an angular advance of 30°. Draw (Fig. 63) AB, BC, CD parallel to OK, OF, OE (Fig. 62) respectively, and representing on some convenient scale the corre- sponding couples given in the schedule. Mark these vectors care- fully with arrows to indicate their way of action. Set out (Fig. 65) A1B1, BICI, CID, parallel to OK, OF, OE (Fig. 64) respectively, BICI, CID1 representing the corresponding couples. AIIS1 is of indefinite length, since the couple due to crank No. 2 is not known. Draw similarly Aobo, BoCo, CoL)0 (Fig. 67) for crank No. 3, leaving Co.L)0 indefinite as to length. There is now a choice of two ways of continuing the work. (1) Assume values for the unknown couples of Figs. 65 and 67. (2) Assume the angles between cranks Nos. 1, 2, and 3. Proceed by the latter method, assuming the angles between cranks Nos. 1 and 2 to be 200°, and between 1 and 3, 100°. IFIG. 61. REVOLVING MASSE5 FIG. 62. <º & P / L^ A1 FIG. 69. FIG. 65. Bl A D1 Y % \ O O 7. FIG, 71. wo W ...A.------ § N83. 5 100 Nº-º--º º ... • * ... • * D Sº {\l C 74 THE BA LA WCIAWG OF EAVGIWE.S. Set out (Fig. 70) the angle between No. 1 and No. 2 cranks to be 200°, and between Nos. 1 and 3 100°. Draw the couple polygon thus— Trace the vectors of Figs. 65 and 67 on separate pieces of tracing-paper. Pin the tracing from Fig. 67 over the vector draw- ing of Fig. 63 by a single pin through A0 on the tracing and D on the drawing. Similarly, pin the tracing of Fig. 65 through Di on the tracing and A on the drawing. Turn the tracings round their pins until AID1 is parallel to crank No. 2, and Co.L)0 to crank No. 3. X, the intersection of the indefinite lines A1B1 and CoID0, defines their length, and therefore the couples for cranks Nos. 2 and 3. Measure these off and enter them in the schedule. Divide each by its appropriate distance from the reference plane, and enter the force so obtained in Column I. It is at once settled that the reciprocating crank-pin mass for No. 2 crank is 1500 pounds, and for No. 3 crank 1275 pounds. Draw a force polygon whose sides are parallel to the couple polygon (Fig. 63), and proportional to the numbers in Column I. (Fig. 71). It fails to close by the vector VA. The valve-gear of No. 4 crank may now be included. To do this, draw ab, be, cd (Fig. 69) parallel to OK, OF, OE (Fig. 68). Make be, cd proportional to the corresponding equivalent masses. The magnitude of ab is as yet unknown. Pin the tracing of these vectors over Fig. 71, d on the tracing being over A on the drawing. Turn it until the indefinite line passes through V. Vb represents the mass of the reciprocating parts of No. 4 crank, and gives the direction of the crank. This crank, is added to Fig. 70. Care must be taken to see that in these vector polygons the arrows are all pointing in the same way. Notice that the valve-gear of No. 4 crank is balanced as regards forces, but not as regards couples. These couples are of small magnitude, and may be neglected. They can be balanced, however— (1) By repeating the whole process and including the couples in the couple polygon, assuming the radii of the two sheaves in question to have the directions found by the first application of the method. (2) By two reciprocating masses arranged in two planes. Pumps worked from the shaft may sometimes be arranged to do this. RECIPROCATING MASSES.—LOWG CONNECTING-RODS. 75 (3) By considering them as couples to be balanced with the revolving masses of the crank-shaft. Assume that it can be done by No. 2 method. 51. Balancing the Crank-shaft. The crank-shaft may be balanced in either of two ways— (1) by the addition of two balancing masses in two separate planes of revolution; (2) by the extension of the arms of each crank to form balancing masses for the revolving masses of that crank. The first case is treated by the general method; the second case by Art. 12. In the first case a crank-shaft in which there are any number of cranks may be balanced by the addition of two masses only. In the Second case there are as many balancing masses as there are crank-arms, but this method has the advantage that the inter- mediate parts of the crank-shaft have not to transmit force from one crank to another, since each mass is balanced in its own plane of revolution. The shaft is thus freed from bending moment due to this cause, so far as the revolving masses are concerned. The addition of balancing masses to the crank-shaft may be avoided altogether if it is balanced in the vertical plane only, leaving the component force and couple in the horizontal plane unbalanced. This may be done in marine engines in those cases in which the ship's hull is stiff enough horizontally to render the disturbances due to a crank-shaft unbalanced horizontally negli- gibly small. To balance an engine in this way, add the revolving masses and the reciprocating masses in each plane together, and proceed by the method of the previous article. Many examples of combining reciprocating and revolving masses into one schedule will be found in the next chapter. Teturning to the example of the previous article, balance the crank-shaft by the first method. The crank angles (Fig. 70) are all fixed by the conditions of balance found for the reciprocating masses. The equivalent revolving masses are all given in Fig. 61. Fill in Schedule 9. 76 THE BALANCING OF ENGINES. SCHEDULE 9. REVOLVING MASSES. Reference plane at No. 4. * Equivalent mass Equivalent mass Number of crank. º lººk when to"R = 1. when coºl? - 1. Feet. Pounds. No. 4 astern ecc. sheave ... tº s ºf – 3:3 70 — 231* No. 4 ahead ecc. sheave ... tº E tº – 3:0 70 — 210* No. 4 crank e tº gº tº e e & E & 0-0 500 *== No. 3 crank gº tº º © & © e & © 4-0 600 2400 No. 3 astern ecc. ... tº ſº tº tº º ſº 7:0 70 490 No. 3 ahead ecc. ... tº gº tº tº tº º 7.3 70 511 No. 2 crank * {º º tº º 0 * * * 10-0 650 6500 No. 2 astern ecc. ... * G \ tº º º 13-0 80 1040 No. 2 ahead ecc. ... tº º q tº q tº 13.3 80 1064 No. 1 crank & © tº e e © tº dº 16-0 500 8000 No. 1 astern ecc. ... • * & © tº 20-0 75 1500 No. 1 ahead ecc. ... tº g tº to tº º 20-3 75 1522 Fig. 72 shows the couple polygon for Schedule 9. It fails to close by the vector VO, which measures 1200 foot-pounds. This couple may be balanced in any convenient way. Obviously, from an inspection of the direction, it would be most convenient to attach a balancing mass opposite crank No. 1. This is 16 feet from the reference plane. The mass required at the crank radius is therefore 75 pounds. Fig. 73 shows the corresponding force polygon. This must be drawn to include the 75 pounds added to balance the couples. The polygon requires the vector V101 to close it, viz. a mass of 12 pounds at the crank radius and in the reference plane in the direction indicated. This could easily be arranged for on crank No. 4. Thus the engine is completely balanced both for reciprocating and revolving masses by the addition of the small masses of 75 and 12 lbs. * Note the way of drawing these vectors, viz. TowARDs the centre of the crank- shaft. The force vectors remain positive. See Art. 28. RECIPROCATING MASSES.—LOWG CONNECTING-RODS. 77 An end view of the crank-shaft centre lines is shown in Fig. 74. FIG. 72. FIG. 73. # 75 POUNDSAT NPI.PLANE à Nº.3 |2 POUNDS AT N94.PLANE No | - FIG, 74. 78 THE BAZAAVC/WG OF EAVGINES. 52, Conditions that an Engine may be balanced without the Addition of Balancing Masses either to the Reciprocating Parts or to the Crank-shaft.-If an engine has four cranks or more, the reciprocating parts may be balanced amongst themselves without the addition of balancing masses in the way already exemplified in Arts. 48, 49, and 50; and this necessarily fixes the crank angles. The distances between the planes of revolution of the revolving masses are practically the same as the distances between the cylinder centre lines, since the crank-arms, etc., are symmetrical with respect to their respective cylinder centre lines, and therefore the mass centre of each revolving mass is in the same plane in which the corresponding reciprocating mass moves; hence, for the re- volving masses to be in balance, the force and couple polygon corresponding to them must be similar to the force and couple polygon belonging to the reciprocating masses, and therefore the revolving masses must be in the same proportion to one another as the reciprocating masses. Or, briefly, no balancing masses will be required if– (1) The mass centres of the revolving and reciprocating masses of each line of parts are in the same plane; (2) The masses of the reciprocating parts are in the same proportions as the masses of the revolving parts; (3) The reciprocating system is balanced by the method of Art. 47. Tor instance, in the example of Art. 49, the masses of the reciprocating parts must be in the following proportions for balance— 2.45 : 3.72 : 4:37 : 3.02 and these, therefore, represent the ratios of the revolving masses if no balancing masses are to be added to the Crank-shaft. This condition rarely obtains in practice, and it is, therefore, necessary to add masses to the crank-shaft in order to balance it. 53. To find the Resultant Unbalanced Force and Couple due to the Revolving and Reciprocating Parts together.—The way to find the unbalanced force and couple for a given system of revolving masses has been given in Art. 31, and the method of estimating the unbalanced force and couple due to a given system of reciprocating masses, moving with simple harmonic motion, RECIPROCA TIAWG MASSES.—ZONG CONWECTING-KOZ).S. 79 has been given in Art. 45. The resultant force or couple at any instant is the resultant of the two vectors representing the force belonging to each system, or of the vectors representing the couples. . Suppose the vector OA (Fig. 75) represents the unbalanced force belonging to the revolving system attached to the shaft, the end of which is shown at O, and OR the unbalanced force of the imaginary revolving system re- placing the reciprocating system operated by the same shaft. The projection of OR on ZZ, that is, Or, is the instantaneous value of the unbalanced reciprocating force (Art. 45). The resultant of OA and Or, that is, OB, is the instantaneous value of the whole disturbing force. It may be shown that the locus of the point B is an ellipse. Notice that OR and OA revolve with the shaft, and that the angle a between them is constant. The couples may be dealt with in a similar way. This need not be further pur- sued, because a method will be 2. given in Chapter VI. by means of FIG. 75. which both the unbalanced force and couple may be found exactly for short rods by a simple geometrical process. 54. Experimental Apparatus.-Fig. 76 shows an apparatus which has been designed by the author to illustrate the principles of balancing reciprocating parts. There are four cranks, all mutually adjustable, three of the flanges of the crank-shaft being divided into degrees for this purpose. Adjustable masses may be fixed to the tails of the respective piston-rods. Of the nine variables concerned in the balancing of a four-crank engine six are susceptible of 80 THE BALANC/VG OF EAVG/NA.S. variation in this apparatus. Suspended from a frame as shown, its motions when running unbalanced exhibit the way a marine engine tries to wobble when running under similar conditions. Properly balanced, it hangs motionless at all speeds, showing only FIG, 76. a little uneasiness when the speed is passing through the natural period of oscillation of the supporting springs. Placed on rollers in the way shown in Fig. 101, it shows the effect on the tractive force due to the unbalanced parts of a four- cylinder locomotive. In this model the revolving parts are so arranged that their balance is not disturbed by any adjust ment that may be made in the reciprocating parts or crank angles. If provided with short connecting-rods, it serves also to illustrate the principles of Chapter W. 55. Balancing Reciprocating Masses by the Addition of Re- volving Masses—The balancing masses found by Art. 30 must, of course, be reciprocated, form- ing, with the given system of reciprocating masses, a system in a balance. If they are added to the system as revolving masses, being attached to the crank-shaft, although they do produce balance amongst the reciprocating masses, they introduce at the same time forces at right angles to the plane of reciprocation exactly equal to the forces they are balancing in the reciprocating system, though differing in phase by 90°. It may be that the unbalanced forces in this direction are less serious than in the original direction, though in general RECIPROCATING MASSES.— LOWG CONNECTING ROD.S. 81 balancing in this way cures one trouble only to introduce another. This case is similar, though opposite, to the method mentioned in Art. 51 of avoiding the addition of balancing masses to the crank- shaft by arranging the reciprocating parts to balance it vertically, leaving it unbalanced horizontally. The method of procedure is alike in both cases, and, in the case of locomotives, is fully illus- trated in the next chapter. CHAPTER IV. THE BALANCING OF LOCOMOTIVES, 56. General Consideration of the Effects produced by the Un- balanced Reciprocating Parts of a Locomotive.—The disturbances caused by the machinery of a locomotive may be divided into those due to the revolving and reciprocating masses respectively. There is no need to discuss the effects due to the unbalanced revolving masses, crank-arms, crank-pins, coupling-rods, etc. These always can and always should be balanced forming a revolving system in equilibrium at all speeds, and affecting, therefore, neither the tractive force nor the rail pressure. The effect due to the unbalanced reciprocating masses may be investigated by reducing them to a reference plane taken centrally, at right angles to the axis of the driving-axle. The unbalanced forces then reduce to a single force and a couple. In Fig. 77 the left-hand crank of a driving-axle belonging to either an inside or outside cylinder engine is shown on the trailing dead centre, and it stands to the front of the reference plane. The right-hand crank is, of course, behind the plane. The reciprocating system consists of two masses, each equal to M pounds say, con- nected to cranks at right angles by rods which are relatively long with respect to the crank radius. The unbalanced force and couple are found by the method of Art. 45. Thus Oab is the force triangle in which Oa and ab are equal, each representing M: the vector Ob, therefore, represents the unbalanced force of the system of revolving masses corresponding to the reciprocating masses. Its projection on the line of stroke, Oa in Fig. 77, Oc in Fig. 78, where the crank axle has turned through the angle 6, 6 being the angle between the 82 THE BAZAAVC/WG OA. A.OCOMOTIVE.S. 83 centre line and the left-hand crank, represents the instantaneous value of the unbalanced force in the centre line of the engine. 2 Majºr lbs. weight, the Since for a given speed Oa = ab = numerical value of this force in terms of the angle 6 is— v/2 Majºr cos (9.4-45°) lbs. weight 9 or, in terms of the revolutions per second of the crank-axle n, M being in pounds and r in feet, it becomes— Unbalanced force in lbs. * = łº, cos (9 + 45°) ap- acting at the centre line proximately . . . (1) Similarly, OAB (Fig. 77) is the couple triangle in which OB R. R. b M z |Line of strokeyº * D * = . * y” g of * ~. & * |- LINE 4 of TRACTion FIG. 77. FIG. 78. represents the value of the unbalanced couple for the corresponding system of revolving masses; its projection, OA in Fig. 77, OD in Fig. 78, therefore, represents the instantaneous value of the un- balanced couple acting in the plane containing the centre lines of the cylinders. This couple oscillates the engine about a vertical axis, or one very nearly vertical when the cylinders are inclined. This axis is always at right angles to the plane of reciprocation. Majºra. foot-lbs., where a Since for a given speed OA = AB = is equal to half the distance between the cylinders, the numerical value of the couple in terms of the angle 6 is— v/2Moºra cos (9 – 45°) 9 84 THE BAZAAVC/WG OF EAWG/WE.S. or, in terms of the revolutions per second n, and the distance between the centre lines of the cylinders, d- ‘85Mn”rd cos (9 – 45°) foot-lbs. approximately . . . (2) The maximum values of the force occur when Ob turns into the line ZZ, that is, when 0 = -45°, or 135°; and the maximum values of the couple when OB turns into the line ZZ, that is, when 6 = +45°, or 225°. There is still another disturbance due to the fact that the centre line of the crank-axle is above or below the line of traction. If the distance between the line of action of the force and the line of traction is t (Fig. 78), the transference of the force to the line of traction is equivalent to (see Art. 24)— (1) An equal and parallel force in the line of traction; (2) A couple whose arm is t. This couple, which is continually varying in magnitude and sign, tends to oscillate the engine in a vertical plane. This would be perceived most with comparatively short tank engines with small wheels. Assuming the distance of the line of traction above the rails to be about 3 feet 4% inches, with a 6 feet 9 inch driving- wheel, this cause of disturbance would be entirely absent, since t = 0. It may be noticed incidentally that the tractive pull of the engine must be transferred from the plane of reciprocation to the line of traction, which transference gives rise to a couple varying in magnitude, but acting to turn the engine about a horizontal axis always in the same direction so long as the engine is pulling. The effect of this couple is to cause a redistribution of the weights on the wheel. If the centre line of the driving-axle is above the line of traction, the load on the leading end of the engine is increased, on the trailing end decreased. On the other hand, if the centre of the driving-axle is below the line of traction, the contrary is the case. Also the turning couples on the crank-axle are necessarily accompanied by equal and opposite turning couples on the engine as a whole. These turning couples are variable, and the variation for each crank and of the total are shown for a particular case in Fig. 91. Since the load is brought on to the main bearings by means of springs, the variations of the turning couples set up Oscillations which are superposed upon the Oscilla- tions due to the inertia forces of the unbalanced machinery. The THE BALANCING OF LOCOMOTIVES. 85 disturbances due to the variation of turning moment cannot be eliminated by the addition of balancing masses; they can be minimized, however, by placing the cylinders as close together as possible. A discussion of this question in connection with marine engines is given in Arts. 130–132. Summarising, if— M is the mass in pounds of the reciprocating parts belonging to each cylinder; T the crank radius in feet ; n the revolutions of the crank-axle per second ; d the distance in feet between the cylinder centre lines; t the distance between the centre line of the driving-axle and the line of traction; the unbalanced reciprocating parts cause— (1) An unbalanced force, the maximum value of which is given by 1:7Mnºr lbs. weight. This force accelerates the whole mass of the train positively and negatively in the direction of travelling. (2) A couple whose maximum value is 85Mn”rd foot-lbs. This couple produces an oscillatory motion about a vertical axis, which, Superposed upon the general forward motion of the engine, causes a swaying from side to side, which, acting on a short engine, may become dangerous at high speeds. The effect of this couple must be judged with reference to the moment of inertia of the engine about a vertical axis through its mass centre. The couple is much less in magnitude for inside cylinders than for outside cylinders, since it varies directly as d. (3) A couple whose maximum value is 1:7Mn°rt foot-lbs. This couple tends to cause oscillation in a vertical plane about a horizontal axis. Its magnitude is usually small, and it disappears altogether if the driving-wheel radius is equal to the height of the line of traction above the rails. 57. Example.—The mass of one set of reciprocating parts of an unbalanced locomotive is 600 pounds, and the cylinders are 2-feet pitch. Find the maximum values of the unbalanced force, and of the couples about a vertical and a horizontal axis, assuming the diameter of the driving-wheel to be 4 feet 6 inches, and the line of traction to be 3 feet 4, inches above the rail, the speed 86 THE BAZAAVC/AWG OF EAVG/AWE.S. being 38.5 miles per hour, and the stroke 26 inches. Find also the values when 0 = 30° (see Fig. 78). The number of revolutions of the driving-wheel per second at the given speed = 4 approximately. M = 600 pounds, and r = 1.08 feet; therefore— (1) Maximum value of the = 1.7 × 600 × 16 × 1-08 unbalanced force = + and — 17,625 lbs. weight + and – 7-87 tons weight This is greater than the average tractive force exerted by the engine. (2) Maximum value of couple about a vertical axis = + and – 17,625 foot-lbs. For outside cylinders d is about 6 feet, so that the value of this couple would in this case be increased to 52,875 foot-lbs. (3) Maximum value of the couple about a horizontal axis is, t being 1125 feet— 17,625 × 1.125 = + and – 19,828 foot-lbs. When 6 = 30°, Instantaneous value of the unbalanced force is— 17,625 cos (30° -- 45°) = 17,625 x 26 = 4582 lbs. weight Instantaneous value of the couple about a vertical axis— 17,625 cos (30° - 45°) = 17,625 x 96 = 16,920 foot-lbs. Instantaneous value of couple about horizontal axis— 4582 × 1.125 = 5155 foot-lbs. This example sufficiently illustrates the effects of the un- balanced reciprocating masses. If the revolving masses are left unbalanced as well, the maximum values are very much increased, and in addition they introduce a couple acting about a longitudinal axis, which causes a variation of the driving-wheel rail-load. * It is instructive to set out the maximum values found above to Scale on a piece of cardboard, dotting in the position of the cranks, in the way shown in Fig. 78. Pin the cardboard to the THE ARA LAMC/NG OF ZOCOMOT/VE.S. 87 drawing-board through O, and bring the edge of the T-square up to the centre. Then turn the cardboard disc, which really repre- sents the reference plane, into various positions: the changing values of the force and swaying couple, represented by the projec- tions of Ob and OB respectively, on to the edge of the T-square, may then be easily studied. 58, Method of Balancing the Reciprocating Masses of a Locomo- tive.—To properly balance the reciprocating masses requires the addition of two more sets of parts reciprocated in the same plane, forming either a two-cylinder engine with two sets of “bob- weights,” as suggested by Mr. Yarrow, or a four-cylinder engine in which the crank angles are found by the principles of Chapter III. At the present time no locomotives in this country are balanced with bob-weights, and no four-crank locomotive has been built in which the crank angles and masses are arranged so that the forces and couples balance, the usual arrangement being to place the four cranks at right angles, in which case the forces are, or rather may be, completely balanced, but the couple cannot be balanced without the addition of balance weights. Four-cylinder engines will be discussed more fully later on. It is almost the universal custom to partially balance the force and couple due to the reciprocating parts in a two-cylinder engine, by masses placed between the spokes of the driving-wheels, these being combined with the masses balancing the revolving parts to form a single pair of balancing masses, or, to use the more familiar term, balance weights. The addition of revolving balancing masses to the reciprocating system introduces forces exactly equal to the forces they balance, acting at right angles to the plane of reciprocation, which, in the present case, causes a variation of the pressure between the wheels and the rail, a variation which in extreme cases is sufficient to double the rail-pressure per wheel at one instant, and lift the wheel clear of the rail in the next, the interval of time correspond- ing to half a revolution of the wheel. This variation of rail- pressure is injurious to the permanent way, to the bridges, and to the engine tyres, and should be kept as Small as possible. The designer has therefore to keep in mind two contradictory conditions. If the reciprocating parts are fully balanced by revolving masses, there is no unbalanced force, and no swaying couple, but there 88 THE BAZAAVC/AWG OF EAWG/WE.S. is a large variation in the rail-pressure. If, on the other hand, the reciprocating parts are left entirely unbalanced (the revolving parts are assumed to be completely balanced), there is no variation of rail-pressure, but there is an unbalanced force and a swaying couple which make it dangerous to run at high speeds. A compromise is usually made, a common practice in this Country being to balance about two-thirds of the reciprocating parts. The effect of the unbalanced part will be examined in detail in Arts. 75 and 76. The next three examples represent typical cases of locomotive balancing. The method followed is to take a set of reciprocating parts and consider them common to different classes of engines. The set of parts taken, dimensions and revolving parts where possible, are those common to a large number of engines on the Lancashire and Yorkshire Railway, the data of which has kindly been supplied by Mr. Aspinall. When revolving masses are used to balance the reciprocating parts of an engine, it is unnecessary to divide the work into two stages, as directed in Art. 47. The proportion of the reciprocating masses to be balanced is to be included with the revolving masses; the revolving balance weight for the two systems is then found at one operation, that is, by one schedule. Following the usual custom, two-thirds of the reciprocating parts are balanced in the typical examples given in Arts, 62, 63, and 64. 59. A Standard Set of Reciprocating Parts (Lancashire and Yorkshire Railway. Cylinder, 18 inches diameter × 26 inches stroke)— 1 piston, 18 inches diameter & tº ... 146 pounds 2 piston-rings tº gº tº tº gº tº gº tº e. • * * 13 , 1 piston-rod and 1 crosshea e & e ... 151 X 3 1 nut ë e e tº º º gº tº º * g tº º 6 55 1 crosshead pin ... tº e t e tº tº tº º 17% , 2 slide-blocks tº g tº e º 'º tº º ve & ſº we 66 ,, Total ... 399% , The connecting-rod weighs 444 pounds, and this mass is to be divided between the reciprocating masses and the revolving 7THE BAZAAVC/AVG OA' LOCOMOTIVES. 89 masses by the method of Art. 46. The position of mass centre is 659 the length, measured from the small end, therefore— ‘659 × 444 = 292; pounds is to be included with the revolving masses, the rest, 151; pounds, being included with the reciprocating masses. This gives finally— Mass reciprocated by the connecting-rod ... 399, pounds Proportion of mass of connecting-rod ... 151%. , Total reciprocating mass per cylinder ... 551 ,, 60. The corresponding Revolving Parts of the Crank Axle are— 1 pair of Crank-arms tº ſº e ... 296 pounds at 13 inches 1 crank-journal ... & is ſº ... 56 , 55 Proportion of connecting-rod ... 292; , > 2 Total revolving mass per crank-pin 644; ,, 25 61. Scales.—In the following examples the distances from the reference plane are expressed in inches. As a consequence, the mass moments, or couples, are given by numbers involving some- times five figures. It will be sufficiently exact for all practical purposes if the scale to which the couple polygons are drawn is chosen so that three significant figures can be read, a fourth being estimated, the fifth being considered zero. The scale of the force polygon should allow three significant figures to be read, 62. Balancing an Inside Cylinder Single Engine.— DATA. Stroke ... $ $ tº tº º e tº gº tº & © & ... 26 inches Distance centre to centre of cylinders ... ... 1 foot 11 inches Distance between the planes containing the mass centres of the balance weights ſº e ge 4 feet 11 , Mass of unbalanced revolving parts per crank- pin reduced to 13 inches radius ... * * tº Mass of reciprocating parts per cylinder at crank radius tº gº º {º º ſº tº tº ſº i.e. º º ... 551 , Proportion of reciprocating parts to be balanced, two-thirds 644 pounds '6L '91'ſ '08 '9IJI A–=—\ I | l d T°478 N —s | | | | | | -T-- ‘SãTTIOIT ºn | \ | SS ‘H. § G-- + D THE BAZAAWCING OF LOCOMOTIVE.S. 91 C B FIG, S1. Masses to be balanced are therefore— Revolving ... e tº e - - - • * > ... 644 pounds # reciprocating ... • * * * * * ... 367 , Total at each crank-pin * - tº ... 1011 ,, Draw the plan and elevation of the crank-axle as shown in Figs. 79 and 80. Notice that the left-hand driving-wheel shows to the front in elevation. Choose a reference plane to coincide with the plane containing the mass centre of the right-hand balance weight, and mark on the drawing the three dimensions i, j, and k. The balance weights are found by the general method. The quantities concerned are shown in Schedule 10. SCHEDULE 10. Inside cylinder single engine. Reference plane at No. 1 (Fig. 80). ! -> Equivalent mass Equivalent mass Distance from at crank radius, moment, Number of crank. reference Or OT plane. centrifugal force, centrifugal couple, when co-R=unity. when toº R-unity. Inches. Pounds. No. 1. R.H. balance weight * - © () 766 O No. 2. R.H. crank ... tº a tº e a • 18 1011 18198 No. 3. L.H. crank ... e * * * 41 1011 41451 No. 4. LH, balance weight ... 59 766 45220 92 THE BAZAAVC/AWG OF EAVGINES. ABC (Fig. 81) is the couple polygon, the closure CA measuring 45,220. This represents the product of the mass of the left-hand balance Weight and its distance from the reference plane, which distance is 59 inches. The mass of the balance weight is therefore 766 pounds. Its angular position in relation to the cranks is at once given by drawing QQ (Fig. 79) parallel to CA (Fig. 81), remembering to draw from the centre of the axle in the direction from C to A. The balance weight is shown in black. The force polygon is Abcd (Fig. 81). Its closure measures 766 pounds, and this is the mass of the balance weight in the right-hand wheel, its angular position being defined by the direction dA (Fig. 81). The balance weight is shown dotted. It is unnecessary to take a new reference plane to check the Work, since the polygons check one another when the masses at each crank-pin are equal, and their planes of revolution and the planes in which the balance weights are placed are symmetrically disposed with reference to the central plane of the engine. Under these conditions the balance weights are equal in magnitude, and their angular positions are symmetrical with respect to the cranks. One balance weight is found from the couple triangle ABC (Fig. 81), the other is therefore known, and the drawing of the force polygon Abcd is therefore only necessary to check the accuracy of the work. The actual mass M0 of the balance weight depends upon the distance R of its mass centre G from the axis. If r is the crank radius, Mo is found from- MoR = Mir = 766r for the present example. Taking r = 13 inches and R = 36 inches, which would be about the practicable distance for a 7 feet 3 inches wheel— M0 = 376 lbs. This should be arranged in crescent form between the spokes, as shown in Fig. 79. 63. Balancing an Outside Cylinder Single Engine,—The re- volving parts in an engine of this class consist of the crank-arm THE BAZAAVC/AWG OF I.OCOMOTYWE.S. 93 formed by the protrusion of the wheel-boss between the spokes of the wheel, the crank-pin, and a proportion of the connecting-rod. The planes in which the mass centre of the crank-arms revolve are not, as in the previous example, coincident with the planes in which the centres of the crank-pins revolve. There are therefore six masses to consider, revolving in six planes, as shown in Fig 83. DATA. Stroke ... tº g tº g º g is g tº & ſº tº ... 26 inches Distance centre to centre of cylinders ... ... 6 feet 13 inches Distance between planes containing the mass centres of the balance weights de & © ... 4 », 11 , Distance between the planes containing the mass centres of the wheel-cranks ... 4 tº e ... 5 , 1} , Reciprocating mass per cylinder tº tº gº ... 551 pounds Unbalanced mass of one crank-arm and the part of the crank-pin therein reduced, to 13 inches radius ... tº º º e º 'º © º e tº gº tº ... 130 ,, Unbalanced mass of the part of the crank-pin and washer outside the crank, 25 pounds; together with 292 pounds of the connecting- rod, both reduced to 13 inches radius ... 317 , Mass at each crank-pin to be considered in the balancing— Revolving ... ſº tº ſº * - © sº º e ... 317 pounds # reciprocating ... e tº a $ 8 º' ... 367 , Total at each crank-pin tº £ tº ... 684 , The plan and elevation of the crank-axle are shown in Figs. 82 and 83. The cranks are arranged as in the previous case; that is, the left-hand crank is to the front and horizontal. Take the reference plane at No. 3 to contain the mass centre of the right- hand balance weight. Fill in Schedule 11. l tº. * © º F==}=l N°2 REF. PLANE | A / U N°3 (. X J I | ! | § tº Q CO So So | | | s FIG, 83. THE BAZAAWC/NG OF / OCOMOTIVE.S. 95 J’ T- . FIG. 84. SCHEDULE 1.1. Outside cylinder single engine. Reference plane at No. 3 (Fig. 83). Equivalent mass Equivalent mass Distance from at crank radius, moment, Number of crank. reference Or OT plane. centrifugal force, centrifugal couple, when toº R=unity. when to R-unity. Inches. Pounds. No. 1. R.H. crank-pin — 72 684 — 4925 No. 2. R.H. wheel-crank — 1'4 130 – 182 No. 3. R.H. balance weight 0 905 O No. 4. L.H. balance weight 59-0 905 53395 No. 5. L.H. wheel-crank 60°4 130 7852 No. 6. L.H. crank-pin (36-2 G84 45280 ABCDE (Fig. 84) is the couple polygon. Observe that the vectors CD and DE are drawn oppositely to the direction of the right-hand crank, because these cranks are on the opposite side of the reference plane to the left-hand crank (Art. 28). The closure EA measures 53,395. The mass of the balance weight is therefore 905 pounds, and its angular position is found by drawing QQ (Fig. 82) parallel to EA (Fig. 84). Abedef is the force polygon (Fig. 84), the closure f4 measuring 96 THE BA ZANC/AWG OF EAWG/WE.S. 905, thereby checking the work. Remember, in drawing the force polygon, that the direction of drawing is always from the axis out- wards to the mass, so that the force vectors cd and de are in the direction of the right-hand crank, and therefore in the opposite direction of the couple vectors CD and D.E. The balancing masses at crank radius are shown in Fig. 82, the left-hand black, the right-hand dotted. 64. Balancing an Inside Cylinder Six-coupled Engine.—(The data for this example correspond with the standard 18 inches by 26 inches six-coupled goods engine of the Lancashire and Yorkshire Railway. Wheels, 5 feet 0% inch diameter.) The new feature in this example is the coupling-rod. Each coupling-rod is to be divided between the three outside crank-pins in the propor- tion that they respectively support its weight since the rod is made with a joint near the centre pin. The proportion in which the division is to be made may be arrived at expeditiously by placing the rod on three knife-edges at its pin centres, each knife-edge being suitably supported on the platform of an independent weigh- ing-machine; or by separately weighing each part of the coupling- rod in the way indicated in Art. 46. In the present example the leading and trailing wheels each take 143 pounds per crank-pin, and the driving-wheel 257 pounds. The total mass of each rod is 543 pounds. Very often the radius of the outside cranks is less than the radius of the inner cranks, so that care must be taken that the masses at the outside Crank-pins are reduced to the inside radius before including them in the schedule. DATA FOR THE DRIVING-WHEEL (Figs. 85 and 86). Stroke ... tº e º tº e & g & & tº º ſº ... 26 inches Tadius of outside cranks tº £ tº tº a tº ... 10 , Distance centre to centre of cylinders ... ... 1 foot 11 inches Distance centre to Centre of coupling-rods ... 6 feet 1; , Distance between planes 2 and 3 containing the mass centres of the wheel-cranks ... ... 5 , 1} , Distance between the planes containing the mass centres of the balance weights & a tº 4. , 11 , Unbalanced mass at each outside crank-pin in planes 1 and 8, made up of 257 pounds of THE BAZAAVCIAVG OF ZOCOMOTIVES. 97 the coupling-rod, and 25 pounds for the crank- pin and washer, in all 282 pounds at 10 inches radius, equivalent at 13 inches radius to tº £ tº & ſº º gº tº º 9 tº e ... 217 pounds Unbalanced mass of each wheel-crank and part of pin in it reduced to 13 inches radius, in planes 2 and 6 tº º ſº tº º tº e ſº tº ... 96 , Unbalanced mass of revolving parts at each inside crank-journal ... tº gº tº tº tº gº ... 644 , Mass of reciprocating parts per cylinder ... 551 ,, Mass to be considered at each inside crank-journal in the balancing is— Revolving ſº tº tº * G tº tº ſº tº tº e G 644 pounds # reciprocating © tº ºt © tº º e is tº 367 , Total per crank ... e tº º ... 1011 ,, |Fill in Schedule 12. SCHEDULE 12. Six-coupled inside cylinder engine. DRIVING-WHEEL. Crank radius = 13 inches. Reference plane at No. 3 (Fig. 86). Equivalent mass | Equivalent mass Distance from at crank radius, Imoment, Number of crank. reference Or OT plane. centrifugal force, centrifugal couple, when co"R=unity. when o’R=unity. Inches. Pounds. No. 1 — 7-2 217 — 1,562 No. 2 — 1'4 96 — 134 No. 3 0-0 494 0 No. 4 18-0 1011 18,198 No. 5 41-0 1011 41,451 No. 6 59-0 494 29,140 No. 7 60-4 96 5,798 No. 8 (36-2 217 14,365 ABCDEFG (Fig. 87) is the couple polygon, in which the closure GA measures 29,140; dividing this by 59, the quotient 494 gives the balance weight for the left-hand wheel. Abcdefgh is the force polygon. The closure ha measures 494, thereby checking the work. H FIG. 85. WHEEL DRIVING FIG. 86. THE BA LA WCIAVG OF LOCOMOTIVE.S. 99 FIG. 87. Notice that the couple vectors EF and FG are drawn oppositely to the corresponding crank directions, because they are on the opposite side of the reference plane to the rest of the cranks. (See Art. 28 for the rules for drawing couple vectors.) Leading-wheel.—The unbalanced masses are entirely revolving, and are therefore to be completely balanced. They consist of the crank-arm, crank-pin, and a proportion of a coupling-rod, in the present example 143 pounds. The masses form a system similar to the system considered in Art. 63. Fig. 83 may be taken to repre- sent the plan of the leading-wheel of the present example. The left-hand outside crank is there shown to the right ; it should, of course, be to the left to correspond with the driving-wheel of Fig. 85, but no confusion is possible, since in whatever position the wheel is placed, the position of the balance weights is found relatively to its cranks. ADDITIONAL DATA. Mass due to coupling-rod ... e in 6 tº º e ... 143 pounds Part of crank-pin outside the crank, and Washer ... 25 5 5 Total in planes Nos.1 and 6 at 10 inches radius 168 ,, 100 THE BALA NCIAVG OF EAWGIWES. Mass of wheel-crank and the part of the crank-pin in it, revolving in planes Nos. 2 and 4 at 10 inches radius tº dº tº n tº a tº ſº tº tº tº º ... 125 pounds Fill in Schedule 13. SCHEDULE 13. Six-coupled inside cylinder engine. LEADING-WHEEL. Crank radius=10 inches. Reference plane at No. 3 (Fig. 83). Equivalent mass | Equivalent mass Distance from at crank radius, moment, Number of crank. reference OT Ol' plane. centrifugal force, centrifugal couple, when o'R=unity. when to"R-unity. Inches. Pounds. No. 1 — 7-2 168 — 1,209 No. 2 — 1'4 125 – 175 No. 3 0-0 317 0 No. 4 59-0 317 18,720 No. 5 60-4 125 7,550 No. 6 66-2 168 11,122 The couple and force polygons corresponding to the schedule are shown in Fig. 89. The balance weights they determine are shown on the elevation re-drawn in Fig. 88. The balancing of the trailing-wheel is the same as the balanc- ing of the leading-wheel in every respect. The radius of the mass centre of each of the actual balance weights may be taken at 1 foot 10 inches. Therefore the actual balancing masses are— Driving-wheel -*; is = 292 pounds. Leading and trailing-wheel = sº = 138 , The angles (measured from the drawings of the polygons), which the respective radii of the left-hand masses make with the hori- Zontal, are— Driving-wheel, 43° below the centre line : Leading and trailing-wheels, 4° below the centre line. THE BAZAAVC/AWG OF LOCOMOTIVE.S. 101 The left-hand side of the engine is shown in Fig. 96, the cranks being shown in their proper relation to one another, and the balancing masses in black. FIG. 88. LEADING AND TRAt LING WHEELS. I ()2 THE BAZAAVCIAWG OF EAWGINES. 65. Wariation of Rail-load: “Hammer Blow,”—The variation of load on the rail caused by the vertical component of the centri- fugal force due to the part of the balance weight concerned in balancing the reciprocating parts is called the “hammer blow.” This description of the effect does not describe what takes place very well, because the variation of load is not sudden, but continuous, except in the extreme Case where the maximum value of the variation is greater than the weight on the wheel; in this case the wheel lifts for an instant, and gives the rail a true blow in coming down. To estimate the variation of load on one rail in any given case, the balance weight concerned in balancing the reciprocating parts must be separated from the main balance weight. The quickest way to do this is to find the balance weight for the proportion of the reciprocating masses balanced, neglecting altogether the revolving masses. The schedule for this problem would be similar to Schedule 10. It is merely necessary to write in for the mass at each crank-pin the proportion of the reciprocat- ing parts to be balanced. A more convenient way is to consider the crank-pin mass unity. Then, in the couple polygon (Fig. 81), AB would represent the dimension j, BC the dimension i. The closure is therefore given by— CA = A/? + jº and the magnitude of the balance weight for unity mass by— CA 1 Then, if M is the mass in pounds of the reciprocating parts per crank-pin and q the fraction of this quantity which is to be balanced, the magnitude, m, of the balance weight required is— 7?? - ºv* + ſº pounds . . . . . . (2) The value of the angle of direction is given by— tan 6 = –4 . . . . . . . . (3 J (3) considering AB to be the initial direction. THE BAZAAVC/WG OF LOCOMOTIVES. 103 Let w be the variation of rail-load, that is, the vertical com- ponent of the centrifugal force due to m, a the instantaneous value of the angle, measured in the positive direction, that is, counter-clockwise between the line of stroke and the radius of the balance weight m, 7 the crank radius in feet, a the angular velocity of the wheel in radians per Second. Then— 2 ** sin a lbs. weight . . . . . (4) Q0 = The sign of w is determined by the sign of sine a ; a positive value indicates a diminution of rail-load, a negative sign an increase. If V is the speed of the train in miles per hour, and D the diameter in feet of the driving-wheel, containing the balance weight— 2 × 5280V * = T3600D Substituting this in 4, and dividing by 2240 to obtain w in tons weight (m is in pounds)— Q 2nny V2 . w = ** sin a . . . . . . (5) A further variation of the rail-load is brought about by the obliquity of the connecting-rods. Considering the R.H. rod, the transmission of force along it is accompanied by a force acting on the slide-bar, and an equal and opposite force acting at the centre of the driving-axle. These two forces, in fact, form a couple instanta- neously equal to the turning couple acting on the crank. In the Ordinary stationary engine this couple tends to turn the engine- frame as a whole. In a locomotive, the main bearings being spring- connected to the frame, and free to move relatively to the frame in the direction in which these forces act, the 'effect is somewhat different. Supposing the engine to be running forward, the force acting on the slide-bar is upwards for the greater part of the stroke. The equal and opposite force acting at the driving-axle causes an increase of the rail-load. The effect of the force on the slide-bar is to lift the leading end of the engine slightly, thereby changing the loads on all the springs and causing a slight diminution of the load on the driving-springs, the amount of which it would be im- practicable to calculate. It is certain, however, that this diminution 104 THE BAZAAVC/AWG OF AEAVGINES. is some fraction of the force acting at the bars because the other springs share the effect, and the total change in load on all the springs must be just equal to the slide-bar force supposing no Oscillations are going on. The variation of rail-load due to the obliquity of the connecting-rod is therefore made up of two parts— the One, a force equal and opposite to the force at the slide-bars acting directly on the driving-axle; the other, a change in the load transmitted to the axle by the driving-springs in consequence of the action of the force at the slide-bars on the frame. These two parts are always of opposite sign, and the second part is always less than the first, so that if +f denotes a force acting upwards at the slide-bars, – f is the force acting at the driving-axle, and if i and k have the same meaning as in Fig. 80 (k now being taken equal to the distance between the wheel treads), the instantaneous value of the variation of the rail-load at the left-hand wheel is given by— J × # minus change of load on L.H. Spring. Similarly for the R.H. wheel. The changes from the left-hand gear added to these give the total results. The magnitude of these quantities depends upon the cut- off in the cylinders, upon the speed, upon the method of springing the engine, upon the strength of the springs, and upon the ratio of the length of the connecting-rod to the crank. The later the cut- off the greater the average value of f, the higher the speed the more uniform the instantaneous values, and the longer the Con- necting-rod the less the values of f. To find the value off for a given crank position, deduce the resultant driving pressure from the indicator diagrams, and take from this the force required for the acceleration of the reciprocating parts. This nett driving pressure multiplied by the tangent of the angle the connecting-rod makes with the line of stroke gives the value off. Multiply f by the distance between the crosshead centre and the centre of the driving-axle, and the product is the turning couple (Art. 130). The nett value of the driving pressure in a particular case is shown in Fig. 95, and the corresponding turning couple by the curve marked L.H., Fig. 91. Compared with the effect of the balance weights, this cause of Variation of rail-load is practically negligible at high speeds. The TAZAZ AAAAAVC/WG OF ZOCOMOTY VES. 105 effects of balance weights increase as the square of the speed, the value of f decreases as the speed increases, in engines of the ordinary proportions. 66. Example—Consider that the example of Art. 62 represents a 7-foot inside single. The value given by equation (1) of the previous article is approximately 0.76; therefore the magnitude of the balance weight required to balance the reciprocating masses is— 0.764M pounds If the whole of the reciprocating parts are balanced, q = 1, and M = 551 pounds; therefore— m = 0.76 × 551 = 419 pounds Let W be 60 miles per hour, and D, the diameter of the wheel, P 22. 9 I. SPEED GO M.P.H. 7' WHEEL | R MASS OF RECIPROCATING PARTS 551 LBS, STROKE 1.08'. FIG. 90. }< 7 feet. Then, from equation (5) of the previous article, the crank radius being 1:08 foot— w = 4 sin a . . . tons weight nearly. When this balance weight is passing through its highest and 106 THE BALANCING OF EAVGIAWES. lowest positions respectively, sin a is +1 and — 1. The load on the rail is decreased in the first case by 4 tons, increased in the second by 4 tons. Supposing the load on the axle to be 15 tons, that is, 7% tons per wheel, at every revolution the load per wheel is alternately decreased and increased by about 54 per cent. If two-thirds of M be balanced, the percentage variation is reduced to 36 per cent. Taking a horizontal base line through 0 (Fig 90) to represent the circumference of the driving-wheel, curves Nos. 1 and 2 respectively represent the variation of the rail-load when the whole and when two-thirds of the reciprocating parts are balanced. The static load on the wheel is represented by the ordinate to the horizontal line PQ. The part of a vertical line cut off by the shaded figure therefore represents the load available for adhesion, at 60 miles per hour, when the point at which it cuts the circumference is in contact with the rail, assuming two-thirds of the reciprocating parts to be balanced. 67. Speed at which a Wheel lifts.--When small wheels are used, the piston speed increases for a given speed of travelling, and the rail-load variation must be carefully considered in the balancing, or the wheels may leave the rail altogether at every revolution, a mistake in design not entirely unknown in practice. The formula, equation (5), Art. 65, may easily be adjusted to find the speed at which lifting takes place. Let W be the static load on the wheel. The rail-load at any instant is given by— W – 70 If w is numerically equal to W, this becomes 0 when the balance weight is passing through its highest position, and 2W when passing through the lowest. Hence, putting W for w in equation (5), Art. 65, and solving for V, sin a being 1– D2W Wo = ––––– . . . . . . . 0 0-00012m.” (6) Vo is now the speed in miles per hour at which the rail-load vanishes when the balance weight passes through its highest position. Taking the data of the example of Art. 66, where W = 7-5 tons, and m = 419 pounds for full balance— Vo = 82 miles per hour approximately THE BALANCING OF LOCOMOTIVE.S. 107. If m = 280 lbs., thus balancing two-thirds of the reciprocating Iſla SS6S- W = 100 miles per hour approximately. These two calculations show that two-thirds is about the greatest proportion of the reciprocating masses which should be balanced in a single engine, and in a coupled engine also, if the balancing mass m is all put in the driving-wheel. Although the rail-load vanishes, slipping may not occur, because the other wheel on the same axle may be able to provide sufficient adhesion at the instant. To detect if slipping is about to take place, the turning effort on the crank must be compared with the couple resisting slipping, this couple depending upon the instantaneous sum of the rail-loads. 68. Slipping.—The driving-wheels tend to slip when the turning effort on the driving-axle is equal to the couple resisting slipping. The forces of this latter couple are, the frictional resistance at the rail and the equal, parallel, and opposite tractive force at the driving- horns; the arm of the couple is the radius of the driving-wheel. The force due to the frictional resistance varies directly as the pressure between the wheel and the rail. If W, is the static load on the two driving-wheels, wi the resultant variation of rail-load for the two wheels, the greatest value of the frictional resistance is about— W, - 101 5 Therefore, if the turning couple on the driving-axle is greater than the couple— (***)R . . . . . . . d) R being the radius of the driving-wheel, slipping will tend to take place. The resultant of the right-hand and left-hand balance weights concerned in balancing the reciprocating parts is equal and opposite to the resultant of the proportion of the reciprocating masses balanced, the latter being considered concentrated at the respective crank-pins. This latter resultant (Art. 56, Fig. 77) is equal to Mºv/ 2 pounds at 45° to each crank. The resultant balance weight 108 THE BALANCING OF ENGIWAES, is, therefore, a mass Mav2 pounds, attached to an imaginary crank at the centre of the driving-axle, placed at 135° to each crank, as shown in Fig. 92. The instantaneous value of the whole variation of the rail-load for the two wheels is given by the pro- jection of the force vector belonging to this imaginary central crank on a line at right angles to the plane of the rails. If 6 is the angle between the left-hand crank (Fig. 78) and the line of stroke, measured counterclockwise, the magnitude of this projection is— * 2n, ** sin (0-225) Ibs, weight. or in terms of the revolutions per second, n, of the crank-axle, M being in pounds, r in feet, and dividing the constant by 2240– wi = 000772Mnºr sin (0-225) tons weight . . . (2) The instantaneous value of the couple resisting slipping is therefore— R w — 00077.0 Mnºr sin (9–1– º foot-tons . (3) 5 • U & This is a maximum when the imaginary central crank is passing through its lowest position, that is, when 0 = 45°; and a minimum when passing through the highest position, that is, when 0 = 225°. For instance, if W = 16 tons, on 7-feet driving-wheels, the mass of the reciprocating parts per cylinder being 551 lbs., of which q = 3 are balanced, stroke 26 inches, the minimum value of the resisting couple when n is 4 per second, corresponding to 60 miles per hour, is— 16 – 00077 × 4 × 551 x 16 × 1-08 sº. 5 } = 7-77 foot-tons The value corresponding to the static load of 16 tons is 11.2 foot-tons. 69. Example.—Tofurther illustrate this point, the actual driving effort or torque is compared with the couple resisting slipping for a complete revolution in Fig. 91, in the case of a Lancashire and Yorkshire 4-coupled bogie express passenger engine running at 65 miles per hour, taking two-thirds of the reciprocating parts to THE BAZAAVC//VG OF COCOMOTIVE.S. 109 th 2 O 8 º }- O O ta. i —?— to.…T `s RESULTANT BALANCE W &y: § * LBS, / WZ 7 | \ ---. 110 THE BAZAAVC/WG OF EAWG/WAZ.S. be balanced in the driving-wheels, which are 7 feet diameter; cylinder, 18 inches diameter; 26 inches stroke. The ordinates of curve No. 1 (Fig. 91) represent the torque or driving-couple acting on the driving-axle, those of curve No. 2 the couple resisting slipping. It will be noticed that the two ordinates are nearly equal for crank position 1. A little more steam and curve No. 1 would have cut curve No. 2, and if this had been a single engine, slight slipping would be the result. In the case in question, the coupled wheels would come into play and prevent it. Between crank positions 7 and 8 there is a large difference in the ordinates, and slipping is not to be feared; the instantaneous load on the rails is increased from 16%, however, to 218 tons. The rail-load corresponding to the minimum value of the resisting couple is 11:2 tons, so that the rail-load under the driving-wheels at 65 miles per hour is continually varying between 112 and 21.8 tons once per revolution, that is, 4.2 times per second. The method of drawing the curves is as follows:— Fig. 92 shows the cranks and the resultant balancing mass, which is 520 pounds. Notice the way the crank positions are numbered round from the initial position of the left-hand crank. The instantaneous angular distance of the radius of the resultant mass from the initial position, 0, of the left-hand crank is given by 0 + 225°, 9 being measured downwards from the horizontal centre line. The left-hand crank shows to the front and the engine is running forward. (a) Divide the crank circle into twelve or more equal parts, and find the corresponding positions of the crosshead graphically. (b) Find the nett driving pressure from the indicator cards which are shown in Fig. 93, by taking the intercepts between the steam-line of one diagram and the exhaust-line of its fellow. The shaded parts of the diagrams show the width to be taken for the left-hand end. These are plotted in Fig. 94, curve No. 1, for both ends, and calibrated to give the total pressure acting on the piston in tons. The numbers on the horizontal axis are those corresponding to the numbers on the crank circle in Fig. 92. (c) The pressures of Fig. 94 are modified by the forces required to accelerate the motion of the reciprocating masses. These are quickly found by Klein's construction (Art. 104). The curve representing them is No. 2 (Fig. 94). The driving pressures on the piston are decreased by the accelerating forces during the first THE BAZAAWCING OF LOCOMOTIVE.S. I 11 part of the stroke, and increased during the second part. The vertical width of the shaded figure gives the instantaneous value of the force operating at the crosshead to turn the crank for any given crank angle. These widths have been re-plotted in Fig. 95. - TOTAL I. H.P. 548, ...T. Boüer Press.160lbs. Cutoff 20% à Speed, 65 M.P.H. É.- Revs. 251 per Mirv. %2 Spring 100. FIG. 93. : FIG. 94. : : FIG. 95. Notice how much more uniform this driving force is made by the action of the accelerating force. In the neighbourhood of the points 5 and 11 the driving force vanishes and becomes negative, that is, the crank is now driving the piston instead of the piston the crank. These are the points at which a change over takes place all through the driving system. The pressure between the connecting-rod brasses and the crank-pin changes from one side. to the other at these points. The slide-blocks leave the top bar I 12 THE BA/A/VC/AWG OF EAVG/AVES. for the bottom bar in forward running, returning to the top bar at positions 6 and 12. These changes are accompanied by a knock if there is any slack at the places where they occur. (d) The crank-effort diagram may be constructed by any of the usual methods from the curve of pressures in Fig. 95. A full discussion of this, and a simple construction for the purpose, are given in Art. 130. The curve marked L.H. in Fig. 91 is the crank effort or torque curve corresponding to the driving pressures of Fig. 95. The curve corresponding to the right-hand crank is assumed to be the same in form; the left-hand curve is, therefore, simply redrawn with an angular difference of 90° to get the crank- effort curve for the right-hand crank. The two are then added to get curve No. 1 (Fig. 91), giving the total crank effort in terms of the position of the left-hand crank. (e) The data for drawing the curve resisting slipping are W = 16, tons; R = 3-5; M = 551 lbs. ; q = 3; n = 4:2; and 7 = 1.08 feet. Expression 3, Art. 68, reduces to— Resisting couple = 11:55 + 378 sin (9 + 225°) the positive sign being used because the angle 0 is measured counter- clockwise and downwards from 0 to the left-hand crank radius. This is represented by curve No 2, Fig. 91. The average crank effort is 5:12 foot-tons. The average resisting couple is 11:55 foot-tons, more than double the average torque. Judging from these figures alone, there would appear to be ample margin to prevent slipping, and yet, as the diagram shows, the engine would be just on the point of slipping if it were not prevented by the coupled wheels. 70. Distribution of the Balance Weights for the Reciprocating Parts amongst the Coupled Wheels—A way of decreasing the variation of rail-load in coupled engines is to divide the balance weight necessary to balance the reciprocating parts, amongst the coupled wheels. The effects of these separate weights on the engine-frame add up to the same horizontal effect as that due to the single balance weight m in the driving-wheel. The variation of rail-load is reduced at the driving-wheel, a proportional variation being introduced at the coupled wheels to which part of the THE BAZAAVC/WG OF / OCOMOTYVE.S. 113 balance weight is transferred. There is also a redistribution of pressures at the horns. To illustrate this method of distribution, consider the example of Art. 64 again. Fig. 96 shows the crank circles drawn out with the balancing masses, shown in black, already found for the com- LEADING. DRIVING. TRAILIN G. 218 LBS, FIG, 9S. plete balance of the revolving parts and two-thirds of the recipro- cating parts, the latter being balanced in the driving-wheel. The balance weight required in the driving-wheel to balance the revolving masses alone is 248 pounds, placed as shown by the dotted circle. The balance weight (m of Art. 65, equation (2)) required to balance the reciprocating parts alone is 279 pounds, placed as shown by I 114 TATE BA LA WCING OF EAVGINES. the full open circle. The black mass of 494 pounds is the resultant of these two. w Draw lines O.Qi and O902, in the leading and trailing wheels respectively, parallel to the radius OQ in the driving-wheel, and place one-third of the 279 pounds, that is, 93 pounds at 13 inches, in each wheel. This is equivalent to 120 pounds at 10 inches, the radius of the cranks for the leading and trailing wheels. Considering the leading or trailing wheel, the 120 pounds due to transferred mass combines with the 317 pounds already found for the revolving masses to form a resultant mass of 218 pounds at 10 inches radius, placed as shown in Fig. 97. Re- combining the 248 pounds balancing the revolving masses of the driving-wheel, with the 93 pounds left of m, a resultant mass of 324 pounds is obtained. The new balance weights are shown in Fig. 97. Balancing is effected to precisely the same extent by them as by the heavier set of Fig. 96; the variation of the rail- load is reduced by two-thirds, though there is now a load variation under each of the coupled wheels. Fig. 98 shows the set of balance weights which will balance the whole of the reciprocating parts (551 lbs.). In this case m, the mass corresponding to the open circle in the driving-wheel of Fig. 96, is 419 pounds, Art. 66 giving 140 pounds at 13 inches radius per wheel to be combined with the masses balancing the revolving parts. The result of this combination is to give 178 pounds in the trailing and leading wheel, and 364 pounds in the driving-wheel, a set of masses weighing less than the sets of Figs. 96 and 97, though balanc- ing the whole of the reciprocating parts, and causing a much less variation of rail-load than the first set. The three sets of balance weights are drawn one under the other for the sake of comparison. This method of distribution is unquestionably the best way of dealing with whatever proportion of the reciprocating parts may be balanced so far as the permanent way is concerned. The variation of the tractive effort may be completely balanced since the whole of the reciprocating masses may be balanced without introducing too great a variation of the rail-load. The division between the coupled wheels may be made in any proportion. Having decided what proportion is to be balanced in the different wheels, the proper balance weights are found directly by the use of a schedule of the style of No. 12. Each wheel, in fact, THE BA LAAWC/NG OF LOCOMOTIVES, 115 to which a part of the reciprocating mass is to be transferred is to be looked upon as a crank-axle coupled by an imaginary connecting- rod at the imaginary inside crank-journals to the real inside crank- journals belonging to the crank-axle, and carrying the part of the reciprocating mass assigned to it at the crank-journals. If an engine without a bogie or small leading-wheel is to be balanced in this way, it would be advisable to assign a less proportion than one-third to the leading-wheel if the engine is to run very fast. 71. American Practice.—Mr. Henszey, of the Baldwin Locomo- tive Works, has kindly furnished the following details of their practice :— All the revolving parts are balanced and two-thirds of the reciprocating parts on single-expansion engines, three-quarters on the Vauclain Compound. The weights balancing the reciprocating parts are distributed equally between the coupled wheels. One- third of the connecting-rod is included with the reciprocating parts, two-thirds with the revolving parts. This is the distribution for a rod whose mass centre is '66 × l, measured from the small end. The coupling-rods are “weighed ” (see Art. 46), to find the mass to be assigned to each crank-pin. The parts are balanced as though their respective mass centres revolved in the same plane, that is, the balance weights are put exactly opposite the cranks. 72. Example—Eight-coupled Engine, Class “E,” Baldwin Company.−Fig. 99 shows the arrangement of the wheels. 2&" +: - 2- - LZ Nº. N. WHEELS 4.8° DIAM. FIG. 99. 2- N Mass of reciprocating parts, including piston, crosshead, one- third of connecting-rod = 1170 lbs. Of this two-thirds is balanced, which, equally distributed between the four wheels, gives 195 lbs. per wheel. 1 16 THE BAZAAVC//VG OF EAWGIWE.S. The mass to be balanced in each wheel is made up as follows:— Wheel numbers. No. 3. No. 4. No. 5. No. 6. Reciprocating parts equally distri- buted tº e tº tº $ tº ... 195 195 195 195 pounds Revolving parts— Two-thirds connecting-rod ... — — 464 — , Coupling-rod & ſº ºf ... 169 214 265 106 ,, "Wrist-pins ... tº $ tº ... 73 90 275 86 , Crank-hubs ... e tº s ... 184 204 272 204 , At 14 inches radius ... 621 703 1471 591 , At 16+inches the radius of the mass centres of the balance weights 531 605 1267 508 , 73. Four-cylinder Locomotives.-The reciprocating masses in a four-crank locomotive may be arranged to balance amongst themselves without the use of balance weights at all. Under these circumstances, always supposing the revolving masses to be balanced, there will be no variation of rail-load, no unbalanced force, and no horizontal swaying couple. The engine will, in fact, be perfectly balanced, neglecting the error due to the obliquity of the connecting-rod. The crank angles involved in balancing four reciprocating masses amongst themselves require the employment of a separate set of valve-gear per cylinder. Considerable mechanical simplicity may be obtained by arranging the cranks in two pairs, the two cranks in each pair being at 180° with one other, the pairs themselves being at 90°. If the reciprocating masses be equal, the force polygon would close, forming a square or a right angle returning on itself; there would be, therefore, no unbalanced force. The couple polygon, however, would not close; there would be, therefore, a horizontal swaying couple left whose maximum magni- tude is by the principles of Art. 56— 1.7Mnºbrcos (9 -- 45°) where b is the distance between the cranks forming a 180° pair. The disturbing effect of this couple will depend upon the moment of inertia of the engine about a vertical axis through its mass THE BALANCING OF LOCOMOTIVES. 117 centre. The longer and heavier the engine, and the more the mass is grouped at the leading and trailing ends, the less the disturbance. If this couple is left in there will be no variation of rail-load because there are no revolving masses in the system applied to balance reciprocating masses; if, however, revolving masses are added to balance this couple, they introduce a variation of rail-load. If the engine will run steadily at high speeds, there is no doubt that the best thing to do under the circumstances is to leave the swaying couple in, so that the engine may run without variation of rail-load. If this couple is left unbalanced, the whole static load on the wheel is always available for adhesion. If, however, there is much swaying, masses may be put in the driving- wheels to reduce it. An example of a successful four-cylinder engine in which advantage is taken of the four sets of reciprocating parts to avoid the use of balance weights is furnished by the four-cylinder com- pounds introduced on the London and North Western Railway in 1897, by Mr. F. W. Webb, for running between Euston and Crewe without a stop with the heavy trains for the North. A description and drawings are published in Engineering, December 3, 1897. The cranks are arranged in two 180° pairs at right angles, and only two sets of valve-gear are employed, one set being arranged to work the two valves of a 180° pair. There is another point in connection with the balancing of this engine which should be noticed. Each inside crank is balanced by prolonging the crank-arms on the opposite side of the axle to form a balance weight. In this way the loading of the axle with centrifugal force between the wheels is avoided. In the usual arrangement, where the balance weights are placed in the wheels, although the axle, as a whole, is thereby freed from dynamical load due to the rotation of the masses belonging to each crank, yet the axle has to transmit the centrifugal force from each revolving mass to the planes in which the balance weights are placed, thereby causing a bending moment on the axle. To emphasise this point, consider the case of a crank-axle of the usual type for an inside cylinder engine where each crank and the part of the connecting-rod included with revolving masses is equivalent to 700 pounds at a radius of 1 foot. At 60 miles per hour, with a 7-foot driving-wheel, the axle is making 4 revolutions per second. The centrifugal force corresponding to this is 6'12 tons 118 THE BA/AMC/AWG OF EAVGINES. weight. At 80 miles per hour the centrifugal force increases to 109 tons per crank. So that at this latter speed acting at each inside crank is a force of 10-9 tons due to the motion alone. This force loads the axle almost as severely as the full steam pressure at starting, so far as the bending moment is concerned. 74. Crank Angles for the Elimination of the Horizontal Swaying Couple.—If four sets of valve-gear be employed, the crank angles may be arranged for complete balance in a large number of ways for a given set of cylinder centre-lines. The only solution which is practicable for coupled engines is that in which the outside cranks are at right angles. Consider the case of a symmetrical engine in which the pitch of the outside cylinders is 6'7 feet and the pitch of the inside cylinders 22 feet. Proceeding by the method of Art. 49 or Art. 37, it will be found that the crank angles must be those shown in Fig. 100, and that each set of inside reciprocating parts must be 2:27 times a set of outside recipro- cating parts. Under these cir- cumstances there would be no unbalanced force, no variation of rail-pressure, and no swaying º; °º flººr couple, except to the extent in- FIG, 100. troduced by the obliquity of the connecting-rods, which in the case of locomotives where the ratio of the rod to the crank is always relatively great, is negligible. Up to the present, no loco- motive in this country has been balanced so perfectly as this, and whether the gain in Smoothness of running and absence of vibration is worth the extra mechanical complication, or whether an engine with such crank angles would work well on the road, are points which can only properly be decided by a practical trial. It should be noticed that an engine with the crank angles of Fig. 100 has only one dead centre at a time. To obtain a good crank-effort curve the work may be distributed amongst the cylinders by the principles of Art. 131. The revolving parts must be balanced independently as a separate system. See Art. 51. INSIDE RIGHT MASS-2-2 OUTSi DE LEFT MASS = 1 . 75. Estimation of the Unbalanced Force and Couple.—If the proportion 7 of the reciprocating masses is balanced, there THE BALA WCIAWG OF LOCOMOTIVE.S. 119 remains (1 – q)M pounds unbalanced. This causes disturbances like those enumerated in Art. 56. The maximum value of the unbalanced force is— 1.7M(1 — q)nºr lbs. weight, from equation 1, Art. 56 the maximum value of the swaying couple is— ‘85M(1 – 7)n°rd foot-lbs., from equation 2, Art. 56 similarly— 1.7M(1 – q)n°rt is the maximum value of the couple acting in a vertical plane. These values are, of course, only true for engines of the Ordinary type, that is, with equal reciprocating masses symmetrically arranged. For any other type, two-cylinder compounds with unequal reciprocating masses, four-cylinder engines with arbitrarily determined reciprocating masses, etc., the general method of Art. 45 must be used to find the magnitudes of the closures to the force and couple polygons. 76. Comparative Tables,—The following schedules have been calculated for the purpose of comparing the different types of engines with regard to their possibilities of balance. Schedule 14 gives the general formulae for the different cases. The expressions in column A. are reduced from the formula— 122 Mºlº. weight . . . . (1) formed by the combination of formulae 4 and 2, Art. 65. The Values of i, j, k, assumed in calculating the quantities stated, are those given in Fig. 80, viz. 18, 41, and 59 inches respectively, for the inside cylinder engines, and those given in Fig. 83, viz. 7-2, 662, and 59 inches respectively, for the outside cylinder engines. Column B. is calculated from— 1.7M(1 – 1)n” . . . . . . . (2) and column C. from- '85M(1 — q)n°rd . . . . . . . (3) The errors of the four-cylinder engines are estimated in the way indicated in Art. 73. 120 THE BA/A/VC/WG OF EAVG/AWE.S. SCHEDULE 14. M = the mass of the reciprocating parts per cylinder in pounds. 2' = the crank radius in feet. m = the number of revolutions of the crank-axle per second. d = the distance between the cylinder centre lines in feet. A. jº |B. C. Proportion of Value or the Maximum tº * º tion of | \, Maximum Type of engine. nº lºº. ºf lºº, per cylinder (q.). P..." * º;. (hammer- blow). Inside single All 0-93 Mº?r Nil Nil Outside single All 1-38M%20° Nil Nil Inside single q = 3 0.62Mn2r 0.57Mººr 0°28Mm?rd Outside single q = } 0-92Mn2). } } 2 ) te sº q = } e 2 Inside 4-coupled { # per whe al 0.31 Mºr J) 5 * tº * Q = # Q | Outside 4-coupled t per whe o 0-46Mºr } } • ? te $º Q - 3. • 9 Mao 2a, Inside 6-coupled º per visa) O-21M172, 2 * | l º *s q = } • 21 Man 2 Outside 6-coupled { per wied) O-31Mºr 5 * 5 * 4-cylinder engine, two 180° pairs of cranks at right angles ... © tº º All Nil Nil 1-73Mn2rd 4-cylinderengine (Fig.100) All Nil Nil Nil The actual values of the different quantities in Schedule 14 are given in Schedule 15, on the assumption that the mass of the reciprocating parts per cylinder is in each case 551 lbs. weight, that the stroke is 26 inches, and that the number of revolutions of the crank-axle is 4 per second. The piston speed corresponding to this assumption is 1036 feet per minute. For the inside cylinder engines the distance between the cylinders, d, is taken, 1.92 feet (Fig. 80); for the outside cylinders, 6-117 feet (Fig. 83). For the four-cylinder engine the distance, d, between the two cranks forming a 180° pair is assumed to be 2 feet. When the recipro- Cating parts are balanced by masses in the driving-wheel alone, the quantities for the coupled engines are the same as for the corresponding type of single engine. THE BALANCING OF LOCOMOTIVES. 121 SCHEDULE 15. Mºn” = 9521. M = 551 lbs. n = 4 revolutions per second. * = 1.08 foot (26-inch stroke). Piston speed, 1036 feet per minute. Type of engine. Inside single Outside single Inside single Outside single e Inside 4-coupled ... Outside 4-coupled Inside 6-coupled ... Outside 6 coupled 4 - cylinder engine, 180° pairs of cranks at right angles 4-cylinder engine (Fig. 1 Reciprocating mass balanced per cylinder. lbs. 551 551 367 367 183 per wheel 183 per wheel 122 per wheel 122 per wheel ! Maximum value of the variation of rail-road ' per wheel in lbs. weight (hammer- blow). i 8,854 13,138 5,902 8,759 2,951 4,379 1,999 2,951 Nil Nil Maximum unbalanced force in lbs. weight. Nil Nil 5426 Nil Nil value of the i : i Maximum value of the swaying couple in foot-lbs. Nil Nil 5,117 16,302 5,117 16,302 5,117 16,302 32,942 Nil The results of Schedule 15 should be considered, together with Schedule 16, to form an idea of the approximate magnitude of the forces and couples corresponding to a given speed. 122 THE BAZAAVC/AWG OF EAWGYNE.S. SCHEDULE 16. Speed corresponding to different diameters of driving- wheel for a piston speed of 1036 feet per minute, and a 26-inch stroke. Revolutions per second = 4. Wheel in feet. Spced in miles per hour. 34°0 38°5 42.6 48-0 51-2 55-5 59.7 64:0 (38'24. | 1)iameter of driving- A peculiarity belonging to locomotives, depending upon the principles of the next chapter for its investigation, may be mentioned here. In this chapter the effect of the obliquity of the connecting-rod has been neglected. As a matter of fact there is no secondary error in the balancing of the forces, nor in the estimation of the magnitude of the unbalanced forces, arising from this cause, because in the case of an engine with two cranks at right angles connected to equal reciprocating masses, the Secondary forces arising from the obliquity of the connecting-rod mutually balance, the secondary force polygon being a line returning upon itself. There is, however, secondary couple error. An engine with four cranks at right angles severally connected to equal reciprocating masses has no secondary force error, the secondary forces mutually balancing one another. In this case the primary forces mutually balance as well, but there is a primary couple and secondary couple left unbalanced (see Schedule 21, p. 196). The primary couple can be balanced, still maintaining the balance amongst |primary and secondary forces, either by the addition of revolving masses, thereby introducing a hammer-blow, or by altering the crank angles and masses in the way explained in Art. 96, though this method carries with it the disadvantage that it is impracticable to arrange the outside cranks at right angles to one another. The Secondary couple cannot be balanced as well in a four-crank THAE BALA WC/NG OF / OCO/07/1/E.S. 123 engine by any practicable arrangement of the cranks. Balanced in the way explained in Art. 74, there is no primary force nor couple error, but both secondary force and couple errors. With the long connecting-rods usual in locomotive practice these secondary errors are negligible. 77. Experimental Apparatus.-Fig. 101 shows a model of an inside four-coupled engine, by means of which the various problems of locomotive balancing may be studied. It is shown resting on rollers. Supported in this way, the effect of the unbalanced masses on the tractive force is separated from the other effects. Unbalanced, the model rolls backwards and forwards when the Lzºº FIG. 101. gear is driven. When the proper masses are added to balance the whole of the reciprocating and the revolving parts, the model stands quite still on the rollers at all speeds of rotation. If the model be suspended by three chains after the manner of the apparatus shown in Fig. 49, the effect of the swaying couple would be seen; if one of these chains be replaced by an elastic link, or a spring, the vertical oscillations would indicate the hammer-blow. CHAPTER V. SECONDARY BALANCING, WHEN the ratio between the length of the connecting-rod and the crank is small, the difference between the true motion of the piston and the motion it would have if the rod were infinitely long, is often sufficiently great to introduce considerable error in the balancing made on the assumption of an infinite rod. The error is called a secondary one, because its chief part is of half the periodic time of the engine, and the maximum value of the force due to it is smaller than the maximum value of the force due to the simple harmonic motion of the reciprocating mass. The two forces are often referred to as the primary and secondary forces due to the reciprocation, and similarly, the terms primary and secondary balancing refer to the balancing of these respective forces. The object of this chapter is to show how to arrange an engine so that the primary and secondary effects of the reciprocating masses may be balanced without the addition of balancing masses. The geometrical ideas of the previous chapters are merged into an analytical method, by means of which a general method is obtained of treating both primary and secondary balancing. The first step in the investigation is to substitute the expression (2), Art. 78, for the simpler expression for the acceleration, equation (2), Art. 40. Notice that the new expression is the old one with the new term Moºr? ! effect. This expression has been used by Mr. Mallock, by Mr. Mark Robinson, and Captain Sankey; it is the basis of Mr. Macfarlane Gray's Accelerity Diagram ; it has been used by cos 2(9 + a) added. This new term represents the secondary 124 SECO/VOA R Y BAZAAVC/AWG. 125 M. Normand, and, more recently, by Herr Schlick. The degree to which it approximates to the real acceleration is examined in Art. 79. The error is small, and its effect is, probably, very much less than the effects of the unbalanced auxiliary engines, or of a propeller, even slightly out of balance. After establishing the general method, it is applied to several actual examples, including the case of partial balancing, given in Herr Schlick's recent paper (Trans. I.N.A., vol. xlii., 1900). The solution for the balancing of a four-crank engine completely is given in Art. 94, and, though it cannot be applied practically, it is used, in Arts. 99 and 100, to obtain new solutions, with respect to the balancing of five- and six-crank engines, which Solutions are shown to satisfy four more conditions in Art. 101. 78. Analytical Expression for the Acceleration of the Reciprocat- ing Masses, including the Secondary Effect.—Let 6 be the variable angle between a fixed line of reference OZ (Fig. 102), the centre line of the engine, Say, and a line of reference OX1, drawn in the revolving reference plane, containing the crank O.P. The circle may be looked upon, in fact, as a crank disc, on which the lines OP and OX1 are scribed. Let abe the constant angle between the direction of the crank OP and the line of reference OX1; p, the angle between the connecting-rod and OZ ; r, the crank radius; l, the length of the con- necting-rod. The distance, ac, of the crosshead, B, from the origin O, is given by the expression— * cos (9 + a) + l cos p FIG. 102. 126 THE BALANCING OF EAVGINES. But since— 7 sin (9 + a) = 1 sin (p cos º-vi-'ºsi"(0-0) . . . . . . (a) 2 “. COs (p = 1 – #sinº (0 + a), approximately “. COs (p = 1 + toos 20+ 2)-1} 7.2 * 7.2 ‘. . = roos (0+ 2 +ºcos 20+ 0 +(-). . . (1) Differentiating twice, with respect to the time, and considering the angular velocity, ! = 0 = 0, of the crank to be sensibly constant, the acceleration of B is— dºw 1962 = — rôº cos (0 + a) — cos 2(9 + a) d;2 l Let M be the mass reciprocated by the point B; then, writing to for 6, the instantaneous value of the unbalanced force, acting on the engine-frame in the line of stroke, which is equal and opposite to the force required for M’s acceleration, is given by— Majºr {cos (0 + a) +}cos 2(0 + a)} . . . . (2) 79. On the Error involved by the Approximation.—It will be observed that the approximate formulae (1) and (2), of Art. 78, depend upon the extraction of the square root of the expression giving cos p, by the Binomial Theorem, to two terms only, it being tacitly assumed that the remaining terms in the expansion may be neglected, without involving serious error. How sur- prisingly small the error is may be seen by comparing the real acceleration for a few crank positions, with the acceleration calculated by the approximate formula (2). Differentiating the true expression for the displacement a, twice, with respect to the time, the true value of the acceleration is given by— r!” cos 2(9 + a) + r" sin” (9 + •) {!” – 7% sin” (9 + a)}} tº- *{cos (6 + a) + (3) SECOMOA R P R A LA WCZWG. 127 The values calculated by this formula are compared with the approximate values calculated by formula (2), for 30° intervals, in the following schedule, for a rod 3} times the length of the crank. It will be noticed that when (9 + a) = 0°, or 180°, the values given by the formula (2) are exact, because sin (0 + a) = 0, and therefore the true expression reduces to the same form as the approximate one. The greatest percentage error in the table is at 90°, and is of the Order 4 per cent. SCHEDULE 17. Crank angle. True acceleration. *:::::::::::. Degrees. -- - --- + 1-286 + 1-286 30 + 1-0.148 + 1-009 60 +0.3571 +0.357 90 — 0°2981 – 0.286 120 — 0-6429 – 0.643 150 — 0-7172 — 0-723 180 — 0°714 — 0-71.4 210 — 0-7172 — 0-723 240 — 0-6429 – 0.643 270 — O-2981 — 0°286 300 +0-3571 +0.357 330 + 1-0.148 + 1-009 360 +1.286 + 1-286 80. Graphical Interpretation of Expression (2), Art. 78.—The first term of the expression, namely— Majºr cos (0 + 2) is equal to the projection OA (Fig. 102), on the line of stroke, of the centrifugal force, due to a mass M, concentrated at the crank radius. Multiply, and divide the second term by 4, giving— 7.2 M(2)(†) cos 2(9 + a) 128 THE BAZAAVC/WG OF EAWGIZVE.S. This is equal to the projection OB, on the line of stroke, of the 2 centrifugal force due to M, concentrated at a crank radius * rotating twice as fast as the main crank. In this way, the unbalanced force caused by the recipro- cation of M may be separated into a primary and a secondary part. The primary part is simply the projection, on the line of stroke, of the centrifugal force, due to the reciprocating mass, supposed transferred to the crank-pin. The secondary part is the projection, on the line of stroke, of the centrifugal force, due to the reciprocating mass, supposed 7° ° 4. radius of the main crank, revolving in the same plane twice as fast as the main crank. The line of reference OX1 (Fig. 102) revolves at the same speed as the crank; the angle a, therefore, remains constant. Similarly, the line of reference OX2 revolves twice as fast as the main crank, and the angle 2a remains constant. transferred to the crank-pin of an imaginary Crank, 73 times the 81. The Effect of the Primary and Secondary Unbalanced Forces with respect to a Reference Plane a Feet from the Plane of Revolu- tion of the Crank.-Fig. 103 shows a crank, and the instantaneous relative position of the imaginary crank causing the secondary forces. The effect of the two cranks, with reference to any chosen reference plane, is conveniently treated by supposing that there are two coincident reference planes: one belonging to the main crank, and revolving as though keyed to the shaft; and one belonging to the Secondary crank, and revolving twice as fast as the shaft. The two planes may be thought of as a pair of infinitely narrow fast-and-loose pulleys. These planes are shown separated in Fig. 103, for the sake of clearness. The principles of the method explained in Chapter II., Arts. 24–28, may now be applied to the consideration of these cranks. Due to the mass M, at the main crank radius r, whose angular velocity is w, there is— (1) A force Moºr, equal and parallel to the centrifugal force due to M, shown by Oa, to scale, in No. 1 reference plane; (2) A couple, whose moment is Majºra, represented by OA in plane No. 1. SECONDAR V AFA LANCIAWG. 129 ”_103-H45） AHVOJNO OES ‘ONISnyo X!NV/?10 AHV NIÐ\/WI ‘80I 'ĐIJI SENV/Teſ 30 N383.438 R I 30 THE BA ZA/VCIAWG OF EAVG/AWES. 22 Due to the mass M, at the imaginary crank radius i. whose angular velocity is 20, there is— 22.2 (1) A force * equal and parallel to the centrifugal force due to M, and shown by Ois in plane No. 2; 22.2 (2) A couple, whose moment is wº g and shown by OS in plane No. 2. 82. Effect of More than One Crank on the Same Shaft.—If there are several cranks on the same shaft, each will be accompanied by its imaginary fellow, and each crank will, therefore, give rise to a set of forces and couples, in a given pair of coincident reference planes, similar to the set stated above. The angles between the set of imaginary cranks will remain constant, although the shaft is revolving twice as fast as the main crank; in fact, they form an imaginary crank-shaft, in which the angles between any pair of the imaginary cranks is always double the angle between the corresponding pair of real cranks. If the effect of each crank is referred to one pair of reference planes, as in Fig. 103, the whole effect will be represented by the vector sums of the several forces and couples. The sum of the projections of the resultant force vectors on the line of stroke will give, at any instant, the value of the disturbing force; and the sum of the projections of the resultant couple vectors, the value of the disturbing couple. Evidently the conditions that there shall be no force and couple are that the four polygons shall separately close. For instance, if there were four cranks on the shaft (Fig. 103), and if it were possible to draw the closed polygons— OABC, Oabc; OSTU, Oistu the engine would be balanced both for primary and secondary forces and couples; because, obviously, their several projections on the line of stroke would be zero for all positions of the polygons, that is, for all the values of 0. 83. The Conditions of Balance.—The conditions of balance, to include the effect of the connecting-rod, are, therefore, completely SECO/VDARY BAZAAWCING, 131 stated thus. Choosing a pair of coincident reference planes any- where along the shaft— (1) The primary force polygon must close. (2) The primary couple polygon must close. (3) The secondary force polygon must close. (4) The secondary couple polygon must i.) 84, Analytical Representation of a Vector Quantity.—Take a pair of axes OX, OY (Fig. 104). Let u represent a vector whose magnitude is OV. The direction V of the vector is determined by Y the two quantities a, and y, q / measured along OX, and parallel A to OY, respectively, having re- gard to the usual convention respecting signs. The end of y, \ remote from the X axis, fixes * J a point, q. The line joining O º to 7 defines the direction of the ! S– X vector completely. O * There are an infinite number —Z JC > e Q of pairs of values of a, and y, º FIG. 104. being constant, which will define the same direction, and the choice of a pair depends upon the particular work in hand. If the direction is to be set out on a drawing, as and y should be chosen to bring q as far from O as possible, to ensure accuracy. For purposes of analytical investigation, it is usually more convenient to choose that pair of values which make— 3.2 + y” = 1 in which case the vector is represented by— OW(a + iy) where i may be looked upon as a symbol of operation, directing that y is to be set out parallel to the axis of Y, from the axis OX towards q. Considering the figure, it is evident that, assuming— a' + y^ = 1 it follows that a = COs a, and that y = sin a ; also, if a = 0, then 132 THE BAZAAVCING OF EAVG/AWE.S. % = 1; and, if y = 0, a = 1, since a " + y” is, by hypothesis, always equal to unity. The magnitude of OW is always considered as a positive quantity. Further, if, in the two vectors, OW(a + iy) and O1V1(21 + iyi), a = 0:1, then + y = + yi; the sign of the y in each case being determined by the conditions of the problem. 85. Relation between the Quantities defining the Directions a and 2a-Let aſ and y (Fig. 105) define the direction Oq, and a1, y1 JC4 QC > IFIG. 105. the direction Ogi. The four quantities are connected by the condition that the angle XOqi is to be double the angle XOq Since— XOqi = 2a and— * wi = cos 2a = cos” a - sin” a or ai = aft — y” since a, and y are respectively, cos a and sin a ; similarly— yi = sin 2a = 2 sin a cos a or yi = 2a:y. .S.A.CO/VOA R Y B.4 ZA/VC//VG. 133 Therefore, if a. and y define a direction a, (a.” — y”) and 20 y are the pair of values of the quantities defining the direction 2a. These are, of course, to be measured along and parallel to OX and OY respectively. 86. Application to the Balancing Problem.--Taking OW(a + iy) to represent a centrifugal force, the magnitude OV is given by Majºr. If it represents a centrifugal couple OW = Maajºr. Hence the side of a primary force polygon is represented by a vector of the form— Majºr(a + iy) the side of a primary couple polygon by— Majºra (2 + iy) the side of a secondary force polygon by— 2a-2 *º-yº tº and the side of a secondary couple polygon by— Mo”a. *{(º-y) + 2y) Adding subscripts to distinguish the different cranks, the statement of the conditions B, Art. 83, may now be transformed into— aſºr (M1(al + tyi) + M2(a:3 + iy2) + M3(~3 + iya) + . . ..} = 0 w”{M1a1(c1+ iy) + Mºſtº (ag + iyº) + . . .} = 0 22.2 * “FM,(º-yº-iºn) + . . .} = 0.) C *M • 2 2 '*) as l * 0 I 101(al — ºft” + i2a1!/1) + . . . . $ºmº It is a property of the quantities of the kind considered that, in any one equation of the above type, all the quantities associated with the symbol i must of themselves form an expression equal to nothing; the remaining quantities forming a second expression, also equal to nothing. 87. The Eight Fundamental Equations.—Using the principle of 134 THE BAZAAVC/WG OF EAWGINES. the previous article, the equations of Group C become as below, 9, r, and l cancelling out— Primary ({Miº. -- M23:2 + . . .} = 0\ (1) forces vanish {Miyi -- M2/2 + . . .} = 0 | (2) Primary ({Miaha, + M222a2 + . . .} = 0 | (3) couples vanish {Miſſiſti -- M2/2a2 + . . . } = 0 (4) D Secondary ( {M1(c.” – ºft”) + M2(cº — yº) + . . .} = 0 (5) forces vanish | {M1(a 1/1) + M2(22/2) + . . .} = 0\ (6) Secondary ({M1(ciº-y;”)al-HM2(a)»”—ys”)ag-- ...} = 0 . (7) couples º {M1(a 1/1)ai + M2(a:2%)ag + . . .} = 0 | (8) It should be carefully remembered that every a, is connected to the y, with the same subscript, by the relation (aft + y”) = 1; So that, when any aſ is known, the corresponding y is known also. A direction is, in fact, completely specified by the value of an as, or, of course, the value of a y. Although the eight equations appear to involve the four sets of unknown quantities in M, a, as, and y, respectively, they are really in the three sets M, a, and as, since the y's can everywhere be expressed in terms of the corresponding 2. The above eight equations express completely the analytical conditions of balance amongst the reciprocating parts, for an engine with any number of cranks, the cylinders being arranged in the way usual in marine work, i.e. all on One side of the crank-shaft, their centre lines being all in the vertical plane, which contains the axis of the crank-shaft. The connecting-rods are supposed to be equal in length, and the masses are the equivalent masses, reduced to the crank radius. 88. On the Relation between the Number of Conditional Equations and the Number of Variables.—In the application of these equa- tions to any particular example, the possibilities of a solution are indicated by the following propositions, quoted from “Chrystal’s Algebra,” pp. 286 and 288, Part I. :- (1) “The solution of a system of equations is in general deter- minate when the number of equations is equal to the number of variables.” (2) “If the number of equations be less than the number of SAECO/VOA R P BAZAAVC//VG. 135 Variables, the Solution is in generalindeterminate” (that is, several Solutions are possible). (3) “If the number of independent equations be greater than the number of variables, there is in general no solution, and the System of equations is said to be inconsistent.” 89. On the Number of Wariables.—It has already been shown (Art. 35) that there are in general 3(n − 1) variables in balancing problems where n is the number of cranks. This formula applies equally to a set of revolving or reciprocating masses. It should be carefully remembered that the variables in M and a are ratios. The value of any one magnitude may always be put equal to unity, if found convenient; for this does not fix the value of a variable in M-it only means that the M’s remaining in the equations represent, not their values in pounds or tons, but their respective ratios to the absolute value of the particular M fixed to unity. Similarly, any one of the a's may be considered unity. 90. On the Selection of the Conditional Equations,—If the eight conditional equations of Art. 87 are satisfied, the solution is independent of the position of the reference plane. The possibili- ties of balancing the reciprocating parts of an engine amongst themselves in which the number of variables is less than eight, are to be investigated by solving a set of equations selected from the fundamental Group D, Art. 87, equal in number to the number of variables concerned in the problem. For instance, the number of variables in a three-crank engine, whose cylinder centre lines are fixed, is 2(n) — 1) = 4. This indicates that four of the eight conditions may be satisfied. The selection cannot be made at will. In the first place, the equations must be taken in pairs, that is, (1) and (2), (3) and (4), etc., must be taken together; and, in the second place, the Solution of the chosen equations must be independent of the position of the reference plane; for there is no axis which can be specified as the particular one about which the engine will tend to turn, consequently any solution which is dependent upon the position of the reference plane is illusory, and is of no practical value. The following theorem shows how this second condition operates to restrict the selection. 136 THE BA LA/VC/WG OF EAVG/AWE.S. Theorem.—In order that the balancing of the secondary couples may be independent of the position of the reference plane, the conditions for the balancing of the secondary forces must be Satisfied as well. Let the functions— f(M,a)=0, represent equations (1)and (2) of Group D, Art. 87 (a) F(M,a,a2,0ſ)=0, ,, , (3) , (4) 5 x , (b) f'(M,a) = 0, 35 52 (5) 35 (6) 55 52 (c) F' (M,a,a2,0ſ) = 0, 25 35 (7) 35 (8) 53 55 (d) Suppose the reference plane moved a distance 2 from its original position, then the values of all the a's change by this amount, becoming tº + 2, a3 + 2, etc. The functions (b) and (d) become— F#M(a + 2),a y} = F(M,a,a2,...) + ºf (M,a),() . . . (e) F'{M(a + 2),a y} = F(M,a,a),(ſ) + ºf'(M,a),(ſ) . . . (f) The condition, that the balancing may be independent of the position of the reference plane, is, that the functions (e) and (f) vanish, when the functions (b) and (d) vanish. The first term of each of the functions (e) and (f) is similar in form to the respective functions (b) and (d), which, by supposition, vanish, leaving— gf (M,a, j) = 0 and— 2f'(M,a),4) = 0 The 2's divide out, leaving functions of the same form as (a) and (6), which must vanish to make (e) and (f) vanish. Hence the theorem. Corollary 1,–It is evident, from equation (e), that the primary couples cannot be balanced independently of the reference plane, unless the primary forces are balanced as well; for, in order that the expression may vanish, each term must separately vanish. Corollary 2.-It is also clear, from equation (e), that if the primary force and couple polygons close for any one position of the reference plane, they will close for all positions of it. Corollary 3.−The unbalanced primary couple is constant for all positions of the reference plane, if the primary force polygon close. For, suppose F(M,a,a2,4) = A, then, from equation (e), if f(May) = 0, FHM (a + 2),(w,j)} = A, for all values of 2. SECOA/AOA R P B A ZAAWC/WG. 137 Corollary 4.—The unbalanced secondary couple, is constant for all positions of the reference plane, only if the secondary force polygon close. This is proved in a similar manner to Corollary 3, by using equation (f). It follows, from the above theorem, that, to obtain any useful practical result, equations (7) and (8) cannot be taken, without at the same time taking equations (5) and (6), and that equations (3) and (4) must be accompanied by equations (1) and (2). Returning now to the selection of the four equations which contain the possibilities of balancing a three-crank engine, whose cylinder centre lines are given, it is evident that, under the operation of the two conditions stated above, the selection can only be made in two ways. These are—equations (1), (2), (3), and (4), or (1), (2), (5), and (6). Any other set of four would be inconsistent with the condition that the equations be taken in pairs, or with the above theorem. This at once shows that a three-crank engine may presumably be balanced— • (1) For primary forces and primary couples, leaving the Secondary forces and couples unbalanced; (2) For primary forces and secondary forces, leaving the primary and Secondary couples unbalanced. The possibilities of balancing the reciprocating parts of an engine of this type amongst themselves, are confined to these two cases, and no arrangement of three cranks is possible whereby anything further can be done, though there remains the practical problem, in all such cases of partial balancing, how to select the data so that the unavoidable errors left are as small as possible. 91. Application of the Method to One- and Two-crank Engines.- Obviously, nothing can be done to balance the reciprocating parts of a one-crank engine without adding another set of reciprocating parts. This evident fact may be used to illustrate the use of the expression giving the number of conditions which may be satis- fied; for, substituting the value 1 for n in expression (1), Art. 89, it becomes— 3(1 – 1) = 0 which means that none of the conditions of balance can be fulfilled. 138 THE BA LA WCING OF EAVG/WE.S. Considering a two-crank engine, the expression becomes— 3(2 – 1) = 3 showing that three of the balancing conditions may be satisfied. The conditions of Art. 90 limit the selection to equations (1) and (2), from the fundamental Group D, Art. 87, because the selection must be made in pairs; therefore, two only can be taken, and (1) and (2) must be taken before (3) and (4), by Corollary 1, Art. 90, and it is evidently no use taking (5) and (6) until (1) and (2) have been satisfied. Hence equations (1) and (2) express all that can be done in the way of balancing the reciprocating parts of a two-crank engine amongst themselves. Taking these two equations to two terms each, they will at Once reduce to values of a; and y, showing that the cranks must lie in the same plane of rotation, exactly opposite to one another, the masses at crank radius being equal. These two cases are given merely to illustrate the way in which the expression giving the number of variables may be used at the beginning of a problem, to define the limits between which a solution is possible. 92. Application to Three-crank Engines.-The number of condi- | N | T la L CON DIT IONS. SOLUTION, No Nº.2. Nº.3. !. / \ / | \ : | | | \ i. | | | | I T º Fº | \! / t Q/2=Ovg = Of FIG, 107. tions which may presumably be satisfied is equal to the number of variables, and this is, by equation (1), Art. 89— 3(3 – 1) = 6 SECO/VOARY BAZAAVC//VG. 139 The conditions of Art. 90 limit the selection to the first six equations of the fundamental Group D, Art. 87. Assume No. 1 crank to be in coincidence with the line of reference. Then a 1 = 1 and yi = 0. Also take the reference plane at No. 1 crank so that a1 = 0. This latter assumption reduces the number of variables by 1, leaving 5. These initial conditions are shown in Fig. 106. To investigate the general conditions, however, take the first six equations and substitute the above values. The equations become— M1 + M203 + M3a's = 0 . . . (1) M2/2 + M3% = () . . . (2) M2a2332 + Maaga's = 0 . . . (3) M242/2 + M3a5% = 0 . . . (4) M1 + M2(22° — ſº) + Mg(aº – ſº) = 0 . . . (5) M23'2ſ2 + M36'3/3 = 0 . . . (6) Eliminating the M’s from equations (2) and (6)– 3/23.3% = 3/23.2% From this, either as = az, in which case 4/3 = y2 (Art. 84), or as is not equal to a 2, in which case the above equation requires that ya = |s = 0, an untenable solution, since this requires that a 2 = as = 1, which is contrary to the second supposition. Taking a 2 = ag, and therefore ſo = 1/3, it follows, from equation (2), that— M2 = Ms Equations (1) and (5), therefore, reduce to— MI + 2Mg2 - 0\ M1 + 2M2(22° tº- !/3%) = 0ſ Eliminating the M's— J'2 that is, substituting (1 — a 2°) for yº”— (2a3 + 1)(a 2 – 1) = 0 140 THE BA / A NC/WG OF EAWG/AWA.S. From this— Q} 2 - $º § 3 = 003 Ol’— Ø2 = 1 = 003 this is untenable, since it involves— 4/2 = 7/3 = 0 and this has been shown above to require that a 2 is not equal to ag. The first value of aº is, therefore, the only tenable one. The values of 4/2 and 4/5 corresponding to this are— ºmº-º-º: 3 3/2 = 3/3 = + VI – H = +\º Substituting the values of as and as in equation (1), remembering that M2 -: M3– M1 = M2 = Ma From equation (2) it is clear that ſ/3 is of opposite sign to ſº, since the M’s must be positive. The crank directions are, there- fore, completely specified by Schedule 18, from which it appears that the cranks are mutually at 120°. SCIIEDULE 18. Crank. 30. Q/. No. 1 1 O No. 2 ... ... ... — $ w/3 sº 2 Q No. 3 ... ... ... — $ _a/3 º 2 Substituting these values, in equations (3) and (4), they become, remembering that the M’s are all equal— – a – as = 0 v/3 A/3 9 as - 3–0. = 0 */ SECO/VOAA’ Y BAZAAVC//VG. 141 |From which— () 0 C/2 Q3 indicating that the planes of revolution must be coincident, as shown in Fig. 107. Three equal masses, therefore, disposed in a plane of revolution, so that their respective radii are mutually at 120°, are balanced for primary and secondary forces. The force triangle corresponding to this is, of course, equilateral. If the planes of revolution are not coincident, couple errors, both primary and secondary, will be introduced to an extent depending upon the distances apart of the three planes. This is, in fact, the case with the three-crank engine, which has been used so much in marine work. Given that the masses are equal, and that the crank angles are mutually at 120°, such an engine would be completely balanced for primary and secondary forces; but there remains a large couple error. Mr. Mark Robinson and Captain Sankey * investi- gated the force errors very carefully in this arrangement, and communicated their results to the Institution of Naval Architects in 1895. By the graphic method they used, no errors could be detected; calculation from an exact formula, however, disclosed a force error of about one lb. weight in a 300 H.P. engine, running at 350 revolutions per minute, the ratio of connecting-rod to crank being 4.77 to 1. 93. Three-crank Engine Cylinder, Centre Lines fixed,—There is another way of looking at the three-crank engine problem which may be noticed. Suppose it given that the cylinder centre lines are fixed. Then the number of independent variables in the problem is— 2(3 – 1) = 4 This shows that only four conditions can be satisfied, that is, that only four of the eight equations of Art. 87 can be chosen. Now, consistent with the conditions of Art. 90, the selection can only be made in two ways, viz. – Equations Nos. (1), (2), (3), and (4) * “On a Method of preventing Vibrations in Marine Engines.” By Mr. Mark Robinson and Captain H. Riall-Sankey. Trans, Inst. Naval Architects. London. 1895. 142 THE BA/AAWC/WG OF EAVG/WES', OI’— Equations Nos. (1), (2), (5), and (6) The latter set have been fully considered in the first part of the preceding article, and they are completely satisfied by an engine with equal masses, and cranks mutually at 120°. The first set merely lead to a solution for complete balancing, supposing the motion to be simple harmonic. The way to obtain the solution of (1), (2), (3), and (4) is as follows, assuming the initial conditions to be the same as before, and as shown by Fig. 106. Eliminating the M’s from equations (2) and (4) of the pre- ceding article— $/20.3% = 3/202/3 a2 is, by supposition, not equal to ag, therefore— !/2 = 1/3 = 0 and therefore— a 2 = 0.3 = + 1 since— oft + y” = 1 Substituting these values in equation (3)— + M202 + Maa3 = 0 If as and as are both positive, a 2 must be of opposite sign to as, in order to leave the above two quantities connected with a minus sign, since the M’s are always positive. Suppose a 2 negative and as positive, then the conditions of balance are stated by the two equations— M1 + Ma = M2, from equation (1) and— M2a2 – Maag = 0 The a's are both of the same sign here, and therefore the cor- responding planes of revolution lie on the same side of the reference plane which is at crank No. 1. Also, since all the y's ,9'ECO/VOAA’ Y BAZAAVC/AWG. 143 are zero, the cranks all lie in the same axial plane. The solution is shown in Fig. 108. If the two afts were of the same sign, the corresponding a's would, of necessity, be of opposite signs, and the two planes of revolution would be on opposite sides of the reference plane. Summarizing, the general conditions of balance for a three- crank engine with fixed centre lines 8.Te— (1) The three Cranks must lie in the Same axial plane. (2) The masses in the two outer FIG. I08. planes of rotation must be such that their moments, with respect to the plane of rotation between them, must be equal and opposite, and the sum of the two outer masses must be equal to the central mass. SOLUTION . M22. Nº 3. 94. Application to a Four-crank Engine. — The number of variables is in general— 3(4 – 1) = 9 In this case, all the conditions expressed by the fundamental group of equations D, Art. 87, may, presumably, be satisfied. Take the reference plane at No. 4 crank so that ai = 0, thus reducing the number of variables by one, and put No. 1 crank in coincidence with the reference line OX, so that ai = 1, and yi = 0. These initial conditions are similar to those shown by Fig. 106, except that the reference plane is now at No. 4 crank, which is not shown there. Substituting these values, and taking four terms, in each of the equations of Group D, Art. 87, they become— 144 THE BA LAAVC/WG OF EAWGYAVES. M1 + M932 + M83's + M42, = 0 . (1) 0 + M24/2 + M3% + Maya = 0 . (2) Midi + M2020:2 + Mgaścs + 0 = 0 . (3) 0 + M222ſ2 + M3a5% + 0 = 0 . (4) M1 + M2(a)" tºº $/3°) + M3(aº º $/3%) + MA(a’ſ” — y;”) = 0 . (5) 0 + M23.3% 2 + M36'3/3 + Møya = 0 . (6) Mia' -- M302(a), — yº) + Mgaś(wº — yº) + 0 = 0 . (7) 0 + M96.93.2/2 + Mgaśćſ/3 + 0 = 0 (8) Equations (3), (4), (7), (8), in Ma above, are precisely similar in form to equations (1), (2), (5), (6) of Art. 92, in M. The solution of (3), (4), (7), (8), in Ma, is, therefore, the same as the solution of (1), (2), (5), (6) of Art. 92, in M. Hence— M1a1 = M20.2 = M3a5 = 1, say . . . . (9) And the cranks are mutually at 120°. Substituting the values of the M’s in terms of the a's, and the values of the ac's and y's from Schedule 18, Art. 92, in equations (1), (2), (5), and (6), they reduce to— 1. 1. 1. M424 = Ta, + 2a, + 2a, T M1(wſ” — y.º.) . . (10) M4/4 = –3. + ¥. = -2M424/4 . . . . . (11) From (10)— 2 2 — a 4* - /4” = 334 that is— (23.4 + 1)(a), — 1) = 0 . . . . . . (12) From (11)— —wº = % . He tº a tº & tº . (13) From (12)— 90.4 = 1 and therefore— !/4 = 0 OT- # £4 = - SECO/VOA R Y BAZAAVC//VG. 145 and therefore— = + Vº Both these Solutions also satisfy equation (13). Considering the first solution, it shows that crank No. 4 is parallel to crank No. 1, and since the right and left hand expressions of equation (11) vanish, y, being Zero, a 2 must be equal to ag. The second solution leads to the same result, in terms of different letters. Hence, from (9)— M2 = M3 . . . . . . . (14) Substituting the values— 90.4 = 1, C/2 = 0.3 in equation (10)— . M4 = + – t . . . . . . . (15) From this it is clear that a must always be greater than a 2, in order to make M4 positive. The relation between the masses is deduced from this equation with the aid of (9), above, from, which— M202 = M10.1 - 1 Substitute these values in the numerators of the terms on the right of equation (15), then— M4 - M2a2 M1a1 02 Cb1 From which— M. -- MI = M. = Ms . . . . . (16) Summarizing, a completely balanced four-crank engine must satisfy the following conditions:— (1) The four cranks must be arranged in three planes of revolution, two of the cranks being in the central plane. (2) The two cranks in the central plane must be at 120° with one another: the two outside cranks must point in the same direction in the same axial plane, which plane is at 120° with each of the cranks in the central plane of rotation. L 146 THE BALA WCIAWG OF EAWGIAWES. (3) The masses in the central plane must be equal. (4) The masses in the outer planes must be such that their moments, with respect to the central plane of rotation, are equal and opposite, and their sum must be equal to one of the equal masses in the central plane. The arrangement is shown in Fig. 109, and, though impracticable to realize, is useful in the solution of other problems. X: "M4. - M1 N R *René ~f Q/; FIG, 109. >i 95. Example.—Let the ratio . be 3. Measuring from No. 4 2 Crank— 0.1 = 3 0.2 = (tg == 1 If Mia! is unity, from (9)— M1 = } and— M2 F. M3 = 1. from (15) or (16)— SECO/VOAA’ Y BAZAAVC//VG. 147 96. Four-crank Engine satisfying Six Conditions. Case I. Schlick Symmetrical Engine.—Conditions 1, 2, 3, 4, 5, and 6 may, presumably, be satisfied, leaving in general three of the nine variables concerned in a four-crank engine to be fixed arbitrarily. The solution of this problem was the main feature of Herr Schlick's paper * at the 1900 Spring Meeting of the Institute of Naval Architects. Choose the line of reference so that it bisects the angle between No. 1 and No. 4 cranks, as OX (Fig. 111), then, a 1 = 24, and therefore yi = — ya. The three conditions which may conveniently be fixed are— (1) That the line of reference shall also bisect the angle between cranks No. 2 and No. 3. This involves that— 932 = 903 and therefore ya is equal in magnitude and opposite in sign to ya. (2) That the ratio M1 : M4 = 1. (3) That the ratio ai : as is given. The first two equations from Group D, Art. 87, become, under these conditions— 2a1 M1 + 22(M2 + Ms) = 0 . . . . . (1) 0 + $/2(M2 – Ma) = 0 . . . . . (2) Considering equation (2), M2 must evidently equal Ma, since ſº is not zero ; therefore (1) becomes— Taking a reference plane to bisect the distance between planes Nos. 1 and 4, so that a1 = — at, the third equation from Group D becomes— M23:2(ag + ag) = 0 whence— (t2 = - 0.3 The consequences of the three assumptions detailed above are, therefore, that M2 = Mg, and a 2 = ag, the two a's being of opposite sign. The cylinder lines are, therefore, symmetrical, with reference * “On Balancing Steam Engines.” By Herr Otto Schlick. Trans. Inst. Naval Architects. London. 1900. 148 THE BA LA/VCIAVG OF EAVG/AVE.S. to the central plane of the engine. Substituting the foregoing equalities in the first six equations of Group D, Art. 87, they become— M13:1 + M M1a1!/1 + M202)/2 = 0 . . . . (5) M1(al" — yº) + M2(2)” – yº”) = 0 (6) Eliminating the M’s from (4) and (6)— a'i(agº — yº”) = 22(21° — yi") . Introducing 1 — ań for y”, this at once reduces to— a 10% = - \ . . . . . . . (7) Eliminating the M’s from (4) and (5)— Ø13/202 = 902?)101 Squaring each side, and substituting — * for a 2, from (7), this <1%/1 reduces to— wit(4a2°) + a'ſ”(a1% * - tº ag”) * a1° = 0 . . . . . (8) If– a 1" — ag” a 1% P = ** as” ” and Q.” = * wi' = –P + VP3+ Q? . . . . . . (9) Also- * = ** from (7) and (4) . . . . (10) 2M1 a 1 can be calculated from (9), when the ratio ai : a 2 is given. Then a 2 can be found from equation (7). The corresponding values of the y's are found from the relation— y” = 1 – a;2 their signs being determined from equations (4), (5), (6), (7). The ratio M1 : Mg is found from equation (10). The relation given in equation (7), above, is that stated in the * cos 7 = ; in Herr Schlick's paper. form COS 2 2 SAE COAV/OAA’ Y BAZA MC/AWG. 149 Case II. Unsymmetrical Engine.—The general solution of the equations (1)–(6), Art. 94, has been published by Dr. Lorenz in a book, entitled Dynamik der Kurbelgetriebe (Leipzig, 1901), from which equations (11)–(14) have been quoted, the symbols having been altered where necessary, to bring them in conformity with those already used in the previous articles. The elimination of the M’s from equations (1), (2), (5), and (6), gives a relation between the angles, which, stated in its trigono- metrical form, is— cos ? tº + cos ?: * = cos B * 3 * . . . (11) 2 2 2 where y2, yi, (3, 8, are pairs of opposite angles arranged in the way shown in Fig. 111. This equation may be reduced to the form— 2 cosº, COS % = cos B g 8 which, if {3 = 8, as in the case of the symmetrical engine, becomes— 'y1 Y2 – 1 cos 3 cos g = 3 a relation already expressed by equation (7) of this article. Returning to the consideration of equation (11), it will be seen that by assuming values for ſ and 8, and thereby fixing the value of (Y9 + yi) since— y2 + yi = 360° – (B + 8) the value of (yo — y1) can be calculated, from which and the previous data y2 and y1 are at once determined. The corresponding masses are to be found by the elimination of the factors of the M’s in equations (1), (2), (5), and (6). If M, is unity, then measuring the angles ai, a2, and as from crank No. 4, Fig. 111— * This condition and similar conditions to those given by equation (12), are given in a paper by Dr. Schubert to the Hamburg Mathematical Society, dated February, 1898. (Mittheilungen der Mathematischen Gesellshaft in Hamburg Band 2, Heft 8.) The geometrical condition between the angles and centre lines, which is the basis of Art. 37, and a solution of the general equations for a five- crank engine, are also given. 150 THE BAZAAVC//WG OF EAVG/AWE.S. e Cl3 te 3 e Cl2 e 3 SIll # Sll] = 0.3 Slrl ºf SlT1 - Cl2 2 2 2 TT 2 Mi” = (12) sin “F* sin (a, -o)sinº “sin (a, - a) 2 gua - a 2 gue – a The expression for M3” is obtained from this by writing the subscripts 1 for 3, 3 for 2, and 2 for 1. For M3” write 2 for 3, 1 for 2, and 3 for 1. Having found the masses, and knowing the angles, the cylinder pitches are to be found from equations (3) and (4) of Art. 94. They reduce to, expressed in the trigonometrical form— as _Misin (az - ai) = - Mi sin (3 + y2) (13) Q1 M3 sin (ag – ag) M3 sin Y2 º e 02 - _MI sin (a8 - ad) *" = M1 sin {3 (14) Q1 M2 sin (as - a2) - M2 sin (— 'y1) In these expressions, the reference plane is taken at No. 4 crank, so that a1 = 0. Thus, assuming the values of the angles {3 and 8, which assumption, of course, carries with it the value of (y1 + y2), the masses and cylinder centre lines can be calculated from the above equations. 97. Examples.—Case I. Symmetrical Engine.—Given that M1 = MA, that the crank angles are symmetrical, and that the ratio ai : a = 6:5 : 2. Find the crank angles, and the ratio of the masses M2 : Mi, so that the engine may be balanced for primary and secondary forces and primary couples. The data necessitate a symmetrical engine, as shown in Fig. 110, i.e. M2 = Mg, and, taking the reference plane at the centre so that a1 = a1, a2 = ag, of necessity. Calculate the values of the quantities P, Pº, and Q”, and solve equation (9) of Art. 96. 6.5” – 2° 1.1 as T STX 32 T 1-195 therefore— P2 - 1'428 6'52 2 — - Z * Q T 4 × 22 2’643 SECOW/DAR Y BAZAAVC//VG. 151 Equation (9) becomes— * a’ = – 1:195 + V1428 -- 2:643 therefore— ac”, = + '823 the negative value being untenable, since a must be real. There- fore, retaining the positive sign, a1 = + '908. FIG. 110. tal 2. # M- 607 M-1 º M-1 M--607 T g| | | k--|---148°. 13'----|->|<-H--113’ & ‘ >k 148'. 13'-----|-> N#4 tal—Nee N#1 | | | I k-2-4-2--> K 6 - 5 + 6 - 5 > Y —r—NCRANK-N° 3. _--T. | _--- AP - 2 t - *P. S}. \, 32- § / N chank Nº. y / \ i * / º t § § & O \ \ \ | * H– –––ºr——k—& \ºv-ºn-y ><----->s-—- -308 | | S- i | V \ t / | | § W. ,” t S$. to * ,” º Qo \ N 2' / ! i N - ^ - © " $’ —Y- i s / Tº-----|----- . . CRAN K N° 4-, | * A& `--—l-–-T CRANK N92. FIG. 111. From equation (7), Art. 96– 1 * = -, -ºs = −551 Then— y = + VI – ’908% = + 421 yº – it v1 – 551* = + 834 152 7 H.A. B.A. Z.A.AWC/AWG OA” AAVG/AVE.S. The three assumptions made include— 3/1 is equal in magnitude and opposite in sign to ya 3/1 52 32 35 3/3 The individual signs of y1 and ye are to be determined from the general equations. From equation (5)— M10.1/1 = – M20.2% M1 and M2 are positive, an and as are of the same sign; therefore 3/1 must be opposite in sign to ya. Arranging the results— a' = +908 yi = + 421, giving the direction of No. 1 crank a:3 = — '551 /2 = – 834, , , ,, No. 2 , Ø3 = — '551 $/3 = +-834, 5 5 5 y 5 x No. 3 X 3 a;4 = + '908 ya = — 421, 5 y 3 x 5 y No. 4 , From equation (4), Art. 96– t M1 : M2 = 0:2 : ai = '551 : '908 = 607 : 1 This is also the ratio MA : M3 And— M = M = 1 by hypothesis Therefore— M3 = Ms = '607 Figs. 110 and 111 show the centre lines and crank angles, set out in their proper relative positions. The crank angles, in degrees, are added between the successive cranks and between the cranks Nos. 1 and 4, and 3 and 2. Case II. Unsymmetrical Engine.—Given that 3 = 100° and that 8 = 90°, find the angles Y1, Y2, the ratio of the reciprocating masses and the cylinder centre lines so that the engine may be in balance for primary and secondary forces and primary couples. The following is a convenient semi-graphical way of solving this problem. Calculate the values of Y2 and Yi from equation (11) of the previous article. Next calculate the value of M1 from equation (12). Now draw an end view of the crank angles, numbering them as in Fig. 111, and set out the two sides of the S/2 COAV/OAA’ Y BAZAAVC//VG. 153 corresponding force polygon which are known, namely, M4 = 1, and M1 from the value just computed from equation (12). Com- plete the polygon by lines parallel to cranks 2 and 3, thus determining the magnitudes of M2 and M3. Now apply the construction of Art. 37 and Fig. 47 to determine the pitches of the cylinders. From the given data— (3 – 8 so *g_* = 5 and therefore— y 2 + y 1 o #–– t = 8 2 5 Hence from equation (11)— COS = COS 5° – COS 8.5° = 0-909039 The angle corresponding to this is 24° 37'. Hence— 72 – 71 = 49° 14' 72 + 71 = 170° 0' Therefore— 72 = 109° 37' 71 = 60° 23' Again, substituting the values of the angles in equation (12)— Mº, sin 80.11 sin 240 34 in 135 in 405. 1 T sin 50° sin 150° sin 104°48' sin 314°24' From this— M1 = 1.2735 Drawing the force polygon, and measuring the sides corresponding to cranks 2 and 3– M2 = 1°685 Applying the construction of Art. 37, Fig. 47– * = 0.386 Qºl * = 0.790 Q1 154 THE BAZAAWCIAVG OF EAWG/WES. 98. Symmetrical Engine. On the Wariation of the Different Quantities in Terms of Pitch of the Cylinders. Two Graphical Methods.—Let b be the ratio of #. or, since the engine is sym- 2 metrical, the ratio of the pitch of the pair of outside cylinders to the pitch of the pair of inner cylinders. Then— _ b” – 1 ––s b2 P * = ′ , and Q 4. The way the different quantities vary for different values of b is shown by the curves in Fig. 112. The values of ac, and a 2 are shown by the ordinates to curves Nos. 1 and 2 respectively. Curve No. 3 gives the ratio § If the pitch of the middle pair of cylinders is considered unity, the value of b is the pitch of the Outer pair. Also, if M1 is considered unity, the ordinates to curve No. 3 give the value of M2. This curve shows how rapidly the ratio #. increases as b increases. The outer cylinders, therefore, should be kept as near the middle cylinders as possible, i.e. b should be as near unity as possible, in order to avoid the necessity of increasing the weight of the reciprocating parts of the inner pair of cylinders too much beyond what would be necessary if designed simply for strength. The angles corresponding to Yi and Y2 between cranks Nos. 1 and 4, and 2 and 3, respectively (Fig. 111), are given by curves Nos. 4 and 5. The scale for these two curves is that on the right- hand side multiplied by 100. For example, suppose the ratio b to be 2. Then, taking the lengths of the ordinates corresponding with 2 to the different curves, a = 83, as = 6, the angle Y = 674°, and the angle 72 = 1063%, ; = 1°38. 1 The crank angles may easily be set out from the curves when the ratio b is given. With the distance unity on the right-hand vertical scale as radius, draw a circle (Fig. 113). Set out along a horizontal diameter the distances a 1 and a 2 taken from the curves, measuring a 2 to the left of a vertical diameter. The intersection of the verticals through p and q with the circumference of the circle determine points r, s, b, u, which fix the crank positions. SAE CONDA R Y BAZA/VC/AWG. I 55 Anticipating Chapter VI., Equation (8), Art. 118 gives the maximum value of the unbalanced secondary couple. Considering 2 - O J. 8 4. • 6 ºº“ l 3 O 3. I. 2 2 : • 4. ; •2 J 2 S 4. b - * - Pitchy of outer Cylº ava Pitchy of middle cyl: FIG. 112, I 56 TAZE BAZAAWCING OF EAVG/NE.S. the factor 2 Baº M1 = unity, the value of the couple for different values of b is shown by curve No. 6 (Fig. 112). The scale for this curve is on the left of the b diagram. This couple increases rapidly as b increases, furnishing an additional reason for keeping the ratio b as near unity as pos- sible. To find the actual value of the couple in any given case, Q. measure off the ordinate to the curve, and multiply by 2M1a2B, S to??? where B = −. gl When b = 2, the Ordinate to the curve is 3:2. If 2a2 = 16, 2a-2 FIG. I.13. and MI = 5, and B = º = 1.5, which is the case when r = 2 feet, l = 7 feet, and the revolutions per minute are 88, the maximum value of the unbalanced Secondary couple is— 1 5 × 5 × 16 × 32 = 384 foot tons When b = 1, cranks Nos. 1 and 2 are opposite and in the same plane of revolution, and cranks Nos. 3 and 4 are similarly cir- cumstanced, the two being at right angles to Nos. 1 and 2. M1 is then equal to M2 and the secondary error is a minimum. The following construction, which is slightly modified from one given by Mr. Macfarlane Grey in the discussion of Mr. Schlick's paper to the Institute of Naval Architects, 1900, gives the relations between the masses, crank angles, and cylinder centre lines for a Symmetrical engine satisfying the conditions of Case I., Art. 96. Draw a circle of any diameter and draw a pair of diameters at right angles AB, CD (Fig. 114). From C with any radius cut the circle in G and the diameter AB in E. Join EC and ED. Join CG, and produce it to cut the diameter AB produced in F. Join FD. Then EDFC is a force polygon such that the angles of the corresponding cranks satisfy the relation of equation (7), Art. 96. To prove this, number the cranks as shown. Consider CE = ED = DH = CG each equal to unity, and draw perpendiculars Gy SECOMDAR Y BALANCIAVG. 157 and HJ, to CD. Then CO represents w1, and Dh = ſ/C represents w2. Also Gg = H/ = y2. Therefore— since Ch × Dh = H/2 (2a1 – 4:3)22 = y2° 2a1a2 — a 2° = 1 – wº i.e. whº = } But the direction of a 2 is negative; therefore— Ø1932 = - & which is the condition of equation (7), Art. 96. The centre lines may be found from the force polygon by the method of Art, 37. In Art. 37 and Fig. 47 it has been shown that angles for which balance is possible may be determined by drawing a pencil C I FIG. 114, FIG. 115. of rays through the traces of the cylinder centre lines, to meet in a point B. In the paper quoted there Mr. Schlick has shown how to find the point B to satisfy the condition of equation (7), Art. 96, for a symmetrical engine. The construction is as follows (Fig. 115):- Set out the centres of the cylinders to scale, ABB1A1. At B erect a perpendicular BC, and make AC = AB1. Produce CB to cut a line drawn at A at right angles to AC in D. Join D and A1, and make DE = DB. Erect another perpendicular at A1 and make A1F = AIE. With radius BC and centre F describe an arc cutting the line AA1 at G. Produce AIT to cut a parallel 158 THA. AEA /A AVC//VG OF EAVG/AVES. to FG through B" in H. Erect a perpendicular at M and make MI = BH. Bisect MI in O. O is the required point. The angles corresponding to this are shown in the figure. 99. Five-crank Engine.—The number of variables in this case is in general— 3(5 – 1) = 12 All the conditions expressed in the eight fundamental equa- tions of Art. 87 may presumably be satisfied, therefore, leaving four of the variables concerned in the problem to be fixed arbi- trarily. The solution of the problem may be derived from the general Solution of the four-crank problem given in Art. 94, without reference to the eight equations of Art. 87. Fig. 109 shows the disposition of four cranks and the corre- sponding masses for complete balance. All the eight equations of Art. 87 are satisfied. A combination of two such systems would also Satisfy all the conditions of balance, for each is in equilibrium; and the combination of two systems, each in equilibrium, must form a single system in equilibrium. Combine two four-crank systems, of the type shown in Fig. 109, to form a single system, in the way shown in Fig. 116, one system being distinguished by the radii of its masses being shown in full lines, the masses forming the system being M1, M2, m1, and m2, the other system being formed of the masses Ma, M4, M5, and me, their radii being shown dotted. It will be observed that the two systems are placed coaxially, with their central double cranks in the same plane of revolution, which plane may conveniently be chosen for the reference plane. The angular disposition of the two systems, relative to one another, is to be such that the four cranks in the reference plane are mutually at 120°. This arrangement is only possible, if one crank of the central pair, belonging to one system, coincides with one crank of the central pair belonging to the other system, as shown in the reference plane (Fig. 116). The set of four masses in the reference plane may now be divided into two groups— (1) A group of three masses, m1, m2, m3, whose radii are mutually inclined at 120°; these are indicated by cross-hatching in Fig. 116. (2) A single mass lettered M5, and shown black; this is drawn SECOAV/DA R Y BA ZA/VC//VG. 1.59 at a slightly greater radius, for the sake of clearness. It is, of course, really coincident in position with m2. If the shaded masses were equal in magnitude, they would form a system in equilibrium amongst themselves, both for the primary and the secondary forces they give rise to. Art. 92 is a proof of this, or it may be proved by substituting these values of FIG. 116. CRANKS Nºş I & 2. CRANKS 3 & 4. YS * >-120- N FIG. I.17. the masses and angles in equations (1), (2), (5), (6), of Group D, Art. 87. They might, under these circumstances, be subtracted from the combined system, without interfering with its equili- brium. Let them be equal, therefore, and let them be subtracted from the combined system; there will be left, in the reference plane, the single mass M5. Now, the masses forming each of the original systems must satisfy the relation expressed in equation (16), Art. 94, which is, I 60 THE BA/AAWC/AWG OF AZAVG//VA.S. considering one system, that the magnitude of each of the two central masses is equal to the sum of the magnitudes of the two Outer masses; the two outer masses are always, of course, in the same axial plane as shown in Fig. 109. Therefore, for the combined system under discussion, the following two equations must hold, the first line referring to the system whose mass radii are shown by thick lines, the Second to the system whose mass radii are shown dotted in Fig. 116:- M1 + M2 = m1 = m2 from equation (16), Art. 94 M3 + M4 = ??? 2 = ???6 ,, 35 25 32 But mi = m2 = mé, by Supposition; therefore— M1 + M2 = M3 + M4 = M5. g e g * º (1) And ſrom the conditions of balance in the original four-crank systems— M1a1 F M202 * * * * * g º e (2) Maas -: M404 a º e º 'º e º e (3) Equations (2) and (3) each express the relation given by expression (9) in Art. 94. From these three equations the values of the masses for given values of the a's may be found in terms of M5. Notice that, in this case, fixing the pitch of the cylinders is equivalent to fixing four of the twelve variables concerned in the problem, and that the remaining eight quantities must be found from the equations. Assuming the pitch of the cylinders to be settled, the magni- tudes of the several masses may be more explicitly stated as follows:— g From equations (1) and (2) above— M1 + M2 = M5 = 1, Say tº e º is a (4) M1a1 -: M242 ſº e º & e º e e (5) Eliminating M2 from (4) and (5)— M = z*z). * tº $ in the same axial plane M2 = 1 — M1 Similarly— at 120° M3 2- 004 * + °49 in the same axial plane M4 2- 1. tº- M3 SECONDA R Y BA LA MCIWG. 161 Example.—Suppose the pitch of the cylinders to be such that a1 = 2 feet, 0.2 = 3 feet, as = 4 feet, aa = 5 feet, each of these dis- tances being measured from the plane of the central crank of the five. Substituting, in the above expressions, and taking M5 = unity, the masses must be proportional to the numbers placed below them in the following rows:– M1 : M2 : M3 : M4 : M5 ; : ; ; ; ; ; ; 1 The crank angles have the same sequence as those shown in Fig. 117. 100, Six-crank Engine.—The number of variables in this case is in general— 3(6 – 1) = 15 A solution of this problem may be derived from the general solution of the four-crank problem given in Art. 94, by the method explained in the previous article, for a five-crank engine, although there are presumably other solutions possible. Combine three four-crank systems of the type shown in Fig. 109, to form a single system, in the way shown in Fig. 118, the masses forming the systems being respectively— M1 M2 7701 ???? M3 M4 7713 770ſ M5 M6 77.5 776 It will be observed that the three systems are placed coaxially, with their central double cranks in the same plane of revolution, which plane may conveniently be chosen for the reference plane. The angular disposition of the three systems, relatively to one another, is to be such that the six cranks in the reference plane are mutually at 120°. This necessitates the arrangement shown in Fig. 118, in which the cranks form coincident pairs. The set of six masses in the reference plane may now be divided into two groups— (1) A group of three masses, m1, m2, m3, shown cross-hatched, whose radii are mutually at 120°. (2) A group of three masses, mg, mº, ms, shown slightly displaced behind the former three, whose radii are mutually inclined at 120°. If the masses in each group were equal, they would form two M 162 THE BAZAAVC/WG OF ENGINE.S. systems, each in equilibrium, both for the primary and the secondary forces they give rise to when reciprocated. (Art. 92 contains the proof of this.) The two systems might, under these circumstances, be subtracted from the combined system, without interfering with its equilibrium. Let them be equal, therefore, and let them be subtracted from the combined system ; there will be left six masses, viz. M1, M2, M3, M4, M5, M6, three on each FIG. 118. / Sº Sº `Tº lºssºiſ. | | jº's | | K----azz ---->k---> * | | | Lºr * * | [* UALA | K OU 6 >|<- CRANKS 1 & 2. FIG. 119, side of the reference plane, forming a six-crank engine, whose cranks are shown in Fig. 119. The following equations are true of each system, respectively, from the conditions of balance of the original systems:— M1 + M2 = m1 = m2 from equation (16), Art. 94 M3 + M4 = ???3 = ??? 4 », 2 3 35 55 M5 -- M6 = ??? 5 = ???6 5 3 33 5 * } } But mi = m2 = me, and mg = mºn, - mºs, by supposition therefore— M1 + M2 = Ma + M4 = M5 + M6 . . . . (1) SECONDARY BAZAAWCING. 163 and, from the conditions of balance of the three original four-crank systems used in the combination— M1a1 -: M202 • * * * * * * (2) M303 -: M404 • a s = * * * (3) M505 -: M606 • , , s = * * (4) Equations (2), (3), and (4), severally, express the relations (9) in Art. 94. This solution adjusts itself most happily to the general condi- tions of design. The rules for designing a six-crank engine of this type may be thus stated— Take a reference plane, and group three pairs of cranks about it, each pair to be in an axial plane and to satisfy the conditions— M1a1 F. M2a2 • * * * * * * (5) M, + M, - a constant = 1, say . . . . (6) z Assuming the spacing of the cylinders to be settled, all the a's are known, and the magnitudes of the several masses may be more explicitly stated as follows :- Cl2 M1 = * + °29 in the same axial plane) M2 == 1 – M1 Ms – –“ % - 49 in the same axial plane at 120° M4 = 1 — M3 M; H –“: * + °69 in the same axial plane ) M6 = 1 – M5 101. Extension of the General Principles of the Foregoing Articles to the balancing of Engines when the Fundamental Expression (2), Art. 78, includes Terms of Higher Order than the Second,—If the extraction of the Square root of expression (a), Art. 78, be con- tinued to more terms, the final expression for the displacement of B (Fig. 102) will contain terms in cos 4(9 + a), cos 6(9 + a), etc. This converging series of cosines, differentiated twice, gives a con- verging series for the acceleration of B (Fig. 102), which, when 164 THE BA LA WCING OF EAVG/AWE.S. multiplied by the mass, gives the values of the unbalanced force acting on the frame. Expression (2), Art. 78, then becomes— Majºr {cos (9 + a) + A cos 2(9 + a) – B cos 4(9 + a) + C cos 6(9-H a) . . . . . (1) where the coefficients A, B, C have the values, c being the ratio – A = ++,+,+,+ B=# F#, c=1.4 t The details of the calculation of the above expression are given in full in an interesting paper by Mr. John H. Macalpine, Engineering, October 22, 1897. Each term in the series may be interpreted in the way ex- plained in Art. 80 for cos 2(9 + a). Thus, the force corresponding to the third term may be considered as the result of the rotation of an imaginary crank, revolving four times as fast as the main crank. The fourth term represents an imaginary crank, revolving six times as fast, and so on. For balancing purposes, therefore, the main crank-shaft of an engine may be looked upon as associated with a series of imaginary crank-shafts, rotating with speeds twice, four times, six times, etc., the speed of the main shaft, about the same axis of rotation ; the angles between the cranks of the imaginary shafts being respec- tively twice, four times, six times, etc., the actual angles between the cranks of the main shaft, the masses carried by each imaginary shaft being in the same proportion as those of the main shaft, the planes of rotation of the series of imaginary cranks, corresponding to any one of the actual cranks, being, of course, coincident. Keyed to each shaft is an appropriate reference plane. The condi- tions that each imaginary shaft may be in balance are, simply, that force and the couple polygons belonging to it shall close. Consider a force vector belonging to one of the main cranks, and let its direction, with the initial line in the reference plane, be a. The series of imaginary cranks belonging to this crank will have SECO/WZOA R V BA/A WC//VG. 1.65 directions 2a, 4a, 6a, etc., with the initial lines in their respective reference planes. These directions may be severally represented analytically in terms of the quantities specifying the direction of the force vector in the reference plane corresponding to the main crank. The way of doing this has been explained in Art. 85 for the 2a direc- tion. By an obvious extension of the method, a vector in the reference plane corresponding to the 4a crank is denoted by— R{(8/* – 8/* + 1) + i4a/(1 — 2/*)} where R is a function of the mass carried by the main crank the length of the rod and the crank radius, but which, for balancing purposes, may be written equal to M, since everything else cancels out when the final conditional equations are formed. Similarly, a vector in the next plane of the series is represented by— - M[{(aº – y”)” – 3(a)" – y”)} + i (6.0/ — 83%)] The eight conditional equations of Art. 87 will be increased by four for every additional imaginary crank-shaft taken in the series. Thus, corresponding to the 4a crank-shaft will be the four additional conditions— SM(8/* – 8/* + 1) SMay(1 – 2/*) SMa(8/4 – 8/* + 1) SMaay(1 – 2/*) () () () - () Similar sets of four more must be added to the conditions, if the 6a crank-shaft is taken in, and so on. The solution of the set of sixteen simultaneous equations, corresponding to the main shaft and the first three imaginary crank-shafts, presents formidable analytical difficulties. They may be used, however, without much trouble, to test the balancing of a proposed arrangement; and, by the introduction of proper coefficients, the magnitude of the unbalanced force and couple corresponding to any crank-shaft in the series may be found. Reverting to the polygons in the series of reference planes, it will be evident that, if a given arrangement of engine is in perfect balance, corresponding to the closed force polygon in the first and second reference planes (see Fig. 103), there will be a closed polygon 166 THE BAZAAVC/AVG OF AZAVG//WE.S. in the third plane, whose sides are proportional in magnitude to the sides of the force polygon in the first plane, but make four times the angle with their initial line that the sides of the first polygon makes with its initial line. There will be a closed polygon in the next reference plane, the directions of whose sides with their initial line is six times the angle of the sides of the first polygon with its initial line. And so on through the whole series. There will, of course, be a similar series of couple polygons. The application of this test to the arrangement shown in Fig. 109 discloses that, not only is the primary and secondary shaft in balance, but that the imaginary shaft, corresponding to 4a, is also in balance for forces. The 6a shaft is out of balance. The 4a shaft is also balanced for couples. Since the 6a forces are not balanced, the 6a couples are not balanced independently of the position of the reference planes. In general, the infinite series of imaginary shafts are in balance, with the exception of the 6a shaft, and all multiples of it. If the reference plane is taken at the centre, the whole infinite series of couples are in balance, but those belonging to the n(6a) series appear again, if the reference plane is taken in any other position; this follows from an exten- sion of the theorem of Art. 90. To prove these statements, take the reference planes at the centre (Fig. 109), and the lines of reference in them, to correspond with the direction of the two outer cranks. The force polygon, in the first plane, is the equila- teral triangle (Fig. 120). In the second plane, corresponding to the 2a shaft, it is also an equilateral triangle (Fig. 121). Fig. 122 shows the triangle for the 4a shaft, still closed. In the next plane, for the 6a shaft, it becomes the line of Fig. 123. Continuing in this way, the 8a, 10a polygons close, opening into a line for 12a. The couple polygon for all planes is a line returning on itself, since the angle of the two outer cranks with the initial line is in each case 0, and therefore all the multiples are nothing. The actual magnitude of the maximum unbalanced force due to the 6a shaft is represented by OC (Fig. 123), and this is evidently = 30A. OA represents to Scale the maximum force due to the mass M. This is given by the value of 2a. * x C lbs. weight SECONDARP BAZANCING. 167 the fourth term in expression (1) of this article, so that it is only necessary to calculate the value of C. As a matter of interest, Å. -4 2--" 2" Fig. 120 & / FIG. 121. f J A i, 24* * --> * / f 4. MAIN SHAFT § B <& `s- ---~~ O A C / gº” \ º # | ; a &” sº § º B § AB * \\ `---- 6 o' FIG. 122. O FIG. I23. ÅA. O however, the whole series is given for the case of a rod equal to three and a half times the length of the crank. It is— Mo” b s º 9 (cos 0 + 291 cos 20 + 006 cos 40 + 0002 cos 60 + . . .) The maximum unbalanced force is then— 2a, 00006 x *ibs, weight 168 THE BAZAAVC/WG OF EAWGINES. If M = 11,200 pounds, r = 2 feet, revolutions per minute = 88, this force is equal to 33-6 lbs. weight, a quantity negligible compared with the unbalanced forces, even from the auxiliary. engines about the ship. From these properties of the arrangement of Fig. 109, it follows that the five- and six-crank engines derived from it are similarly, in balance. Thus, an engine balanced by the rules of Art. 99, or a six-crank engine designed from Art. 100, would be in balance for forces and couples up to those of the 6a class. There can be no doubt but that these are the kinds of engines to use to avoid vibration. The four-crank engine cannot approach them in this respect, since, even in the best arrangement, that of Art. 96, couples of the 2a class of considerable magnitude are left un- balanced. Moreover, the crank angles of the five- and six-crank engines fit in so well with the other conditions of design. A uniform crank-effort diagram can be obtained with ease, and there are no awkward starting angles. 102. General Summary.—(a) One or any number, m, of rigidly connected masses revolving in the same plane may be balanced, by the addition of one mass in that plane, whose magnitude and position are found by Art. 14, forming with the given masses a system of n + 1 masses in balance; or, by the addition of two masses, each in a given separate plane of revolution, forming a system of n + 2 masses in balance. These two masses may be found by the general method of Art. 28, or by using first Art. 14 to find the mass in the plane of the given masses and then divid- ing it between the given planes, inversely as their distances from the plane of revolution of the given masses. (b) One or any number, n, of rigidly connected masses, each revolving in a different plane, may be balanced by the addition of two balancing masses in any two given planes of revolution, forming with the given masses a system of n+2 masses in balance. These masses are to be found by the general method of Art. 28. (c) It is generally better to reduce the masses to a common radius, the crank radius in engine problems, before filling in the schedules. This is done by the method of Art. 33. Any balanc- ing mass “at crank radius” may be placed at any radius, provided it is changed so that the product Mr remains constant. (See Art. 12 and Fig. 12.) SECOWDAA' P BA/A/VC/WG. - 1.69 (d) Assuming that the masses forming a reciprocating system, as in a multi-cylinder engine, move with simple harmonic motion, any number, n, may be balanced by the general method of Art. 28 (see Art. 44), the balancing masses found being reciprocated in the common plane of reciprocation. The system then consists of n + 2 reciprocating masses. The problem of balancing the reciprocating parts of an engine generally presents itself in this form—given n cylinders to find the masses of the corresponding reciprocating parts and the crank angles so that the system may be in balance without the addition of balancing masses of any kind. Typical solutions for four masses, and for four masses and their valve-gear, are given in examples Arts. 48 to 50. (e) In some cases, locomotives in particular, after finding the balancing masses for the reciprocating parts, in order to avoid the practical objection to reciprocating them, they are added to the system as revolving masses. Under these circumstances, the balancing masses introduce an unbalanced force and couple equal to those they are balancing, in a plane at right angles to the plane of reciprocation. (See Art. 65 for an illustration.) (f) Secondary balancing can be partially effected in four-crank engines and completely effected in five- and six-crank engines. The following general directions show how to set about balancing an engine, and the possibilities of balancing each type are briefly indicated :— (g) Two-cylinder Engine.—Put the cranks at 180°, and make the reciprocating masses equal. This ensures balance for primary forces. The primary couples and the secondary forces and couples cannot be balanced. Keep the two planes of revolution as close as possible to minimize the couple error. Three-cylinder Engine.—The cranks must all be in the same axial plane as shown in Fig. 28, and the masses proportioned so that the middle one is equal to the sum of the outer two, and the moments of the outer two about the plane of the middle one are equal. In this arrangement the primary forces and couples are balanced. The secondary forces and couples are not balanced. If equal masses at crank radius are operated by cranks at 120°, the primary and secondary forces are balanced (Art. 93), but the primary and secondary couples are unbalanced. To minimize the couple error, keep the planes of revolution as close together as possible. 170 THE BA LA WCIAVG OF EAWG//WES. Four-cylinder Engine.—Set out the centre lines of the cylinders, and choose any three of the remaining seven variables (see Art. 35), and apply the method of the example of Arts. 48 or 49. Do not fail to check the work as directed in Art. 47. To include the valve-gear, proceed as in the example of Art. 50. A four-cylinder engine balanced in this way is completely balanced for primary forces and couples, but is unbalanced for secondary forces and couples. If the cylinder centre lines are set out symmetrically as shown in Fig. 110, lay down the centre lines, and then calcu- late the ratio of the pitch of the extreme cylinders to the pitch of the inner cylinders, which has been called b. Find this number on the horizontal scale of Fig. 112, and read off the quantities ai, as from curves Nos. 1 and 2 respectively, and set out the crank angles in the way illustrated in Fig. 113. Then read off the mass ratio §: from curve No. 3. An engine Con- structed from these angles and masses will be in balance for primary and secondary forces and primary couples. There will be a secondary couple error, the variation of which is shown by curve No. 6. The amount of the error in foot-tons is found by multiplying the ordinate from the diagram by the product of the pitch of the inner cylinders in feet and the mass in tons of 2002 one of the outer reciprocating masses, and by 12"ſ. where m = the revolutions per second, and l and r are the length of the connecting-rod and the crank radius respectively. This error is smaller the smaller the value of b; therefore keep the distance between the Outer pairs of cylinders as Small as possible re- latively to the pitch of the middle pair. The balancing of a sym- metrical engine in this way was described by Mr. Schlick in his paper to the Institute of Naval Architects, 1900. For an example, see the s.s. Deutschland, Engineering, November 23, 1900. To include the valve-gear, find the angle between the two inner cranks either from equation (9), Art. 96, or from Fig. 1.12; assume two equal masses for the inner cranks, and then apply the method of Art. 50, taking a reference plane at one of the outer cranks. The final angles will be slightly different from those required for the balancing of the secondary forces. See exercises 43 and 44 at the end. If the angles 3 and 8 (Fig. 111) are not equal, the conditions SAE CO/WDAR Y BAZAAVC/AVG. | 7 | for balancing the primary and secondary force and the primary couple necessitate an unsymmetrical engine, the details of which are to be found by the method illustrated in Art. 97, Case II. Five- and Six-cylinder Engines.—Set out the cylinder centre lines. Then the crank angles must be placed in 120° pairs, two pairs and a central crank for the five cylinders (Figs. 116 and 117), and three pairs for six cylinders, as in Figs. 118 and 119. The masses. are then calculated by the formulae at the ends of Arts. 99 and 100. In these engines the primary and secondary forces and couples are completely balanced, and further, as shown in Art. 101, the fourth period forces are balanced also. In fact, five- and six-crank engines arranged in this way are the most perfectly balanced engines of the usual multi-cylinder type that it is possible to Construct. The Crank-shaft.-The crank-shaft is in general an unbalanced revolving system in which the planes of revolution and the crank angles are fixed by the reciprocating system which it operates, It can be balanced by either of the methods of Art. 51. If the ratio of the revolving to the reciprocating masses is the same for every cylinder in the engine, no balancing masses will be required for the crank-shaft. CHAPTER VI. ESTIMATION OF THE PRIMARY AND SECONDARY UNEALANCED FORCES AND COUPLES, 103.—IN this chapter it is shown how the unbalanced force and couple may be estimated for an engine whether any attempt has been made to balance it or not. Methods have been given in Art. 31 for dealing with a set of revolving masses, and in Art. 53 for finding the resultant force and couple due to a revolving and reciprocating system together, the reciprocation, however, being simple harmonic. The method about to be explained combines that of Art. 31 with, in the first place, a construction which gives the accelerating force acting on the piston accurately for any length of connecting-rod, and, in the second place, with an approximate graphical method, requiring rather less work, based upon the formula (2), Art. 78. Finally, it is shown how the equations of Art. 87 may be used to calculate the maximum values of the primary and secondary components of the resultant force and couple. 104. Klein’s Construction for finding the Acceleration of the Piston.-This construction is theoretically accurate. It is, in fact, the graphical solution of expression (3), Art. 79. It is simple, quickly applied, and gives the result in a convenient way. The construction is given in the Journal of the Franklin Institute, Vol. CXXXII., September, 1891.* Let OK (Fig. 124) be the crank, KB the connecting-rod, BO the line of stroke. The acceleration of the piston masses is the same as the acceleration of the point B, representing the cross-head. It is required to find the acceleration of the point B in terms of the crank angle 6, assuming the crank to rotate uniformly. * See reference on page 234. 172 UWBA CANCAE D FORCES AAWD COUPLE.S. 173 Produce the connecting-rod BK to meet a perpendicular to BO, the line of stroke, in W. On BK as diameter draw a circle. From K as centre, with radius KV, draw an arc cutting this circle in QQ1. Join QQ1, producing the line if necessary to cut the line of stroke in A. Then AO represents the acceleration of the cross- head B, to the same scale that KO, the crank radius, represents |FIG. 124. FIG, 125. the acceleration of the crank-pin K. The latter is uniform and is equal to wºko. For instance, suppose the drawing to be made to a scale of 1% inches = 1 foot, and that AO scales 2-2 feet. Then the acceleration of B, from B towards A, is 2.2 × 60°. If the crank is making 88 revolutions per minute, w” = 84.5, and therefore the acceleration of B is 1859 feet per second per second. 105, Proof–Let BX be the acceleration which is to be found. It may be resolved into two components, BZ along the rod, ZX at right angles to the rod. If the magnitude of either of these components is found, the scale of the acceleration triangle BZX is fixed, and the values of BX, and the component ZK, may be scaled off. Now, BZ is equal to the radial acceleration of B 174 THE BAZAAVC//VG OF EAVGZAVES. relatively to K, plus the component acceleration of the crank-pin R along the rod. Draw OT at right angles to BV and qu parallel to BC. If KO represents the radial acceleration of the crank-pin, KT is clearly its component along the rod. It will be proved immediately that q K is the acceleration of B relatively to K. Therefore gT represents the whole acceleration of B along the rod, and qu is evidently the acceleration of B in the line of stroke, which is again equal to AO. It remains to show that q K is the acceleration of B relatively to K. Let W be the velocity of the crank-pin K. This is known in magnitude and direction. U the velocity of the cross-head B, known in direction only. v the velocity of B relatively to K, known in direction, since it must always be at right angles to the rod BK. v = the vector difference (U – V) (Art. 6) This is a case of B (Art. 8). Set out ab (Fig. 125) to represent V, and ac in the direction of the velocity U. Draw be at right angles to the rod BK, thereby fixing the point c and the magni- tudes of the velocities v and U. Comparing this triangle with the triangle OVK (Fig. 124), it will be perceived that the two are similar, OK corresponding to ab and OV with ac. Therefore KV represents v to the scale on which OK, the crank radius, represents V. Thus the first operation in Klein's construction gives the velocity triangle KOW in which the radius represents the velocity of the crank-pin. Under these circumstances KO also represents y = @*r, the radial acceleration of the crank-pin. Again, the triangles BKQ and QKq are similar, BK corre- sponding with QK. Therefore— g|K : KQ = KQ : BK 2 But KQ = KV = v. Therefore q K = Er the radial accelera- tion of B relatively to K. Hence the construction. The way this construction may be used to investigate the state of balance of an engine is most easily explained by applying it directly to an example. UAVAEA ZA/VCED FORCES AAWD COUPLE.S. 175 106, Data of a Typical Engine.—The following data may be taken to apply to a modern engine of about 12,000 I.H.P. They are given in Schedule form at once to avoid repetition, and for con- venience of reference. The engine is supposed to be in full gear ahead. The H.P. and L.P. forward reciprocating masses include an allowance for the air-pump gear operated by their respective Cross-heads. The ahead gear includes the mass of the valves, valve spindle, motion block, half the link, a proportion of the eccentric rod. In the two L.P. gears the mass of the balancing pistons is included. The astern gear includes a proportion of the eccentric rod and half the link, and reciprocation of the rod is supposed to take place in the main plane of reciprocation, an assumption which involves a negligibly small error. The rest of the valve-gear is included with the revolving masses. The angular advance of all the eccentric sheaves is assumed to be 45°. In an actual engine they would differ for the H.P., Int., and L.P. cylinders. SCHEDULE 19. RECIPROCATING MASSES. 100 revolutions per minute, Crank radius, 2 feet. Ratio crank to rod, 1 : 3’9. Mass at Mass Pºe radiº*m. Mass moment or Moment Crank. reference trifugal ratios. ..º. ratios. plane. force, when o°R=l. o?R=l. Feet. Tons. 4, L.P. Aft © tº º tº a º 0 6-6 1-1 | 0 - Ast. Sheave ... tº e & 4-7 0.1 - 0°47 - ...ºmº Ah. Sheave ... te e " 5'4 0-6 -- 3:24 -º- Ah, sheave ... Q & tº 10:6 O-6 -*. (3-36 *-*. Ast. Sheave ... e e o 11-3 0.1 - 1-13 -*- 3. L.P. Forward ... e - e. 16-0 7:0 1-166 112-0 1 Ah, sheave ... tº tº º 23-6 0.6 -* 14-16 •mme Ast. Sheave ... • * * 24°3 0-1 - 2.43 --- 2. Int. e Q & tº tº º tº º º 29-0 (3-3 1.05 1827 1-631 Ah. sheave ... 9 * * 36-6 0.6 *s 21.96 -º- Ast. Sheave ... e 4 tº 37-3 0.1 - 3-73 -* 1. H.P. ... e t → tº e - 42-0 (3-0 1-0 252-0 2-25 176 THE BA/AAVC/AVG OF EAWG/AWE.S. SCHEDULE 20. REVOLVING MASSES. | } e Mass at Mass Distance crank - - I moment or I Crank. º, ºn tº jº, *: plane. force, when º co”R=1. wº- ºr, e. Feet. Tons. 4. L.P.A. ... e e º a - e. 0 4'41 1 *- --> Ast. Sheave ... tº $ tº 4-7 0-2 - 0-94 - Ah. sheave ... e - º 5-4 0-2 -*- 1.08 - Ah. Sheave ... e - e. 10:6 O-2 -ms- 2-12 -s Ast. sheave ... tº e tº 11-3 O-2 *- 2.26 - 3. L.P.F. ... tº $ tº © - e. 16-0 4'41 1 70-56 1 Ah. Sheave ... & e ºf 23-6 0-2 *- 4.72 - Ast. Sheave ... - - - 24.3 0-2 -º- 4-86 - 2. Int. ... * * > tº º º 29-0 4'41 1 127-89 1.81 Ah, sheave ... is tº a 36-6 O-2 - 7.32 *º Ast. sheave ... - - - 37.3 O-2 - 7°46 - 1. H.P. ... e e - © tº tº 42-0 4'41 1 185°22 2.62 107. Calibration of a Klein Curve to give the Accelerating Force.—The changing values of the force acting on the frame due to any one of the set of reciprocating masses may be conveniently exhibited by a curve plotted on a crank base. Consider the H.P. cylinder of the example. The ratio of the crank to the rod is 1: 3-9. Choose a suitable scale, and apply Klein's construction at, say, twelve equidistant angular positions of the crank. Let XX (Fig. 126), taken any length, be divided into the same number of equal divisions. Set out the accelerations found by the constructions, vertically above or below the corresponding angles on XX, as the case may be. AA, for instance, represents the acceleration of the reciprocating parts when the crank angle is 30°. Mark off XR to represent the crank radius. Draw a curve through the ends of the ordinates. All the intermediate values of the acceleration may be scaled off this curve. Since the mass accelerated is constant, the ordinates are also proportional to the accelerating force. The scale is at once fixed for a given speed from the fact that XR UAVAEA /AAVCAE/O APORCES AAWD COUP/LA.S. 177 represents wºr, the acceleration of the crank-pin, the corresponding Majºr force in tons is therefore At 100 revolutions per minute w” is very nearly equal to 110. From the schedule, M, for the H.P. cylinder, is 6 tons, and r is 2 feet, so that XR represents a force of 41 tons. Draw Xr at any angle, and set off 41 to a suit- able Scale, marking off the points 10, 20, etc., at the same time, to correspond to 10-ton intervals; join T.R, and draw parallels to it from 10, 20, etc., thereby reducing the 10-ton intervals to the 60 V. 50 40 30 20 10 X 70 : 20 30 4. 0 FIG. 126. Scale of the drawing at the points where these parallels cut the line XR. The curve is properly calibrated by drawing horizontals through these points, repeating them and subdividing the intervals as much as may be required. The curve shows that the force at the top, dead centre, is about 51 tons, and on the bottom centre 31 tons, that when the crank is at right angles to the line of stroke it is 10 tons; in fact, the disturbing force for this particular cylinder is known for any assigned position of the crank belonging to it. 108. Derivation of Curves to represent the Forces due to the Other Reciprocating Masses in the Engine, the Ratio of Crank to Rod being Constant.—When the ratio of the crank to rod is a constant for all the cylinders, the Klein curves are all similar. Further, if the strokes are the same, the only variable in the expression for the value of the force is M. If, therefore, a series of similar curves be drawn, the maximum ordinates being in the ratio of the several masses, the curves will severally represent the magnitudes of the disturbing forces belonging to each cylinder. Applying this N 178 THE BAZAAWCING OF EAVGIAVES. to the example, Schedule 19, if lengths XT, XU, XV are set out So that— XS : XT : XU : XV = 6 : 6'3 : 7 : 6-6 = 1 : 1:05 : 1-166 : 1-1 and curves be drawn through the points T, U, V similar to the curve through S, from the appropriate curve of the set so obtained the force corresponding to the motion of the parts belonging to any one of the cylinders may be read off for any given position of the Crank belonging to that cylinder. The proportional increase in the ordinates of the curves are quickly found by the use of a pro- portional compass. 109. Combination of the Curves for their Phase Differences.— In order that the simultaneous values of the forces corresponding to an assigned angular position of one of the cranks may be read off, the curves must be combined for the phase differences of their cranks relatively to it. In Fig. 128, the four curves corre- sponding to the cranks 1, 2, 3, 4 of Schedule 19 are combined in their phase relation to the H.P. crank. Thus when the crank is at 30° with the line of stroke, the ordinates ab, ac, ad, ae respectively represent the instantaneous value of the forces acting on the frame due to the reciprocation of the parts belonging to cylinders Nos. 1, 2, 3, and 4. Their algebraic sum— ab + ac – ad – we = — af is the instantaneous value of the unbalanced force acting on the engine-frame. The ordinates for a number of positions are quickly added by means of a pair of dividers, and through the series of points so obtained, of which f is one, the error curve may be sketched in. It will be observed that the maximum error is about 10 tons. The combination for difference of phase may be made in the following way. Draw an end elevation of the cranks (Fig. 127), marking on the drawing their angular distances from the H.P. crank. Consider the intermediate crank No. 2. Assuming the four curves, through the points S, T, U, V (Fig. 126), to have been drawn according to the instructions of Art. 108, trace the curve belonging to the intermediate cylinder, that is, the one through T, and mark on the axis XX the point corresponding to the angular distance of its crank from the H.P. crank, in this case 270°. Place the tracing over a drawing of the H.P. curve so that UAVBA LA/VCED FORCES AAWD COUPLES’. 179 the angle marked on the XX of the tracing coincides with the zero of the H.P. axis XX, and so that the XX of the tracing coin- cides with the XX of the drawing. Then prick or rub the curve through on to the drawing. Do this for each curve in turn, obtaining finally a set like those shown in Fig. 128. The calibration of the diagram is obtained, of course, from the calibra- tion of any one of the curves, since they are all drawn to the FIG, 127. . H.P. Nº. 2. I. P. N° 4, LPA : FIG. I28. same force scale. For instance, XR on Fig. 128 is the XR of Fig. 126, and this has been shown to represent 41 tons for the example under discussion. The force-error curve shows at a glance the variation of the unbalanced force acting through a complete revolution of the crank-shaft. The component of the unbalanced force due to the revolving parts in the plane of recipro- cation must be added to this to get the whole force in that 180 THE BA/A A/CING OF EAVG/AWA.S. plane. In the present example the force polygon for the re- volving parts is a closed square, so that there is nothing to add to the curve shown. The work entailed by strictly following the directions of this and the preceding two articles may be consider- ably curtailed in ways which will be obvious after one trial of the method. 110. Calibration and Combination of Klein Curves to obtain the Unbalanced Couple belonging to the Reciprocating Masses of an Engine.—If a is the distance of the centre line of a cylinder from the reference plane, the magnitude of the couple in foot-tons is given by the *. × acceleration The only variable in this expression is the acceleration. But this is given by a Klein curve. Therefore the ordinates of the curve represent to some scale the changing value of the couple in terms of the crank angle. To fix the scale it is only necessary to observe that the length XR, representing the radius of the crank, now stands for the couple Moto” This is equal to 1722 foot-tons for the quantities corresponding to the H.P. cylinder, Schedule 19. The curve marked 1 (Fig. 130), which is merely the curve of Fig. 126 recalibrated, thus represents the couples at any instant due to the H.P. parts taken with reference to a plane at the L.P.A. cylinder centre line. Curves Nos. 2 and 3 are drawn so that their maximum ordinates have the ratios to No. 1 curve given in the last column of Schedule 19, and so that their phase differences are those given by Fig. 127. The thick curve marked “reciprocat- ing masses” is the resultant curve of the three found by taking the sum of their ordinates at points along the base in the way already explained. The drawing of this curve finishes the problem so far as the reciprocating masses are concerned, for corresponding to any assigned crank angle of the H.P. crank, the instantaneous value of the unbalanced couple can be read off. Thus when the crank angle is about 45°, the couple acting due to the motion of the piston masses is about 1600 foot-tons, changing to about 1700 foot-tons, whilst the H.P. crank turns through about half a revolu- tion, the corresponding time interval being 0-3 second. Serious as UAVBAZAAVCAE D POA’CES AAWD COUP/LA.S. 18| this couple is, it is not by any means the whole unbalanced couple, for there must be added to it the component of the un- balanced couple due to the revolving masses. FIG. 130. 1722 FIG. T. 29. 111. Addition of Couple due to Revolving Parts.--To find the effect of the revolving parts, set out the couple polygon correspond- ing to the data for the main cranks given in Schedule 20. It will be found that the vector sum of the couples, that is, the line joining the origin to the end of the last vector, or the “closure" reversed, scales 170, and makes an angle of 312° with the H.P. crank (Fig. 129). The projection of this line, as the system revolves, on the plane of reciprocation gives the instantaneous value of the component couple. These projections, set out on the crank base (Fig. 130), form a cosine curve whose maximum ordinate represents the actual value of the couple. If the diagram is cali- brated in foot-tons for a given value of a) and r, then the maximum ordinate of the cosine curve must be taken so that it represents— Hy 2a, 1700° foot-tons This curve is shown dotted in the figure. The thick curve marked 182 THAE PA /A A/C/AVG O F EAVG//VE.S. “total” is obtained by adding the ordinates of the dotted curve and the curve marked “reciprocating masses.” It will be noticed now that the unbalanced couple changes from a positive value of about 3000 foot-tons to a negative value of the same amount in about half a revolution. A ship with a similar set of engines to those specified in the schedules vibrated so much under the action of the unbalanced couple that they had to be altered. 112. Acceleration Curve corresponding to the Approximate Formula (2), Art. 78.-It has been shown (Art. 79) that the differ- ence between the true acceleration and that given by the approxi- mate formula is small. It follows that the difference between the acceleration curve constructed from this latter formula and the Rlein curve, which realizes the exact formula, will be negligibly small also. The method of building up the approximate accelera- tion curve from a primary and a secondary curve is exhibited in Fig. 131. XX and XR are taken equal in length to the correspond- FIG. 131. ing lines in Fig. 126, so that the two curves may be compared. Referring to formula (2), Art. 78, the form of the acceleration curve is given by the quantities in the brackets, since the factor Moºr is constant. Measuring angles from the H.P. crank, a becomes zero. The first term in the brackets is represented graphically by a cosine curve whose maximum ordinate is XR (Fig. 131), the second term by a cosine curve whose maximum ordinate is the 7° 7 the periodic time. In fact, the values of the ordinate of the two fraction ; of XR, in phase with the other curve at O, but of half ONBAZAAVCED AſORCES AAWD COUPLE.S. 183 component curves are given respectively by the projections of the main crank and its imaginary fellow, revolving twice as fast, on the line XR. The positions of the two cranks corresponding to 30° of the main crank are shown to the left of the figure. Adding these two curves together, the thick curve is obtained, which will be found to differ only slightly from the true curve of Fig. 126. The approximate curve may, therefore, be used as the basis for esti- mating the unbalanced force and couple for most practical purposes. Majºr 9 The calibration of the curve is fixed as before from XR = Mao” g A well-known and convenient way of finding the ordinate for this approximate curve is illustrated in Fig. 132. Let 6 be any 2 Ol' , as the case may be. FIG, 132. crank angle, B0 the line of stroke, OK the crank. Draw two circles, each with radius º, so disposed that they touch one another externally at the centre O, and that the line joining their centres coincides with the line of stroke. From the points K and 6 drop perpendiculars to the line of stroke. Then— wO + fe represents the acceleration of the piston to the same scale that KO represents wºr, the acceleration of K. 184 THAE BA ZA/VC/WG OF EAWG/AWAE.S. The intercept fe is a vector quantity (c is the centre of the circle nearest the cross-head) always directed from f towards c.; but when the crank cuts the other circle, the direction of the intercept is from the centre d towards f, the foot of the perpendicular. Also a0 is a vector always directed towards O. In the position shown in the figure the quantities happen to be oppositely directed. When the crank is at 90°, the point f coincides with O, and the intercept fe is then equal to the radius of the circle, that is, º The proof of this construction is simple. If KO represents wºr, ex then fo represents *(; COS 20) since the exterior angle ace of the triangle c0e, is equal to the two interior and opposite angles, each of which is equal to 0. The projection aO represents toºr cos 0; hence the vector (aO +fc) represents wºr(cos 0 + ſcos 20), a form identical with that of the expression giving the accelera- tion. The advantage of this method is that it can be applied without having to draw the connecting-rod in at all. 113. Process of finding the Primary and Secondary Components of the Resultant Force and Couple Curves, The resultant of curves like Fig. 131, might be found in the way that has just been described when exact curves like Fig. 126 are used. A much simpler method is available, however, since the component vectors of all the curves are known. The total unbalanced force may be represented by— o°rS4M cos (9 + a) + M; cos 2(9 + a)} To find the effect of the first terms alone, set out the M’s in order parallel to the crank directions in the usual way, to find their vector sum. Set them out again, only this time draw parallel to the cranks of the imaginary shaft, that is, a shaft in which all the crank angles are doubled. This second sum, multi- plied by f is a vector representing the effect of the second terms in the expression. These resultant vectors may now be conceived attached, the one to the crank-shaft, the other to the imaginary shaft, which is revolving twice as fast. The respective projections of OAVBAZAAVC/E/D FO/CC/ES AAW/D COO/P/A.S. 185 these two vectors on the plane of reciprocation will then represent the instantaneous values of the ordinates of the two curves which are the components of the resultant curve, in the same way that the two thin curves of Fig. 131 are the components of the thick curve, only now there will in general be a phase difference. This method avoids the necessity of drawing any but the two component curves to obtain the resultant curve. A similar process may be used to find the unbalanced couple curve. The method is illustrated by applying it to find the unbalanced couples of the example previously considered. 114, Application to the Couples of the Example.-The vector OC (Fig. 133) represents the sum of the couples in Schedule 19, found by drawing a polygon relatively to the main cranks of the engine. LPF FIG. 133. An end view of the imaginary crank-shaft is shown in Fig. 134. The angle between the H.P. and L.P.A. crank is now 180°, between the L.P.A. and the L.P.F. 180°, and between the L.P.F. and Int. 180°. OD (Fig. 133) is the vector sum of the couples, multiplied by i. found from a polygon whose sides are drawn parallel to the cranks on the imaginary shaft (Fig. 134). The curves corresponding to the rotation of these two vectors, the latter twice as fast as the former, are shown in Fig. 135, the one marked b corresponding to the vector OC, the one marked a, to the vector OD. Their sum is shown by the thick line. Comparing this line with the corre- sponding one in Fig. 130, marked “reciprocating masses,” very 186 THE BA/AWC/WG OF EAVG/WE.S. little difference will be observable to the Scale of the illustrations. The sum of the couples belonging to the crank-shaft system is shown by OE (Fig. 133). The curve corresponding to this may be drawn on the diagram, as in Fig. 130, and then added to the 3 000 2000 7000 1.000 2000 30 00 IFIG. I.35. resultant curve for the reciprocating masses as before. Or the sum of the vectors OC and OE may be taken, and a curve drawn for their resultant. This curve added to curve a will give the curve representing the total unbalanced couple acting in the plane of reciprocation. The principle of the method explained in the last two articles is the principle of Mr. Macfarlane Gray’s “ Accelerity Diagram. 115. Valve-gear and Summary-The ratio of the eccentricity to the length of the eccentric rod is usually so small that the effect of the rod's obliquity may be neglected, and the motion of the valves, therefore, treated as simple harmonic. Also considering the engine in full gear ahead, the effect of the idle eccentric rod * “On the Accelerity Diagram of the Steam Engine.” By Mr. J. Macfarlane Gray. Trans. Inst. Naval Architects. London, 1897, UNBALANCED FORCES AND COUPLES. 187 may, without sensible error, be supposed to take place in the main plane of reciprocation. On this understanding the unbalanced forces and couples due to the valve-gear are to be found by the principles of the preceding articles, and added to the total curves given for the main cranks. The several masses belonging to the reciprocating and revolving parts of the valve-gear, all reduced to the crank radius of 2 feet, are given in Schedules 19 and 20 for the example under consideration. Instead of drawing two polygons, one for the reciprocating and one for the revolving masses, one only need be drawn for the purpose in hand, using for the equivalent mass the sum of the reciprocating masses and the revolving masses given in the schedules. The angular advance of the eccentrics for both ahead and astern sheaves is 45°. An end view of the eight eccentric cranks concerned can therefore be drawn. It will be found that the force polygon closes, and that the vector sum of the couples is a line 29(Ma) units long, making an angle of 1124° with the H.P. crank. It is shown in its proper phase relation and Scale to the other vectors in Fig. 133 by OV, its length being given by the radius of the small inner circle. The cosine curve corresponding to it is shown by chain-dotted lines marked valve- gear in Fig. 135. It will be observed that in this particular case the unbalanced couple due to the valve-gear is not so great as the unbalanced Secondary couple due to the piston masses. In other cases it might be greater. This shows that if an engine is balanced for secondary forces or couples the balancing can hardly be considered satisfactory unless the valve-gear is balanced also. This can be done conveni- ently by the addition of balancing masses to the crank-shaft, it being unnecessary to distinguish between the reciprocating and revolving parts of the valve-gear, since the projection of the centrifugal force due to the proportion of the masses balancing the reciprocating parts on a horizontal plane will cause little effect about a vertical axis. The processes of this and the preceding two articles may be summarized thus— (1) Make a schedule in which the equivalent mass M repre- sents the sum of the revolving and reciprocating masses appropriate to each crank in the system, including the eccentrics. (2) Find the vector sum of the forces and the corresponding sets of couples. I 88 THE BAZAAVC/AVG OF AEAVG/AVE.S. (3) Make a schedule of the reciprocating parts alone belonging to the main cranks of the engine, and draw an end view of the imaginary crank-shaft in which all the main crank angles are doubled. Using the same force and couple scale as in (2), find the vector sum of the forces and couples for these angles, remembering that the lengths scaled, as the sum or closure reversed, is to be 7° ſ vectors found under (2). (4) Combine these vectors, two for the forces and two for the couples, in the way shown in Fig. 135, by the curves a and b for the couples. multiplied by in each case, to reduce them to the scale of the 116. Calculation of the Maximum Ordinates of the Components of the Resultant Force and Couple Curves, Extension to any Number of Terms in the General Series. The “vector sums ” of the forces and couples may be calculated directly from the general equations D, Art. 87, the coefficients being introduced which divide out when those equations are equal to 0. Consider the primary force polygon. Equation (1), of the set D, gives the value of the sum of the a components of all the disturbing forces, equation (2), the sum 2a, of the iſ components, º being equal to unity. Calling the first Sum X, and the second Y, the magnitude of the sum is— V/X3TY2 This is, in fact, the length of the closure to the polygon. The direction of this is given by- tº Ti Y X = Cl Care must be taken to give the proper signs to the numerator and denominator of this fraction in order to fix the quadrant in which the vector lies. The magnitudes of the four sums are given in the following general form for the sake of reference. The Summation sign indicates that a number of terms equal to the number of cranks in the problem is to be taken of the form to which the sign is prefixed. Magnitude of the maximum unbalanced primary force in lbs. weight— O/WBA/CAAVCED FORCES AAWD COUPLES. 189 2 ºr swº. . . . . . (1) Direction— _ >(My) 9 tan a = >(Mo) ſº g & e & e & (2) Magnitude of the maximum unbalanced primary couple in foot-lbs.— *{(>May 1 (SMay).” 3) ſ {(>Max, (>May)*}” . . . . . (3 Direction— ..., a >(May) tan ſ3 *=s >(Maa) * * ſº & sº gº g (4) Magnitude of the maximum unbalanced secondary force in lbs. weight— 60%-2 9 Y 2 * 21% # Itswoº – 1)}* + (X(M2ay)}*]* . . . . (5) Direction— *-* >(2May) tan y – >{M(23-1)} & * e tº & & (6) Magnitude of the maximum unbalanced secondary couple in foot-lbs.— *HSM42 – 1)} + x(2Maº H- #Iſswage-by-seva)) . . . () Direction— _ >(2Maay) S tan 8 = >{Ma(239 – 1)} ' ' ' ' ' ' (8) 117. Application to the Example of Art. 106.-The values of the a's and y's to be substituted, taking the H.P. crank for the axis of X, are (Fig. 127)— a'i = 1 J/1 = a 2 = 0 $/2 = - 1 a:3 = - 1 3/3 = 24 = 0 V4 = 1 The values of the M’s are given in Schedule 19. 190 THE BA/A WCZWG OF EAWG/AVES. Substituting these values in equations (1) and (2) of D, Art. 87— >Ma: = –1 Substituting these sums in (1) of the previous article— 1'04 × o°r g Maximum primary force = lbs. Weight app. Substituting them in (2)— Direction relatively to the H.P. crank is tan" 03 0.3 1 Similarly, the secondary force error, by substitution in (5) and (6), reduces to— tº 2a, 026 × 0°r lbs. weight and the direction is coincident with the H.P. crank, since >(2Macy) is zero. The set of expressions for primary effects may be used to find the maximum values of unbalanced forces and couples due to the revolving masses. It will be found, using the numbers from Schedule 20, that the sum of the forces is zero. Again, substituting the proper values from Schedule 19 in equations (3) and (4) of D, Art. 87– >Mao: = + 140 SMay = –1827 and these sums substituted in (3) of the previous article give— Maximum primary couple foot-lbs. 230 X o°r J Substituting them in (4)– – 1827 + 140’0 the corresponding angle being 308°. This vector is shown by OC (Fig. 133). Direction relatively to H.P. crank is tan-" UWBALANCED Forces AWD cover Es. 191 For the secondary couples, substituting the proper values in equations (7) and (8) of D, Art. 87– >Ma(2a9 – 1) = 1813 >Ma 20 y' = 0 Substituting these values in (7) of the previous article— • *. 22, Maximum secondary couple is * * * foot.Ibs. g The direction of the vector is coincident with the H.P. crank. It is shown by OD (Fig. 133). Similarly, the couple due to the revolving masses may be found by using the proper quantities from Schedule 20 in expressions (3) and (4) of D, Art. 87, and (3) and (4) of the preceding article. The valve-gear may, of course, be treated in the same way. Thus the maximum values of the component curves can be calculated directly for any given arrangement of cranks and magnitudes of parts. From these values a very good idea can be formed of the engine as to the balance. Proceeding further and calculating the directions, all the data are obtained for Fig. 133, and the resultant curves may be built up from these vectors in the way already explained. 2. 118. General Formulae for Typical Cases.—Let A represent or. g Q 6,2,.” gl Type 1–Three-crank engine, the cranks all being in the same plane, the masses being proportioned by Art. 93, so that the primary force and couple polygons are closed, but the secondary force and couple polygons are open to an unknown extent. The values to be used in the general equations of Art. 116 are, taking the reference plane at the central crank— and let B represent wi = 1 !/1 = 0 a 2 = –1 3/2 = M1a1 = Mgaº [92 THE BAZAAVC/AWG OF EAWG/AWE.S. The closure for the secondary force polygon reduces to— 2PM(*** lbs. weight . . . . . (1) 0.3 in terms of the pitch of the cylinders; or— 2B(M1 + Ms), or 2BM2 lbs. weight . . . . (2) in terms of the masses. The length of the closure to the secondary couple polygon depends upon the position of the reference plane, since there is secondary force error (Theorem 1, Art. 90). If the reference plane is taken at the central crank, the secondary couple polygon closes, and there is, therefore, no error; if it be taken at an outer crank, the closure reduces to 2BMa foot-lbs., the M and a being respectively the mass and the distance, corresponding to the crank farthest from the reference plane. Type 2.—Three-crank engine, arranged as in Art. 92. Taking the reference plane at the central crank, the quantities to be substituted in the general equations, Art. 116, are— a 1 = 1 y1 = 0 a's = – * = -\º M1 = M2 = Ms In this case the primary and Secondary force polygons are both closed, the remaining two polygons being open. The closure for the primary couple polygon reduces to— AMVa} -- a” + ayas foot-lbs. . . . . (3) and for the Secondary couple polygon to— BMVa” + a” + alag foot-lbs. . . . . . (4) If the cylinders are equally spaced, so that a1 = ag, these two closures reduce respectively to— AMa V3, and BMav3 foot-lbs. . . . . (5) The magnitudes of both the primary and secondary couple OAVEA CAMCAE/D AWORCES AAWD COUPLE.S. 193 errors are constant, because both the primary and Secondary force polygons close. Type 3.—Symmetrical four-crank engine, arranged as in Art. 96, illustrated by Figs. 110 and 111. The general data in this case, the reference plane being at the centre, are— 031 = 0.34 $/1 = -$/4 - 302 = - 0.3 2/3 = - ſo (t1 = - (tº (02 = - 0.3 The only open polygon is that representing the secondary couples; the other three are closed, by the conditions of the problem. The closure reduces to— 4B(M10121ſ1 + M90922/2) foot-lbs. . . . . (6) Referring to equation (10), Art. 96, it will be seen that— 3 M2 £1° = 2M By means of this relation, and— &º- Midly, M2/2 Cl2 = , from equation (5), Art. 96 and— Øo F -º-, from equation (7), Art. 96 2001 2 the ac's and y's may be eliminated from the above expression for the error; for, substituting these values for a 2, a2, ...and putting v/T – º for yi, and multiplying and dividing by 21, where necessary, to obtain the alſº form, the expression becomes— 2Bo(MVº-1+MV. - 1) tool-lºs 901 31° and this, substituting # for a 1°, gives— 1 --- 2B40ſ, + Mºv/*-1} foot-lbs, . . . (7) 194 THA. B.A. Z.A/VC//VG OF EAVG/AVE.S. an expression of the same form as that given by Herr Schlick in the paper already quoted. If b = #. this expression may be written— 2 mºmºsºm-ºsmºs---sºm 2Boyſ (1 + ;)( * – 1) foot-lbs. . . . (8) a convenient form for use when 2a2, M1, and B are each put equal to unity (see Art. 98, and Fig. 112). Type 4.—Four-crank engine with cranks at 90°, and equal reciprocating masses. Take the reference plane at the centre, so that a1 and as are positive; and as and a 4 are negative. The data are— a 1 = + 1 !/1 = 0 a 2 = –1 !/2 = 0 a 3 = 0 $/3 = +1 a;4 = 0 !/4 = - 1 M1 -: M2 - M3 = M1 Substitute these values in the proper equations, and the following results will be obtained:— Primary force polygon— Length of closure = 0 Primary couple polygon : the closure reduces to— AMV/(an – as)” + (a| – as)” foot-lbs. which is equal to— AM(a1 – ag) V2 foot-lbs. . . . . . . (9) if the distances between each pair of cranks, which are at 180°, are equal. The secondary force polygon closes, and the secondary couple polygon reduces to BM(a1 + q2 + as + au). - This is equal to— 2BM(a1 + ag) if the distances between each pair of cranks which are at 180°, are equal. OWBAZAAWCED FORCES AAWD COUPLE.S. 195 119. Comparative Examples—It is interesting and instructive to compare the disturbing effect, due to different arrangements of an engine, on the assumption that the magnitudes of the lightest set of reciprocating parts is the same in each case, and that the engine cylinders are spaced a given distance apart. For this purpose, assume the following data:— Mass of the lightest set of reciprocating parts = 5 tons Crank radius = 2 feet. Cylinders, 16 feet pitch Ratio of crank to rod = 1 : 3:5 Revolutions per minute = 88 Then— w = 9.2 approximately o” = 84-5 2 *" = 525 = A #. o” = 1.5 = B gl The force errors may be compared with the maximum disturb- ing force, due to the reciprocation of the lightest mass, with simple harmonic motion, which, in the case under discussion, is R. .2a. º: = 26:25 tons weight. The different results are set forth in Schedule 21. The general formulae for the errors will be found in the first horizontal row, corresponding to each type. The second row gives the forms the formulae assume when the several pitches of the cylinders are equal. The figures in the third row are the numerical results corresponding to the above data. SchEDULE r Primar Pri Type. force *::::. cºsºr. M2 Tons. Foot-tons. O 0 -d. --> * - - - - - - - - - - - -D - * *** = ** - - - - - - - - - - - Me M1 CRANKS AT 120°. 0 AM WDP-E d? --Df. : | T T T AMD w/3 a --D - - - - - - - . 728 SYMMETRICAL ARRANGEMENT. 0 0 K -a, ----> -- a -- --> i. -- D ---------- --- > CRANKS IN 180° PAIRs AT 90°. k--- c ----k-d--> O AM WDº-E & i | | | | | — TH F- AMD w/ 2 k--------1 … . 594 k--D ---> 5- and 6-crank engines arranged 0 O as in Figs. 105, 106, 107, and 108. Revolutions per minute, 88. A = 5:25. B = 1°5. Disturbing 21. Secondary Secondary force error. couple error. Tons. Foot-tons. 2BM; or 2B(M, +M) Variable. 4BM, Reference plane at centre, error = 0, at No. 1 = 480. 30 O BMA/D*-E d” --T)d BMD w/3 208 O B(Miaſtiſ, -- Mºſtºy.) 2M BD (M, + M) vſ. * - 1; 1.28M, DB 465 O BM(L + c) BM(L + c) 480 0 O force due to the lightest mass of 5 tons is 26:25 tons weight. 198 THE BAZAAWCIAVG OF EAVG/NE.S. Although the magnitudes of the unbalanced force and couple of a given engine of any one of the types considered may readily be inferred from Schedule 21, yet nothing can be predicted about the behaviour of the support or foundation of the engine from this knowledge alone. On some supports the engine may run without causing vibration enough to be troublesome, even though there is a large unbalanced couple. Place the same engine on another support, and the vibration it causes may be exceedingly great. Again, a support may be sensitive to a Small unbalanced force, and at the same time undisturbed by a large unbalanced couple, and vice versá. It is always necessary, therefore, when considering the effect likely to be produced by an unbalanced engine, to have in mind the general principles governing the vibration of supports under the action of an external force or couple. A short account of the main features of this subject is given in the next chapter. CHAPTER VII. THE WIBRATION OF THE SUPPORTS, 120. Preliminary.—The foundation or support of an engine or machine is in general an elastic system susceptible of vibrating in a variety of ways. If any part of the system is displaced from its position of equilibrium by the action of an external force which suddenly ceases to act, the system, in returning to its position of rest, overshoots the mark, and begins to vibrate in one of the ways peculiar to it. The energy of the vibration is gradually dissipated in heat, and by communication to the surrounding medium. If the application of the disturbing force is repeated at regular intervals, that is to say, if a periodic force acts on the system, it is compelled to vibrate in a way foreign to it, the period of this forced vibration being equal to the period of the force producing it. This period is in general different to the period of any of the natural modes of vibration of which the system is susceptible if disturbed and then left to itself. If the periodic time of the dis- turbing force should happen to be equal to the periodic time of any one of the many natural modes of vibration peculiar to the system, a large disturbance is produced, even though the magni- tude of the force producing it is extremely small. The work done by the force in displacing the system from its position of rest appears as the energy of the vibration; if the next applica- tion of the force is so timed that it begins to act just at the time that the vibration caused by the first application is about to repeat itself, it is able to communicate another small amount of energy to the system without interfering with the amount communicated by its first action. If this timed action of the force be continued 199 200 TA/A2 A3 A LA A/C/AVG OA' AAVG//VE.S. indefinitely, energy is gradually accumulated in the system to such an extent that the consequent vibration may be sufficient to break it down. - An unbalanced engine or machine applies a periodic force and couple to its supports, compelling them to execute forced vibra- tions, which may be of small amplitude and little consequence, though the force and couple acting may be large. If, however, the support possesses amongst its many natural modes of vibra- tion, one whose period is equal or nearly equal to the time of revolution of the engine or machine bolted to it, then the ampli- tude of the force vibration is large, and out of all proportion to the force producing it. In this way the engines of an electric light station, though bolted to large concrete blocks, may set up vibra- tions in the ground in which the block is embedded, which may be transmitted all round the neighbourhood. The hull of a steamer may be thrown into violent vibration to the discomfort of the passengers and the injury of the ship. An unbalanced machine bolted to a shop floor may shake the whole building, though the actual value of the force and couple may be insignifi- cant. A locomotive may be thrown into violent and dangerous Oscillation if the swaying couple coincides in period with the natural period of Oscillation of the engine On its springs. A carriage will ride roughly if the speed of the train is such that the interval of time between the blows from the fish-joints is equal to a periodic time of one or more of the loaded carriage springs. The precise investigation of even the simpler problems of this kind is difficult. A great deal of instruction, however, may be obtained from the detailed consideration of the natural mode of vibration of the simplest kind of support and its behaviour under the action of a periodic force. 121. Natural Period of Vibration of a Simple Elastic System.— Let a mass M be supported by a steel bar whose mass is negligibly small compared with M ; and let the bar rest in two closed V’s (Fig. 136). The bar is the support whose behaviour is to be examined. One of the objects in the investigation of the vibration of such a system is to find an expression which will give the displacement of the point c, relatively to its position of rest, at the end of any time after the commencement of the vibration. The assumption is made that the system is only free TAZZ VIAPRA 77OAV OA. THE SUPPORTS. 201 to vibrate in a vertical plane, and that the displacements are not so great that the bar or spring is strained beyond the elastic limit. The magnitude of the displacement will be fixed by one variable, which will be denoted by y. This symbol will, therefore, always stand for the distance of the centre of the mass M from its position FIG. 136. of rest. The equations are formed from the familiar fundamental form— Force = mass × acceleration In the problems under discussion the force and the acceleration are continuously variable and functions of the time. Newton’s method of indicating differentiation with respect to the time is used, so that— y = the displacement of c from its position of rest ſ = % = the instantaneous value of c's velocity C sº s d?, * 3. * $/ = # = the instantaneous value of c's acceleration It may be noticed that the problem is just the converse of that in Art. 78. There the displacement a is fixed for a given crank angle, which, of course, is a function of the time, by the mechanism itself, and it is desired to find an expression for the acceleration. This is accomplished by two differentiations. In the problems about to be discussed the acceleration is stated in terms of the force, and 202 THE BA/A A/C/AWG OF EAVG/AWAE.S. the displacement is required in terms of the time. This involves two integrations to get to the displacement equation. In what follows the equations will be stated and their solutions given, because the interest lies in their solution and not in the method of obtaining it. The equations all belong to the type known as linear differential equations with constant coefficients, and rules for their solution are to be found in any work on differential equations.” The solutions may always be verified by substituting the value of y in the original equation and differentiating. Returning to the problem, let F be the force in absolute units, acting at c, which produces the deflection y. This force varies as the deflection, and therefore is the force which must be applied to produce unit displacement of one foot. This may be found experimentally or by calculation from the formula— _48EI - - - for the case under consideration, E being Young's modulus, and I the moment of inertia of the section of the bar about the neutral axis, l the distance between the W’s. Let be represented by pu. Then when the deflection is y, the accelerating force is given by Mj. This force is also equal to – uſ, the minus sign being introduced because the force acts to oppose the motion. Hence the equation of motion is— Mī) + uſ = 0 . . . . . . . (1) of which the solution is— */ = a cos (qt — a) . . . . . . (2) The two constants, a and a, are determined by the initial conditions of the motion, and q = *. M Thus, if the mass of M be 50 pounds, and it requires a force of * “Calculus for Engineers,” Perry, London, 1897. “Integral Calculus,” Edwards. London, 1894. THE WIBRATION OF THE SUPPORTS. 203 10 lbs. weight = 10g poundals, to produce a deflection of 0-0071 feet, u, the force which will produce unit deflection, is— a . x 322 = 45,852 poundals 0-0071 q therefore is 30:14. To find the constants a and a, the initial conditions of the motion must be specified in some way. Suppose that the system be held at 0:1 foot displacement, and that the time be counted from the instant it is let go, then y, the displacement, is 0-1 foot When t is nothing, and the constant a is 0, since the angle is nothing when t is nothing; substituting these values of t, y, and a in equation (2), a = 0:1. Substituting the values of a and q in equation (2), the displacement for any given time t is— y = 0:1 cos 30:14t feet The displacement curve for this free vibration is drawn and calibrated in Fig. 137. FEET •7 FIG. I57. The cosine of an angle repeats itself exactly if the angle be increased by 2it. Reverting to equation (2), all the circumstances of the motion will be the same, therefore, when— (qtº — a) — (qti — a) = 2ir from which— tg * -º-º: tl - T - 2 Tr ; 204 THE BA /AAWCIAWG OF EAWG/AWE.S. giving on substitution of the value of q, the well-known formula— T = 2rv/* Al T being called the periodic time of the vibration. The system under consideration will, therefore, execute vibra- tions in periodic time— sia = 0.208 seconds or its frequency will be 4-8 vibrations per second. This vibration is the principal one natural to the system, and whatever be the magnitude of the force producing the initial displacement, providing always that it does not strain the system beyond the limit of its elasticity, the vibrations following its removal will always have the same periodic time, though the amplitude will depend upon the magnitude of the force starting the motion. 122. Damping.—The successive ordinates denoting the maximum amplitude of the vibration in Fig. 137 are shown equal. This could only be true if none of the energy of the vibration were lost. In any actual system, part of the energy is gradually frittered away in heat, partly through the imperfect elasticity of the bar, partly through the frictional resistances between the surface of the system and the medium in which it is vibrating, another part being communicated to the environment, so that eventually the system is brought to rest. In many cases the loss through frictional resistance is very great relatively to the other causes of loss, and its effect in modifying or damping the amplitude of the vibration is considerable. In some cases it is artificially increased by means of dash-pots and such-like apparatus. Frictional resistance is equivalent to a force acting to oppose the motion, and its magnitude may be assumed propor- tional to the velocity of the mass M in the simple case under discussion. In the consideration of the motion of the system (Fig. 136), the effect of the resistance at the surface of the bar may be neglected. The frictional force at any instant may 7A/A WIBAEA 77OAV OA' THE SUPPORTS. 205 therefore be written — 87, 8 being a constant determined by experiment. The force acting to retard the motion is now— – 87 – uſ and the equation of motion is— Mý + 8 + ay = 0 . . . . . . (3) the Solution of which is— y = Ae-#" cos(v/(g” – 4b*) – ał . . . . (4) where b = * and q -V* A and a being constants determined by the initial circumstances of the motion. If 45° is less than 0°. so that the quantity under the root is positive, the motion is oscillatory, otherwise the displaced mass M returns gradually towards its position of equilibrium, but does not overshoot it. Fig. 138 is a copy to scale of a displacement curve which AAAAA* FIG. 138. was automatically drawn by a damped system of the simple kind under consideration, and in which the mass was 683 pounds and the support was a spiral spring. Equation (4), with proper constants, may be taken to represent this curve. It is an interesting exer- cise to determine the values of these constants from the curve. This incidentally indicates a method of experimentally finding 8 and u. The time of a complete vibration is represented by twice the distance AC or A1C1. This was observed to be 1:46 seconds. The isochronous character of the motion can be verified by measur- ing the series of intercepts made by the curve along the axis. At the point A, the angle is such that its cosine is unity, whatever 206 - THE BA LA WCINVG OF EAWGIAWE.S. may be the value of the individual quantities in the brackets of equation (4). Similarly, after 0.73 seconds, the angle has changed by r, and its cosine is — 1. Therefore the ratios— C1D1 = CD, etc., -: Ae-?” 6 —#0(ta-ti) A1B1 AB Ae-#bt" Measuring several pairs of successive ordinates the ratio was found to be 0-91. And to — ti = 0.73, therefore— — 0.36b = loge 0.91 from which– b = 0°262 The mass being 6'83 pounds— = 0.262 × 6'38 = 1.67 This means that the frictional resistance to motion is 1-67 poundals when the velocity is unity. Again, choosing the origin of co-ordinates at a place corre- sponding to the maximum excursion of the point from its position of rest, a = 0. The coefficient of t is found from the consideration that for a change in the time 0-73 seconds the change in the angle is tr. Hence— 0.73 Vº-Tº = r In this b = 0.262. Solving for g”— q* = 18'54 and since this = §: and M = 6:38 pounds— pu = 118 poundals meaning that a force of 118 poundals will displace the system 1 foot from its position of rest. To find A, measure a pair of simultaneous values of y and the angle. One pair of values is y = 094 feet, when t = 0. Substi- tuting these in equation (4), it at once reduces to— A = 094 THE WIBFA TION OF THE SUPPORTS. 207 Hence— y = 0.094e–91* cos (4:3t) feet 123. Vibration of the System under the Action of a Periodic Force.—Suppose now that the mass M (Fig. 136) is an engine, Self-contained, whose crank-shaft makes n revolutions per second. Let there be an unbalanced force in it whose maximum value is E poundals. Whether this be from revolving or reciprocating parts, the unbalanced force in the vertical plane, which is all that is under consideration at present, will have the value— E cos pt where p = 2irm. As in the previous case, suppose the frictional forces resisting motion to be expressed by 8). Then the sum of the forces acting to cause motion is— – 87 – uy + E cos pt and the equation of motion is— Mý + 8) + uy = E cos pt . . . . . (5) The solution of this is— y = *(u-o . . . . . . (6) where the value of e, which fixes the phase of the motion, is given by— tan s = gº p” ' ' ' ' ' ' (7) In these expressions b = 8. and q = º = 2irm.1, P = E. M’ M 5 M’ When q = p, the revolutions of the engine per second are equal the number of Oscillations per second natural to the system, since p = 2mm and q = 2itni. Under these circumstances, tan s = in- finity, and therefore s = 90°, so that sin s = 1 and cos s = 0. Equation (6) then becomes— ,, . P sin pt v==;e . . . . . . (8) 208 THE BA LAAWCIAVG OF EAWGINES. If b is small, y becomes large; in fact, y tends to infinity as b tends to zero. This shows that when b is Small, and when q = p, the engine could set up vibrations of sufficient amplitude to break down its supporting rod altogether. This equation is one of great importance, and it should be carefully studied, trying the effect on y of different values of b, and gradually approaching values of p and q. The effect of the frictional resistance in modifying the maximum amplitude of the forced vibration is illustrated by the diagram (Fig. 139). Considering a system similar to Fig. 136, whose natural --~ *-* * * *** *** ----- * *-* *- - - - *, *. ---. —— - ——'. ( — , — - L. ------TETEE-FT 280 281 232 283 284. 285 . .286 || 287 288 289 290 .297 292 293 Revs. per Trun. 286-46 - FIG. 139. number of vibrations per minute is 286:48, suppose the speed of the engine, assumed contained in the mass M, to be gradually increased from 283 to 291 revolutions per minute. Further, suppose P = 1, and that b has the different values shown in the figure against the corresponding curves (Fig. 139). The maximum amplitude of the forced vibration, calculated from equations (7) and (6) of the present article, are shown by the Ordinates of the curves, for the different speeds and for the different values of b. When the engine is running at 283revolutions per minute, themaximum values of the amplitudes are practically the same for all values of b taken, and they are relatively insignificant in amount. At 286 revolutions per minute the effect of the different values of b is distinctly shown. At the synchronizing speed, b exerts its maximum influence. It THE WIBAEA TVOAV OA' THAE SUPAORTS. 209 should be remembered in considering this diagram, that if b = 0, the maximum amplitude of the forced vibration at the Synchroniz- ing speed is infinite. Beyond the synchronizing speed the curves rapidly drop again to insignificant values of y. This indicates how it is that a dangerous speed may be run through to higher speeds at which the disturbance becomes negligible. These curves, though only illustrating an example of the simplest kind, show how necessary it is to include the frictional resistance into any actual problem to get even an approximate result. At the synchronizing speed the phase difference is 90°. Below 285 and above 288 revolutions per minute the phase difference is small; it increases as b increases. When b = so, the phase differences corresponding to different speeds are shown below. Revs. per min. Phase difference. 284. 3° 48' 285 5° 3' 286 16° 5'1" 286'3 41° 14' 286°48 90° Synchronism. 286-6 — 53° 9' 286-8 – 26° 16' 287 – 17° 34' 288 — 7° 9' 292 — 19 39' Thus within a few revolutions of synchronism either way the phase difference is practically nothing. 124. Natural Vibrations of an Elastic Rod of Uniform Section.— The vibrating system of the last three articles consists of a single mass whose motion is controlled by a steel bar. The mass of the bar itself has been neglected. Consider now that the system consists simply of a steel rod, so that the forces called into play when any point of it is displaced from the position of equilibrium act on the mass of the bar only. This is a much more complex system to deal with than the first one. The bar possesses several natural modes of vibration. The first three, corresponding to the gravest periodic time, and the next two higher, are shown in Figs. 140, 141, and 142. The points marked with the dots are the nodes, or places of rest for that particular vibration, though they P 210 THE BA LA WCING OF EAWGIAWE.S. need not be points of absolute rest, since all these separate modes of vibration may exist simultaneously. The respective numbers of vibrations per second corresponding to successive modes of division are very nearly in the ratios of the squares of the odd FIG. 140. | | k-004 4–1. --- * * - - - - - - k* - - - - - - as sº ºw 288 - k •261 ----------!- FIG. 142. numbers. The next three lines show the characteristics of these natural modes of vibration. Number of nodes ... 2 3 4 5 6 7 Frequency ratio ... ... 3° 5* 72 92 112 132 Ratios of periodic times 1 0:36 0183 0.111 0.074 0.053 The gravest frequency may be calculated from the formula— 22:4k /E 70, - 272 p & b º & * o & (1) TAZE WIBA'A TION OF THE SUPPORTS. 211 where p is the density of the material, E Young's modulus, l the length in feet, k the radius of gyration of the Section about an axis perpendicular to the plane of bending. 125. On the Point of Application of a Force and the Vibrations produced.—If the rod considered in the last article be supported in such a way that it is free to vibrate in any of its natural modes, the application of an external force, whose periodic time is equal to or is a multiple of any of the natural periods of the bar, is capable of forcing vibrations of considerable magnitude of that period, corresponding with the case of Art. 123. It is a fundamental principle that the point displaced by the action of the force cannot be a node in the subsequent vibration, so that all modes of vibration requiring the point of application for a node disappear. Advantage is taken of this principle in fixing the point at which the hammer shall strike a piano string. A point is chosen which would form a node in a mode of vibration inharmonious with the principal modes of vibration of the string. The act of striking the string at this point eliminates those particular modes from the note. Suppose now that the period of the applied force acting on the bar is equal to the gravest mode of vibration natural to the bar. If the force be applied at either of the corresponding nodes, it cannot induce the vibration, though applied at any other part of the bar large forced vibrations will result. If a couple be applied at the bar, it can force vibra- tions of its period if applied at a node of that period, but not if applied midway between a pair of nodes. These two principles will perhaps be more clearly understood if the matter is considered in another way. When the bar is vibrating in any one of the modes peculiar to it, every point is constrained to move in a certain way. A point midway between a pair of nodes moves vertically in a straight line, and the element of length surrounding this point has no freedom to turn. The point forming a node is not free to move, but the element of length in its neighbour- hood is free to turn about an axis through the node. Motion in a straight line is produced by the action of a force, turning about an axis by the action of a couple, and neither can produce the effect of the other. Hence a periodic force in agreement with a natural period of the bar is unable to force the corresponding 212 THE BA/A/VC//WG OF AAVG/AWAES. mode of vibration if it is applied at a node belonging to that mode. Neither can a periodic couple in agreement force the corresponding vibration, if it is applied midway between a pair of nodes corresponding to the system. If the engine of Art. 123 have a period of revolution in agreement with one of the natural mode of vibration of the bar under consideration, it would still be possible to prevent it forcing the corresponding vibration on the bar by properly choosing its position of attachment to the bar, supposing it had either an unbalanced force or an unbalanced couple. These principles may be studied practically by suspending the model already described in Art. 54, from a flexible bar, itself supported in hooks hanging freely from a cross-bar, so that their distance apart may be varied quickly. At a synchronizing speed of the model the part of the bar projecting beyond the frame may be thrown into vibration with nodes which can be made to dis- appear by an increase or decrease of the speed. The model can be adjusted to give either a force or a couple. 126, Longitudinal and Torsional Vibrations,— In addition to the lateral vibrations already considered, the bar is susceptible of vibrating in the direction of its axis and about its axis, these being called longitudinal and torsional modes of vibration respec- tively. Corresponding to each kind, there is a fundamental vibration with one node at the centre and a series of higher vibrations with two, three, etc., nodes. The nodes and the corresponding ratios of the frequencies and periodic times are given in the following lines:— Nodes g 1 2 3 4 5 Ratios of frequencies 1 2 3 4 5 Ratios of periodic times 1 5 333 25 .2 The frequency of the gravest period is given in the case of longitudinal vibrations by— 1 /E and for torsional vibrations by— 1 . /C * = 5 v . . . . . . . . . (3) TAZAZ VIAEA-ATION OF THE SUPPORTS. 213 where p is the density, E Young's modulus, C the modulus of rigidity, l the length in feet. Ea'ample.—Let l be 10 feet; E, 30,000,000 × 144 × 32 poundals per square foot; 0, 490 pounds per cubic foot. Intro- ducing these values in equation (2), n, the frequency of the gravest longitudinal mode of vibration, is 842 per second. The gravest frequency of the torsional vibrations would be less than this in the ratio of VE to v/C. For hard steel, E: C about in the ratio 1 : 4, consequently the ratio of the frequencies is 1 : “53, or the gravest torsional vibration corresponding to one node at the centre is 530 per second. The gravest vertical vibration may be calculated from equation (1), Art. 124, when the diameter of the bar is given. Suppose it to be 2 feet diameter, then k, the radius of gyration, will be '05. Substituting this and the data of the example in the equation, the gravest frequency is found to be 30 vibrations per second. The higher frequencies follow the ratios given in Art. 124. Sum- marizing the results— Vertical. Longitudinal. Torsional. Gravest frequency ... 30 842 530 Frequency of next mode 83 1684 1060 / Vibrations Frequency of third mode 163 2526 1590 ( per second Frequency of fourth mode 270 3368 2120 A periodic force or couple agreeing with the periodic time of any of these modes of vibration, and applied in the proper manner to the bar, is capable of exciting large vibrations of that period. 127. Simultaneous Action of Several Forces and Couples of Different Periods,--The effect of the simultaneous action of a set of forces is indicated by the following quotation from Lord Rayleigh’s “Sound,” Vol. I. p. 49 : “From the linearity of the equations it follows that the motion resulting from the simultaneous action of any number of forces is the simple sum of the motions due to the forces taken separately. Each force causes the vibration proper to itself, without regard to the presence or absence of any others. The peculiarities of a force are thus in a manner trans- mitted into the system. For example, if the force be periodic in time t, so will the resulting vibration. Each harmonic element of 214 THE BA/AAVC/AWG OF EAVGINES. the force will call forth a corresponding harmonic vibration in the system. But since the retardation of phase s and the ratio of the amplitudes are not the same for the different components, the resulting vibration, though periodic in time, is different in character from the force. It may happen, for instance, that one of the com- ponents is isochronous, or nearly so, with the free vibration, in which case it will manifest itself in the motion out of all proportion to its original importance.” From this it appears that the vibra- tion for each periodic force is to be found as though it alone acted. If the displacement of any point from its position of rest be examined, its magnitude and position at any instant will be the vector sum of the several displacements which would be caused by each force acting separately. • A knowledge of the principles of the preceding articles of this chapter will often indicate a way in which troublesome vibrations of foundations or supports may be minimized. 128. Possible Modes of Vibration of a Ship's Hull and the Forces present to produce them,-A ship's hull is an elastic structure susceptible of vibrating in all the different modes which have been considered for the solid bar, although the positions of the nodes and the corresponding periodic times of vibration are very different. Exact mathematical treatment is impossible. The different parts of the hull are loaded differently at different times, and since the loads have to share the vibrations, they must be included even in an approximate treatment of the problem. Any resulting state of vibration may be analyzed into the following Components:— (1) Vibrations in a vertical plane after the manner of the rods in Figs. 140 to 142, though the nodes will not be in the same relative positions as there shown. (2) Vibrations in a horizontal plane. (3) Longitudinal vibrations. (4) Torsional vibrations. The two latter types approximate much more closely to the modes of the rigid bar than the former two. Their frequencies are, however, so great relatively to the frequency of the engine that synchronism is a remote contingency. The periodic forces acting to throw the hull into vibration are– (1) The unbalanced force and couple due to the motion of the THE VIBA'A TWOAV OF THE SUPAORTS'. 215 reciprocating parts of the engine. These force oscillations in a vertical plane. (2) The unbalanced force and couple from the revolving parts of the engine. The horizontal and vertical components of these force vibrations in a horizontal and vertical plane respectively. . (3) The variation of turning moment on the propeller shaft. This tends to force torsional oscillations. (4) Corresponding to the variation of turning moment, there is a variation of the thrust on the propeller. (5) The variation of thrust due to a partially immersed pro- peller. (6) Want of symmetry in the propeller, and slight variations of the pitch of the blades. With all these exciting forces in simultaneous activity, each producing a forced vibration proper to itself, the hull is thoroughly searched for any of its natural modes of vibration of corresponding periodic time. Any that are found are immediately exalted in amplitude above all the rest of the co-existing forced vibrations, and as a result the ship is thoroughly uncomfortable to voyage in. This agreement in periodic time may take place at relatively slow speeds—for instance, a ship may be in violent vibration at half speed, and quite comfortable at full speed, or the oscillations may be unbearable at 80 revolutions per minute, and insignificant at 90. It has been shown by experiment, and may be predicted from theoretical considerations, that the unbalanced forces from the engines are the most important agents in producing forcing Oscillations. The magnitude of the force or the couple may be large, and in addition the engine may be placed in just that part of the ship, relatively to the nodes, most favourable for forcing the oscillations. There is a peculiarity in the vibration of twin-screw steamers which may be mentioned here. The two engines cannot be run at exactly the same speed. One is continually gaining slightly on the other. Suppose each engine to force a vibration. Then two vibrations of very nearly equal period will be impressed on the hull. The combination of two such vibrations gives a resultant vibration of varying amplitude. One instant the resultant is equal to the sum of the components, at another instant equal to the difference of the component amplitudes, the number of maxima or minima per minute being given by the difference between the 216 THE BAL.1 WCING OF EAVG/WES. number of revolutions of the engines per minute. The * is similar to that of beats in music. If, for example, the speeds of the two engines are 80 and 81 revolutions per minute, and the maxi- mum amplitudes of the vibration due to each separately are equal, each being a inches, there will be once per minute an amplitude of 2a inches, decreasing gradually to 0, and then increasing again to the maximum 2a. This peculiarity may be illustrated if two curves like the one in Fig. 137 are added to get a resultant curve, the base of one curve, however, being taken slightly longer than the base of the other. 129. Experimental Results, Mr. Yarrow * has given direct experimental confirmation of the foregoing principles by means of a Series of costly and beautiful experiments, in which the actual vibration of the hull of a torpedo-boat was measured under different circumstances by a “Vibrometer.” To separate the effect of the propeller from the effect of the engines, vibra- tion diagrams were taken, firstly with the boat under weigh, secondly with the boat at rest and the propeller removed, thereby entirely eliminating whatever vibration it caused. The recorded vibrations were practically alike when the engines ran at the same speed in the two experiments, and this agreement continued in over a hundred similar experiments at different speeds, showing that the effect of the propeller was small, and that the real cause of vibration was to be looked for in the engines. Experiments were then made on a first-class, 23-knot torpedo-boat, 130 feet long, 13 feet 6 inches beam, carrying 20 tons. The reciprocating parts of the three-crank engines were balanced by means of bob-weights (Art. 47), so that the reciprocating system (neglecting the valve- gear) really consisted of five cranks. It was found that 248 revo- lutions per minute corresponded with a natural mode of vibration of the hull. The experiments were made under three conditions of balancing, in all of which the speed was kept at 248. The amplitude of the vertical vibration at the stem was #4 inch, when no balancing of any kind was used, the cranks being at 120°, and the engines being of the usual design. This was reduced to #} when balance-weights were attached to the crank-shaft, properly placed to balance the revolving masses only, and to ºr when bob-weights * “On Balancing Marine Engines, and the Vibration of Wessels.” By Mr. A. F. Yarrow. Thans. Inst. Naval Architects. London, 1892. THE VIBAEA TWO/V OA' THE SUPPORTS. 217 were added to balance the reciprocating parts. Thus at a critical Speed, the proper balancing of the engine, neglecting the effect of the obliquity of the connecting-rod, reduced the maximum ampli- tude of the vibrations in the ratio of 27 to 7. The hull being moored and the propeller removed, instantaneous photographs were taken of the ripples on the surface of the water caused by the vibrating hull for each of the three conditions above stated. The ripples indicated the nodes in the hull and the places of maximum vibration, and their decreasing amplitude showed the decreasing amplitude of the hull’s vibrations as the balancing of the engines Was improved. The photographs are published in Engineering, April 5, 1892. In Some observations made by Mr. Schlick on the twin-screw despatch vessel, Meteor, belonging to the Imperial German Navy,” the maximum amplitude of the vertical vibration just near the stern post was ſº, inch, and the maximum horizontal vibration ſº, at a speed of 186 revolutions per minute. At 220 revolutions per minute the amplitudes of both directions were comparatively Small, the maximum amplitude being observed at 175 revolutions per minute. At 120 revolutions the vibrations practically dis- appeared. Mr. Schlick attributed the lateral vibrations recorded by his instrument (described in Trans. I.N.A., 1893) to the effect of torsional oscillations, the stiffness of the ship horizontally being too great to have a natural period corresponding to the speed of the engine. Mr. Schlick f has devised the following simple formula from his experimental observations, designed to give the number of vibrations per minute of the gravest natural mode of vibration of the hull:— N = vibrations per minute. D = the displacement in tons. L = length of the hull in feet. I = the moment of inertia of the midship Section, in the calculation of which the Several areas constituting the Section are to be expressed in Square inches, and the respective distances of their centres of gravity from the neutral axis in feet. * “On an Apparatus for Measuring and Registering the Vibrations of Steamers.” By Herr Otto Schlick. Trans. Inst. Naval Architects. London, 1893. f “Further Investigations of the Vibrations of Steamers.” By Herr Otto Schlick. Trans. Inst. Naval Architects. London, 1894. 218 THE BAZAAVC/AWG OF EAWG/WE.S. Wessels with very fine lines, such as torpedo-boat | q = 156,850 destroyers g is tº ge tº ſº. tº tº º e tº e gº tº e Large transatlantic passenger steamers with fine | q = 143,500 lines tº e & © e tº tº ſº tº tº $ tº • * > Cargo-boats with full lines ... tº s tº e is tº ... q = 127,900 Then— where for— Mr. Schlick” also states that for ships with very sharp lines, like cruisers and despatch-boats, the after node is from 0.23L to 0.25L from the after perpendicular, and the forward node from 0.31L to 0.36L measured from the fore perpendicular, these corre- sponding to the gravest period of vibration of the hull. The number of vibrations corresponding to the next mode of vibration is sometimes only twice the number of the gravest kind. Referring to Art. 124, it will be seen that for the solid rod the ratio is much higher. Mr. Mallock f has suggested a method by means of which the natural period of vibration and the position of the nodes in a proposed ship may be approximately found from the behaviour of a plank shaped so that its width is everywhere proportional to the moment of inertia of the corresponding section of a wood model of the hull under consideration, and loaded so that the weight at any cross-section is proportional to the weight at the corresponding cross-section of the model. In a recent paper to the Institution of Naval Architects, Mr. Schlick f describes some interesting experiments carried out on the S.S. Deutschland to ascertain the cause producing the vibrations of the hull at the synchronizing speed. The instrument recording the vibrations was placed at the ex- treme after end, and at the synchronizing speed of 67 revolutions * “On Vibrations of Higher Order in Steamers and on Torsional Vibrations.” By Herr Otto Schlick. Trans. Inst. Naval Architects. London, 1895. : + “On the Vibration of Ships and Engines.” By Mr. A. Mallock. Trans. Inst. Naval Architects. London, 1895. it “On Some Experiments made on Board the Atlantic Liner Deutschland during her Trial Trip, June, 1900.” By Herr Otto Schlick. Trans. Inst. Naval Architects. 1901. THE V/AEA'A TWOAV OAP TAZAZ S UPA'OR T.S. 219 per minute the maximum amplitude of the vertical vibrations recorded was about 5% inch, a small amount for a ship 662 feet long and 37,000 horse-power. The curve indicated a simple vibration of the same periodic time as the engine. The engines were balanced, by Mr. Schlick's method, so that the primary forces and couples in both engines are presumably completely balanced, and there is only a secondary couple un- balanced. Electrical apparatus connected each crank-shaft to the drum of the recording instrument, so that an indication was made when the respective forward cranks of the engines were vertical. A time-line was also drawn, so that from the lines on the drum the revolutions of the engines could be exactly computed, and the position of the cranks relatively to the vibration curve fixed at any instant. It has been shown, in Art. 123, that in the immediate neighbourhood of synchronism, the phase difference between the force and the vibration it causes is about 90°, being exactly 90° at the critical speed. Using this principle, Mr. Schlick inferred that the vibrations produced in the vertical plane were caused by a difference of resistance amongst the blades of the propellers, which difference is attributed to slight differences in their pitch. 130. Turning Moment on the Crank-shaft.—When the forces and couples due to the motion of the parts of an engine have been balanced, there still remains a couple acting on the frame equal and opposite to the turning couple on the crank-shaft. This couple may have a periodic variation sufficiently great to cause vibration. In the case of a ship, whatever be the turning moment or couple exerted by the engine on the propeller shaft, there is of necessity an equal and opposite couple acting on the hull of the ship, which, if the couple is uniform, holds the hull steadily in a position imperceptibly inclined to the vertical. There is a proper position of equilibrium corresponding to every value of the turning couple. A periodic variation of the couple causes an oscillation of the hull about the position of equilibrium corresponding to its average value, that is, to the average value of the turning moment on the shaft, and is thus able to force the hull into torsional vibrations, which may be insignificant or important according as the periodic time of the variation approaches the period belonging to one of the hull’s natural modes of torsional vibration. The 220 THE BA LA A/C//VG OF EAVG/AVE.S. cause of the oscillations is removed if the turning effort on the shaft is made uniform. Fig. 143 shows an engine-frame in diagrammatic form ; the forces in thick lines are those acting on the frame in Consequence of the driving pressures exerted by the fluid pressure P. The couple acting on the frame is represented by R × OB for the crank position shown. Considering the question in more detail, let M be the mass of the reciprocating parts, and a their acceleration. Then, if p is the steam-pressure or gas-pressure per square inch in the cylinder, p1 the back pressure, and A the area of the cylinder in square inches, the total resultant pressure acting on the piston and equally on the cylinder cover is (p – pi)A. Of this, the part º in which the acceleration a can be found by Klein's construction (Art. 104), is required for the acceleration of the reciprocating masses; the remainder— (n-p)A-º-r is the pressure acting to produce the turning of the crank. (The continuous variation of P is shown in Fig. 95 for the case of a locomotive running at 65 miles per hour. The indicator cards are shown in Fig. 93. The ordinates of curve No. 1, in Fig. 94, represent the values of (p — pi) A, and those of curve No. 2, º for one revolution of the crank.) The forces and couples due to the motion of the parts can be balanced by the methods already given; the finding of the effect of P may therefore be considered as a statical problem. Assuming, therefore, that all the inertia effects are balanced, P is kept in equilibrium at the cross-head B (Fig. 143), by the force Q, along the connecting-rod, and the slide- bar reaction R (acting to the right as shown by the dotted force R). The values of these three forces are given by the force triangle abc (Fig. 144), where ab represents P, and ca, and be represent respectively the force Q and the reaction R. The equal and opposite value of R, viz. cb, is the force acting on the frame from the cross-head. The connecting-rod applies the force Q to the crank-pin K in the direction shown, and its effect with respect to the axis O of the crank-shaft is equal to (Art. 24)— 7'HE VIB/8A TVOAW OF THE SUPPORTS. 221 Z à % % % % * % % % % % % % % % % % % % % % % % % % P % % & % % Y % == N º FIG. 144. º sº sº §§ (L N N # P| \Q Y b C R FIG, 145. d FIG. 143. 222 THE BA LA WCZWG OF EAVGINA.S. (1) An equal and parallel force acting at O. (2) A couple, Q × Ok, acting to turn the crank, Ok being perpendicular to the rod BK. The force Q, represented by de in the force triangle déf (Fig. 145), causes pressure on the main bearing, and its horizontal and vertical components are evidently equal to R and P of Fig. 144. The forces acting on the frame at the cylinder end are cb, that is be reversed, at the bars; and ba, or ab reversed, at the cylinder cover. Thus R at the main bearing and R at the slide- bars form a couple acting on the frame from the gear. The vertical downward component, df = P at the main bearing, is balanced by the vertical upward force ba = P acting on the top cylinder cover. To show that the couple R × OB is equal to the turning couple on the crank, it is only necessary to express Q in terms of R, and OB in terms of the angles 6 and q. The angle OKg is equal to (6 -- p), therefore Ok = OK sin (9 + q) = BO sin p. And Q = R cosec p, and hence the couple Q × Ok = R cosec p X OK sin (9 + (p). But BO = OK sin (0 + (p) cosec p, therefore R × OB represents the magnitude of the turning couple on the crank-shaft. Again, draw Og at right angles to the line of stroke; then, since P = R cot q = R × º the product P × Og = R × OB, also repre- sents the moment of the couple. Also, if Y represents the resolved component of Q at right angles to the crank, that is, the tangential force, Y × OK is another product representing the couple. Collect- ing these results, any one of the four products— Q × Ok = R × OB = P × Og = Y x OK. . . (1) may be taken to represent the moment of the turning couple, and therefore the equal and opposite couple acting on the frame, as may be most convenient for the problem in hand. It will be observed that the factors of the first three are all variable, but that Y only is variable in the fourth. Therefore, the variation of Y represents the variation of the couple. A convenient construc- tion for finding this is as follows:– Set out Ka (Fig. 143) to represent the value of P, measuring from K along the crank radius; draw acy parallel to Og, that is, THE VIBAEA TWOAW OF THE SOAPA’OA’T.S. 223 horizontal; then acy represents the value of Y. To prove this— Ra: : a y = KO : Og that is— P: a y = KO : Og therefore— aſy × KO = P × Og But P × Og is equal to the moment of the couple from (1), therefore a y = Y. This method may be very quickly applied to find the value of the turning couple for any position of the crank, and the variations of the couple may be exhibited by plotting Y vertically over a straight line drawn to represent the circumference of the crank circle. Such a curve is usually called a crank-effort diagram. The curve may ob- viously be drawn so that the vertical ordi- nate represents the couple Y × OK, in which case the length of the base-line represents 2tr. It is a matter of in- difference whether the constant multiplier OR, the crank radius, be introduced vertically or horizontally, the form of the curve and its area remain the same. The 6. T F r z- \{TN * / ſ\ 8-2 / *\º k- N O \alatºr º 4-8 / Péſº Nº.3 / | N i / & Nºt liso. * DY —5.7 Y.8%. 4.8 binae-53 cos 28 Nº FIG, 146. curve marked L.H. (Fig. 91) represents the turning couple acting on the left-hand crank of the locomo- tive under consideration in that article, resulting from the varying values of P shown in Fig. 95. There the base represents 27, and consequently the ordinates represent the couple Y × OK. Curve No. 1 (Fig. 146) shows the crank-effort curve for an engine, 6 feet 224 THE BA LA WCIWG OF EAVGIZVES. stroke, running at 60 revolutions per minute, cutting off at about 20 per cent of the stroke, and in which the reciprocating parts weigh 84 tons. In this case the base represents half the circumference of the crank-pin circle, so that the ordinates represent the tangential force. The vertical scale must be multiplied by 218 to find the tangential force in tons, the figures on the diagram being proportional only. Fig. 147 shows the crank-effort curve # of a gas-engine, and 3 may be taken as typi- -l cal for many explosion motors. The cycle 12. t º iº 4Tr. F" 3 of operation In the FIG. 147. cylinder requires four strokes to complete it; Y has very great values in the first stroke, and in the following three becomes relatively insignificant. Each of the crank-effort curves mentioned represents also the couple acting to turn the frame. It will be noticed what a large variation there is between the maximum and minimum values. The actual values may be read off the curve (Fig. 91). The maximum value in Fig. 146 is over 100 foot-tons and the minimum just under – 10 foot-tons. In Fig. 147 the variation is from 1700 foot-lbs. to — 332 foot-lbs. Expressing this in another way, the maximum value in the first case is 2-3 times the mean, which is 2:56 foot-tons, and the minimum is negative; in the second case the maximum is nearly twice the mean, falling to a negative quantity; and in the third case, where the mean is 186 foot-lbs., the maximum is 9:2 times the mean. These variations take place in a fraction of a second in each case. There is therefore, in each case acting in the plane containing the cylinder centre line, and at right angles to the axis of the crank-shaft, a couple, whose variations are approximately periodic, and which is therefore capable of forcing vibration in the frame of the same period as itself about any axis parallel to the axis of the crank-shaft. Notice that the plane in which this couple acts is at right angles to the plane of the couples of the unbalanced moving parts of the engine. THE WIBAEA 7/ON OF THE SUPPORTS. 225 131. Uniformity of Turning Moment.—In a single-cylinder engine there will always be a large variation of turning moment of the order shown in the cases just discussed. Where there are more cylinders than one, the cranks may be so arranged that the combination of the crank-effort curves corresponding to each results in a more uniform curve. In motor-cars, where the cylinders are often put in the same plane with cranks at 180° or parallel, the variation is of the type shown in Fig. 147. Con- sequently, even if these engines were properly balanced for the moving parts, there will always be acting on the frame a variable couple tending to cause oscillations. Speaking generally, syn- chronizing oscillations of large amplitude are not to be feared, because the speed of the engine is so high that the supporting springs have no grave periods to correspond to it. There are, however, forced oscillations of small amplitude, and to get rid of these the engines must be arranged, not only for balance amongst the moving parts, but so that the turning effort is much more uniform than is usually the case. The result of combining two crank-effort curves when the cranks are at right angles is shown in Fig. 91, by curve No. 1. The varia- tion is now much less, but takes place twice as fast. The ratio of the maximum to the mean is now reduced to 1:46, and, instead of the minimum being negative, it is positive and 0:45 of the mean. The crank-effort curves and their resultant curve for the s.s. Kaiser Wilhelm der Grosse are given in Engineering for April 8, 1898, p. 434. There are four cylinders, and the ratio of the maximum to the mean is 1:19, and the minimum to the mean '66. In three- or four-cylinder engines with cranks at 120° and 90° respectively, the usual equal division of work amongst the cylinders is the best to get a uniform turning moment on the shaft. The same rule would give equally good results applied to the five- and six-crank engines of Arts. 99 and 100. Applied to four-crank engines in which the cranks are arranged at angles specially found for balancing the reciprocating masses amongst themselves, the equal division of work amongst the cylinders results in a turning effort in which there may be considerable variation during a stroke. Dr. Lorenz" has shown how the * “On the Uniformity of Turning Moments in Marine Engines.” By Dr. Lorenz. Trans. Inst. Naval Architects. London, 1900. Also “Dynamik der Kurbelgetriebe.” By Dr. Lorenz. Leipzig, 1901, Q 226 THE BA LA WCIAVG OF EAWGIAVES. division of work may be made under these circumstances, to secure a better approach to uniformity of effort. The rule applies to engines of any number of cranks with any angles between them. The process of finding the rule is given in full, to show exactly what assumptions are made to obtain the results. The form of the crank-effort curve corresponding to one cylinder, depends upon the cut-off, back pressure, mass of the reciprocating parts, the length of the connecting-rod relatively to the crank, and the speed of the engine, yet, when the average speed and the rate of working are constant, the curve repeats itself at every revolution, or neglecting the effect of the obliquity of the connecting-rod, at every stroke. In other words, the turning effort is continuous and periodic in the time occupied by half a revolution, and it may therefore be represented by a Fourier series. That is to say, if Y is the moment of the turning couple at any instant corresponding to a crank angle 6, which is dependent upon the time, the value of Y is expressed by— Y = A0 + A2 cos 26 + A4 cos 40 . . . + B2 sin 20 + B, sin 46 . . . where A0 is the average height of the crank-effort curve, and A2, B2, etc., are numerical coefficients. The odd values of the angles do not appear in the series because the curve is periodic in half a revolution. There are a variety of ways of finding the values of the coefficients in the series corresponding to a given curve, the quickest and most convenient being to use some form of harmonic analyzer. It is an essential feature of Dr. Lorenz's method that terms above 20 must be discarded, consequently the value of Y is assumed to be given by three terms only of the series, or— Y = A0 + A2 cos 20+ B2 sin 26 . . . . (1) To show to what extent this value of Y differs from the true value in a typical case, the author analyzed the tangential force shown by the crank-effort curve marked No. 1, in Fig. 146, into the above harmonic constituents, finding the coefficients by means of a Henrici analyzer. The value of A0 is found by measuring the area of the curve with a planimeter and calculating the average height in the usual way. The result is— Y = 8.2 + 48 sin 20 — 5:7 cos 20 THE VIERA TION OF THE SUPPORTS. 227 The component curves are shown in Fig. 146, and are numbered Nos. 2 and 3. Their sum, including the mean height, is shown by the chain-dotted curve No. 4. A comparison between this curve and the true curve, No. 1, with which it is assumed to coincide, shows to what extent equation No. 1 is able to realize the actual conditions of any given case. It does so nearly enough to use as a basis for obtaining a working rule. To find the turning moment on the propeller-shaft for a multi- cylinder engine, the ordinates of the crank-effort curves corre- sponding to each cylinder are added together after the curves have been adjusted relatively to one another for their phase differences, in precisely the same way that has been explained for the force curves in Art. 109. Or stating the process more generally, if there are n cylinders, the turning effort on the propeller-shaft is represented in terms of the crank angle of any one assigned crank by the proper combination of n curves of the type No. 1 (Fig. 146); or, assuming No. 1 curve to be represented nearly enough by No. 4 curve, by the proper combination of n curves of No. 4 type. But the resultant curve in this latter case may be found by com- bining its components separately, thus finding the components of the resultant curve. Now, the sum of the components of the mean heights will be a straight line parallel to the axis under all circum- stances, assuming uniform speed. The resultant components of the curves of the types Nos. 2 and 3 give two resultant components which are variable. If, however, it were possible to arrange that these two resultant components were separately zero, then the resultant turning effort would be constant, since under those cir- cumstances it would be represented by the sum of the average heights of the n crank-effort curves. Dr. Lorenz shows how the separate sums of the n component curves of the types 2 and 3 (Fig. 146) may be made nothing if the n crank-effort curves are similar. This assumption of similarity cannot be exactly realized in four-cylinder balanced engines, because the inertia correction for each set of reciprocating masses is different, and therefore, even if cut off, etc., and all the other circumstances of a stroke could be kept constant through all the av cylinders of the engine, there would always remain the different inertia cor- rections to destroy the assumed similarity of the crank-effort diagrams. Dr. Lorenz finds the rule analytically as follows:— 228 THE BA/A MCIWG OF EAWGI WES. Let Y1, Y2, Ys, etc., be the turning moments on cranks Nos. 1, 2, 3, etc. Also let 6 be the variable angle between an initial line of reference revolving with the crank-shaft and a fixed line, the fixed vertical centre line of the engine, and let al, ag, as, etc., be the respective crank angles measured from this revolving line of reference. (Fig. 102 illustrates this: OX1 is the revolving line of reference and OM is one crank, OZ being the fixed line from which the variable angle is measured.) Then for crank No. 1– Y = A0 + A2 COS (20 + 2a1) + B2 sin (26 +2a1) or expanding the cosine and sine and rearranging the terms, Y = A0 + COs 20ſ A2 cos 2a1 + B2 sin 201) – sin 200A2 sin 201 – B2 cos 2a1) The total turning effort on the shaft is then— XY = XA0 + cos 20>(A2 cos 2a + B2 sin 20) — sin 20X(A9 sin 2a – B2 cos 2a) If this total effort is not to be influenced by the variations originating from the double angle 20, then both equations— > (A2 cos 2a + B2 sin 20) = 0 >(A2 Sin 20 — B2 cos 2a) = 0 must be separately fulfilled. Assuming a similar form of indicator diagram for all cylinders, the coefficients A and B (dropping the subscript 2) are proportional to the average turning effort Y., of each crank, or if a. and b denote constant quantities— A = a Y., and B = b Y 7??, 2/2 Then the conditions are— axY, cos 2a + b XY, sin 20 = 0 7??, a XY, sin 20 — blº V, cos 2a = 0 which for finite values of a and b require that— XY, COs 2a = 0 XY, sin 20 = 0 Interpreted graphically, this means that it must be possible to draw a closed polygon whose sides are respectively proportional to THE WIBA’ATION OF THE SUPAORTS. 229 the average turning effort exerted by the cranks, and the directions of whose sides are parallel to directions which are double the actual crank angles. 132. Example,_To illustrate this method, consider the example in Chapter III., Art. 48. Find what must be the distribution of work amongst the cylinders in order to obtain the most uniform turning moment. Draw an end view of the crank-shaft showing the crank angles (Fig. 148). Measuring angles from crank No. 1, draw another end view in which the angles are all doubled (Fig. 149). Draw AB (Fig. 150) parallel to No. 1, and BC parallel to No. 4 in Fig. 149, taking them any lengths. Close the quadrilateral by two JD O No || ! { Nº.3. FIG. 148. FIG, 149. FIG. 150. * } lines, CD and AD, drawn parallel respectively to cranks Nos. 3 and 2 (Fig. 149). Then the distribution of work should be in the proportion— AB : BC : CD : DA for the cylinders Nos. 1, 4, 3, 2 respectively. A parallel da gives another set of lengths, a B, BC, Cd, da, which satisfy the necessary conditions, and there are obviously a great many other ways in which they may be satisfied. Fig. 151, taken from Dr. Lorenz's paper already quoted, is the crank-effort diagram of the S.S. Medjerda. The line AB represents 230 THE BA/AAWCING OF ENGINES, the circumference of the crank circle, the angles between the cranks being indicated by the figures below it. The work done in each cylinder is written against the corresponding vertical for the respective cylinders. It will be found that the horse-powers are Very nearly in the ratio of the sides of the double-angled polygon corresponding to the angles given. In this case the maximum is C 2 N 2-TNS. 2-t\ /TN D NLY Yºlº NZ N ſº 0: S.S. I : MEDJERDA : º: lf) § § O O w; § 3 § 3 Oſ) - lil º (W) ſº 0. U) Oſ) 0. * § # ir § 0- 0. S2 5 – 5 > > * 2 -1 9 A. | B Re---------- 7 O 7+---------->|<------ 94 - 25+-----.>|<----64.5°---->|<------34-25" ----> FIG. 151. 1:21, and the minimum 85 of the mean. Of course, if the condi- tions stated in obtaining the rule were exactly fulfilled, the curve would be a straight line corresponding with CD, the mean value. 133. Short-framed Engines.—Although the forces acting on the frame due to a set of reciprocating and revolving masses in balance, form a system of forces in equilibrium, the individual forces of the system cause elastic deformation of the frame at the places where they act. If the frame is flexible these local deformations, being of course periodic, may conceivably be of sufficient amplitude to cause vibration of the supports. Generally, engine-frames are stiff enough to limit the elastic deformations to negligibly small amplitude, and a frame may be indefinitely stiffened if these elastic deformations become troublesome. In balanced engines of the usual type there are forces belonging to the system acting on the frame, at the intermediate bearings of the crank-shaft. Mr. Macalpine has proposed a form of engine * in which none of the forces forming the system acting on the frame act at any of the main bearings. In this engine, which is balanced in the longitudinal vertical plane and in a horizontal plane, the Trans. Inst. Naval Architects, 1901, and Engineering, July 12, 1901. TAZAZ VIERA TION OF THE SUPPORTS. 231 cylinders are placed in pairs across the shaft. The cross-heads of a pair are connected by a rocking beam, and one of the pair of cross-heads is connected to a crank by the usual form of connecting- rod. The two sets of reciprocating parts forming a pair are made equal in mass, and being moved in opposite directions by the rocking beam, the forces in the arrangement proposed are very nearly perfectly balanced, but there remains a couple in the plane of the two cylinders tending to force torsional oscillations. Mr. Macalpine's contention is, that it is preferable to have couples in a plane across the ship of relatively great magnitude, to having forces and couples of exceedingly small magnitude in the longi- tudinal vertical plane, since in the former case the periodic time of the gravest torsional oscillation of the hull is so Small compared with the periodic time of the engine that trouble from torsional vibrations is not to be feared. In the Wigzell engine the cylinders are all placed athwart the frame in one plane, and balance amongst the reciprocating masses effected so that there is no torsional couple left. The frame is very short, and the only forces acting on it are those due to the angle of the connecting-rods. A full description of this engine will be found in Engineering, September 7, 1900. There are three cylinders, each containing two pistons, which move oppositely to one another in each case. The three piston-rods coming through the lower cylinder-covers are connected to one central crank by a tri- angular connecting-rod. The corresponding three piston rods taken through the upper cylinder-covers are similarly connected to two outer cranks by triangular connecting-rods connected to the upper set of cross-heads by coupling-rods. An incidental advantage of this arrangement is that there is practically no pressure between the crank-shaft and the main bearings. What pressure there is, is chiefly due to the weight of the crank-shaft system alone. CHAPTER VIII. THE MOTION OF THE CONNECTING-ROD. 134, THE motion of the connecting-rod is one of periodic acceleration, and therefore the forces required to produce the motion have also a periodic variation. These forces ultimately appear as reactions on the frame, and, being periodic, may cause vibration. The effect of the rod may be divided into two distinct parts: first, it disturbs the simple harmonic motion of the reciprocating parts; secondly, the forces required for its accelera- tion appear as forces on the frame tending to cause oscillation. Chapter V. shows how the first effect is dealt with in the balancing of the engine, and a rule has been given in Art. 46 for dividing the mass of the rod between the revolving and reciprocating parts of the gear to eliminate the second effect, it being tacitly assumed that the accelerations of these two masses require reactions on the frame equal to those required by the motion of the actual rod. The object of this chapter is to investigate the motion of the rod, and show to what extent the rule given is a valid one. 135. Dynamical Principles on which the Investigation is based.— Let C (Fig. 152) be the mass centre of a body, free to move in the plane of the paper and acted upon by a system of co-planar forces whose resultant is R. This force is equivalent to— (1) An equal and parallel force acting at the mass centre. (2) A couple whose moment is R × CZ = L, say. The effect of the force is to accelerate the motion of the mass centre in the direction of its line of action. The effect of the couple is to accelerate the angular motion of the body about an 232 THE MOTION OF THE CO/WWECT/WG-ROD. 233 axis through the mass centre, at right angles to the plane contain- ing the force R and the point C. It is a fundamental dynamical principle that the acceleration of the mass centre C is the same as if the whole mass of the body were concentrated at the one point C, and that the angular acceleration of the body about C is the same as if the axis through C were fixed. Hence, if R be given in position and magni- tude, the instantaneous acceleration of the mass centre, and the angular ac- celeration about the mass centre, can <> be found when the mass of the body and its moment of inertia about the perpendicular axis through C are re- FIG. 152. spectively given. The acceleration of any point in the body is then determined, being the vector sum of the acceleration of C, and the acceleration of the point about C. The connecting-rod problem is precisely the converse of this. Two points in the rod are compelled to move in a definite manner, the one in a circle guided by the crank-pin, the other in a straight line guided by slide-bars, and if the acceleration of the first point be given, the acceleration of the second can be readily found. But if the accelerations of two points in a body moving in a plane be given, the acceleration of every point can be immediately deduced. From the acceleration of the mass centre and the mass of the rod, the force R producing this acceleration can be at once calculated. Similarly, the angular acceleration of the rod about the mass centre can be found when the accelerations of two points are given, and hence the couple L may be inferred when the moment of inertia about the mass centre is known. The problem naturally divides itself into two parts: first, the determination of the acceleration of the mass centre and the magnitude and direction of the force R; secondly, the determination of the position of R relative to the mass centre so that it causes a couple equal to L. In the geometrical method of finding R, which will be explained first, the first construction gives the acceleration of the mass centre, after which there are several ways of finding the position of R relative to the mass centre, depend- ing upon the artifice of concentrating the mass of the rod at two points, so that the two masses thus concentrated form an equivalent 234 THE BAZAAVC//VG OF EAVGZAVES. dynamical system to the actual rod. Then the two forces which must act to produce the instantaneous acceleration of this two- mass system, which has, of course, the same acceleration as the actual rod, must have R for a resultant. The magnitude and direction of R is found by the first construction. The aim of the second is to find a point on the line of action of R. This point is discovered by the intersection of the directions of the two forces producing the instantaneous motion of the equivalent two-mass system. 136. Graphical Method for finding the Acceleration of the Mass Centre of the Rod.*—Assume the angular velocity of the crank to be sensibly constant. Let OK (Fig. 153) be the crank, KB the connecting-rod, and B0 the line of stroke. Let C be the mass FIG. 153. centre of the rod. Draw CQ parallel to BA. The acceleration of the point K is a "KO. The acceleration of the point B may be found by Klein's construction (Art. 104). This acceleration is shown by AO in the figure. From a reference to Art. 6, it will be understood that AK is the acceleration of B relatively to K. QK is then the acceleration of C relatively to K, since QK : AK = CK : BK. The whole acceleration of C is the vector sum of its acceleration relatively to K and the acceleration of K, that is, the vector sum of QK and KO = Q0. It is evident that there is nothing to restrict this reasoning to the point C. Therefore, the acceleration of any point on the rod may be found by projecting the point on to KA by a line parallel to BO, and joining the point so found to O. Any point thus projected divides K.A. in the * Many of the following graphical methods are given in “New Constructions of the Force of Inertia of Connecting-rods and Couplers and Constructions of the Pressures on their Pins.” By Professor J. F. Klein. Journal of the Franklin Institute, Vol. CXXXII., September and October, 1891. THE MOTYOAV OA' THE COMAVECT/WG-FOD. 235 same ratio that it divides KB. In fact, KA is the rod drawn to a smaller scale, and in such a position that the line joining any point on KA to O, represents the acceleration of the corresponding point in the actual rod. On account of this property, KA has been called the acceleration image of the rod. QO is to be measured to the scale on which KO represents the crank radius; then the actual value of the acceleration is wºQO. If M is the mass of the rod, the magnitude of the force R, is— y w°Q0 lbs. weight acting at C in a direction parallel to Q0 from Q towards O. This force applied at the mass centre, would produce the instantaneous acceleration of the mass centre which it actually undergoes. 137. Equivalent Dynamical System.—The conditions to be Satisfied by the concentration of the mass of the rod at two points are— (1) The sum of the two masses must be equal to the mass of the rod. (2) Their mass centre must coincide with the mass centre of the rod. (3) Their moment of inertia about an axis through the mass centre at right angles to the plane of motion of the rod, must be equal to the moment of inertia of the rod about the same axis. Let mi, m3 be the two masses into which M is divided, distant respectively di and d2 feet from the mass centre of the rod, and let /º be the radius of gyration of the rod about the axis at the mass centre. The three conditions stated above are expressed by the equations— m1 + m2 = M . . . . . . (1) "mid1 - m2d2 – () s º e +- º e (2) midi" + m2d2° = Mk” . . . . . . (3) From (1) and (2)— Md., 771.1 = all-Ed, ' ' ' ' ' ' ' (4) and— Mdi (5 T. . . . . . . . (5) 77b2 F 236 THE BA /.4 AVC/AVG OF EAVG/NES’. Combining these with (3)— didº = k” . . . . . . . . (6) This is the relation governing the position of the masses. Their magnitudes are, as shown by equation (2), inversely as the mass centre divides the distance di + do. Observe that k is a mean proportional between di and d2. Hence, if the position of one mass and k be given, the position of the second mass can be found by the following construction:— Let C be the mass centre of the body, which is symmetrical about the line sh, shown in Fig. 154. Let S be the given position P FIG, 154. of one mass. Draw CP at right angles to Hs, its length repre- senting to scale the radius of gyration k. Join S and P, and draw PH at right angles to sp. H then marks the position of the second mass. It is evident that there is nothing to restrict the position of the given point s, consequently an infinite number of pairs of points, s and H, can be found. It may be noticed that if the body be suspended from an axis through the point s, shi would be the length of the simple equivalent pendulum, and that H would be the centre of percussion relatively to 8, and s the centre of percussion relatively to H. Returning to the connecting-rod, the value of k must be known, or the position of the two points s and H, before the division of the mass can be made. The best way to arrive at the position of a pair of points of a finished rod is to let it oscillate about some selected point s, the axis through the centre of the small end usually, and adjust the TAZAZ MOTIO/W OF THE CO/WWECT/WG-ROD. 237 length of a plumb-line until it swings in unison with the rod. The length of the plumb-line, measured from the point about which it swings to the centre of the bob, gives the distance shi, that is, d1 + do. The position of the mass centre C can be found by balancing the rod on a knife-edge; thus the individual values of d1 and d2 are known, and k can at once be calculated. Ea’ample.—The length of a plumb-line adjusted to swing in unison with the connecting-rod belonging to an inside-cylinder, four-coupled passenger engine, the rod swinging about an axis at the small end-centre, was found by experiment to be 5-9 feet. The distance of the mass centre, the position of which was found by balancing the rod on a knife-edge, was 492 feet from the axis of suspension. Therefore— di = 4.92 feet, d2 = 0.98 feet Hence— /* = 4.82, and k = 2:19 feet 138. Constructions for fixing a Point in the Line of Action of R.—Having found a pair of points, s and H, on the assumption that k is known, the object of each of the following constructions is to find the respective lines of action of the forces acting on the concentrated pair of masses. The intersection of these two lines of action fix a point X on the resultant, defining thereby its line of action, since its direction is known from the construction of Art. 136. Construction 1 (Fig. 155).-Choose the point s to correspond mass-rººms--- * * * * *-* * -- ~ — —t FIG, 155. with B, and find H by the method of Fig. 154. Repeat the con- struction of Fig. 153 to find KA, the acceleration image of the rod. The direction of the acceleration of B is along the line of stroke. To find the direction of acceleration of H, refer it to the line KA, then h() gives the required direction. Therefore, the 238 THE BALA WCING OF EAVG/NWES. intersection X, of a line parallel to h9 drawn through H, with BO fixes a point in the line of action of R. The line through X parallel to Q0, R's direction, is the line of action of the resultant R. Its perpendicular distance from C is such that it causes the couple L, producing the angular acceleration of the rod. Construction 2 (Fig. 155).-The motion of the rod may be analyzed into a translation of the rod parallel to itself, every point moving with the acceleration of B, and an angular motion about B. The force to produce the first acceleration must be applied to the system at the mass centre C in a direction parallel to the line of stroke. The force to produce the angular acceleration about B must be applied at H in the direction of acceleration of H relatively to B, that is, in the direction h A. The intersection of a line through H, parallel to ha, with CQ fixes a point, XI, in the line of action of R. A line through X1 parallel to QO is therefore the line of action of R. w Construction 3 (Fig. 156).-Choose s to correspond with the big end-centre K, and apply the construction of Fig. 154 to find H. The direction of acceleration of K is along the crank radius. Project H on to the image of the rod KA, to find h0, the direction of acceleration of H. A line through H, parallel to h(), intersects the crank produced in a point, X2, on the line of action of the resultant R. Construction 4 (Fig. 156).-The motion of the rod may be analyzed into a motion parallel to itself, where every point moves in a circle of radius equal to the crank radius, and an angular motion about K, the crank-pin. To produce the first, a force must be applied at the mass centre C in a direction parallel to the crank KO. To produce the angular motion about K, a force must be applied at H in the direction of the acceleration of H relatively to K, that is, in the direction h R. Therefore, the intersection X5 of a line through H. parallel to lik, with a line through C parallel to the crank radius KO, fixes a point in the line of action of R. Construction 5–The point 8 may be taken in any position on the centre line of the rod, and a point, H, to correspond with it found by the construction of Fig. 154. The directions of their accelerations are found at once from the acceleration image, from which a point, X, can immediately be fixed. Any one of these constructions may be used in combination with the construction of Art. 136 to find R. The combination THE MOTYON OF THE CONNECT/WG-ROD. 239 with Construction 1, and, in fact, the whole question of the inertia loading of links, is discussed in Professor Klein's paper already quoted. The combination with Construction 2 is due to Professor Dunkerley and Mr. J. B. Peace. The use that may be made of FIG. I56. it to find the inertia loading of a connecting-rod is shown in an article by Professor Dunkerley in Engineering, June 2, 1899. The whole question of acceleration and velocity images is treated in a general way in Professor Smith’s “Graphics.” 139. Combined Construction,-For convenience of reference, the .# steps in the combination of the construction of Art. 136, with construction No. 1 of the previous article, are restated. To find the force R (Fig. 155) for any given crank angle : mark the points C and H on the rod; find AO, the acceleration of B, by Art. 104; join KA; draw CQ and Hh, respectively parallel to B0. Then the intersection of a line through H parallel to ho with the line of stroke fixes a point, X, in the line of action of R. 240 THE BA/A WCINVG OF EAWG/WE.S. A line through X parallel to Q0 is the line of action of R, and R’s magnitude is— *oo lbs. weight 140. Effect on the Frame and on the Turning Moment exerted by the Crank-Since the force R, under whose action the instan- taneous acceleration might be produced, can be found for any crank angle by the construction of the previous article, the effect of the rod's motion on the frame is reduced to the statical problem of finding the effect of R on the frame. Let R be the resultant force (Fig. 157) for the crank angle shown. Referring R to the crank-pin, it is equivalent to— (1) An equal and parallel force R acting on the crank-pin. (2) A couple whose moment is R × gi acting on the rod. The couple is actually applied to the rod by a pair of parallel forces acting respectively at the cross-head and crank-pin in a direction at right angles to the line of stroke BO, since this is the only direction in which a force can act from the frame at the slide-bars, neglecting friction. Draw Kp at right angles to BO, then the magnitude of each force of the couple is— R × gi Bp So that acting at the crank-pin there are two forces, S, acting always at right angles to the line of stroke, and a force equal and parallel to R. Referring these forces to the main bearing, O, each is equivalent to an equal and parallel force acting at the bearing and a couple acting on the crank. The whole couple acting on the crank is thus— F. S, Say R × Oj + S × Op This is the extent to which the turning moment exerted by the crank is modified by the motion of the connecting-rod. The resultant force R, therefore, requires that— (1) A force, S, acts at the slide-bars from the bars to the rod. (2) Forces S and R act at the main bearing O. These are the forces which must act from the frame on the gear to produce the acceleration of the rod. The forces acting from the gear on the frame are therefore equal and opposite to these, and THE Morrow of THE cowlyzcz/WG-Rop. 241 are shown below in Fig. 158. Notice that S at the bars and –S at the main bearing form a couple. The effect of R on the frame is thus equal to— (1) A force, equal, opposite, and parallel to R, acting on the main bearing. (2) A couple S x OB. FIG. 157. v S Q _--~~~~ K * A 2^ / / | & N C * * / | sº B : | | p O N * s \ fºx \ | / | R R.S. S * f \— FIG. 158. Another way of stating the effect is— (1) The resultant OR1 of R and S at the main bearing. Call this F. (2) A single force S at the slide-bars. The forces F and S may be found directly by a simple con- struction (Fig. 159). The rod is at any instant in equilibrium under the action of three forces, viz. the forces acting through its ends and R reversed. These three forces must therefore meet in a R 242 THE BAZA NCIWG OF EAVG/AVES. point. The force acting at the slide-bars must be at right angles to the line of stroke, and the line of action of R is known; hence, draw a line at right angles to the line of stroke through B, to meet R produced in the point V. Join V.K. Set off WW1 equal to R. Draw WIV, parallel to VB. Then V2W is the whole force acting at the crank-pin, from the rod to the pin, and VIV2 is the force from the rod to the slide-bars. The force V2W at the crank-pin is equivalent to an equal and parallel force at the main bearing and a couple WW2 × Oa. This couple is equal and opposite to that V V2 N. V, FIG, 159. previously given, which modifies the turning moment of the crank, and the force F is the same as the resultant of R and S (Fig. 158). When the construction is carried out on the figure used to find the value of the resultant R, that is, Q0 in Fig. 153, the point V is only required to fix the direction WW2, since, if a perpendicular to the line of stroke be drawn from Q to cut a line through O parallel to WW2, the point of intersection fixes the value of F and S. The triangle formed in this way is shown in Fig. 158, where OR is the resultant, OR1 the direction V2W, and RR1, perpendicular to the line of stroke which fixes the point R1, defining thereby the lengths of OR1 and RRI, that is, F and S. 141. Examples.—Fig. 160 shows the frame reactions at the main bearing and the slide-bars respectively for the rod of an inside cylinder locomotive. The specification of the rod is as follows:— THAZ MOTIOM OF THE COMWAVECT/WG-AEOD. 243 Length, centre to centre, 6'81 feet Distance of mass centre from the small end, 493 feet Distance of the centre of percussion H from the small end, 6:0 feet - Mass of rod, 454 pounds The method of finding points on the curves was first to find R by the construction of Art. 139, and then to apply the construction FIG. 160. k-------------L = 6.3_CRA NKS ----------------- F-> S →- * B I G e-** - - - - - - - - - - • 72, IL IC END | | * • &3 L ---------------------- > | | FIG. 162. t | 41 N° 3. | …TT s | | Tö1 & 3. 4. 5 6 7 8 .S. I6 FIG. I.61. of Fig. 159 to find S and F, for twelve different angular positions of the crank. Curve No. 1 is the locus of the point corresponding to R (Fig. 158), and curve No. 2 the locus corresponding to point R1 (Fig. 158). The vertical distance between the curves represents RRI (Fig. 158), that is, the force S, whose equal and opposite acts at the bars. Curve No. 3 (Fig. 161) represents the changing values 244 THE BAZANCING OF ENGINES. of S, set out at the slide-bars. Thus, when the crank is at OK (Fig. 160), the length of O4 measured to curve No. 1 represents the magnitude and direction of R, the length of O4, the magnitude and direction of OR1 = F, the whole force on the main bearing, and the intercept, 441, equal to 44, on curve No. 3 (Fig. 161), represents the force at the slide-bars. The actual magnitudes of these forces are to be found by measuring the lengths of the lines respectively 5 7 CURVE N9 I. * 5 6 4–ſºr Hās; gº *T." Tºs; 3-H. 3 4 § & º FIG. 163 k------------------ L - 4 - 5 CRANKS Tº "... sº— * *:::: ſ 7 L ----------------- >: C * * * ------------------------------- •93L > FIG. 165. N 9.3 . T, T2 sº <3. J 9. 7 8 3-#Hz FIG. 164. representing them on the diagram, to the scale on which OK represents the crank radius, and multiplying the lengths so found, in feet, by the mass of the rod and the square of the angular velocity of the crank, dividing by g to get the result in lbs. weight. For example, if the crank radius OK represents 1:08 feet, O4 measures 0.846 feet, O41 measures 0-76 feet, and 441 measures 0.114 feet. At 240 revolutions per minute w” is 632. THE MOTION OF THE CONNECTING-ROD. 245 Therefore— O4 = M. × 0.846 = 7538 lbs. weight = R _M O41 = ;” × 0.76 = 6770 lbs. weight = F 44 = ** x 0.114 = 1016 lbs weight = S (ſ p Figs. 163 and 164 show sets of curves for the rod of a torpedo- FIG. I66. 5 6 4. 8 3 CURVE N9 I. 2 4. 5. 7 tº II. His 6 7 N 1 5 3 NX10 77 *_ 4. S Nº. 2 1 7 3. s M1 2. o 12, 172, * Tº T3 4. *T-AE-F-I-IVB- | | k L - 4 CRANKS ------------------- > |-e-Ha-º. * ſº Fº 168. ~I O 1 2. 3 4. s 6. 17 8 9 10 1ſ 12, FIG. I.67. boat destroyer, and Figs. 166 and 167 curves for a rod of uniform section. These latter curves are merely of theoretical interest, since the big and Small ends of the rod must always be of con- siderably greater section than the body connecting them. In each case the dynamical peculiarities of the rods are indicated by the centre-line drawings (Figs. 162, 165, and 168). Comparing Figs. 246 THE BAZAAVC/WG OF EAWGINES. 160, 163, and 166, it will be noticed that the nearer the point H, the centre of percussion relatively to the small end, is to the big end, the smaller is the maximum vertical distance between the curves Nos. 1 and 2, which indicates that the couple, of which the vertical distance represents one force, is small also. If H coincide with the big end, the line of action of the resultant R passes through the point O in every position of the crank; and if in any position of the crank R. pass through K, the couple vanishes. This is the case in Fig. 163 at No. 10 crank position. 142. Balancing the Rod.—Suppose the form of the rod to be such that, when s is taken at the small end, H falls on the crank-pin centre. The two masses forming the equivalent system are then concentrated at these two points, their magnitudes being inversely as the mass centre divides the rod; this follows from expressions (4) and (5) (Art. 137). If the masses are treated independently, the one considered to be attached to and moving with the cross- head reciprocating masses, the other attached to and moving with the crank-pin, the forces required for their acceleration must have R for their resultant, since the masses are equivalent to a dynamical system of which R is the resultant accelerating force. If these forces are exactly balanced, it is clear that R at the main bearing is balanced, since in all positions of the gear the line of action of R passes through the centre of the main bearing. There still remains the couple S × OB acting on the frames, consequent upon the existence of the couple R × gi (Figs. 157, 158). The frame- couple only vanishes when this couple vanishes, that is, when the line of action of R passes through the crank-pin centre. . The rod is seldom of such a form that s and H fall at its centres. But it will be noticed that their position has no effect on either the magnitude or the direction of the force R. These points only determine the line of action of R relative to the mass centre, and therefore affect only the magnitude of the couple R × ig, and ultimately the frame-couple S × OB; and since S is always at right angles to the line of stroke, it has no component in the line of stroke. Hence, so far as the unbalanced forces in the line of stroke are concerned, it is only necessary to consider R at the main bearing for any form of rod. If the mass of the rod is distributed between the crank-pin and cross-head inversely as the mass centre divides the rod, the assumption is tacitly made that THE MOTIO/V OA' THE CONNECT/WG-ROD. 247 these two separate and independent masses form an equivalent dynamical system ; so far as the forces in the line of stroke are concerned, there is no error in the assumption; at right angles to the line of stroke, however, this assumption involves an error in S depending upon the position of the point H. No attempt is usually made to balance this couple on the frame. - The author suggested a way of dividing the mass of the rod between the crank-pin and cross-head whereby the resultant of the 6 4. A &º & Q sº Q $- --> & * dº IV $- 24, III dº IV --------§§'s a nº * Ar 10 ¥ J.-->4.T i 8 __>~ 6 & e _2~ II q. - → N 2 tº a gº & No z I __ IOX \ % sº \ 12 O O BH = • 66 L H C = • 5 L L == • 4 CRANKS FIG. 169. forces required to accelerate the two masses is approximately, but very nearly, coincident with the force OR1, the whole force on the main bearing, during the whole revolution. The mass at the crank-pin is found from the expression— M × BC X BH RB2 M being the mass of the rod, BC and BH the respective distances of the mass centre and the centre of percussion from the small-end centre, KB the length of the rod. The remainder of M is placed at the cross-head. The curves of Fig. 169 show to what extent the resultant force at the main bearing, due to the acceleration of the two masses found from this formula, differs from the true value of 248 THAE BA ZAAVC/WG OF EAWG/AWE.S. OR1. The thick curve (No. I.) shows the locus of the point corresponding to R1 of Fig. 158, and is, in fact, curve No. 2 of Fig. 166. The thin curve (No. II.) shows the locus of the end of the force acting on the main bearing due to the accelera- tion of the masses divided in the way suggested, and the dotted curve (No. III) shows the locus of the end of the force due to the acceleration of the cross-head and crank-pin masses divided inversely as the mass centre divides the rod. In the position 8, for example, the length of O8 measured to curve No. I. is the actual force on the main bearing, the length of O8 measured to curve No. II. is the force consequent upon the mass division suggested, the length of 08 measured to curve No. III. is the force consequent upon the inverse mass division. This shows that a better agreement at the main bearing can be obtained if desired, leaving the force S at the slide-bars unbalanced. There is an error introduced in the line of stroke by this method, but it should not be forgotten that in many cases the mass added to balance the reciprocating mass can only be arranged to do so in a very approximate manner, and that therefore a small error in the line of stroke in the mass assumed concentrated at the cross-head is of no consequence. The method of division suggested may sometimes be found useful when it is desired to relieve the main bearing of the whole force OR, due to the acceleration of the rod, and when the point H is some distance from the big end. 143, Particular Form of Balanced Engine,—The rod is some- times balanced by opposing to it a similarly formed rod of equal mass, moving similarly but in opposition. Fig. 170 shows the arrangement. The crank is prolonged so that there are two cranks at 180°, and the mass centres of each rod move in the same plane. The forces acting on the frame are two sets like those of Fig. 158. They are indicated in Fig. 170, one set being shown in dotted lines. It will be seen that the forces at the main bearing mutually balance, leaving a force S at each slide-bar, which together form an unbalanced couple of moment S × B1.B2. To realize this arrangement practically, one connecting-rod must be divided into two, each, half the mass of the original rod, and formed similarly to it, and each being placed at equal distances on opposite sides of the central plane of motion. It should be noticed also that in this arrangement equal reciprocating THE MOTYO/W OF THE COMAWAECT/AWG-A’OD. 249 masses of any magnitude, placed at B2 and B1 respectively, balance each other exactly. The effect of the angle of the connecting-rod is entirely eliminated in the line of stroke, since B2 and B1 move with exactly equal and opposite acceleration. Suppose that instead of splitting one rod, the two are displaced relatively to one another along the axis of the shaft so that their FIG. 170. respective planes of motion, 1 and 2 (Fig. 171), are a feet apart. It will be apparent that there will be now couples R × a and S X a, acting on the frame derived from the forces at the main bearing, and a couple, S × B2B1, the diagonal distance between the cross-heads, derived from the pair of equal and opposite forces acting at the bars. If a similar system of two cranks is arranged 250 THE BAZAAVC/AVG OF EAVG/AVES. anywhere along the shaft in the manner shown in planes 3 and 4 (Fig. 171), the couples arising from the forces at the main bearing mutually balance. The forces at the four cross-heads form two couples acting respectively in the planes of which Biba and Baba are the horizontal traces. Their axes are in the respective directions XX and YY, and their magnitudes are equal. The vector sum of these couples, found by the triangle (Fig. 172), is represented by the line AC. This couple is always equal in magnitude to twice the projection of the distance B1R2 = Baba, on the line B1 (Fig. 171), which equals BiB2 (Fig. 170) multiplied by S, and it tends to rock the frame about the axis of the crank- shaft. There is always, in addition to this, the couple equal and opposite to the driving couple (see Art. 130), and the couple due to the acceleration of the reciprocating masses by the rod, tending to rock the frame about the crank-shaft. These are usually great in comparison with the couple due to the connecting-rod. Mr. W. G. Wilson has designed a motor-car engine on these principles, and its smoothness of running at all speeds, the maximum being 1500 revolutions per minute, is remarkable. 5. k-------- va. – SO j B, * 144, Analytical Method of finding R and L. Let M be the mass of the rod (Fig. 173); 6, the crank angle measured from the initial direction OX; o, the constant angular velocity of the crank; q, the angle made by the rod with the initial direction OX; a, the angle made by the resultant force R with OX; OK = a, the crank radius in feet; FCC = b, the distance of the mass centre C from the crank- pin K ; KB = l, the length of the rod; k, the radius of gyration about the mass centre; a, ], the co-ordinates of the mass centre C. THE MOTIO/V OA' TAZE CONAVECT/WG-ROD. 251 Use the Newtonian notation for representing differentiations with regard to the time, viz. – da: e * - d?” à, for º, ø, for “... d; ** if: d”q, 72° etc. $ for In all that follows, b = 0, the angular velocity of the crank is assumed to be sensibly constant, so that 0 = 0. Suppose the force R transferred to the mass centre C, the transference giving rise to the couple L (see Art. 135). The component of R parallel to the X axis is R cos a. The acceleration of the mass centre parallel to the X axis is 3. Therefore— R cos a = Mā; . . . . . . . (1) Similarly— R sin a = Mjj . . . . . . . (2) and— L = Mk.” i . . . . . . . (3) The direction of R is evidently given by – tan a = § 30 Its magnitude can be calculated from either (1) or (2). The position of R relatively to the mass centre is given by— L * =& -2 The constraint applied to the rod by the crank-pin and slide- bars is such that, for all positions of the gear, whether the rod is arranged to work to the left or to the right of the crank as shown in Fig. 173, by full and dotted lines respectively— sinº = -}sin 9 . . . . . . . . (4) Therefore— * l cos q = + VP – a sin” 0 . . . . . (5) the positive sign to be taken if the rod is to the right of the crank, the negative sign if to the left, because, in the first case, cos p is 252 TA/AE BAZAAVC/WG OF ENGWAVES. positive during the whole motion of the crank; in the second case, it is always negative. In what follows the minus sign will be retained in the radical corresponding with the arrangement of gear shown in full lines in Fig. 173. From equation (4)– q = sin- (-4sin 0) . . . . . . (6) The angular velocity of the rod— (p = ôa cos 0 v//? – a” sin” 0 The angular acceleration— (7) 6Pa(!” – a”) sin 9 :-. . . . . . . (8) (!” – a” sin” 9); If the rod is arranged to the right of the crank will be negative and j positive. Using the value of p given by equation (8) in (3), and Writing o, the constant angular velocity of the crank, for 0, the couple L is given by— o°a(l’ – a”) sin 9 . . . . . (9 (? – a” sin”0)? | tº (9) from which its magnitude can be computed for any given value of 0, the crank angle. It will be noticed that at the dead centres, where sin () is zero, the couple vanishes. Again, the position of the mass centre relatively to O, the origin, is found by taking the sum of the two vectors, a, 6, and b, p ; or by taking the sum of their horizontal and vertical components. Thus, æ is always the vector sum of a cos 0 and b cos p, and y the vector sum of a sin 6 and b sin ºp. Then— a = a cos 0 + b cos p . . . . . . (10) Velocity of C parallel to X— L = –Mº * # = — aſ sin 6 — by sing . . . . . (11) Acceleration of C parallel to X, remembering that b = 0– ă = — aff” cos 0 – biºcos 4 - bi sin p . . . (12) THE MOTIOAV OAT THE COMAWAZCTIAWG-RO D. 253 Similarly— y = a sin 6 + b sin p . . . . . . (13) Velocity of C parallel to Y- # = affcos 0 + bi cos 4. . . . . . . (14) Acceleration of C parallel to Y-- & # = — aff'sin 9 — b% sin q + bó, CoS p . . . (15) Therefore— * — aff” cos 0 – biºcos p - bi sin q and— R Mž My (17) cos a sin a 145, Values of j, q, and 3, §, at the Dead Centres.--When () = 0, or 180°, cos 0 = + 1 and sin 6 = 0. Substituting these values in equation (7)— # = *-*. hen 6 = 0 and— e 00) O q = – , = - , when {} = 180 A substitution of the value of the sine in equation (8) shows that— j = 0 Again, substituting these values of , and $ in equation (12), writing to for 6, and noting that q = 180° when 0 = 0 or 180°, so that cos q is negative— 3 = -aw (1 tº- º when 0 = 0 . . . (18) and— * = a(1 + %) when 0 = 180° . . . (19) § vanishes for both values of 6. 254 THE BALA/VCIWG OF ENGINES. 146. The Acceleration of the Cross-head in the Line of Stroke may be found by putting l for b in equation (12). Let }, represent the acceleration in the line of stroke; then– à = —aff cos 0 – li” cos ? – li sin p . . . (20) If the values of j, q, from equations (7) and (8), and the values of sin q = –% sin 6 and cos (p = +}v$2 – 2 sin” 9, be substituted in this equation, it reduces to— * al” cos 20 + a” sin" ! (21) (!” – a” sin” (); e tº e *1 = —alºcos {} |. in which the upper sign is to be taken if the rod is to the left of the crank, and the lower sign if to the right. The acceleration of the cross-heads at the dead centres may be found by writing l for b in equations (18) and (19), giving, when 0 = 0 . . . . . . . . . . . (22) £1 = —ao (1 º- %) and when 0 = 180° . . . . . . . . . . (23) Q1 = aw (1 + %) 147. Example.—Given that— a = 1 foot b = 1.2 feet = 3 feet 2 = 0.81 and that the rod is arranged to the left of the crank in the way shown in Fig. 173, find the magnitude, direction, and position of R, and the value of the couple L, when 0 = 120° and at the dead centres, in terms of the mass of the rod M and the angular velocity w. If the values of , and j in terms of 0, given by equations (7) and (8), be substituted in equations (12) and (15), giving the value of 3 and j, 6” will appear as a factor of every term. Hence, in THE MOTION OF THE CONNECTIVG-AZOD. 255 computing the values of these expressions, 6 may be taken equal to unity; P = 0° must then be introduced as a factor in the final result. Hence, wherever 9 appears, take it equal to unity. From the given data— q = 196° 46' from equation (6) q = –0.1740 radians per second from equation (7) # = –0.2920.” radians per second per second from equation (8) 㺠= 0:4340” feet per second per second from equation (12) # = –0-52000” feet per second per second from equation (15) Then— tana = -0.520° 0.43400° from which— a = 309° 55' Also— Mæ M% = tº = tº , f e 2 gº. = sin. trom equations (1) and (2) Therefore— R = 0-68Moj” in absolute units of force The value of the couple L, found by substituting the value of q, in equation (3), is— – 0.292Mkºo,” = –0.236Moo? The perpendicular distance, hig (Fig. 157), of R from the mass centre, to cause this couple, is found from— & 2 hig = # = º = 0.35 feet approximately R must be placed so that the couple is negative, i.e. clockwise. At the dead centres, using equations (18) and (19)— = - 0-96Mo” when 0 = 0 R = 1.96 May” when 6 = 180° L vanishes for both angles. 256 THE BAZA MCYAVG OF EAVG/AVES. The acceleration of the cross-head, found from equations (20) or (21), is— 0-330° At the dead centres, using equations (22) and (23), the accele- ration is— – 0.660° when {} = 0° and— 1-330° when 6 = 180° EXERCISES, GRAPHICAL work is connected with most of the following exercises. The student is recommended to get a set of scales in which the foot is divided decimally instead of into inches. The glass scales made by Zeiss, divided into centimetres and tenths, are very useful, and are more accurate than the ordinary type of boxwood or ivory scale. The scale divisions are engraved on the side of the glass, which is put in contact with the paper; parallax in reading the scale is thereby avoided. * To set out an angle, as a (Fig. 104), measure out the distance a; equal to unity, using as large a scale as possible. Then set out y at right angles to OX, equal in length to the trigonometrical tangent of the given angle, the value of which is to be found from a table of tangents. Conversely to measure any angle, set out a equal to unity, to as large a scale as possible, and draw y at right angles. Measure !), and look out the corresponding value of the angle from a table of tangents. In the following examples many of the results are given approximately, angles being given to the nearest degree, and magnitudes to the nearest whole number. Unless otherwise stated, an angle is always to be measured in the way shown in Fig. 6, from a horizontal initial line. Direction will be indicated in some cases by subscript figures or signs; thus— 1280, 603809, ABue mean, respectively, a vector quantity whose magnitude is 12 and 257 S 258 7"HAE BAZAAVC/AWG OF EAWG/AWE.S. whose direction is 30° with the initial direction; a quantity of magnitude 603 whose direction is 340° with the initial line, measured of course counter-clockwise; a vector of magnitude AB inclined 6 degrees to the initial line. 1. Find the sum of the vectors given in Schedule 1, setting them out in any four of the possible twenty-four different orders. 2. Show that, if the magnitudes of the three vectors are equal— AB02 + BC1200 + CD2400 = 0 and that— AB900 + BC1802 + CD270° = AB1800 and that— A Boo + BC600 + CD600 = 2,65410 3. Assuming the following set of vectors to represent forces acting at a point, find the force required to maintain equi- librium— 120°, 24.300, 361900, 101900 Answer.—41’ 7269. 4. Find the velocity of a point in the rim of a driving-wheel turning in the positive direction, belonging to a locomotive running at 60 miles per hour— (1) When the point is in its highest position. (2) When in its lowest position. (3) When the radius of the point is at 120° with the initial line. Diameter of the wheel, 7 feet. *. (Take the vector sum of the velocity of the wheel-centre and the Velocity of the point in the several cases, supposing the wheel-centre to be fixed.) Answers.- (1) 1761800 feet per second. (2) 0, (3) 1701959 feet per second. FXAEA’C/S E.S. 259 5. Find the difference of the vectors— (1) 1230° gº 301800. (2) 301802 – 1230°. Answers.--(1) 40°36′. (2) 40.81sg”. 6. A bicyclist rides at 12 miles per hour due north. Find the direction from which the wind appears to him to be blowing, and find the component velocity of the wind he must actually ride against when the actual velocity of the wind is— (1) 12 miles per hour blowing from the E. (2) 12 5 5 2 3 2 3 S. (3) 12 3 5 5 x y 2 N. (4) 12 5 5 5 * 2 3 N.E. Answers.-- (1) Direction N.E. Northerly component, 12 miles per hour. (2) No wind. (3) N. Northerly component, 24 miles per hour. (4) 23° to E. of North. Northerly component, 20.5 miles per hour. 7. The crank of an engine makes 5 revolutions per second in the clockwise direction, and the crank-pin is 9 inches radius, the connecting-rod is 3 feet long. Find— (1) the velocity of the cross-head pin relative to the frame, (2) the velocity of the cross-head pin relative to the crank pin, (3) the velocity of the crank-pin relative to the cross-head pin, when the crank angles are respectively 30, 90, and 120 degrees, measured from the initial direction, the cross-head being arranged to the left of the crank-shaft (as shown in Fig. 153). (The velocity of the cross-head relatively to the crank-pin is given by— Vector difference (velocity of cross-head minus velocity of crank-pin) The direction of relative velocity is at right angles to the connecting-rod ; hence the directions of the three sides of the vector triangle are given and the magni- tude of one of them from which the various problems above can be solved.) 260 THE BAZAAWC/AVG OF EAVGINES. Answers.-- Angles ... ... * * * 30° 90° 120° Velocity of cross-head pin rela- Feet per second. tive to frame ... tº tº wº * e ºs 9:2180° 23:551800 23180° Velocity of cross-head pin rela- tive to crank-pin tº º 'º. * * * 20-6272 O° 12102. Velocity of crank-pin relative to Cross-head ... * * * * * * 20-6970 0° 1228% 8. Defining acceleration as the change of velocity per second, prove formula (2), Art. 9. (The change of velocity in time dt is the vector difference of the velocity at the end and at the beginning of the interval. When a body is moving uniformly in a circle, the magnitude of the velocity at the beginning and at the end of the interval is the same, but the direction is different. Drawing a triangle to find this difference for a small angular displacement of the radius d6, which takes place in time dt, it will be observed that in the limit, the vector difference may be expressed by Vd6 = ord{}, in a direction towards the centre of the º ſe 0.6 circle. The change per second is therefore wr: = 0°r, dź and the force corresponding to this acceleration is— Majºr poundals 22, Moºr lbs. weight. J 9, Find the two vectors, a, 3, required to make— Ot + 3 + 1430 + 2000 + 100 = 0 having given that the magnitudes of the two vectors are re- spectively— (1) 20 and 14.9. (2) 12:5 and 32-7. Az XERCISES. 261 Answers.--(1) 202800 and 149230°. (2) 12:5300° and 32°7210°. Oſ– 12:51610 and 32.72592. 10. A clack-box is bolted to an angle-plate on the face-plate of a lathe for boring out the valve seating. The mass of the box and the angle-plate and attaching bolts is equivalent to 50 pounds at 3 inches radius. What mass must be added to the face-plate at 1 foot 6 inches radius to effect balance 2 (The seating cannot be bored truly unless the work is balanced.) Answer.—8-33 pounds. 11. A rope-wheel, weighing 1 ton, driving the main shaft of a mill at 150 revolutions per minute, caused the bearing nearest to it to heat. The distances of the right-hand and the left-hand bear- ings from the centre of the wheel are respectively 1 foot 9 inches and 5 feet. The shaft was disconnected, and it was found by experiment that the wheel was out of balance (the rim was not turned inside) to the extent of 34 pounds at 2:8 feet radius. Find the distance of the mass centre of the wheel from the centre of the shaft, and the dynamical load on the bearings. Assuming a co- efficient of friction of 0-1, find the horse-power required to over- come the friction of the bearings due to the dynamical load alone. The diameter of the shaft is 5 inches. What fraction is the dynamical load of the static load 2 (Horse-power required is given by— Dynaml. load on bearing x coefficientoffriction × rad. of jour.inft. × 60 550 Answers.— (1) # inch nearly. (2) Dynamical load on right-hand bearing, 540 lbs. weight; on the left-hand, 189 lbs. weight. (3) 0.44 H.P. (4) 32% per cent. 12. The crank-arms and crank-pin of a crank-shaft are equivalent to a mass of 700 pounds at 1 foot radius. The shaft is supported in two bearings, 5 feet centre to centre, and the centre of the crank 262 THE BAZAAVC//VG OF EAVGIZVE.S. is 1.5 feet from the left-hand bearing. The diameter of the shaft at the journals is 8 inches. Find the dynamical load on the shaft and on the bearings for a speed of 240 revolutions per minute, and calculate the rate, in horse-power, at which work is dissipated in heat at each bearing, assuming a coefficient of friction of 0-05. Answers.- Dynamical load on shaft, 13,708 lbs. weight, load on bearings, left hand, 9595-6 lbs. weight; right hand, 4112-4 lbs. weight. H.P. loss, left-hand bearing, 7:30; right hand, 3:12; total, 10:42. 13. Draw the bending moment diagram and the shearing force diagram for the shaft of the previous question due to the revolution of the unbalanced mass, assuming the shaft to be a straight one. State the numerical value of the maximum bending moment. Answer.—Maximum bending moment, 172,720 inch lbs. 14, Find the single mass at 3 feet radius which will balance the mass of question 12. Find also the magnitudes of two masses which will effect balance when they are placed in the same plane of revolution as the disturbing mass, at radii of 4 feet and 5 feet respectively, these radii being inclined to the radius of the disturbing mass at 160 degrees and 220 degrees respectively. Answers.--(1) 2334 pounds. (2) jº pounds at 4 feet ...; 55 pounds at 5 feet radius. 15. What is the direction of the axis of the couple exerted on a double-ended wrench, to tap a nut with a right-hand thread 2 Answer.- A line drawn parallel to the axis of the tap with the arrow- head on it pointing from the wrench towards the point of the tap. 16. Add the couples— 140°, 20309, 10900 Answer.— 37:233°, meaning that the axis is inclined 33 degrees to the initial line, and that the moment of the couple is 37°2. Aº XERCYSA.S. 263 17. A three-legged table touches the ground at points which joined, form an equilateral triangle, ABC, of 2-feet side. A force of 7 lbs. weight acts vertically on the table at a point D, 12 feet from A and 14 feet from B. Find the pressure at the points of Support. -- (Take A for the origin. Consider the lines AB, AC, AD, to be the horizontal traces of three vertical planes. Referring the given force to the origin A, there will be an equal and parallel force at A, and a couple 7 x AD foot-lbs. acting in the plane of which AD is the trace. This couple must be balanced by couples in the planes of which AB and AC are the horizontal traces. Therefore set out the axis of the couple in plane AD, 7 × AD units long, and complete the triangle by lines drawn at right angles to the remaining traces, the intersection of these lines will determine the length and Sense of the two unknown axes. Measuring these off and dividing respectively by the arms AC and AB, the forces of the two couples at the supporting points are known.) Answer. Force at A = 2:23 lbs. weight } } at B == 1'4 32 j } at C = 3:37 ..., 2 ) } } 18, Find the two masses which will balance the mass of question 12, when they are placed at 4 feet and 5 feet radii respectively— (1) In planes of revolution, the first 1 foot to the left of the plane of the given mass, the second 2 feet to the right. (2) In planes distant 1 foot and 3 feet respectively to the right of the given mass. Answers, (1) Mass in plane to the left 116-6 pounds at 4 feet radius, the radius being at 180° with the radius of the given mass. Mass in plane to the right 46.6 pounds at 5 feet radius, the radius being at 180° with the radius of the given mass. (2) Mass in plane nearer the given mass, 262-5 pounds at 264 THE BAZAAVC/AWG OF EAWG/WE.S. 4 feet radius, the radius being at 180° with the radius of the given IOOla SS. Mass in further plane 70 pounds at 5 feet radius, the radius being at 0° with the radius of the given mass. 19, Draw the bending moment and shearing force diagrams for the shafts of the two balanced systems of question 18, due to the dynamical loading alone, when the speed is 250 revolutions per minute. 20. Five pullies, equally spaced at 2 feet apart, are keyed to a shaft which is supported on bearings 12 feet apart. The pullies are out of balance to the following extent :— No. 1, 5 pounds at 1 foot radius. No. 2, 6 2 3 2 feet ,, No. 3, 7 , 1 foot , No. 4, 2 , 2 feet ,, No. 5, 6 3 x 1 foot , The angles between the several mass radii and the mass radius of No. 1 pulley are respectively 45, 90, 120, and 240 degrees. Find the two masses which will balance the system— (1) When placed in Nos. 1 and 5 pullies at 1 foot radius. (2) 22 35 2 3 2 25 4 2 3 22 22 Answers.- (1) } 3-8430so in No. ; (2) { 9:150 in No. : 15-252250 in No. 1. 22:72205 in No. 2. The angles are measured from the direction of No. 1 radius. 21. Find at what speed the maximum value of the unbalanced force of a locomotive is 4 tons, assuming that the revolving masses are balanced and that the reciprocating masses, which weigh 600 pounds per cylinder, are unbalanced. Diameter of wheels, 4 feet 6 inches. Stroke, 2 feet. (Use formula in (1), page 85.) Answer-20.2 miles per hour. 22, What is the speed in question 21, if the revolving parts, which weigh 700 pounds per cylinder, and are at 1 foot radius, are unbalanced as well ? Answer.-13.7 miles per hour. EXERCISES. 265 23. Calculate the values of the swaying couples in questions 21 and 22, assuming the cylinders to be inside and 2 feet centre to centre. (Use formula in (2), page 85.) Answer.—4480 foot-lbs. in each case. 24, What is the speed in question 21, if two-thirds of the reciprocating masses are balanced and all the revolving masses 3 Answer.-35 miles per hour. 25, Calculate the speed in the previous question when the diameter of the driving-wheel is 7 feet. Answer.—54.4 miles per hour. * 26, Calculate the values of the respective swaying couples for a speed of 60 miles per hour and their respective periodic times, for the following engines, in each of which the revolving masses are balanced and the mass of the reciprocating parts per cylinder is 600 pounds, two-thirds of which is balanced. Stroke, 2 feet. (1) A 7-foot inside single, cylinders 2 feet pitch. (2) A 7-foot outside single, cylinders 6 feet pitch. (3) An 8-foot outside single, cylinders 6 feet pitch. (4) A 5-feet inside cylinder tank, cylinders 2 feet pitch. Answers—(1) 5440 foot-lbs. 0.25 seconds. (2) 16320 , 0.25 , (3) 12495 , 0.286 (4) 10662 , 0.179 , 27. Two engines are built with similar sets of reciprocating parts, one as a 7-foot outside single in which the cylinders are 6 feet pitch, the other as an inside cylinder tank engine in which the cylinders are 2 feet pitch. The revolving masses and two-thirds of the reciprocating masses are balanced in each case. For what diameter of the driving-wheels would the swaying couple acting on the tank engine be equal to that acting on the single engine, when both engines are running at the same speed ? Answer.—4 feet diameter approx. 28. Assuming that the tractive force exerted by an engine 266 THE BALANCING OF ENGINES. varies inversely as the speed, and that the tractive force is 2 tons when the speed is 30 miles per hour, and also that there are 200 pounds reciprocating mass unbalanced per cylinder, find the speed at which the maximum value of the unbalanced force becomes equal to the average tractive force. Wheels, 7 feet diameter. Stroke, 2 feet. Answer.—44-6 miles per hour. 29. Find the balance weights for the inside single engine speci- fied by the following data :— Stroke, 26 inches. Cranks at right angles, left-hand crank lead- ing. (In Fig. 79 the right-hand crank is leading.) All the revolving and two-thirds of the reciprocating masses are to be balanced. Distance centre to centre of the cylinders ... 2 feet 4 inches Distance between the planes containing the mass centres of the balance weights tº º ſº ... 4 , 11% , Mass of the reciprocating parts per cylinder ... 612 pounds Mass of the revolving parts per cylinder ... 720 ,, Answer.- Left-hand wheel. 880 pounds at 13 inches radius at an angle of 160 degrees measured from the left-hand crank, counter-clock- wise, when facing the left-hand wheel. 30. Find the balance weights for the inside cylinder four- coupled engine specified by the following data:— w Stroke, 24 inches. Inside cranks at right angles, right-hand crank leading. Outside cranks, 11 inches radius, placed oppositely to the corresponding inside cranks. All the revolving and two-thirds of the reciprocating masses to be balanced. The mass of each coupling-rod to be divided equally between the driving and trailing wheels. The balancing mass for the reciprocating parts to be divided equally between the driving and trailing wheels. Distance centre to centre of cylinders 2 feet Distance between planes of motion of wheel-cranks tº e & tº gº & ... 5:166 feet Distance between planes of motion of coupling-rods ... § - ſº, ... 6'27 2 3 A: YAZACC/SA.S. 267 Distance between the planes containing the mass centres of the balance weights ... ſº tº ºt tº º ſº ... 494 feet Mass of reciprocating parts per cylinder 642 pounds at 12 inches Inside revolving masses per cylinder 723 , , 12 , Mass of each coupling-rod ... ... 224.5 ,, » 11 > Mass of each wheel-crank in driving and trailing wheels tº tº wº ... 117-6 ,, , 11 , Mass of part of crank-pin outside crank, together with the pin and washer for each outside crank ... ... 38'2 2 3 , 11 . Answer.— Left-hand driving-wheel. 493 pounds at 12 inches radius at an angle of 37 degrees, measured from the outside crank, counter- clockwise, when facing the left-hand wheel. Trailing-wheel. 145 pounds at 12 inches radius at an angle of 144 degrees, measured from the outside crank, counter-clockwise, When facing the left-hand wheel. 31. Find the balance weights for the engine of the previous question when the two-thirds of the reciprocating masses are balanced entirely in the driving-wheels. Answer.- Left-hand driving-wheel. 651 pounds at 12 inches radius at an angle of 34 degrees, measured from the outside crank, counter- clockwise, when facing the left hand wheel. Trailing-wheel. 268 pounds at 12 inches radius at an angle of 1753 degrees, measured from the outside crank, counter-clock- Wise, when facing the left-hand wheel. 32. Calculate the maximum value of the hammer-blow for the driving-wheel in the two preceding examples when the crank-shaft is making four turns per second (corresponding to 60 miles per hour with a 7-foot wheel), and hence find the maximum and minimum load on the rail, supposing the static load to be 7% tons. (Use expression (1), Art. 76.) * Answers.- For example 30, where one-third is balanced in the driving- wheel, hammer-blow = 3185 lbs. weight. 268 THE BA LA WCING OF EAWGINES. Maximum load on the rail, 892 tons weight; minimum, 6-08 tons weight. For example 31, where two-thirds is balanced in the driving- wheel, hammer blow = 6370 lbs. weight. Maximum load on the rail, 10:34 tons weight; minimum, 4.66 tons weight. 33. Assuming the mass of the reciprocating parts shown in Fig. 50 to be 1000 pounds, and the crank radius to be 1 foot, find the accelerating force when the crank makes 480 revolutions per minute, for the crank angles, 0, 60, 90, 120, and 180 degrees. Answer, –78440, -39220, 0, -1-39220, -1-78440 lbs. weight. 34. Find the maximum values of the unbalanced force and the unbalanced couple in terms of the revolutions per second, due to the reciprocating masses of a four-crank engine in which the cylinder pitches, reckoning from the left, are 10 feet, 12 feet, and 10 feet respectively, the corresponding masses, reckoning from the left, being 2, 3, 4, and 2 tons, and in which the crank angles are, reckoning from the left, between cranks 1 and 2, 90 degrees; between cranks 2 and 3, 90 degrees; between cranks 3 and 4, 90 degrees. Stroke, 4 feet. Answer.— 2 Unbalanced force, 1765; lbs. weight. Unbalanced couple in plane of reciprocation about an axis at the centre of the engine, sº foot-lbs. 35. Find the unbalanced force and couple using the data of the previous question when the sequence of cranks is changed to the following, reckoning from the left: angle between cranks Nos. 1 and 2, 180 degrees; between cranks 2 and 3, 90 degrees; between cranks 3 and 4, 180 degrees. 2 Answer.—Unbalanced force, 1785. lbs. weight. 2 Unbalanced couple at the centre, tº: foot-lbs. A.XAZRCISE.S. 269 36. The mass centre of a connecting-rod, l feet long, is 0.8l from the Small end. Its mass is 850 pounds. Find the masses which must be included with the revolving and reciprocating parts which it connects, in order that the effect of the rod may be balanced when these masses are balanced. Answer.—680 pounds with the revolving mass. 170 32 55 reciprocating mass. 37. Reckoning from the left in order, let the letters A, B, C, D denote the cylinders of a four-crank engine. The distances between them are, 5 feet between A and B, 8 feet between B and C, 6 feet between C and D. The revolving masses corresponding to A, B, C, D are respectively 1, 1}, 1}, and 1 ton at crank radius. Given that the angle between the cranks of cylinders B and C is 105 degrees, and that the reciprocating masses of cylinders B and C are respectively 2} and 2 tons, find the remaining crank angles and masses so that the reciprocating parts may be in balance amongst themselves, neglecting the obliquity of the connecting-rod. Find also the masses which must be added to the crank-shaft at cranks A and B to balance it. Answer.— measured in Angle between cranks C and A, 97 degrees 32 35 ,, A , D, 58 > 5 | 55 33 , D , B, 99% 3 x order. Reciprocating mass at A, 1615 tons. 55 ,, at D, 1343 , Revolving mass attached to crank-shaft at A, 0.071ste tons. 55 55 33 at D, 0.1589% , The subscript directions being measured from crank B towards crank C. 38. Taking the data of the previous example, find the remaining crank angles and reciprocating masses, including the revolving masses with the reciprocating, so that the reciprocating and revolv- ing masses are together in balance in the plane of reciprocation, but the revolving masses are unbalanced in the plane at right angles to it. 270 TAA BALA WCING OF EAVGINES. Answer.— Angle between cranks C and A, 96% degrees & º |-> IIl A. }} D, 57; 53 order. 33 : 2 35 D 33 B, 101% Reciprocating mass at A, 1.69 tons. at D, 1:19 tons. 35 2 3 } } 2 3 5 5 5 y 39. Taking the data of question 37, balance the reciprocating masses amongst themselves, and find what the corresponding revolving masses should be so that they may be in balance without the addition of balance weights, having given that the revolving mass at A is 1 ton. Answer.— Revolving masses must be in the same ratio as the reciprocating IQ 8 SS6S. Revolving mass at B, 1394 tons. at C, 1.238 ,, at D, 0.833 , 2 3 } ) y 5 5 § 40. The reciprocating masses of a four-crank engine are respectively 5, 7, 6, and 5 tons, taken in order. Find the cylinder centre lines having given that the pitch of the extreme cylinders is 39 feet, and that the remaining two are to be arranged sym- metrically with respect to them, neglecting the obliquity of the connecting-rod. Find also the corresponding set of crank angles. (This problem may be solved by taking advantage of the geometrical properties of the combined force and couple polygons, Art. 37, Fig. 47, to calculate the ratio º the 0.1 distances at and a2 being measured from a central refer- ence plane, as in Fig. 110, and then drawing the couple triangle ABā or CBC, from either of which the force polygon may be completed. Referring to Fig. 47, page 50, the condition of symmetry requires that— Ad Co. AD CD and therefore the line joining d to c is parallel to AC, Aº XERCISE.S. 271 Hence— 20. Be = 2 x Ba = *—Ba . . . (I € de X a 1 + a 2 - (1) Since if AC represents 2a1, de represents a1 + q2, and ef represents 202. Again— Bd = cD = CD(1 - *. *) (t1 since— cC = %DC - (* gº)po Substituting the value of Bd, in (1)– Be = **CD Cºl. Similarly— * Bf - *AD (tºl Again considering the triangle ABC, let O be the central point of AC, and let the angle BOC be 0. Then— AB” – BC* = 2a1OB cos 6 Be” – Bf4 = 2a2OB cos 9 Dividing and substituting the values of Be and Bf- AB2 *Eº B62 — 02 — M12 gºe M.” $ CD3 – AD? Tº TM3 – Mº After the value of this ratio has been calculated from the four given masses, either of the couple triangles ABd, or BeC, may be drawn. Completing the force polygon all the crank angles are determined. The lengths to be used in drawing the triangle ABſ are— AB = 2a1 M1 Bd = (a1 + a2)M2 Ad = (a1 – ag)M3 * This relation is given in Herr Schlick's paper “On Balancing Steam Engines,” Trans. Inst. Naval Architect8, 1900, 272 THAE AEA LA WCW/VG OF EAVG/AWAE.S. Answers— > - 1: - Q2 21 Ratio a T 52 Pitch of inner cylinders 1575 feet Angle between cranks 1 and 3, 118 degrees (see Fig. 48) 55 55 5 x 3 and 2, 81% 55 55 ,, 2 and 4, 123 , 55 33 > y 4 and 1, 37} 5 y X 3 41,–Referring to the engine of which the data are given in Art. 50, find the change which must be made in the masses belonging to Nos. 2, 3, and 4 cranks, and the change in the direction of crank No. 4, relatively to crank No. 1, in order that the reciprocat- ing parts may be in balance amongst themselves, when the valve- gear is neglected. Answer.— No. 2 mass, increased 100 pounds, 6-66 per cent. of the Original Iſla SS. No. 3 mass, increased 125 pounds, 9.8 per cent. Of the original Iſla SS. No. 4 mass, increased 75 pounds, 7.2 per cent. of the original Illa SS. Angle between cranks 1 and 4, 47; degrees, being a change of 2. degrees. 42.—Referring to the engine of which the data are given in Art. 50, assume that the masses of the reciprocating parts corre- sponding to cranks Nos. 1, 2, and 3 are given, being respectively 1000, 1500, and 1275 pounds, and find the two solutions which are possible when the valve-gear is included.* (This is case C of Art. 8. Neglecting the valve-gear, the solutions differ only in the respect that the sequence of angles in the one is opposite to that of the other. Or, holding the drawing of the angles from one solution in front of a looking-glass, the reflection is the other solu- tion. Including the valve-gear, there are two different solutions, one of which in the case in question is of course the solution of Art. 50.) * I am indebted to Professor Dunkerley for this extension of Art. 50. Az XAEA’C/SA.S. 273 Answer.— 2nd solution. Angles between cranks 1 and 4, 4 and 2, 2 and 3, are respectively 44% degrees, 118 degrees, and 93% degrees, measured counter-clockwise. Mass at No. 4 = 1095 pounds. 43. The cylinders of a four-crank engine are arranged symmetrically. The pitch of the outer pair is 35 feet, and of the inner pair 15 feet. The mass of each set of reciprocating parts belonging to the outer cylinders is 6 tons. Find the crank angles and the inner masses, so that the reciprocating masses may be in balance for primary and secondary forces, and primary couples. (Use equation (9), Art. 96, and check the work by Fig. 112.) Answer.— Let A, B, C, D indicate the cylinders taken in order. Angle between cranks A and D, 61 degrees 42 minutes. 55 35 , B , C, 108 , 48 , Reciprocating mass corresponding to cylinders B and C, 8:844 tons per cylinder. - 44. Details of the valve-gear belonging to the engine of the previous question are given below. Distance from Mass at crank Crank A. radius. Feet. Tons. Ahead sheave — 5:0 0-6 Astern sheave —4:3 0-1 Crank A O-0 Crank B 10-0 8°844 Ahead sheave 14-3 0.6 Astern sheave 15-0 0-1 Astern sheave 20-0 0-1 Ahead sheave 20-7 0.6 Crank C 25-0 8-844 Crank D 35-0 Astern sheave 39-3 0.1 Ahead sheave 40’0 0-6 Assuming that the angle between cranks B and C is 108 degrees 48 minutes, and that the corresponding masses are 8,844 T 274 THE BAZAAVC/WG OF AZAVG/AWE.S. tons per cylinder, as found in the previous question, find the remaining angles and the masses corresponding to cylinders A and D, including the effect of the valve-gear. - (Take a reference plane at crank A and include the couples belonging to the valve-gear of crank A, by assuming that the direction of crank A is that found in the previous question.) Answer.— Angle between cranks A and D, 63 degrees 53 minutes. A. 5 y C, 95 35 56 35 B x 5 D, 91 X 3 23 55 35 2 3 55 33 25 22 Mass at A, 6-3 tons. ,, at D, 6-0 ,, 45. An opposite pair of crank angles, [3 and 8, in a four-crank engine, have the values (3 = 110 degrees and 8 = 90 degrees. Find the remaining angles, the ratio of the reciprocating masses, and the pitch of the cylinders so that the engine may be in balance for primary and secondary forces and primary couples. (Find y1 and yo from equation (11), Art. 96, and M1 from equation (12), Art. 96; then find M2, M3, ag, as, either by calculation or by the graphical method of Art. 37) Answer.— y1 = 44 degrees 12 minutes 'Y2 - 115 5 5 48 35 If M4 = 1, then— M1 = 2.3155 Ma = 2-9563 If the reference plane is taken at No. 4 crank, so that a1 = 0, and all be put equal to unity— a 2 = 0.8326 as + 0.6236 46. Given that the crank radius of the symmetrical engine detailed in question 43 is 2 feet, that the connecting-rod is 7 feet A.Y.E./CC/S E.S. 275 long, and that the engine runs at 80 revolutions per minute, find the maximum value of the unbalanced secondary couple. (See Art. 118, type 3, or the formulae in Schedule 21.) Answer.—385 foot-tons. 47. Given that the crank radius of the unsymmetrical engine detailed in question 45 is 2 feet, that the connecting-rod is 7 feet long, that the pitch of the extreme cylinders is 35 feet, that M4 is 6 tons, and that the speed is 80 revolutions per minute, find the maximum value of the unbalanced secondary couple. (The most expeditious way of doing this is to find the closure of the secondary couple polygon graphically. This tº tº to???? measures 312 units. Multiply this by *#) Answer.—385 foot-tons. 48. A steel tube, 6 feet long say, is firmly fixed at one end in a vertical position, and loaded at the free end with a mass of 10 pounds. It is found by experiment that a force of 10 lbs. weight, applied at the centre of the mass at right angles to the length of the tube, produces a displacement from the position of equilibrium of a tenth of a foot. Find the time of vibration of the system, and the displacement of the centre of the mass from the position of equilibrium at the end of 10 seconds, having given that the displacement is 0.3 feet when the time is nothing, and that a is nothing when the time is nothing. Neglect the mass of the tube. Answer.—Time of vibration = 0-35 seconds. Displacement at the end of 10 seconds is — 028 feet. 49. Find the displacement at the end of 1 second for the system of question 48 when there is a frictional resistance acting, which is proportional to the velocity, having given that the resistance to motion is 3 lbs. weight when the velocity is 1 foot per second. Answer.—0:00004 feet. 50. Suppose a periodic force whose periodic time is 0.4 seconds, º ... 10 e and whose maximum value is g lbs. weight, to act on the mass of 276 THE BAZAAWCING OF EAWG/NWAE.S. the system of question 48. Find the maximum amplitudes when the frictional resistance to motion at unit velocity is respectively 3 lbs. weight and; lbs. weight. Find the maximum amplitudes in the two cases when the periodic time of the force is 0.35 seconds. Answers.- 0.0059 feet and 0.0066 feet when the periodic time is 0.4 seconds. 0-1306 feet and 0.637 feet when synchronism takes place. 51. The reciprocating masses of a single-cylinder engine weigh 6 tons. The cylinder is 70 inches diameter and 4 feet stroke, and the connecting-rod is 7 feet long. When the connecting-rod sub- tends a crank angle of 60 degrees, the difference between the forward and the back pressure in the cylinder is 30 lbs. per Square inch. Find the turning moment on the crank when the speed is 80 revolutions per minute. (Calculate the acceleration of the reciprocating masses by equation (2), Art. 78 (a is zero in this case), or use Rlein's construction, Art. 104. Calculate the length of Og (Fig. 143), and use the appropriate product from equation (1), Art. 130.) Answer.—83-6 foot-tons. 52. Find the ratio in which the total horse-power should be distributed between the four cylinders of the example, case 2, Art. 97, in order to obtain a good crank-effort curve, having given that the horse-power of No. 2 cylinder is to be equal to the horse- power of No. 3 cylinder. Answer.— Calling the horse-power of cylinders Nos. 2 and 3 each equal to unity— Horse-power of No. 1 cylinder = 0-74 5 5 }} 3) 4 >> - 0:6 Aº XERCISE.S. 277 53. Find the ratios in the previous example, if the horse- powers in cylinders 1 and 4 are each equal to unity. Answer.—Horse-power in cylinder No. 2 = 1. 55 55 32 No. 3 54. The mass centre of a connecting-rod 7 feet long from centre to centre, is 4.9 feet from the small end. The rod, sus- pended so that it is free to swing about an axis through the small- end centre, swings in unison with a plumb-line 6-3 feet long. Find k, the radius of gyration about a parallel axis through the maSS Centre. Answer.—2-61 feet. 55. The rod of the previous question is suspended so that it is free to swing about an axis through the big-end centre. Find the length of a plumb-line which will swing in unison with it. Answer.—5.36 feet. 56. The connecting-rod specified in question 54 drives a crank 2 feet radius. The mass of the rod is 1610 pounds. Find the force required to produce the instantaneous motion of the rod when the rod subtends a crank angle of 60 degrees, and the crank revolves uniformly at 80 revolutions per minute. Answer.— 5305 lbs. weight in a direction inclined 307 degrees with the line of stroke (the crank radius in the line of stroke, representing the initial direction, points away from the cross-head), and cutting the line of stroke at 0-71 feet measured from the centre of the crank-shaft towards the cross-head. 57. Taking the data of the previous question, find the instan- taneous values of the reactions on the frame. Answer.— (1) A force of 5305 lbs. weight inclined 127 degrees with the line of stroke (the initial direction pointing away from the cross- head) acting at the axis of the crank-shaft. (2) A couple whose moment is 4930 foot-lbs., the equal forces of which are each 634 lbs. weight. T 3 I N D E X. A. Accelerating force, action on the frame, 12, 56, 58; approximate expression for, to include effect of the obliquity of the connecting-rod, 126; exact expression for, to include obliquity of connecting- rod, 126; expression for, infinitely long rod, 56; for uniform motion in a circle (Schedule 2), 10, 11 ; on connecting- rod, 233, 237–240; series for, on piston masses, 164 Acceleration of commecting-rod, 232; angular, 251; at dead centres, 253; resultant force causing, 237–239 Acceleration of mass centre of commecting- rod, analytical calculation of, 250 ; graphical constructions for, 234 Acceleration of piston masses, approximate construction for short rods, 182; approxi- mate formula for short rods, 126; curves calibrated for unbalanced couple, 180 ; curves calibrated for unbalanced force, 177, 178; exact construction for, 172; exact expression for short rods, 126, 254; expression for long rods, 56 Addition of couples, 26 Addition of vector quantities, 2 Analytical method of investigating condi- tions of balance, short connecting-rods, eight fundamental equations, 134; five- crank engine, 158; four-crank engine, 143–153; fundamental theorem, 136; general method, 122–134; six-crank engine, 161; three-crank engine, 138; two-crank engine, 137 Axial plane, 22 Axis, central, reduction to, 39 Axis of a couple, 25 Axis, reaction on, 11 B Balancing an engine without the addition of balancing masses, 78 Balancing connecting-rod, 62, SS, 246 Balancing, higher orders than the second, 163; primary and secondary, 124 Balancing reciprocating masses, 80, 87; eight-coupled locomotive, Baldwin Com- pany, 115; inside cylinder single loco- motive, 89; inside cylinder six-coupled locomotive, 96; outside cylinder single locomotive, 92 Balancing reciprocating masses, long con- mecting-rods, example, including valve- gear, four-crank engine, 71; examples, four-crank engines, 67, 69 ; general form of the problem, 65 ; investigation of method, 58; locomotive, 87; number of variables, 66; statement of general rule, 64 Balancing revolving masses, any number of given masses in different planes of revolution, 34; carriage-wheels, L. and N. W. R., 20; crank-shaft, 75, 117; general conditions of balance (equations (1) and (2), p. 34), 36 Balancing revolving masses by recipro- cating masses, 75 Balancing revolving masses in one plane of revolution, any number of masses, 16; One mass, 13; two masses, 15 Balancing revolving masses in three planes of revolution, 30 Balancing reciprocating masses, short con- necting-rods, 124. See “Analytical ‘’ Bob-weight, 65 C Central axis, reduction of unbalanced force and couple to, 39 Centrifugal couple, 22; vectors, primary and secondary, 133 ; vectors, rule for way of drawing, 35 Centrifugal force, 12 ; effect of, with reference to a point, 29 ; vectors, primary and secondary, 133; vectors, rule for way of drawing, .35 Centripetal force, 12 280 IAWDAE.X. Closing a vector polygon, quantities deter- mined by, 7 Closure, 37 Common or crank radius, reduction of masses to, 41 Condition for no turning moment, 27 Condition that vector sum may be zero, 3 Conditional equations for primary and secondary balancing, 134; selection of, 135; theorem controlling selection of,136 Conditions of balance, a system of revolving masses (equations (1) and (2), p. 34), 36, 42; five-crank engine, short connecting- rods,160; four-crank engine, long connect- ing-rods, 78; four-crank engine, short connecting-rods, 145; four-crank sym- metrical engine, short connecting-rods, 148, 155; four-crank unsymmetrical engine, short connecting-rods, 149; four revolving masses, 47; primary and secondary, 130, 134; reciprocating masses, long connecting-rods, 61 ; re- ciprocating masses, short connecting- rods, 130, 134; six-crank engine, short connecting-rods, 163; two-crank engine, short connecting-rods, 138; three- crank engine, short connecting-rods, 143; three revolving masses, 46; two revolving masses, 46 Connecting-rod, action on frame, 240– 250; action on turning moment, 241; analytical investigation of motion of, 250; balancing, 62, 88, 246; motion of, 232, et seq.; opposing motion of two rods, 249 Couple, addition of, 26; arm of, 22; axis of, 25; causing angular acceleration of connecting-rod, 250, 252; centri- fugal couple, 22; closed couple polygon, 27; curves for unbalanced engine, 181, 186; definition of, 22; equivalent, 24; modifying crank-effort due to motion of connecting-rod, 241; moment of, 23; rules for way of drawing vectors repre- senting a couple in the balancing pro- blem, 35 Crank-effort diagrams, 223; analysis of, into harmonic elements, 226; four-crank marine engine, 230; gas engine, 224; locomotive, 109,223; method of drawing, 222; single-cylinder engine, 223 Crank-effort, ratio of maximum to mean, 225, 226; uniformity of, 225 Crank-shaft, balancing, 75, 117; turning moment on, 219, 225 D Damped vibrations, 204 Data, Lancashire and Yorkshire engines, 88, 89, 96 Data, selection of, 43, 66, 135 Definition of, arm of couple, 22; axial plane, 22; axis of couple, 25; balanc- ing reciprocating systems, 57; balancing revolving systems, 13 ; direction, to include sense, 6 ; mass moment, 14 ; moment of mass moment, 36; plane of revolution, 13; reciprocating system, 57; reference plane, 30; secondary balancing, 124; turning moment of a couple, 23; vector quantity, 1 Displacement vectors, 4 Division of horse-power amongst cylinders for uniform turning moment, 228, 229 Dunkerley, Professor S., 272 Dynamical load on shaft, 12 Dynamical system, equivalent, 235 E Effect, any number of forces acting simul- taneously, 27; centrifugal force with respect to a point on axis, 29; connect- ing-rod on frame, 232, 240–250; of a force on a rigid body, 27; unbalanced forces on supports, 12, 56, 199, 214, 216; unbalanced reciprocating parts of a loco- motive, 82, 102, 120 Errors, arithmetical computation of maximum primary and secondary com- ponents, 188; formulae for typical cases (Schedule 21), 191–197; four-crank engines specified in Schedules 19 and 20...175–186, 189; four-, five-, and six-crank engines when balanced for primary and secondary effect, 167; general method of finding for given engine, 187; involved by use of approxi- mate expression for acceleration of piston masses, 126 Equations, fundamental, for primary and secondary balancing, 134 Equivalent dynamical system, 235 Equivalent mass at crank radius, 41 Exercises, 257 Experimental apparatus, four-crank reci- procating systems, 80 ; locomotives, 122; revolving systems, 52 Experimentally testing the balance, 19 F Five-crank engine, 158, 166, 168 Force, accelerating, 10, 11, 56, 126, 164, 233 Force, centrifugal, and centripetal, 11, 12 ; effect of centrifugal, on axis of rotating body, 12, 27, 29; effect of, on a rigid body, 27 MAVOAX. 281 Force required to constrain circular motion (Ex. 8, p. 262), 10 Foot-lbs., 25 Foot-pounds, 25 Forced vibrations, 207,211, 214 Four-crank engine, balanced for primary, secondary, and tertiary effects, 166; con- necting-rods opposed in pairs, 249; crank- effort diagram of, 225, 230; examples of balancing (S.H.M.), 66, 69, 71; primary and secondary balancing of, 143 ; primary and secondary forces balanced, and primary couples, 147; Schlick symmetrical engine, 147, 150, 154; unsymmetrical engine, balanced for six conditions, 149, 152; unbalanced forces and couples from, cranks at right angles, 175–186, 189 Four-crank revolving systems, properties of, 50, 51 Frame, action of connecting-rod on, 233, 237–240 Frame, action upon, by accelerating forces, 12, 56, 58 r G Geometrical properties of a pair of force and couple polygons, 44–46 Geometrical solutions of particular pro- blems, four-crank systems, 50, 157 Gray, Mr. Macfarlane, 48, 124, 156 H Hammer-blow, 102, 105, 120, 121 Henszey, Mr., 115 Horse-power, rule for division of, amongst cylinders, 228, 229 I Indicator cards, L. and Y. locomotive at 65 M.P.H., 111 K Klein's construction for, acceleration of connecting-rod, 234, 237; acceleration of piston, 172 L Lateral vibrations (or vertical), 210 Locomotive, adhesion modified by balance weight, 105, 107, 108, 112; American practice, 115; balancing reciprocating parts,87; comparative tables, 1193 con- necting-rod, 88; crank-axle, 89, 117; crank-effort curve, 109; distribution of balance weights amongst coupled wheels, 112; effect of obliquity of connecting- rod, 103; effect of unbalanced recipro- cating parts, 82, 118; elimination of horizontal couple, 118; examples of balancing, 89, '92, 96, 99, 112, 115; experimental apparatus, 122 ; four- cylinder engine, 116 ; hammer-blow, 102, 105, 120, 121; horizontal swaying couple, 83, 85, 118, 120, 121, 123; indicator diagrams, 111; nett piston pressure, 111; secondary balancing, 122; slipping, 107; speed at which driving- wheel lifts, 106; unbalanced force, 83, 85, 118, 120, 121, 123; variation of rail-load, 102, 105, 120, 121 Longitudinal vibrations, 212 Lorenz, Dr., 149, 225 M Macalpine, Mr. J. H., 164,239 Magnitude of unbalanced force. balanced ” Mallock, Mr. A., 124 Mass centre, 18 Mass moment, 36 Moment of mass moment, 36 Motion in a circle, acceleration for (Ex. 8), 262; force to constrain, 10 See “ Un- N Natural period of vibration, concentrated mass, 200; Schlick's formula for ship’s hull, 217; ship's hull, 214, 216, 218; s.s. Deutschland, 218; S.S. Meteor, 217 ; torpedo-boat hull, 216 Nodes, application of force or couple at, 211; of a rod vibrating laterally (verti- cally), 210; of a rod vibrating longi- tudinally, 212; of a rod vibrating torsionally, 212; ship's hull, 218 Normand, M., 125 O Obliquity of connecting-rod, effect on rail, 103; approximate formula for piston acceleration, including, 126; exact for- mula for piston acceleration, including, 126; formula for piston acceleration, neglecting, 56 2 8. 2 AVDA2A. P Polygon, two vector quantities determined by closing, 7 Pound, signification and abbreviation of, 25; foot-lbs., 25; foot-pound, 25; lbs.-feet, 25; lbs. weight, 25; pound, 25; poun- dal, 11, 25; poundals-feet, 25 Product of inertia, 36 R Radial connector, 12 Reciprocating parts, method of balancing, locomotive (see “Balancing”), 87; stan- dard set of, 88 Reciprocating system, balancing of 57; balanced by revolving masses (see “Balancing ”), 80; definition of, 57; in- vestigating balancing conditions of, long rods, 58 ; investigating balancing con- ditions of, short rods, 125, et seq. Reduction of masses to a common or crank radius, 41 Reference plane, definition of, 30 Reference plane, position of, and couple polygon, 49 Reference plane, position of, and number of variables, 44 Relation between force and couple polygon belonging to a balanced system, 44 Revolving masses. See “Balancing” Revolving systems, two-, three-, and four- crank, 46–50 Robinson, Mr. Mark, 124, 141 Rules, for checking accuracy of work, 35; general, for balancing an engine, long rods, 64; general, for balancing two-, three-, four-, five-, and six-crank engines, 169–171; way of drawing centrifugal couple (moment of mass moment, or product of inertia) vectors in balancing problems, 35; way of drawing centri- fugal couple (moment of mass moment, or product of inertia) vectors in general, 25; way of drawing centrifugal force (mass moment) vectors, 32 S Sankey, Captain, 124, 141 Scalar quantity, 1 Scales, 89 Schedules, list of, (1) a set of vectors, 7; (2) different ways of expressing force constraining circular motion, and there- fore centrifugal force, 11; (3) typical example of a revolving system, 37; (4) quantities derived from a force polygon drawn at random, 48; (5) standard form of schedule, 64; (6) reciprocat- ing masses, four-crank engine, 67; (7) reciprocating masses, four-crank engine, 70; (8) reciprocating masses, four-crank engine, including valve-gear, 72; (9) revolving masses, four-crank engine, in- cluding valve-gear, 76; (10) data of inside cylinder single engine, 91; (11) data of outside cylinder single engine, 95; (12) data of six-coupled engine, driving-wheel, 97; (13) data of six- coupled engine, leading and trailing wheel, 100; (14) expressions for un- balanced force, hammer-blow and sway- ing couple, 120; (15) expressions for unbalanced force, hammer-blow and swaying couple when piston speed is 1036 ft. per min., 121 ; (16) speed and diameter of driving-wheel, piston speed 1036 ft. per min., 122; (17) errors of approximate formula for piston accelera- tion, 127; (18) crank directions for three-crank engine, 140; (19) data of reciprocating masses, typical four-crank engine, 175; (20) data of revolving masses, typical four-crank engine, 176; (21) primary and secondary force errors, typical engines, 196 Schlick, Herr Otto, 52, 125, 147, 217, 218 Schubert, Dr., 52, 149 Secondary balancing. See “Analytical" Secondary unbalanced force, effect of, 128, 130 Shaft, dynamical load on, 12 Short-framed engines, 230 Simple harmonic motion, balancing for (Chaps. III. and IV.), 58; relation to circular motion, 57 Six-crank engines, 161, 168 Slipping, locomotive driving-wheel, 107, 109 Smith, Professor R. H., 239 Subtraction of vectors, 5 Summary, general, 168; method of finding errors, 187, 188 - Swaying couple, locomotives, 83, 85, 118, 120, 121, 123 Symmetrical engine, determination of, from given centre lines geometrically, 157 Symmetrical engine, determination of, from given force polygon geometrically, 156 Symmetrical engine, Schlick four-crank, errors of, 155, 193, 196 Symmetrical Schlick four-crank, 147, 150 Synchronism, 208 Synchronizing speed, Schlick formula for, 218; S.S. Deutschland, 218; torpedo-boat, 216 AVZ) E.X. 283 T Taylor, Mr. D. W., 61 Three-crank engine, 138, 141; balancing of (see “Balancing”); errors of, 62, 196 Torpedo-boat, vibration of, 216 Torsional vibration, 212 Turning moment on crank-shaft, 219; curves representing (see “Crank-effort diagram ”); effect on frame, 219; effect of motion of connecting-rod on, 240; uniformity of, 225; ratio of maximum to mean, 224, 225 Two-crank engine, 137, 168 |U Unbalanced force and couple (see “Errors”); force and couple, general investigation, 172; force and couple, reciprocating system, long rods, 61; force and couple, revolving system, 37; force and couple, revolving and reciprocating parts, long rods, 78; forces, primary and secondary, effect of, with respect to a given reference plane, 128; reciprocating parts of loco- motive, 82, 118; revolving co-planar system, 18 V Valve-gear, included in balancing an engine, 71–75 Variables, number of, and conditional equations, 134; number of, in a given problem, 65, 135 Vector quantities, 1 ; addition of, 1 ; analytical relation between, defining directions a and 22, 132; analytical re- presentation of, 131 ; condition that the sum of, may be zero, 3; determined by closing a polygon, 7; direction of, 6; subtraction of, 5 Vector representing, centrifugal couple (moment of a mass moment), 25 ; centrifugal force (mass moment), 14; displacement, 4; primary and secon- dary forces and couples, 133; process of finding primary and secondary un- balanced force and couple, 184, 185, 188 Vibration and natural modes possible to a ship's hull, 214; formula for gravest vibration, 218; twin-screw steamers, 215 Vibration, experimental results, S.S. Deut- schland, 219 ; S.S. Meteor, 217; torpedo- boat, 216 . Vibration, lateral (vertical), of a solid rod, 209, 213; longitudinal and torsional, of a solid rod, 212, 213; of support, and point of application of force, 211; under action of several periodic forces, 213 Vibration of a concentrated mass, curves exhibiting, 203, 205, 208; damped by friction, 204; damped by friction and under action of periodic force, 207 ; natural period of, 200; synchronizing with period of supports, 208, 209 W Webb, Mr. F. W., 117 Wigzell engine, 231 Work, division of, amongst cylinders, 228, 229 Y Yarrow, Mr. A. F., 65, 87, 216 THE END. PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES. M. R. E. D WA R D A R N O L D'S LIST OF Seientifie and Jeehmiedl 3ooks. HUMAN EMBRYOLOGY AND MORPHOLOGY. By A. KEITH, M.D., F.R.C.S. Eng., Lecturer on Anatomy at the London Hospital Medical College. With 250 Illustrations. Demy 8vo., I2s. 6d. net. FOOD AND THE PRINCIPLES OF DIETETICS. By Robert HUTCHISON, M.D. Edin., M.R.C.P., Assistant Physician to the London Hospital and to the Hospital for Sick Children, Great Ormond Street. Fourth Impression Illustrated. Demy 8vo., 16s. net. ‘It is seldom we take up a book on dietetics which is at the same time so readable and so scientific as this is. It is the author's intimate touch with the actualities of life which gives to this book much of its vivacity, and lightens the load of the scientific facts which are found on every page.’—Hospital. THE PHYSIOLOGICAL ACTION OF DRUGS. An Introduction to Practical Pharmacology. By M. S. PEMBREY, M.A., M.D., Joint Lecturer on Physiology in Guy's Hospital Medical School; and C. D. F. PHILLIPS, M.D., LL.D., Examiner in Materia Medica and Therapeutics in Aberdeen University. Fully Illustrated. Demy 8vo., 4s. 6d. net. ‘Will undoubtedly be most useful to everyone undertaking the study of Experimental Pharmacology.’ —Dublin Journal of Medical Science. PHOTOTHERAPY. By N. R. FINSEN. Translated by J. H. SEQUEIRA, M.D. With Illustrations. Demy 8vo., 4s. 6d. net. CONTENTS.—I. The Chemical Rays of Light and Smallpox—II. Light as an Irritant—III. Treatment of Lupus Vulgaris by concentrated Chemical Rays. A MANUAL OF HUMAN PHYSIOLOGY. By LEONARD HILL, M.D., F.R.S., Lecturer in Physiology at the London Hospital Medical College. With 173 Illustrations. xii-i-484 pages. Crown 8vo., cloth, 6s. “Anyone who conscientiously works through this book will not only gain a sound knowledge of physiology, but will be grounded in a thoroughly scientific method of observation and deduction. . . . The book is accurate, simple, and scientific, and in many places philosophical, and we can heartily recommend its perusal.’—Lanceſſ., A PRIMER OF PHYSIOLOGY. By Dr. LEONARD HILL, Author of ‘A Manual of Human Physiology.’ [In the Press. A.ONDON : EDWARD ARAVO/I), 37 BEDAWORD ST REAE 7, S7RAAWD. 2 Mr. Edward Arnold's List of Scientific. Books. LECTURES ON THEORETICAL AND PHYSICAL CHEMISTRY. By Dr. J. H. VAN 'T HoFF, Professor of Chemistry at the University of Berlin. Translated by Dr. R. A. LEHFELDT. In three volumes, demy 8vo., Illustrated, 28s. met; or obtainable separately, as follows : Vol. I. Chemical Dynamics. I2s. net. Vol. II. Chemical Statics. 8s. 6d. net. Vol. III. Relations between Properties and Composition. 7s. 6d. net. ‘It is everywhere evident that the material has been wrought into form by a powerful thinker, who sees deeper and more clearly into his subject than any of his contemporaries.”—Nature. A TEXT-BOOK OF PHYSICAL CHEMISTRY. By Dr. R. A. LEH- FELDT, Professor of Physics at the East London Technical College. With 4o Illustrations. Crown 8vo., cloth, 7s. 6d. SUMIMARY OF CONTENTS.–Introduction, Physical Summary—Chap. I. Deter- mination of Molecular Weight—Chap. II. Physical Constants in Relation to Chemical Constitution — Chap. III. The Principles of Thermodynamics — Chap. IV. Chemical Dynamics of Homogeneous Systems—Chap. V. Chemical Dynamics of Heterogeneous Systems—Chap. VI. Application of Thermodynamics to Chemical Equilibrium—Chap. VII. Electro-Chemistry—Index and Biblio- graphy. r PHYSICAL CHEMISTRY FOR BEGINNERS. By Dr. Ch. M. VAN DEVENTER. With a Preface by J. H. VAN'T HoFF. Translated by Dr. R. A. LEHFELDT, Professor of Physics at the East London Technical College. 2s. 6d. CONTENTS.–Chap. I. Definitions—Chap. II. Fundamental Laws of Com- bination—Chap. III. Behaviour of Gases—Chap. IV. Some Points of Thermo- Chemistry—Chap. V. Solutions—Chap. VI. Photo-Chemistry –Chap. VII. The Periodic System. THE ELEMENTS OF INORGANIC CHEMISTRY. For use in Schools and Colleges. By W. A. SHENSTONE, F.R.S., Lecturer in Chemistry at Clifton College. With nearly 150 Illustrations and a Coloured Table of Spectra. xii + 506 pages. Crown 8vo., cloth, 4s. 6d. LABORATORY COMPANION. For use with Shenstone’s “Inorganic Chemistry.’ By W. A. SHENSTONE, F.R.S. viii-H 117 pages. Crown 8vo., cloth, Is. 6d. A FIRST YEAR’S COURSE OF EXPERIMENTAL WORK IN CHEMISTRY. By ERNEST. H. Cook, D.Sc., F.I.C., Principal of the Clifton Laboratory, Bristol. With 26 Illustrations. viii-H 135 pages. Crown 8vo., cloth, Is. 6d. AN EXPERIMIENTAL COURSE OF CHEMISTRY FOR AGRI- CULTURAL STUDENTS. By T. S. Dymond, F.I.C., Lecturer on Agricultural Chemistry in the County Technical Laboratories, Chelmsford. With 50 Illustrations. 192 pages. Crown 8vo., cloth, 2s. 6d. ‘Mr. Dymond is much to be congratulated on the very barefully constructed scheme of work which he has now published. . . . . . The ideas of the author have been gradually perfected by practice. We can cordially commend the book.’—Nature. LOAWDOAV: EDWARD ARNOLD, 37 BAE DA'ORD STREET, STRAAWD. Mr. Edward Arnold's List of Scientific Books. 3 THE STANDARD COURSE OF ELEMENTARY CHEMISTRY. By E. J. Cox, F.C.S., Headmaster of the George Dixon Higher Grade School, Birmingham. With 90 Illustrations. 350 pages. Crown 8vo., cloth, 3s. Also obtainable in five parts, limp cloth. Parts I.-IV., 7d. each ; Part V., Is. ‘A capital book both for the teacher and practical elementary student.”—Journal of Education. THE BALANCING OF ENGINES. By W. E. DALBy, M.A., B.Sc., M. Inst. C. E., M.I.M.E., Professor of Mechanical Engineering and Applied Mathematics in the City and Guilds of London Technical College, Finsbury. With 173 Illustrations. Demy 8vo., Ios. 6d. net. PHYSICAL DETERMINATIONS. Laboratory Instructions for the Determination of Physical Quantities connected with General Physics, Heat, Electricity and Magnetism, Light and Sound. By W. R. KELSEY, B.Sc., A.I.E.E. Crown 8vo., cloth, 4s. 6d. TRAVERSE TABLES. With an Introductory Chapter on Co- ordinate Surveying. By HENRY LOUIS, M.A., A.R.S.M., F.I.C., F.G.S., etc., Professor of Mining and Lecturer on Surveying, Durham College, Newcastle- on-Tyne; and G. W. CAUNT, M.A., Lecturer in Mathematics, Durham College of Science, Newcastle-on-Tyne. Demy 8vo., 4s. 6d. net. THE CALCULUS FOR ENGINEERS. By JoHN PERRy, M.E., D.S.C., F.R.S., Professor of Mechanics and Mathematics in the Royal College of Science, Vice-President of the Physical Society, Vice-President of the Institution of Electrical Engineers, etc. Fourth Edition. Crown 8vo., cloth, 7s. 6d. ‘The author has been successful in retaining all that liveliness and originality of illustration which distinguishes him as a lecturer. The volume abounds in humorous remarks, but the wit is never at the expense of sound advice and instruction, of which the student who would teach himself stands in need.’—Electriczazz. MAGNETISM AND ELECTRICITY. An Elementary Treatise for Junior Students, Descriptive and Experimental. By J. PALEY YORKE, of the Northern Polytechnic Institute, London. With nearly 150 Illustrations. Crown 8vo., cloth, 3s. 6d. ‘The author notes clearly the fundamental facts and laws of Magnetism and Electricity. The explanations are lucid, and the illustrations have a freshness not usually seen in text-books. It will probably be largely adopted in schools.”—Scientific American. ELEMENTARY NATURAL PHILOSOPHY. By ALFRED EARL, M.A., Assistant Master at Tonbridge School. With numerous Illustrations and Diagrams. Crown 8vo., cloth, 4s. 6d. AN ELEMENTARY TREATISE ON PRACTICAL MATHEMATICS. By JOHN GRAHAM, B.A., Demonstrator of Mechanical Engineering and Applied Mathematics in the Technical College, Finsbury. Crown 8vo., cloth, 3s. 6d. ELECTRICAL TRACTION. By ERNEST WILSON, Wh.Sc., M.I.E.E., Professor of Electrical Engineering in the Siemens Laboratory, King's College, London. Crown 8vo., 5s. STEAM BOILERS. By GEORGE HALLIDAY, late Demonstrator at the Finsbury Technical College. With numerous Diagrams and Illustrations. 400 pages. Crown 8vo., 5s. ZOAVDON: EDWARD ARAVOLD, 37 BAEZ) RORD STREAE 7, STRAAWD. 4 Mr. Edward Arnold's List of Scientific Books. AN ELEMENTARY TEXT-BOOK OF MECHANICS. By R. WoR- MELL, D.Sc., M.A. With 90 Illustrations. Crown 8vo., cloth, 3s.6d. *...* Solutions to Problems for Teachers and Prizate Students. 3s. 6d. § A TEXT-BOOK OF ZOOLOGY. By G. P. MUDGE, A.R.C.Sc. LOND., Lecturer on Biology at the London School of Medicine for Women, and the Polytechnic Institute, Regent Street. With about 200 original Illustrations. Crown 8vo., cloth, 7s. 6d. A MANUAL OF ALCOHOLIC FERMENTATION, AND THE ALLIED INDUSTRIES. By CHARLEs G. MATTHEws, F.I.C., F.C.S., etc. Fully illustrated. Crown 8vo., cloth, 7s. 6d. net. A MANUAL OF THE NATURAL HISTORY AND INDUSTRIAL APPLICATIONS OF THE TIMBERS OF COMMERCE. By G. S. BoulgeR, F. L.S., F.G.S., A.S.I., Professor of Botany and Lecturer on Forestry in the City of London College, and formerly in the Royal Agricultural College. [In the Press. * WORKS BY PROFESSOR C. LLOYD MORGAN, F.R.S. ANIMAL BEHAVIOUR. By C. LLoyd MoRGAN, F.R.S., Principal of University College, Bristol, author of ‘Animal Life and Intelligence,” etc. With numerous Illustrations. Large crown 8vo., Ios. 6d. OUTLINE OF CONTENTs. –I. Organic Behaviour—II. Consciousness—III. The Mental Faculties of Animals—IV. Heredity and Evolution—V. Instinctive Behaviour—VI. Intelligent Behaviour—VII. Some Types of Animal Behaviour —VIII. Social Behaviour—IX. Emotional Behaviour—X. The rôle of Con- sciousness in Evolution. HABIT AND INSTINCT. By C. LLOYD MORGAN, F.R.S. viii-H 352 pages. With Photogravure Frontispiece. Demy 8vo., cloth, 16s. The earlier chapters contain the author's original observations upon the young of many species of birds and mammals. These are followed by an interesting discussion upon Animal Habits and Instincts, and the work ends with chapters on ‘The Relation of Organic to Mental Evolution,’ ‘Are Acquired Habits Inherited P’ ‘Modification and Variation,” and ‘Heredity in Man.’ “An admirable introduction to the study of a most important and fascinating branch of Biology, now for the first time based upon a substantial foundation of carefully observed facts and logical induction from them.”—Professor ALFRED RUSSEL WALLACE. PSYCHOLOGY FOR TEACHERS. By C. LLOYD MORGAN, F.R.S. xii-H 251 pages. Crown 8vo., 3s. 6d. CONTENTS.—States of Consciousness—Association—Experience—Perception —Analysis and Generalization—Description and Explanation—Mental Develop- ment—Language and Thought—Literature—Character and Conduct. “The author of this book speaks with special emphasis and authority, and his remarks will be found of exceptional value to students.”—Sir J. G. FITCH, in the Preface. ANIMAL SKETCHES. By C. LLoyd MoRGAN, F.R.S. viii-H 312 pages, with 52 illustrations (many of them full-page). Crown 8vo., cloth, 3s.6d, LOAVDON: ED WAA D ARWOZA), 37 BAEA) FOA'D STRAEAET, STRAND. 3 9015 01185 6484 - sº y º sº. - … . tº ºr ºf Y.Y. º. ," º * . . . . . .'; sº sº . . . - ...? ...? º : *:::: * * * tº . . . . . º. º * * *.*.*.*. * : *- : * * * * * * * * * * - - - ºr *** * * * * - . . . . . . . . . º “...º. º.º. - §º º z; † : * * *.*, * * * * **** * * : . .3 y \ - ~ w s' & 3. - - * : * : * * * * * : * : * : * : * * : * g ºr - ºf ſº fºr gº º * : * : * * * * º : º, . ; º º," *.*.* º ſº. º.º. º. ºº & k > * : * º .# 3. … ºf *...*, *, *, *, *, *, *, *, * : * * : *...* *** * * *; i. º. ºº ºf $.” ** ** * * * * *ś'º.º. º ; : * * º º, º.º. ºº $ $1.3 ºr sº :* - ºr sº tº is ſº *...* ; : * ~ - § 3.5% - ... ? § 3. ğ ºgº. . … * * : * > * - * * * & sº º ºr ; - º'...'. * : *, * ~ * " * - . … . . . . . ºf & - - ** * d **, *...ºf . i. fºº # ºf * * * : * * * * * * * * : * : *- : * * * * * . ... & Jºº - º . . . . * … . . * * v - * *: 3-ºxº * : * * * * * *...* : * * ; : *- : * * : * ~ * * * * * * * ºr tº sº wº a º f : * : *. * * * * * * : * * * *... ºr ; ; , ºr ºf ºf . ; ; ; º,”,” ; : Tº fº ºne ºr tº: * a & sº º º * * * **, ºr: . . . . . . . . . ;-º. º. º. * , ºf a ºf , sº . . . .” sº ... . . . . º. ºf Tº º sº a º ºf º * * : * * * .*.*.*.** * * * * : : * , *...* * * ; : * * * * * * * * tº sº. ºr r < º º, 'º' 㺠: * * : * *.*.*.*.*.*.* * * * * §:4. 2. % º- Fº * : * : * * * * * * * * * * * * º. 4 º' : - ſº ºf 4 × : * * * : * : * * * e º '... 3 º v E. : * : * * * * * * * * * * * * * * * * sº sº sº sº ºr gº iſ . ºf:: * : º ** * * * * * * - & K . . . . * * * * * * * * * * * * * * * * * * ~ º -** tº ś , ; ; ; § ** * * * º tº . . . . . - * * * … “º gº º, sº gº tº # * r * * * * * * * * * * *.*, *:: * : * * *, *, *, * * * * **, ºf # & tº # 4 & 5 * * * * * * * tº sº ºr * * * * r * * # * sº, a º ºs. sº I - , * * * * § 3. … w * * * ~ * : * * * * * * * g : : , ; ; * * * * sº ºf § ** * : ; ; º ; : * ºf - is . . . ; rºſ. # 4 × ... < & * * * * * * * * * * * * * - s' ºs s...} : ***, * : *- : * : ; sº tº ºw º, º gº º & g º #4 & gº ºf º - * . * . . . . . . . . ; *, * * * .* : * **, 1 #, sº - # * : * **, * * * : * * * º * : * : * : * sº ; : . . . . * * * *, * * * * **, r* % 5-9 ğ ; : . . .3 º * , sº it is ׺ gº. 3 & 13 × º ; sº P 3. * * * * > . . . ..","..'s º K. tº sº. - - º - ; : * : *. §§ - ? § # , º. fºr. º > ~ I " ..." - º: ºº : • * -> tº #: ; sºº &. . .” . *::4 ºs ºf ...º.º.º.º.º.º. & º º 'º. #: , ; º * . . . . . . . . . . . . . # *... . ; tº ºf 8 * . - - sº ..º.º. ; : * : ... a . . . . ; º - - - 3 *, * v #: ºº #. ºººº Yºº º Pº ; : , ; ; ; ; - º Aº w . . . . . . . ; : ... sºr: * i... * * * § 3.4% º “ . . . . ." * * * * * * * , #.º. º.º.º.º. # ***.*.*.* :: *.*.* : * ºx: - g Nº. . . . . . . . . . . . . - s: " . . . . tº gº fºr g º - - º * * * * . . ; * * * * * * * * * º: § ºº: s: º º *: ; ; ; ; ; ; sº : * * * * # , º, . . . . . .” *.*.* **, ºr † : - 3. sº ºf ºr e - sº tº F- ºr :* º & º' Jº, ºr 3 ſº º f * : sº tºº, ; : * §: ; § #2 * : * ºf . ; ; : º : § § 5.