LIB RARY (^CONGR ESS, ©§ap^.!? Itipgng^ ¥)n. Shelf >}Ä5.to> UNITED STATES OF AMEEICIA. THE GRAPHICAL STATICS MECHANISM. A GUIDE FOR THE USE OF MACHINISTS, ARCHITECTS, AND ENGINEERS; AND ALSO A TEXT-BOOK FOR TECHNICAL SCHOOLS. BY GUSTAV HERRMANN, PROFESSOR IN THE ROYAL POLYTECHNIC SCHOOL AT AIX-LA-CHAPELLE. TRANSLATED AND ANNOTATED BY I A. P. SMITH, M.E NEW YORK: D. VAN NOSTRAND, PUBLISHER, 23 Mueuay and 27 Wirren Streets. 1887. SU Copyright, 1887, By W. H. FARRIXGTON. RAND AVERY COMPANY, ELECTKOTYPERS AND PRINTERS, BOSTON. PREFACE. Since the appearance of Culmann's work, which marked an epoch in the history of graphical statics, the graphical method has attained pretty general dissemination in engineering circles ; its advantages over the analytical have been recognized more and more, and its further development kept constantly in view. It is universally applied in the design- ing of stationary structures, — such as bridges, — for determining the requirements of the individual parts. In machine design, also, the graphical method gives valuable aid in finding the moments to which the machine parts are subjected, and in determining di- mensions. Accordingly courses in graphical statics have been introduced in all technical schools. The graphical method has also been adapted with advan- tage to certain departments of dynamics, as in Radinger's " Dampfmaschinen mit hoher Kolbenge- schwindigkeit," and Pröll's " Versucht einer graphischen Dynamik." * * Radinger' s Steam-engines with High Piston Speed, and Pröll's An Attempt at Graphical Dynamics. iii IV PREFACE. In all these determinations, however, friction and the special hurtful resistances to motion have not been taken into account. Heretofore all attempts to ascertain these hurtful resistances in machines, and to determine the efficiency which is dependent upon them, and is of such great importance in practice, have been confined to the analytical method, which is often awkward and at times utterly inapplicable. No method is as yet known to me by which the frictional resistances and efficiency of any desired mechanism can be graphically determined. In my lectures on machinery in the polytechnic schools of this place I have endeavored to show the relations existing between the forces in mechanism in a simpler form than that offered by the analytical method. Out. of that endeavor has grown the following treatise, which in reality amounts to nothing more than a wider application of the long known but little used angle of friction. My object in the present treatise was principally to facilitate study for the students of the technical schools, upon whose time and industry increasing demands are made from day to day ; perhaps the work may also be of interest and value to those more advanced. GUSTAV HERRMANN. Aix-la-Chapelle, May, 1879. TRANSLATOR'S PREFACE. The following translation was undertaken from a belief that a knowledge of Professor Herrmann's work on — w^e might almost say discoveries in — this subject should not be bounded among English and American engineers by an ability to read German or French; the treatise having already been translated into the latter language. The original has been fol- lowed faithfully. There is one word that perhaps needs explanation. The German expression im Sinne seems to have no technical equivalent in English, and has been literally translated "in the sense of," and coupled with " direction," the latter being used loosely. "Acting in the sense of a force Q" means producing a similar result to Q. Thus, suppose a force P applied to a crank tangentially to the crank-circle at every point of revolution. The load Q is suspended from a rope wound on the windlass drum turned by the crank. Then if the rope is being wound on to the drum, the force P acts at every instant in an opposite " sense " to Q, while there is only one point in each revolution at which it acts in an opposite direction. VI TRANSLATOR'S PREFACE. To complete the definition, suppose the windlass to turn stiffly in its bearings, and the weight Q to be so small that when the crank is released the weight will not run down. To lower the weight, then, a force (P) would have to be applied to the crank to turn it in the opposite direction. This force (P) would then act in the same " sense " as Q at every instant, though it would have the same direction as Q but once in each revolution. " Sense," then, has reference simply to the effect produced through no matter what amount of intervening mechanism, and is entirely distinct from " direction" in its stricter meaning. A few of the terms of Professor Kennedy's transla- tion of Reuleaux's " Kinematics " have been employed, because there are no other well known equivalents, but they will be easily understood even by those who are not as yet familiar with that work. The great advantage which the method presents is its simplicity. By the use over and over again of a few easily mastered principles, the most complicated problem may be solved. No knowledge of higher mathematics is required in its mastery, and no hand- ling of lengthy and involved algebraic formula is necessary in its use. A few lines are drawn in accord- ance with easily understood rules, and the result stands out so clearly on the paper that every bright mechanic ought to understand it at a glance. This, TRANSLATOR'S PREFACE. Vll with the additional merit of rapidity and accuracy, should soon render its use common in every class- room and drafting-office. The respectful thanks of the translator are due to Professor J. F. Klein of Lehigh University and to Mr. T. M. Eynon, recently of the same institution, for their critical revision of the manuscript. A. P. SMITH, M.E. United States Patent Office, Washington, D.C. CONTENTS. PAGE § 1. — The Efficiency of Mechanisms 1 § 2. — The Equilibrium of Mechanisms 14 § 3. —Sliding Friction 23 § 4. — Journal Friction . 39 § 5. — Rolling Friction 76 § 6. — Chain Friction 84 § 7. —Stiffness of Ropes 102 § 8. —Tooth Friction 110 § 9. — Belt Gearing 122 § 10.— Examples ...'*'* 140 Concluding Remarks . 150 ix THE GRAPHICAL STATICS OF MECHA § 1. — THE EFFICIENCY OF MECHANISE The object of every mechanism is to transfer 3 in a certain way, from one particular machine another. In this transfer the magnitude and di of the motion may remain unchanged, or the re: motion may vary in either or both of these res and the rate of variation in either magnitude 01 tion may be constant or not. In the steam-e for instance, the straight-line motion of the pis communicated continuously to the cross-head piston-rod unchanged in magnitude and while this right-line motion generates a motion i crank-pin, through the agency of the connectin whose direction varies constantly, and whose magi] differs in its relation to the magnitude of the head's sliding motion at every instant. In the ca two cylindrical toothed wheels of equal diameter, ^ ing together, the rotation of one is transformed ii rotation of the other, equal, but in an opposite direc The ratio between the lengths of the paths thn which the two pieces move depends upon the : which exists between the force P working upon driving member of the machine, and the resistanc 1 HE GRAPHICAL STATICS OF MECHANISM. 3 driven member encounters. If no hurtful ?s or hinderances, such as friction, appeared in )ii under consideration, the mechanical work of Qg-force, for any period of time, would exactly t done during the same period in overcoming ance presented to the driven member, under the on of uniform motion. This follows from the wn principle of virtual velocities. If p and q he distances over which the points of applica- P and Q travel along the directions of these md if we let Pp and Qq — i.e., the product of ito distance — represent the mechanical work r the respective forces, we have the equation Pp = Qq, or Pp - Qq = 0. jr the supposition, on the other hand, that one . of the machine parts does not have a uniform according to the law of kinetic energy, again ill hurtful resistances, the equation becomes Pp - Qq = L, L represents the amount of work which the I mass M has either absorbed or given out in the *ed time. According as the above value of Pp — positive or negative, the moving masses have been crated or retarded ; and from Pp — Qq z= there - r s evidently a uniform motion during the assumed i. From the preceding it follows, that, always Aug hurtful resistances, in no machine construction hatever form is there either a loss or gain in lanical work ; for if, in the case of an inequality THE EFFICIENCY OF MECHANISMS. 6 between Pp and Qq, the masses M are accelerated, this acceleration represents a storing-up of mechanical work, which, later in the period, is available to its full amount. But there can be, in reality, no movement between two bodies in contact without certain hinderances aris- ing ; which hinderances, since they always have the char- acter of an impediment to the desired motion, are gener- ally known as hurtful resistances, in opposition to the useful resistances in the overcoming of which the useful work of the machine consists. In every case, such hurt- ful resistances arise only where one body has a motion relative to another: as, for instance, the friction of a journal turning in a bearing which is stationary, or rotates at a different speed ; the resistance of the air to arms or wings in rapid motion, etc. These resisting forces disappear with the cessation of relative motion between the two bodies assumed. Therefore, we must admit that no loss of work occurs when tensile or compressive forces act on a solid body, so long as the elastic limit is not exceeded, since their effect is only to increase or diminish the distance between the molecules of the body, — there being no contact between these molecules, — and the contraction of the body gives out again the work used in expanding it; at any rate, experience has not yet proved the contrary. That no one may bring forward the example of the resistance due to the stiffness of ropes as an exception to the above generalization, it should be remarked, that, when a rope is bent around a pulley there occurs a relative motion between any two threads or strands which are at different distances from the axis of the pulley. A certain amount of mechanical work, Ww, is necessary to overcome the hurtful resistances TT during a period 4 THE GRAPHICAL STATICS OF MECHANISM. of motion in which the distance w is traversed. This is not only lost as far as the work proper of the machine is concerned, but exercises a further and evil influence, in that it is transformed into molecular work, whose effect is evident in the wear of the moving parts. Taking into account the hurtful resistances the equa- tion becomes Pp — Wto = Qq + -L, or Pp — Wiv = Qq, if, in the last equation, we suppose the useful work Qq to include the positive or negative quantity L repre- senting the work done in the acceleration or retardation of the masses M. The last assumption, which will be taken as the basis of all that follows, is always admissi- ble, because, as before remarked, ail work stored up in the acceleration of the masses will be given out again in the performance of useful work when the velocity of the masses sinks to its previous value. But these regular or periodic accelerations and retardations characterize the state of permanency (or normal running condition) of the machine, which is the only state or condition taken into consideration in the present case. From what precedes it follows without further ex- planation, that in all machines, without any exception whatever, the useful work done is less than the work expended by the driving-force in producing motion, by an amount exactly equal to the work required to over- come the hurtful resistances. The smaller these latter are, the more completely will the machine accomplish its object ; that is, the work of the driving-force will be transferred to its point of useful application with the greater economy. We may therefore regard the ratio THE EFFICIENCY OF MECHANISMS. 5 of the accomplished useful work Qq to the work Pp expended by the driving-force as the efficiency of the machine ; and the ratio Qq_ Pp- 71 is known by this name, or by that of " useful effect." The efficiency of a machine — which will always be represented by rj hereafter — is, according to what pre- cedes, less than unity in every case except the ideal one, in which all hurtful resistances, TF, disappear. The value of r) will evidently grow smaller as the number and magnitude of the hurtful resistances increase; and, since the latter occur wherever there is relative motion, it follows that, in general, the simpler a machine is in its construction the more economical will it be in work- ing, since every additional moving piece reduces the efficiency by the addition of a new hurtful resistance. Tt is hardly necessary to remark that the value of the useful effect depends not only upon the number, but upon the magnitude, of the hurtful resistances, which magnitude varies greatly according to the kind of resistance offered. Thus there are not a few machines whose efficiency, on account of the resistances incurred, sinks to a small percentage ; in fact, the value of 77 may even become equal to zero or negative. The numerator of the fraction __l = y becomes zero in all cases where Pp a weight Q is moved in a horizontal plane, — for instance, in the rolling of an object along a horizontal track, in the turning of a crane, of a turn-table, etc., — because the distance q passed over in the " sense, 1 ' or direction, of the load Q (i.e., the line of direction along which it (j THE GRAPHICAL STATICS OF MECHANISM. would -tend to move if left without any support what- ever) is equal to zero. In such cases the entire work of the driving-force P is employed in overcoming the hurtful resistances W; and we have for these cases, for such mechanisms, the equation Pp = Ww. If, further, we suppose that in a certain case Ww>Pp, there follows, from the general equation, Pp — Ww = Qq, a negative value for Qq ; i.e., in order to render motion possible in such a case the force Q must work in a direction, — y, opposite to its usual " sense " or direc- tion.* The force Q is no longer a resistance to be overcome through the agency of the machine: it is, on the contrary, to be regarded as one of the forces necessary to produce motion, and working in the same " sense," or direction, as P. There can no longer be any question of useful work, and machines of this nature are never used to produce motion : they find their application in those cases where it is required to hinder an undesired motion. Every screw the pitch- angle of whose thread is not sufficiently great to cause it to turn backwards under the influence of the load upon its nut, is a machine with negative efficienc}' for the case of backward motion ; and the same is true of the differential jail ley as generally constructed. The * Thus, if it was a weight to be lifted, the attraction of gravitation would have to be reversed, or overcome by an opposite and greater force before motion would occur. — Trans. THE EFFICIENCY OF MECHANISMS. 7 clamps, so common in practice, which never release their hold under any strain whatever, and of which we shall treat more particularly hereafter, may always be regarded as machines with negative efficiency. We may define efficiency in another way which is often used, and in many cases is more convenient. If in any mechanism where, as before, a driving-force P expends the work Pp in order to move a load Q along the path q, we neglect the hurtful resistances, there would evidently be necessary to accomplish the useful work Qq, under this supposition, only a driving-force P , of a less value than P, but working along the same distance /?, at the same point of application, and in the same direction. In the case of this force P we have of course, through the neglect of all hurtful resistances, the equation from which the expression for the efficiency becomes v — pp - Pp — p- If, for brevity, we call P the theoretic, and P the actual, driving-force, the efficiency becomes the ratio of the theoretic to the actual driving-force ; and this value accordingly represents that percentage of the driving- force which serves to overcome the useful resistance. If, on the other hand, under the assumption of fric- tionless motion we regard the force P as working along the path p, it would overcome a resistance Q along the path q, which theoretic resistance Q would have a larger value than the useful resistance Q actually overcome ; 8 THE GRAPHICAL STATICS OF MECHANISM. and, from the equation Pp = Q q, we should have the efficiency, __Qq_ __Q± _ Q_ *- Pp~ Q q~ Q > that is to say, equal to the ratio which the actual useful resistance Q bears to the theoretic one Q . If, in the case of any particular mechanism with the useful load Q, the driving-force becomes less than P, there can be no motion in the " sense " of P; the mechanism remains at rest. The same thing continues until the driving-force approaches the theoretic P in value. If, then, there were no hinderances to motion to be taken into account, the slightest further diminution of P below the value P would result in a backward or reverse motion of the mechanism, the load Q becoming the driving-force. But since certain hurtful resistances, which may be represented by ( IF), oppose the backward motion, that backward motion can only occur from that moment when the force _P, through further diminution below P , has sunk to a value (P), for which the relation Qq — ( W)w = (P)p holds. The force (P) which is just able to prevent the backward motion, or running down, of the mechanism, is smaller than the theoretic driving-force P by an amount which increases with the magnitude of the resistance (IT) offered to the backward motion of the mechanism. Thus, in speaking of the efficiency for backward motion, we will hereafter understand the ratio « = From (,) = i|) = it follows that (P) =2 0, and therefore since (P)jp + QW)w — Qq. On the other hand, for forward motion, Pp — Qq + Ww. If the hurtful resistances (TT) for the backward motion are equal to those W for the forward, the substitution of the former for the latter in the last of the above equations would give Pp = Qq + (Tf> = Qq + Qq=z 2Qq ; from which Pp 2Qq THE EFFICIENCY OF MECHANISMS. 11 The assumption that W ■= (TF) is a sufficiently close approximation, since the hurtful resistances are the result of both P and Q* Moreover, since TT> ("FT), it follows that 2Qq < Pp, and consequently that the efficiency rj must always be less than |. Therefore we may lay down the principle that every self-locking machine has an efficiency of less than fifty per cent, a conclusion which experiments with screw hoisting-gear and differential pulley-blocks confirm. Machines met with in practice consist generally of a collection of elementary mechanisms, so that the driving- force P r necessary to overcome the useful resistance Q of the first mechanism must be regarded as the useful resistance Q' of the second mechanism, for the overcom- ing of which in the second mechanism another driving- force P" is required, and so on. Take, for example, an ordinary warehouse hoist : here the rope with its drum is one mechanism, in which the load hanging from the hook on the end of the rope is the useful resistance $, to overcome which a force P f must be applied at the circumference of the toothed wheel connected with the drum. As regards the toothed gearing which forms the second mechanism, the force P f becomes the useful resistance Q\ which must be overcome by a force P" applied at the circumference of the second large gear * The force of this reasoning lies in the fact that W\ (W) :: friction of (P and Q) : friction of (P) and Q, in which Q is generally much the larger factor; and hence the differ- ence between IF and (W) is much less than it would be if they were dependent solely upon P and (P), and the relation W : ( W) :: friction P : friction (P) existed. — Trans. 12 THE GRAPHICAL STATICS OF MECHANISM. upon the axis of the pinion. In the same way, this force P" is the resistance Q" for the third mechanism, represented by the second pair of gears and the crank. To overcome this resistance Q", a force P is necessary at the crank-handle. When, as in the case of a crane, the load Q does not hang directly from the drum, but the rope, or chain, is first led over one or more pulleys, each of these pulleys is to be regarded as a separate mechanism. In the same manner, every machine, however complicated, can be resolved into its simple elementary mechanisms. Such a resolution greatly simplifies the determination of efficiency of machines, inasmuch as the number of common mechanisms is small, while the diversity between complete machines is almost endless. One general law may be established for the efficiency of a machine consisting of any desired number of ele- ments. If we let Q again represent the load, and q the path described by it in a slight movement, and let P v P 2 . . . P n denote the driving-forces for the separate mechanisms, and p v p 2 . . . p n the corresponding paths of these forces, we may understand by rj v rj 2 . . . rj n the efficiencies of the individual mechanisms. For the first machine, we have -&- = Vl . For the second, in which P x is the useful resistance, and P 2 the driving-force, we have THE EFFICIENCY OF MECHANISMS. 13 and so on for each separate mechanism. From the multiplication of all these equations, there results QV P iPi PjiPn -Pi.-iPi.-i - v v v v or Qq jr = v = v t . v 2 • % • • • %. That is to say, the efficiency of a machine composed of any number of mechanisms is equal to the product of the efficiencies of all the separate mechanisms. From the fact that the efficiency of a simple mech- anism is always less than unity it follows, as before remarked, that in general the useful effect of a machine decreases as the number of its constituent elements increases. Since, furthermore, the above product cannot be negative as long as all the factors are positive, it follows that a machine can only be self-locking when this property belongs to at least one of its elementary mechanisms. 14 THE GRAPHICAL STATICS OF MECHANISM. § 2. — THE EQUILIBRIUM OP MECHANISMS. Although mechanisms, by their very nature, can only effect their object while in motion, or by virtue of the same, yet, for the ascertaining of the relation existing between the various forces, we may always assume as a basis that condition of equilibrium which corresponds to the limit where the slightest increase of the driving- force would produce a motion in the "sense*-" or direc- tion, of that force. In what follows P will again represent the driving-force and Q the useful resistance. Neglecting for the present any acceleration of the masses, we will suppose a uniform motion in which, at each instant, the work of the force during a small portion of time is just sufficient to overcome the useful resistance Q after the hurtful resistances W have been disposed of. It will then easily appear in what way the accelerating force working upon the mass M in the case of variable motion can be ascertained. The exterior forces P and Q working upon any mechanism, call forth certain internal forces, or re-ac- tions J2, between the members of the machine wherever two parts come in contact. These re-actions are to be regarded as two equal and opposing forces occurring at every surface of contact. Every pair of forces thus arising at the same point is, therefore, in equilibrium. We must imagine such re-actions wherever two bodies come in contact, whether the bodies move relatively to each other or not. We can, therefore, in every case neglect the bodies in-contact and think only of the THE EQUILIBRIUM OF MECHANISMS. 15 re-actions offered by those bodies. Under this supposi- tion, any member of a machine which is acted upon by certain exterior forces P and Q, and which is supported at certain points by neighboring bodies, must be under the influence of the exterior forces P and Q, and of the re-actions Ä, which are sufficient to replace the imagined supports, in order to be in the supposed limiting con- dition of equilibrium. The conditions of equilibrium furnish us, in general, with a means by which from the known elements, — direction and magnitude of indi- vidual forces, — we may ascertain the unknown. In the majority of cases the intensity of the re-actions of the supports is unknown ; of the exterior forces, there is, as a rule, one element — the direction, or intensity, of one force — unknown at first. As regards the direc- tion of the re-action replacing a support, it is deter- mined empirically by the condition that it shall be inclined to the supporting surface at a certain determi- nate angle whose magnitude depends upon the nature of the two bodies in contact, as to smoothness, hardness* etc. The hurtful resistances to motion, W, which, as previously remarked, arise only at the point of contact between two bodies (i.e., at the supporting surfaces), depend on the nature of the material, and of the sur- faces constituting the supports. The size of the angle at which the surfaces of contact will be cut by the direction of the re-action existing between them de- pends closely, as will be shown in what follows, upon the amount of hurtful resistance generated between the surfaces. If we suppose, in the next place, that no hurtful resistance W exists, — a condition of affairs which, of course, never occurs in practice, — the angle formed by 16 THE GRAPHICAL STATICS OF MECHANISM. the direction of re-action and the supporting surface would be a right angle ; in other words, when there are no hurtful resistances a supporting surface can only re-act, at each of its points, in the direction of a normal. With the assumption which thus entirely ignores all hurtful resistances, it is a simple matter to determine the proportion of power and load at every point in a machine ; and, for this purpose, a graphical method can be used to good advantage. A few examples will serve to illustrate this procedure. Let J-, B, and (Fig. 1, plate I.) be the centres of the three pins on a bell-crank, the middle one of which, 6 7 , turns in the fixed bearing C v while the end pivot A is enclosed by the eye or end bearing A x of the rod A X A V and the other pivot B is attached in the same way to the rod B X B V If, now, a force Q acts upon the lower end A 2 of the rod A l A 2 in a certain direction, the rod A 1 A 2 must, according to the preceding principles, be in equilibrium under the influence of the force Q, and the re-action M 1 replacing the pivot or journal A ; and this latter re-action, being perpendicular to the surface of the journal, must pass through its centre. Two forces, however, can only be in equilibrium when they are equal, and work in opposite directions along the same straight line ; from which it follows immediately that the line of direction of the force Q must pass through the centre of A, or, if it does not, there wiH be a turning of the rod A 1 A 2 about the journal A until this condition is fulfilled. In the same way, it follows that the direc- tion of the force P acting on the rod B X B 2 must pass through the centre of JS, and that the journal B must exert upon the rod B X B 2 a re-action R 2 which shall be equal and opposite to P. The rods A X A 2 and B X B 2 will THE EQUILIBRIUM OF MECHANISMS. 17 be subjected to tension by the forces Q, M 1 and P, P 2 , respectively ; i.e., there will be called forth internal elastic stresses in the material of each section of the rods, which will be in equilibrium one with another, and with the exterior forces at the ends of the rods. These interior strains are of great importance in determining the dimensions of cross-sections of the separate machine parts, but have no direct influence upon the conditions of equilibrium of the machine. We shall not, however, go into the determination of dimensions here, or in the remaining portion of the treatise. For all that the reader is referred to the well-known works on machine construction and the resistance of materials. Regard- ing now the lever ABC alone, we have the external forces Q and P acting upon it at the points A and B ; and we can also suppose the bearing C x to be replaced by a re-action P 3 , whose line of direction passes through the centre of C. For the condition under consideration these three forces must be in equilibrium. This can only occur when the three forces intersect at the same point ; and therefore the re-action P 3 , exerted by the bearing C\ upon the journal C, must also pass through the intersection of the lines of the forces P and Q. Moreover, the relative intensities of the three forces, P, $, and P 3 , are easily determined if we let OD == Q, the load, and complete the parallelogram ODFE, whose other side falls upon OP, and whose diagonal coincides with 00. We have then in OE the force P, and in OF the pressure of the journal O upon its bearing, and the equal but opposite re-action P 3 of the bearing C x against the journal C. The relative intensities of P, Q, and P 3 correspond to the oft-mentioned limiting condition in which the 18 THE GRAPHICAL STATICS OF MECHANISM. slightest increase of Q or P would cause motion in the "sense/ 1 or direction, of the force so increased, as can be at once seen from the figure ; for, if we increase Q until it is equal to OD\ the diagonal 0F\ which repre- sents the pressure of the journal upon its bearing goes, to the left of the centre Ö. This points to a motion of the lever in the "sense," or direction, of Q. On the other hand, an increase of P to the value 0E f gives a journal pressure OF", in consequence of which there would be a right-handed revolution of the lever. As a further example, the slider-crank mechanism (Fig. 2, plate I.) may be adduced. The force acting from the piston-rod AA' upon the pin A of the cross- head is transferred through the connecting-rod A 1 B 1 to the crank-pin B; and, under the supposition of entirely frictionless motion, the force T must pass through the centres of A and B since it is perpendicular to the surfaces of contact of the journals A and B with their bearings. Since the forces P and T have different directions, the pin A cannot be in equilibrium under their influence alone : a re-action It 1 exerted by the guide jPjDg upon the cross-head J) is necessary. For the condition of equilibrium this force acting jierpen- dicularly upon the supporting surface D^D^ must pass through the intersection A of the piston-thrust P and the connecting-rod resistance T. From this we can easily find the forces R x and T by drawing AF equal to P, the piston-thrust, and completing the triangle AFGr, in which FGr is parallel to R v i.e., perpendicular to the guide D X D T If, now, the axis C of the crank meets a resistance Q at the distance CV7,— as though a gear-wheel, with the radius CJ was on the axis C and meshed with another gear-wheel JK, whose resistance THE EQUILIBRIUM OF MECHANISMS. 19 would be represented by Q, — it must be in equilibrium under the influence of the connecting-rod thrust acting at jB, the resistance Q acting at J, and a re-action R. 2 of the bearing O v The direction of the latter again coin- cides with the line joining and (7, if represents the intersection of Q and T. By constructing upon the line AG, already determined as the value and direction of the force T, the triangle AGH, whose sides AH and GH are parallel to the direction of the resistance Q, and of the re-action R v respectively, we have in GH the value of the re-action iü 2 , and in AH that of the resistance Q overcome at J, at the instant under consid- eration. It will be seen that the ratio of P to Q varies for every instant, and that, with a constant piston- thrust the amount of resistance Q overcome at J will vary between zero at the dead points and a maximum at some intermediate position. When, therefore, as in the ordinary case, the resistance which the gear JK presents to the motion of the crank is constant, this resistance must have a mean value between zero and the maximum value of Q ; and there will result an acceleration or retardation of the masses (of fly-wheel) according as the value of Q, determined as above, ex- ceeds, or falls short of, this average value of the resist- ance between the gear-wheels. This peculiarity of slider-crank mechanism is well enough known to render a further discussion of it here superfluous. In the same way as in the two examples shown we can obtain in every case the ratio of the force P to the resistance Q. In nearly all cases the graphical method offers great advantages over the analytical on account of its simplicity and plainness. The analytical method frequently leads to involved expressions, espe- 20 THE GRAPHICAL STATICS OF MECHANISM. cially when it is attempted to bring the hurtful resist- ances into the calculation ; when, in other words, it is no longer a question of determining P or $ , but of P or Q. Since the economic value, or efficiency, of any machine depends directly upon the magnitude of the hurtful resistances connected with it, it is evident that the determination of the ratio actually existing between the forces when these resistances are taken into account is of vastly more importance in practice than any deter- mination of the merely theoretical forces. We have been accustomed in the past to the use of only the analytical methods. The graphical methods for the determination of the friction in, and hence the effi- ciency of, mechanisms have hitherto been rarely em- ployed ; at any rate, in all the text-books only the formulse for these determinations have been given. How complicated such investigations often became even in the simplest machine as soon as any exact calculation was attempted, is well known. Thus, for example, we could only obtain an expression for the journal friction of a bell-crank, as in Fig. 1, plate I., or of a pulley where the ropes were not parallel, through a long radical, — a circumstance which compelled the assumption of parallelism in all cases of pulley friction, even when there was a marked inaccuracy in the same. For the same reason it is the custom to assume an infinite length of connecting-rod in all cases of crank- gear, in order to render less unwieldy the expressions in which the angle of crank to piston-rod occurs. From the well-known advantages which the use of the graph- ical method offers in the designing of machine elements, as in the determination of the moment to which axles, cranks, etc., are subjected, arose the idea of finding an THE EQUILIBRIUM OF MECHANISMS. 21 expression for friction by the same method. From that idea has sprung the following treatise. In the course of the same, it will be shown that we can obtain a graphical determination of the actual proportion exist- ing between the forces in the same simple and sure way as was done in the preceding examples where all friction was neglected; and it is evident, that, by comparing the values obtained for P or Q with those of P or # > we have immediately an expression for the efficiency, — ■ ' , = £.= £ The methods used do not differ from those indicated in the determination of the theoretical forces ; and in nearly every case the drawing of force polygons suffices to attain the desired result. The main question will therefore be to express the separate hurtful resistances graphically. The solution of this problem will be at- tempted in what follows. The hurtful resistances in mechanisms which are to be taken into consideration are few in kind and consist, if we neglect the resistance of the medium in which they work, only of friction, which may occur as sliding and rolling friction, journal friction, chain friction, and the friction of toothed wheels. The stiffness of ropes may be considered as equivalent to chain friction. In the following pages these simple resistances will be taken up one by one. The resistance of air or water is here neglected because in ordinary machines it may be left out of account as insignificant, and in most cases is not regarded. In individual cases and in particular machines where such neglect is not allow- 22 THE GRAPHICAL STATICS OF MECHANISM. able, where (as in ventilators) the moving of this medium is the principal object, its resistance should be determined by the rules of hydraulics. It has ceased to be a hurtful, and become a useful resistance. SLIDING FRICTION. 23 §3. — SLIDING FRICTION. A load which presses upon a horizontal surface (7(7 (Fig. 3, plate I.) with its weight Q = AB calls forth, while at rest, an infinite number of re-actions from points in that surface, whose resultant is equal to the weight Q, and opposite in direction. This re-action passes, therefore, through the same point I) in the supporting surface as the weight Q (the term " weight " being here used in its meaning of a resultant of the forces of gravitation acting upon each particle of the body ^4). Now, it is known that a force P = pQ, acting parallel to the plane (7(7, is necessary to produce a horizontal sliding of the body A along that plane, where fx repre- sents the co-efficient of friction. If, now, a force P, represented in the figure by A C, acts upon the body J., the body is under the influence of the two forces P and Q, which must be in equilibrium with the re-action R of the supporting surface ; it being always remembered that we are assuming that limiting condition of equi- librium where the slightest increase of P would cause motion. This condition of equilibrium requires, there- fore, a re-action R = EA of the supporting surface equal to the resultant of P and Q, and opposite in direction. From the equation P = ^Q is determined the angle $ = EAB which this resultant makes with the normal AB to the supporting surface, since /* = — = tan ; 24 THE GRAPHICAL STATICS OF MECHANISM. and this is called the angle of friction for the materials composing the bodies A and GG. If we suppose the force P to increase gradually from zero to the value AC, the re-action R, given forth by the supporting surface, would gradually deflect from its original direc- tion BA to HA; and for all positions between these two the conditions of equilibrium would be satisfied. In this way the point of intersection D of the re-action Avith the surface would move from D to F; but in every position it is to be regarded as the point of application of the resultant of all the elementary sur- face re-actions. These relations evidently hold what- ever the direction of the horizontal force P. For example, it is true that when P has the direction AC V the re-action of the supporting surface coincides with the line E X A. By a complete revolution of the force P in the horizontal plane the re-action would describe a conic surface concentric to AB. This is called the cone of friction, and limits the space within which the re-action of the surface GG may exist with- out motion resulting. We must therefore regard the supporting surface as re-acting against the supported body in certain directions whose angles of intersection with the normal are less than the angle of friction ; motion commencing from that moment at which this angle of intersection exceeds, by the slightest amount, the angle of friction. We may employ this property of surfaces in the graphic representation of sliding friction under the following rule : — If two bodies having plane surfaces in contact under- go relative motion along that plane of contact, we may completely replace each of these supporting planes by a re-action which is inclined to the normal at an angle SLIDING FRICTION. 25 equal to the angle of friction, and so situated that its component parallel to the plane of contact will work in the direction of the motion which the supporting sur- face has relatively to the body supported. The necessity and correctness of the last part of this can easily be seen from the figure. If, for instance, the supported body A is moved by the force P in the direc- tion from A to (7, the relative sliding of the support, or bearing, GG to the body A is evidently in the opposite direction CA; and the re-action R, by which GG is replaced, has, according to previous demonstration, the direction EA, whose component CA works in the direc- tion of the sliding of GG upon the body A. By virtue of this general law we can easily obtain in every case the value of the sliding friction called forth by the relative motion of one body upon another. If, for instance, in the case of the slider-crank gear (Fig. 2, plate I.), we wish to determine the influence of the friction existing between the slipper-block D of the cross-head and the cross-head guide, we have only to draw the re-action R v passing through A, in the direc- tion E X A) making the angle E X AE = <£ with the normal to the cross-head guide. If, then, we draw through F the line FG X parallel to E X A, we have in AG X the actual thrust T of the connecting-rod ; while the value pre- viously obtained by neglecting friction was AG = T . The thrust of the connecting-rod has been reduced, through the sliding friction of the cross-head i>, by the amount GG X ; and we have the value AG, T „ AG T V for the efficiency of the right-line motion in the slider- 26 THE GRAPHICAL STATICS OF MECHANISM. crank gear. It is unnecessar}^ to explain that the value of this sliding friction and its dependent efficiency varies for every position of the gear. By the use of the angle of friction in the manner discussed above, the efficiency of a great number of machines may be easily determined, as will be shown in a few examples. Let ABC (Fig. 4, plate I.) be the rack of an ordinary jack, upon whose lug, or claw, A the load Q = Io l acts vertically downward ; while at the pitch- line DD of the rack, the force P acts vertically upward. In consequence of the fact that the load Q acts in one direction only, the rack is continually pressed against the casing at B and C; and we may regard it as being supported by the resultants b and c of the re-actions R x and 7t 2 . The four forces Q, P, R v and R 2 must be in equilibrium, which can only be the case when the resultant of any two is equal to that of the other two, and opposite in direction. If, therefore, o x is the inter- section of Q and R v and o 2 is that of P and R v the line o x o 2 is the direction of the resultant of Q and R 1 as well as of P and R 2 . From the given load Q, we can now determine the forces P, R v and R 2 by drawing the force polygon. Make Io x = Q; draw III parallel to R v and intersecting o x o 2 ; and then construct the tri- angle II III o x by drawing II III parallel to P, and III o x parallel to R 2 . We have, then, in II III the force -P, which must be applied at the pitch-line of the rack to lift the load Q. Without friction, Po = Q, and therefore the efficiency of this mechanism - ^2 V P is known. SLIDING FRICTION. 27 The figure has been constructed with a co-efficient of friction ^ = 0.2, and a corresponding angle of fric- tion = 11° 20', and gives P = 122.3 for Q = 100; therefore 100 122.3 = 0.817. It should be here remarked, that in this and all following cases the numerical results are not taken from the lithographed plates, but from the original drawings of the author, which were on a much larger scale. In drawing the direction of the re-action it is not necessary to know the angle of friction in degrees and minutes ; a knowledge of the co-efficient of friction is entirely sufficient, and both accuracy and simplicity recommend the construction of the angle of friction, through the geometric method, from its tangent == (jl rather than taking it direct from the table. The effi- ciency rj as determined above is, of course, only the efficiency of the rack in its casing. In order to get at the efficiency of the entire jack we must take into account the friction of the gears and of their journals, in the manner hereafter to be explained. For the case of backward motion in the jack, that is, when the load Q is the cause of motion, the same construction holds, with the assumption that the re- actions (.ßj) and (i£ 2 ) are inclined to the opposite side cf the normal by an amount equal tQ the angle <£, and we have the force polygon / 2 3 o v shown in broken and dotted lines in the figure, where the diagonal 2o 1 is parallel to (oj) ( with the horizontal. Taking the points of application of these re-actions at b and c, the centres of the sliding surfaces, we have again, in the line con- necting the point of intersection o l of Q and Ii 1 with the intersection o 2 of P and R v the position of the re- sultants of these two pairs of forces. If, then, we make Io x equal to the total load, and draw I II parallel to R v and o x III parallel to R v and through the point of 30 THE GRAPHICAL STATICS OF MECHANISM. intersection II the vertical II III, we have in the last the necessary driving-force P, and in II /and o 1 III the re-actions R x and H T The force Z =: o x II is the one which acts directly upon the platform. For back- ward motion the necessary brake force (P) is given by the line 3 2 of the diagram 1 2 3 o v From the propor- tions given in the drawing, with a co-efficient of friction ix — 0.16, we have, for Q — P = 100, P = 105.9 ; therefore V = 105.9 = °- 944 ' and (P) = 93.5 ; therefore 0» = ~ = 0-935. It can easily be seen that the friction grows less, and the efficiency increases, as the horizontal distance be- tween the forces P and Q becomes less, and that between the sliding surfaces B and C greater. In practice, there- fore, the height BO should not be taken too small, and the line of force P should be brought as near as possible to centre of gravity A by bending outwards the iron D to which the hoisting-rope is attached. For the same reason, in lifting a weight it should be placed as near the guide EE as possible, while in descending the strain can be partially taken off the brake by placing the weight far away from the guide. The influence which the ratio of the distance between P and Q to the distance between the guiding surfaces SLIDING FRICTION. 81 B and C has upon the relative magnitudes of P and Q can be seen in the simple clamping device shown in Fig. G, plate L, a mechanism commonly used in saw- mills to hold the block upon the carriage. The log is held firmly upon the carriage, or table, simply by the wrought-iron clamp AB which rests upon it and runs loosely up and down the fixed standard DE. If from any cause, as the upward stroke of the saw, a force Q acts vertically and tends to slide the clamp AB upward along the cylindrical standard DE, there is a resultant tendency to rotate on the part of AB which presses the outer edges b and c of the eye hard against the standard. The latter re-acts at these points with the forces R x and i? 2> which, under the supposition of an actual slip of the bushing, act downward at an angle to the hori- zontal, and oppose such slipping motion. The force R x acts from b toward 0, and B 2 from toward . SLIDING FRICTION. 35 Also the re-action (i2 2 ) of the casing which holds in equilibrium the forces Q and (Ü^) must come from the opposite side of the casing in the direction from ((7) to A. Further, the position of the re-action (i2 8 ) of the support H is fixed by the condition that it must pass through (0), the point of intersection of the re-action (iüj) with the force (P), which acts along the line of P, but in an opposite direction. To complete the force polygon for backward motion we draw A 3 parallel to (P), and 2 3 parallel to (i£ 3 ) ; and in the former we have the value of the force (P) necessary to withdraw the wedge from under the bearing. Since this force (P) acts from A toward 3 in the same "sense," or direc- tion, in which the load Q tends to move the wedge it is evident that a backward motion of the wedge cannot result from the action of Q alone, and we must regard the efficiency (jf) = ^— - as being negative. With a value /a = 0.16, and a taper of 1 in 9 for the wedge, we get from the drawing, for Q = 100, P » 11.1, P = 45, and therefore V = ^ = 0.247, also (P) == -19.8, Hence W - TiT = - 1 ' 78 ' It may be remarked here, to avoid repetition, that, as in Fig. 7, the lines of force and the force polygon will 36 THE GRAPHICAL STATICS OF MECHANISM. be drawn in full lines for the forward motion, those for backward motion in broken and dotted lines, and those for the theoretical force P in broken lines. All con- struction lines will be simply dotted. The resistance which the foot-journal of an upright shaft encounters from the plane surface FF (Fig. 8, plate I.) upon which it rests may be deduced by the methods of sliding friction. We may suppose the ele- ments of friction which arise at every point in the sur- face of contact to be concentrated in the circumference of a circle A x A 2 , whose radius A A 2 = §r, r being the radius of the journal A F. If we imagine the load Q to be replaced by two equal forces CA, each equal to | Q, which act at diametrically opposite points A x and A 2 of the circumference, we can replace the re-action of the bearing by two forces R l and R 2 at these points A x and A 2 . These forces will be inclined to the axis AC by an amount equal to <£, the angle of friction, and will lie in planes perpendicular to the plane of the forces ~ passing through A x and A 2 . The horizontal projections A X E X and A 2 B 1 represent the resistances of this species of pivot friction concentrated at A x and A 2 . To over- come these an equal and opposite couple with forces E x A x = DiA 2 = P must be applied at the extremities of the lever-arm A 1 A 2 . When, as is always the case in practice, the driving-force P is applied only at one point the journal will press against one side of the bushing in which it runs on account of its one-sided working, and there will result a certain amount of neck- journal friction beside the pivot friction already taken into account between the end of the shaft and its sup- porting surface. The determination of this neck fric- SLIDING FRICTION. 37 tion corresponds to journal friction, which will be dis- cussed in another chapter. In the same way we can ascertain the amount of friction in the thread of a screw. Suppose SS (Fig. 9, plate I.) to be a screw provided with both right and left handed threads, as occurs in certain forms of coup- ling for railway-cars ; and suppose that each of the nuts M t and M 2 is drawn outward with a force Q = AB; Ave may then obtain the force necessary at the lever N to turn the spindle of the screw in the following way : Each of the two nuts is regarded as acting upon the screw in two diametrically opposite points of a central helix (or pitch line) whose diameter d is equal to the arithmetic mean 1 "t" — 2 of the inner and outer helices 2 of the screw. And letting A represent that point of the first pair, and C that point of the second pair, nearest the observer, we have in the two lines AO and CO drawn at the angle $ from the normals A0 and C0 to the direction of the screw-thread Aa and Cc, the direction of the re-actions at A and C. Now draw from the point of intersection the line OJ parallel to the axis and equal to ^Q. Then in the line KL drawn perpendicular to JO we have the value of the force P which is in equilibrium with the two re-actions of the nuts M x and M 2 against the screw. Since the same construction holds for the two other points diametrically opposite to A and 6 y , it follows that for the turning of the screw a couple of forces, each equal to P = KL, is necessary ; the arm of the couple being twice the radius r of the helix in which the action of the nuts upon the screw is supposed to be concentrated. Without friction we have the force 88 THE GRAPHICAL STATICS OF MECHANISM. P =£ 2f L i the re-actions being assumed in the direc- tion of the normals O A and C. For a co-efficient of friction /x = 0.1, and a pitch of screw n = ^ we have, with %Q = 100, and P ~ 37.8, P = 16.67, ^^ = 0.441. The construction remains the same when the pitch of the two screws is different, as in differential screw-gear- ing ; or when the pitch of one screw equals zero, as in the much-used tension mechanism (Fig. 10, plate I.) where the thread of one screw is merged into a ring and swivel in which end-journal friction only occurs. Here, as before, we make OJ = J$, and have in KL one force of the couple necessary to turn the nut M ; it being understood that the line of re-action OK is drawn at an angle = the angle of friction to the axis of the rod S v Without friction the force P would be given in JL Q . For a pitch n = T V, and a co-efficient of fric- tion ix = 0.1, the construction gives, for \Q = 100, P = 28.9, P s 8.33, and v = ^ = 0.288. The application of the turning-force at one side only of the mechanisms (sketched in Figs. 9 and 10, plate I.) causes a certain amount of neck friction which may be determined by methods to be explained in the following chapter. JOURNAL FRICTION. 39 § 4. — JOURNAL FRICTION. If a cylindrical journal (Fig. 11, plate I.) is pressed bj' a force BE = Q perpendicular to its geometric axis into a bearing A v that bearing re-acts at B with a force equal to Q and opposite in direction, the same as any other supporting surface would do. If the journal revolves in the direction of the arrow a certain force is necessary which does not go through the axis. Let BG be such a force acting at i?, and of such value P that it is just sufficient to hold the journal in equilib- rium, and by the slightest increase to cause a turning. Under this supposition the journal is in equilibrium under the two forces P and Q, and of the re-action R exerted upon it by the bearing. This last re-action must be equal and opposite to OJ, the resultant of P and Q. The bearing A i acts upon the journal at the point If with the force BK == JO. By that we do not mean that the re-action is actually exerted at the point K, but that the resultant of all the re-actions of the elements of the bearing against the journal passes through the point K of its surface. . We have to assume that the resistance to turning offered by the bearing is friction at the point K. This friction is exerted in the direction of motion of the supporting surface, as previously laid down in a general law, and has a value //JV; N being the normal pressure at if, and /x the co-efficient of friction. If we resolve the re-action BK into components at right angles, one normal and 40 THE GRAPHICAL STATICS OF MECHANISM. the other tangent to the surface of contact at K, we have in the component &fiT the normal pressure, and in the tangential component LS the friction al resistance T V at the circumference of the journal. Since, now, — — = J SK fj, — tan (f) we find that the angle LKS or OKA which the re-action makes with the radius drawn to its point of application equals the angle of friction , where r is the radius of the journal. The same value for this arm TA will be obtained in every case wherever the turning-force P is applied ; and if Ave suppose the force F to occupy, one after another, all possible posi- tions about the axis A the re-action of the bearing will in each case be tangent to a circle described about the centre A with a radius AT = p = r sin . We can therefore regard the journal bearing, in its action upon the journal, as entirely replaced by a re- action which is tangent to a circle drawn about A with the radius r sin ; and the direction and situation of this re-action will be known as soon as any other condi- tion is settled, as, for instance, in the present case, that it must pass through the point 0. This circle of a radius p == r sin , which for brevity will be called the Friction circle through analogy to the nomenclature, fric- tion angle, friction cone, etc., oilers a convenient means JOURNAL FRICTION. 41 for the graphic expression of journal friction. Since, under the supposition of an entirely frictionless motion, the re-action of the bearing passes through the centre of the journal, we may regard it as being tangent to the friction circle, which has become equal to zero in this case. It is evident from the figure that the second tangent drawn from to the friction circle, shown in the dotted line OIF, corresponds to a revolution of the journal opposite to the arrow, and that in this case, when the turning-force acts in the direction OB^ the point of application of the bearing against the journal is to be assumed at W. Either If or IF may represent the point of support of the journal, according to circumstances. If we further assume that the journal A is fixed, and that the bearing is acted upon by the forces P and Q like the hub of a wheel running loosely upon an axle, we must then regard the point K A or W x as the point at which the hub is supported by the fixed axle ; the point K 1 corresponding to a left-handed revolution like that indicated by the arrow, and the point W 1 to an opposite revolution. The amount and direction of the re-action R are not affected by these changes, there being merely a transfer of the point of application from if to if x , and from W to W x . On the contrary, if we suppose the force P and Q, and consequently their resultant, to act in the opposite directions OF\ 0H\ and OJ\ the point of support will fall upon K 1 , while the journal will have a revolution in i\ right-handed * direction opposite to the arrow. We * In general right-handed revolution is to be understood hereafter as meaning revolution in the direction of the hands of a watch. 42 THE GRAPHICAL STATICS OF MECHANISM. must determine, therefore, in every case, which of the two tangents is the line of direction for the re-action in that case. This determination is rendered much less difficult here, as in the case of sliding friction, if we keep resolutely before us the principle that each of the two pieces exerts upon the other a re-action which coin- cides in direction with the motion of that piece relative to the other. Even if both parts, journal and bearing, have absolute motion, as is the case in all link connec- tions, it is still not difficult to determine the relative motion of one part with respect to the other. Later remarks will serve for the further elucidation of this law. To determine graphically the friction circle of a jour- nal, we have only to draw any radius AB (Fig. 12, plate I.) of the journal, and lay off the angle of friction := BAC. When this angle is not given directly, but only the co-efficient of friction /x, draw BO perpendicu- lar to AB and equal to /x . AB. Then draw BE through -Z), the point of intersection of the hypothenuse AG with the circumference of the journal, parallel to AB. We have then, in the circle drawn about A tangent to the line DE, the desired friction circle of a radius AE = p = r sin . Rods, or links, provided with two pins, or eyes, con- necting them with other machine parts, often occur in mechanisms. Such a rod, or link, as AB (Fig. 13, plate I.) would, in the absence of friction, simply transmit force from journal to journal along the line AB con- necting their centres. A force applied to either journal, AC iov instance, would call forth in the other, BD, an opposite force with which it would be in equilibrium. Since, when friction is neglected, the pressure of the JOURNAL FRICTION. 43 journal upon its bearing in the rod can only be perpen- dicular to the surface of contact, both these equal and opposing forces must necessarily pass through the cen- tres A and B. This would also occur in reality when there is no motion of the journal relative to its bearing, as, for instance, in the joints of a linked chain which, loaded with a weight, is drawn vertically upwards. When, on the contrary, a turning of the journal rela- tive to its bearing occurs friction enters into the consid- eration of the motion ; and according to preceding prin- ciples force can only be transferred from journal to bearing along lines which are tangent to the friction circle. Therefore, in the present case, a transfer of force between the journals AC and BD can only take place along one of the four possible tangents to both friction circles AE and BF. We can easily determine which of the four tangents is to be regarded as the line of re- action in any particular case by the application of the rule previously given. With it we have only to decide in what direction, whether to the right hand or to the left hand, the turning of the journal occurs in its bear- ing, and in what direction the journal acts upon the rod ; i.e., whether the latter is in tension or compression. The action of the link on a pin or journal must then have such a direction that, in consequence of this action, the eye of the link will assume relatively to the pin the rotation which actually does take place. This rule holds also for the action of the pin on the link ; for not only the direction of the force, but the direction of rotation, will be reversed in this case. In Fig. 13 the four tangents are denoted by the figures 1, 2, 3, and 4, for brevity. For still greater clearness there are shown in Fig. 14, plate I., four sepa- 44 THE GRAPHICAL STATICS OF MECHANISM. rate mechanisms in which a rod AB of the type under discussion unites two swinging levers MA and NB. In these four cases, I., II., III., and IV., the heavy arrow drawn at one lever denotes not only the motion of that one lever, but of the whole system ; the other lever is therefore always a resisting body. By the little arrows drawn at each joint is shown the motion of the rod, or link, relative to that lever with which it is there united. We may regard the journal as forming a part either of the link or the lever, since, as before explained, such assumption has no effect on the direction and amount of the re-action, merely shifting the point of application to another position on the same line. If we suppose the link AB to be in tension in the cases I. and III., and in compression in II. and IV., it will be readily seen that tangents 1, 2, 3, and 4 in Fig. 13 correspond respectively to cases L, IL, III., and IV. in Fig. 14. This correspondence of Fig. 13 to Fig. 14 holds also for the opposite motion on condition that the driving-force is applied at the same lever, for then both the nature of the force acting in the link and the direction of the turning will be reversed. In a similar way we can determine the direction of re-action of a journal upon its bearing in every particu- lar case. This is in reality all that needs explanation or demonstration in the method of calculating journal friction, since the further graphical determinations con- sist simply of the application of well-known principles in regard to the resolution and composition of forces. The method above referred to may be shown in a few examples for the sake of clearness. Let ABC (Fig. 15, plate II.) again represent a bell- crank for which we are to determine the force P that JOURNAL FRICTION. 45 must be applied to the arm B to lift the load acting upon the pin A of the arm CA. Drawing the friction circles for the journals A, B, and C we have, in the tangents ao and bo parallel to the lines of force Q and P, the lines of direction for these forces. Because of the turning of the bell-crank to the right as shown by the arrow, the rod grasping the journal A has a left- handed turning about that journal, and the tangent oa must be drawn, according to the principles previously established, touching that side of the friction circle farthest away from C; so that one might say that the arm of the resistance to be overcome is increased by journal friction. It follows, in the same way, that the line of direction for the re-action of journal B against its rod, which also has a left-handed turning, must be tangent to the inner side of the friction circle, so that the arm of the force P is shortened by journal friction. The re-action of the supporting bearing against the journal C must pass through o, the intersection of oa and ob. There being a turning of the bell-crank to the right hand the re-action R of the bearing to the journal C will lie along the line oe, passing through o and tangent to the friction circle of 0. Since P, P, and Q are in equilibrium we have, by making ol ~ Q and completing the parallelogram ol II III the force P in oIII and in oil, the re-action of the bearing equal and opposite to the journal pressure. Without friction we should obtain, as in Fig. 1, plate I., the force P = III ; i.e., if the direction of the forces passed through the centres of the pins and OI — Q. In the case under consideration, with a co-efficient of friction /* = 0.1, the drawing gives, for Q = 100, P = 91.3, P = 87.8, 46 THE GRAPHICAL STATICS OF MECHANISM. and r, = ^0 = 0.962. For another example we will choose the ordinary slicler-crank train (Fig. 16, plate II.) as used in steam- engines. The downward pull P of the piston is exerted by the piston-rod JTupon the pin A of the connecting- rod. We shall endeavor to determine how great a resistance Q at the distance CE from the shaft can be overcome by this piston force in the position of the mechanism shown by the drawing, friction being taken into account. By the force P we do not represent the entire pressure of steam upon the piston, but simply that really acting upon the cross-head after piston and stuffing-box friction have been deducted. The pressure variations caused by the acceleration and retardation of the piston will also be taken into account in the value of P. The line of this force P must pass through the centre A of the cross-head pin, since this pin is rigidly fixed to the cross-head and piston-rod, and no relative turning can occur between them. There is such turn- ing, however, between the pin A and the end A 2 of the connecting-rod A 2 B 2 . The line of direction of the force S acting in the same will therefore lie along the tangent ab to the friction circles of the two journals A and B. (Which of the four possible tangents is to be here taken is shown by the little arrows which indicate the direc- tion of turning of the connecting-rod ends about the journals A, B, and enable us to apply the principles explained in Figs. 13 and 14, plate I.) The cross-head pin cannot be in equilibrium under the influence of the two forces S and P alone, which have different direc- tions. Equilibrium requires a third force which must JOURNAL FRICTION. 47 be exerted by the guide D in the re-action li 1 . The direction of this re-action is fixed by the condition that it must be inclined at the angle to the normal to the guide, and its situation by the requirement that it must pass through the intersection o x of the forces P and S. Accordingly the guide D exerts upon the cross-head a re-action along the line do 1 . Further, the force S will be transmitted without loss by the connecting-rod from the pin A to crank-pin B along the line ha ; and we find, as in the case of the bell-crank (Fig. 15), that the force iS'and the resistance Q acting at E are held in equilib- rium by a re-action R 2 exerted by the bearing upon the shaft C. We have the direction of the latter in the line o 2 c which is drawn from the intersection o 2 of the forces S and Q tangent to the friction circle of the shaft 0. To determine each force we have only to draw the force polygon in which Io 1 — P,I II is parallel to o { d, II III is parallel to o 2 c, and o 1 177 is parallel to Q or FE, The line o^II ■=. Q gives the resistance acting at E, and II o x gives the tension S in the connecting-rod, while II I represents the re-action of the guides, and II III the pressure upon the bearing of C; the determination of the latter forces being of especial importance in pro- portioning the parts under consideration. For the determination of the theoretical resistance Q = o x III^ it is only necessary to draw the re-action R 1 normal to the guide 7>, and the directions of aS^ and R 2 through the centres of A, 7?, and (7, as is shown in the force polygon drawn in broken lines. With the assumption of a co- efficient of journal friction /x r= 0.1, and sliding friction at the guides /x z= 0.16, the drawing gives, for P = 100, Q = 48.4, Q = 52.5 ; 48 THE GRAPHICAL STATICS OF MECHANISM. therefore v = Q- = 0.922. It is, of course, self-evident that for any other position of the mechanism there would be a different efficiency. If we turn now to the diagram for the oscillating engine (Fig. 17, plate II.) it will be observed that the force S acting in the piston-rod B X D is tangential to both journals B and C\ as in the preceding case, and falls along the line be. The demonstration of this fact is as follows : If P is the force exerted b}^ the steam pressure upon the piston E, and which acts along the line of centres BC, this force must be in equilibrium with the other forces which act upon the piston-rod. There are besides P only three forces to be considered : the re-action >S' of the crank-pin B, the only known con- dition of which is that it must be tangent to the friction circle of i?, and the re-action ll x exerted by the stuffing- box D 1 against the piston-rod at the point rf, together with that, R 2 , of the cylinder against the piston E at the point e. These re-actions are inclined at an angle to the normal to the geometric axis of the cylinder, and act from the cylinder toward the piston and piston-rod respectively. If we now regard the relative motion of cylinder to piston-rod only we may imagine the piston and piston- rod to be held fast, while the cylinder with its bearing moves a short distance along the piston-rod under the pressure of the steam upon the cylinder-head. Here also the various forces acting upon the moving cylinder must be in equilibrium. These forces are the following: First, the pressure of the steam upon the cylinder-head, JOURNAL FRICTION. 49 which is of course equal to the steam pressure upon the piston, and acts along the line of centres CB, but in the opposite direction, and is therefore to be denoted by — P ; next, the re-actions which formerly acted upon the piston and piston-rod at e and d now act at the same points, but in opposite directions, against the cylinder and stuffing-box, and are to be denoted by — R 1 and — i? 2 ; finally, there is the re-action Z exerted by the bearing C x against the trunnion-journal (7, the only known condition of this re-action being that its line of direction must be tangent to the friction circle of 0. Since, then, the condition of equilibrium exists between the four forces P, R v R v and S\ and also between — P, — R v — P 2 , and Z, it is evident that P, R v and R 2 will balance — P, — R v and — R 2 respectively, thus leaving S = — Z; i.e., the forces /S r and Z are equal and act in opposite directions. The force &\ therefore, must coincide with the tangent cb to both friction circles. For the condition of equilibrium between the forces P, R v R 2 , and S, it is necessary that the resultant of any two, as P and P x , shall be equal and opposite to the resultant of the other two, S and R 2 . By joining o v the intersection of P and P 1 , with o 2 , the intersection of # and P 2 , we get in o l o 2 the direction of these result- ants. If we make C I = P, and draw C II parallel to JK 1 , ///parallel to o 2 o v /////parallel to Jc, and I III parallel to P 2 , the line II III gives us the value of the force S which tends to pull the crank-pin around. Of equal value, as has been demonstrated above, is the re- action which the bearing C\ exerts upon the journal C of the trunnion. The further determination of the acting forces is carried out in the usual manner. If a resistance Q acts at the radius AF from the shaft we 50 THE GRAPHICAL STATICS OF MECHANISM. have in the line o s a drawn from the intersection o s of Q and S tangent to the friction circle of A the direction of the re-action i? 3 exerted by the bearing upon the shaft journal A ; and to determine the value of this re- action and of the resistance Q we have only to resolve II III into U IV and III IV parallel to the directions o s a and o s F of the re-action and resistance Q respec- tively. Without friction we should have P = S, and the triangle CI IV would give immediately the value of Q = I IVq, CI being resolved in the directions of Q and OA. For a co-efficient of journal friction fx = 0.1, and of friction in the stuffing-box and cylinder a = 0.16, the drawing gives, for P = 100, Q = 61.0, Q = 64.2, and = -£ Q, 0.950. It should be remarked that the friction here consid- ered as existing between the piston and cylinder, and between piston-rod and stuffing-box, is only that arising from one-sided or lateral pressure. The friction winch is caused by the pressure of piston and stuffing-box packing must be estimated in other ways, and deducted directly from the piston pressure.* We have, therefore, * The " one-sided or lateral forces " to which the author here refers are those resulting from the non-coincidence of the opposing forces S an 1 P. They arc. in other words, the re-actions which keep the piston-rod and cylinder in the line BC, their tendency being to arrange themselves along the line of tension be. If Ave imagine the piston, piston-rod, and cylinder to be made of some elastic material which would bend at the slightest application of a deflecting force, they would so arrange themselves when the steam pressure P was applied that JOURNAL FRICTION. 51 in the value of y\ given by the figure, not the efficiency of the steam-engine, but only the efficiency ot those parts composing the mechanism. The same slider-crank motion which lies at the basis of the oscillating engine finds frequent application in planing-machines for the production of a quick return motion. Fig. 18, plate IL, represents an arrangement of this kind, where the slide HGc bearing the tool M is moved back and forth in a prismatic guide by the link F 1 F 1 which receives its reciprocating motion from the lever D l E l oscillating about the fixed journal D. The oscillation of the latter is produced by the crank AB, whose crank-pin B works in the block C sliding along the slot C v of the oscillating lever F X I) V The resistance Q offered by the material to the cutting-tool M produces the re-actions M 1 and li 2 at the points g and h ; and these three forces, Q, M v and Ii v must be in equilibrium with the force 8 acting in the reciprocating link E X F V The the line passing through the centres of piston and cylinder-head would coincide with the line be of tension between trunnion and crank-pin. There is another class of "one-sided or lateral forces" which the author has omitted to mention; namely, those arising from the oscilla- tions of the cylinder. Taking the case when the crank is upon either dead-point, with the engine running at a high rate of speed, Ave have the entire mass of the cylinder, by no means small in practice, in rapid motion in one direction; passing to the other dead-point we find it in equally rapid motion in the opposite direction. Between the two the inertia of this mass moving at this rate of speed has twice to be over- come byre-actions of the nature of Tt 1 and R>, but vastly greater than II C and I III. The resultant of these re-actions applied at the crank- pin is of the nature of those accelerations and retardations to which reference is made in the first chapter, and have no effect on the work done, since the work stored up in the quadrant on one side of the dead- point is given out in the quadrant on the other side; but the friction at the points e and a, or their opposites, represents a loss of energy never given back again. — Trans. 52 THE GRAPHICAL STATICS OF MECHANISM. direction of the force S is given by the tangent ef to the friction circles of E and F. If we therefore connect o x , the intersection of Q and _B X , with o 2 , the intersec- tion of S and i2 2 , we have in the side II III of the force polygon o x I II III, the value of the force S acting in the link B 1 F 1 , under the supposition that o x I ' = Q, 1 II is drawn parallel to B x , II III parallel to ef, and o x III parallel to B 2 . The crank-pin B acts upon the sliding bearing C, and through it upon the guide C 1 of the oscillating lever, in a direction o 3 c which must be tangent to the friction circle of B and inclined at the angle to the normal to the slot C v The line of this re-action T is therefore o s c, and similarly the line do s drawn from the intersection o s of T and S tangent to the friction circle of D is the direction along which the journal D re-acts against the lever B l B 1 . If, further, the driving-force P is applied at the end if of the radius AH from the shaft A we can get the direction of the re-action exerted upon the crank-shaft A by its bearing, in the line o^a drawn tangent to the friction circle of A from 6> 4 , the intersection of T and P. Completing the force polygon by resolving the force # = II III into II IV Mid IV III, parallel to o%c and o s d, and drawing IT ["parallel to o 4 a, and IV ^parallel to o 4 ff, we have in V IV =. P the force which must be applied at H to overcome the resistance Q = o x I at 31. Assuming a co-efficient of journal friction /x = 0.1, and of sliding friction ^ ^= 0.16, the construction gives, for Q = 100, P = 84.1, P = 66.6, and v = 5) z= 0.792. JOURNAL FRICTION. 53 In the case of a steam-engine with single beam (Fig. 19, plate II.) the investigation is pursued as follows: The steam pressure P acting upwards upon the piston-rod AB coincides with the geometric axis of the cylinder, and passes through the centre of the jour- nal A. On the other hand is the thrust S of the rod, i.e., the re-action of the beam upon A, which must lie along a line tangential to the friction circle at A. These two forces, S and P, cannot be in equilibrium, since they do not act along the same line. For this a third force is necessary, which is found in the re-action Il 1 of the stuffing-box against the piston-rod. By drawing through a middle point b in the stuffing-box the direc- tion of M 1 at an angle with the normal we have, in the intersection o x of R i with P, the point through which the thrust S must pass; this force then acts along the line o x a. If we make o Y I ' = P, and draw through I a parallel to P x , we have in o x II the force S exerted upon the journal A of the beam. It may not be uninteresting to remark that the stuffing-box P a is exposed, according to the above demonstration, to a certain side thrust II I This side thrust is not the result of an inaccuracy in the parallel motion, as is the case in approximate motions like that of Watt, for the Evans motion here represented is well known for its absolute accuracy. This side thrust is directly the result of the journal friction occurring at A. For a clearer understanding of this fact we may imagine the piston-rod to be bent to the left by a force equal to the journal friction of the beam applied at its upper end, and tending to revolve toward the right. Such a force would evidently call forth in the stuffing-box the re- action which we have been considering. It is also 54 THE GRAPHICAL STATICS OF MECHANISM. evident that the side thrust R x will act in an opposite direction and from the other side of the stuffing-box when the piston makes a down-stroke, since at each change of motion the direction of the journal friction at A is reversed. On the other hand, the piston-rod thrust or pull S in upward and downward stroke remains always in the line of the tangent o x a, since at each reversal of motion, i.e., at each dead-point, the direction of turning of the journal A, as well as the direction of the force S, is reversed. We may therefore regard the effect of journal friction at A as being in every case to diminish the lever-arm of the force S. Beside the force S three other forces act upon the beam ; they are the re-actions of the radius-rod DE, of the connecting-rod FG, and of the guides K X K V We find the directions of the force L in the radius-rod, and T in the connecting- rod, according to previous rules, in tangents to the friction circles at D and E, and at F and G. Recol- lecting that the radius-rod is under compression while the connecting-rod is in tension, and noticing the direc- tion of turning at each journal as indicated by the arrows, we decide upon cle and fg as being the desired tangents. The sliding-block C x exerts upon the journal C a re-action which must be tangent to the friction circle, and upon the guide K X K 2 a re-action which must be inclined at an angle $ to the normal ; these two con- ditions fix the position of R 2 as coinciding with the line he. In Fig. 19a the sliding-block C x is shown in detail. We first observe that during up-stroke the lower guide I^K.2 exerts the re-action, and during down-stroke the upper guide K^K^ is under pressure. In further ana- lyzing the re-actions exerted upon the sliding-block O x we must remember that it has a double reciprocating JOURNAL FRICTION. 55 motion ; that is, it makes a complete stroke out and back for every half-stroke of the piston. Beginning then at the lower dead-point with the beam in the posi- tion CA 1 we have, during the first half of the up-stroke, a motion of C x toward the right, and right-handed turn- ing of the journal (7, which gives k 1 c 1 as the direction of the re-action R 2 . When the beam reaches the posi- tion CA 2 the radius-rod DU is parallel with it, and any further motion will, by the action of the radius-rod DE, cause G x to move to the left. During the second half of the up-stroke, therefore, k 2 c 2 is the direction of the re-action R 2 . At the beginning of the down-stroke, when the beam is in the position CA^ the sliding-block C x again reverses its motion and moves to the right; but it now presses against the upper guide K x f K 2 \ and the direction of turning at the journal has also been reversed, and the re-action lies along the line from Jc s to c 2 or Jc 2 . Similarly, during the last half of the down- stroke, there is sliding toward the left, and left-handed turning of the journal ; so that k 4 c v or k v is the line of re-action. In every case, therefore, the re-action is so applied as to lengthen its lever-arm and render the force #less effective, which is entirely in keeping with the obstructive action of friction. In order that the four forces S, L, T, and R 2 acting upon the beam shall be in equilibrium the resultant of the two forces L and R 2 intersecting at o 2 , and the resultant of the two forces T and S intersecting in , yör, and ao- will intersect in one and the same point : but the intersection of ßr (the line of the force T~) and ao- (the line of the force >S r ) is the "desired point o 3 . There- fore the line o 2 w gives the desired direction o 2 o s . The point w can easily be located by drawing any line, aS, making a8 equal to 1 . Taking these re-actions at e and d inclined at the angle (/> to the normal we have again, in the line joining the point of intersection o x of B Y and Q with that o 2 of R 2 and S, the direction of the resultants of these pairs of forces. If we therefore make o x I — Q, and draw / II parallel with o x o^ and II III parallel to be, and I III parallel with R 2 , we have in III II the thrust aS' exerted in the eccentric-rod. If, further, the motion of the shaft A is caused by a force P applied at the end F of the lever-arm AF we have again in ao« the direction of the re-action B s exerted by the bearing against the shaft A ; and by resolving III II into III IV and II IV parallel to ao z and P we get in IV lithe value of the force P which must be applied 58 THE GRAPHICAL STATICS OF MECHANISM. to overcome the resistance Q. For /* == 0.1 and /x = 0.16 the drawing gives, for Q ~ 100, P = 36.2, P = 22.8, and v = ^ = 0.630, a small value for 77, due to the large radius of the friction circle of the eccentric B. An interesting application of the crank-train is found in the ordinary Blake crusher shown in Fig. 21, plate III. The crank-rod BC is here connected with two links DE and FG, forming a knee, by the pressure of which the materials fed in at L are crushed, through the intervening plate JH swinging from the centre II. It is evident that as turning occurs at both ends of each link the pressure can only be transmitted through them along the tangents de and fg to the friction circles. Furthermore, the force S of the crank-rod or pitman which is tangent to the friction circle at B must also pass through the intersection o 2 of the forces jPand T x ; i.e., act along the line o 2 b. And also the re-action R x against the journal H of the swing-plate must pass through o v the intersection of the thrusting-force T in the link DE with the working-resistance Q. If the motion of the crank-shaft A is caused by a force P applied at iT wo have, according to well-known methods, the re-action of the bearing against the shaft in the line o 3 a tangent to the friction circle of A. Then draw the force-polygon as follows: Make o 1 I equal to Q, the crushing-resistance of the material; draw / II parallel to öjA, and IIo l is the thrust T sustained by the link With a co-efficient /x = 0.1 Q = 100, P = WA, P and JOURNAL FRICTION. 59 BE. In III II we have the thrust T x in Fa, if III II is drawn parallel to fy, and öjZZZ parallel to 6o 2 . Finally, by resolving IIIo 1 = #, the strain in the pitman, into Oj/F parallel to o 8 a, and III IV parallel to P, we have in III IV the value of the force P which must be applied at K to crush the material at L. The broken lines again indicate the construction by which the theo- retic force P = HI Q IV is obtained. the drawing gives, for = 12.3, rj = ^ = 0.80. Another example of toggle-joint mechanism is the hand-punch shown in Fig. 22, plate III. Two bent levers, A 1 CD l and B 1 CE 1 , are here connected by a hinge or bolt at (7, so that when their ends D t and E x approach each other under the action of the screw FGr the head HJis forced down, and the punch L forces the metal under it through the die K. To produce this result the screw FGr is turned by a long wrench applied at its square head F, and its right and left threads draw the nuts D 2 and F 2 together with a certain force P, as was the case in the coupling shown in Fig. 9, plate I. On account of the turning of the levers the nuts are connected to them by the journals D and E, from which it follows that the force P by which they are drawn toward each other acts along the tangent de to the two friction circles of D and E. If we now suppose one of the levers (the under one, for instance) to be at rest it must be in equilibrium under the forces acting 60 THE GRAPHICAL STATICS OF MECHANISM. upon it. The forces consist of the driving-pressure P acting along de, and the two re-actions R 1 at and R 2 at B. For the first re-action R 1 exerted by the upper lever A 1 CD X we have the direction ac tangent to the two friction circles of A and G 7 , the lever A 1 CD l turning toward the right hand ; while the direction of R 2 must pass through o v the intersection of P and R x , and be tangent to the friction circle of B. It lies, there- fore, along the line o x b. If, then, w T e make o x I = P, and resolve it parallel to the two re-actions by drawing I II parallel to o x b, we have in IIo x the re-action JR 1 exerted by the upper lever upon B x OE x , and in I II the re-action R 2 offered by the journal B. The head HI must, in its turn, be in equilibrium under the influence of the force R 2 acting along bo x , the resistance Q which the metal offers to punching acting along the axis of the punch Z, and the two re- actions R s and i2 4 of the guide-bushing H^^ If we assume h and i at a sufficient distance (say 5 mm., or \ inch) from the edges as the points of application for the re-actions Ii s and R 4 which are drawn at the angle of friction <£ to the normal, the line connecting o 2 , the intersection of JR S and R 2 , with o 3 , the intersection of 72 4 and $, furnishes us with the means of completing the force polygon in the usual way. Resolving II I — R 2 into III I parallel to R s and II III parallel to o 2 3 , and then II III — o 2 o s into III IV parallel to R ± and II IV parallel to Q, we have in IV II = Q the resistance which can be overcome by the application of the force J? upon the nuts D 1 and JE V It is also evident that the side IV Q II of the force polygon drawn in broken lines normal to the surfaces, and passing through the journal centres, is the value of the theoretic' resist- JOURNAL FRICTION. 61 anee # which should be overcome by the same force P. With the usual values /* = 0.1 and /x x = 0.16, the draw- ing gives, for P = 100, Q = 194, Q = 350, and V = -^ = 0.554. Vo It will be seen that the ratio of P to # would remain the same whether we use the arrangement shown, or whether we suppose the force P to act only upon one arm B l U 1 ^ while the nut D 2 is replaced by a cylindrical eye and spindle so arranged that no sliding can occur in the direction of the axis of the spindle, as in the swivel, Fig. 10, plate I. The ratio between P and Q would not be changed, because if the driving-force P was applied only at JE X there would arise at i) 1 an opposite equal re-action which would be transmitted by the swivel-ring to D. The only difference between the two arrangements is that by the movement of both levers the space traversed by the head HJ for any given portion of a revolution of the screw FGr is double that which would result if only one lever moved while the other was held fast. The work done by the turning- force for this portion of a revolution is, of course, twice as great in one case as in the other. The above remarks apply exactly only in the case of frictionless motion ; for with the arrangement giving a re-action — P of a iixed point, as in the swivel, a new friction enters which must be determined in the same way as in Fig. 10, plate I. The investigation of the present mechanism has not included the determination of resistances arising in the 62 THE GRAPHICAL STATICS OF MECHANISM. screw ; such a determination would be made in the manner shown in Fig. 9, plate I. The amount of resistance Q which can be overcome by a certain force P, as well as the efficiency of the mechanism where knee-joints are employed, depends upon the angle which the centre lines of the links form- ing the joint make one with another. It will be readily seen that this resistance Q becomes greater, and the efficiency ?] smaller, as this angle approaches 180 de- grees. If we assume this last value, or one which differs from it by an infinitely small amount, as in Fig. 23, plate III., a force P acting upon the journal C would be able, in the absence of friction, to overcome an infinite resistance M=^w =cc )- On account of journal friction, however, these lines of force are to be found in the tangents ca and cb to the friction circles ; and we therefore find the actual forces in the parallelogram acblii we let cl = P. The greatest resistance Q which can be overcome is therefore given by the equation Q = P tan ^#, where 2w is the obtuse angle acb of the two directions of pressure. The efficiency rj = -¥- in this case where Q = oo is equal to zero. If we further suppose tlie knee to be in the condition of backward motion, i.e., if we assume that the tendency of the re-actions at ^1 and B is to force the joint C out JOURNAL FRICTION. 63 to one side or the other, it is evident that such backward motion can only begin at that instant in which these re-actions coincide in one straight line. If, therefore, we draw such a case (Fig. 24, plate III.) in which the tan- gents ac and be to the friction circles fall upon the same straight line we have the limiting position at which the knee is self-locking. The angle 2(w) = ACB, which differs from the angle 2w in Fig. 23 only by an inappre- ciable amount, determines on both sides the limiting position within which a backward motion, i.e., an open- ing of the press by the re-action offered by the material within its jaws, is impossible. As Qtv) depends on the proportions of links assumed, the limits within which the mechanism is locked become greater as the distance from journal to journal becomes smaller, and as the radius of the journals increases. The knowledge of these proportions is of special importance in the design- ing of mechanisms in which the knee-joint is employed to grip an object and hold it fast, as in certain forms of vise. The methods heretofore employed in determining the efficiency of machines can also serve the purpose of determining friction as applied to useful ends in many machines and processes. Thus friction serves to produce the necessary tension in all spinning-machinery, and is employed also in sewing-machines and water-frames or throstles. Let Fig. 25, plate III., represent the ordinary spindle with the Arkwright flyer 00 which rotates with the rapidly moving spindle. The thread F passing through the stationary glass eye at D with a certain velocity u, and leading to the loose spool L after several turns about the arm of the flyer, serves as a driver to the 64 THE GRAPHICAL STATICS OF MECHANISM. spool, which is caused by the thread to revolve in the same direction as the spindle and the flyer. The spool holds back on account of the frictional resistance offered to it, and at each instant the portion of thread running out is unwound by this difference between spool revolu- tion and flyer revolution. The friction of the spool by which the tension in the thread is determined occurs principally in two places, — at the circumference of the spindle as journal friction, and as pivot friction where the under surface of spool at Gr rests upon the bobbin- frame EE which slowly rises and falls. This friction must attain a certain value in order that the tension of the thread shall be sufficient for a certain amount of twist, and in order that the thread shall not belly out between B and I) under the influence of centrifugal force, and become entangled with the thread of the neighboring spindle. The tension S of the thread may be determined as follows: If 6r = HA is the weight of the spool with the quantity of yarn already upon it, it produces friction upon a ring-shaped portion of the surface of the bobbin-frame EE. We may therefore suppose the bobbin-frame EE to be replaced by re- actions which are uniformly distributed over an average circle of contact whose diameter is EE* all these re- actions making the angle <£ with the normal in the direction prescribed by the motion. As was shown in * The radius p of the circumference of contact is = where r, and r 2 represent respectively the radii of the outer and inner circumference of the supporting surface. JOURNAL FRICTION. 65 the case of pivot friction (Fig. 8, plate I.) we can here suppose all re-actions to be concentrated in two diamet- rically opposite points, a\ and a 2 . From the parallelo- gram AJHK we get in AJ and AK the re-actions exerted by the bobbin-frame, and in their horizontal components a x i and a 2 k the frictional resistance offered by these re-actions to the revolution of the spool. If the thread draws the spool in the direction fa at a cer- tain moment with a tension S the spindle re-acts with equal force along a line ab drawn parallel to the direc- tion fc, and tangent to the friction circle of the spindle. The question then is of the equilibrium of the spool under the influence of the two frictions a x i and a 2 k, the tension S, and the re-action of the spindle along the line ab. Joining the intersection o^ of the thread- tension S with the friction a 2 k, with that o 2 of the friction a x i with the spindle re-action, we have the ten- sion $ given in value by the line IIo 2 if Io 2 = a x i and I II is drawn parallel to o x o 2 . So far in these investigations it has been tacitly assumed that the journal friction was only exerted upon one bearing. This is never the case in practice. Every shaft has at least two supporting points or bear- ings, and at these the forces P and Q will call forth certain pressures and re-actions of a value proportional to the distance of the point of application of the forces from the bearings. If we suppose the shaft supported by the bearings A t and A 2 (Fig. 26, plate III.) to encounter a resistance Q acting through a wheel or pulley placed at C, with a radius q — AG, we can resolve this fpvce into two components parallel to Q, and having the same lever-arm q, lying in planes passing through the points A x and A v and normal to the shaft. 66 THE GRAPHICAL STATICS OF MECHANISM. The forces are determined by the well-known rela- tions Qi = q aa~ and Q * = q ja: 12 12 and can easily be found by construction. If the same construction is carried through for the determination of the driving-force acting at each bear- ing we have in P x and P 2 the forces which, lying in the planes through A x and A 2 , act upon the parallel lever- arms AH = p to overcome the resistances Q x and Q 2 . It follows that under the supposition of equal journal radii and equal co-efficients of friction, i.e., with equal friction circles, at A x and A 2 these forces P 1 and P 2 must be in the same ratio one to another as Q Y to Q 2 ; so that if P x and P 2 were compounded in one result- ant P it would have to lie in the same plane, passing through 6 y , in which Q acts. Therefore we should obtain by this method the same value of P which has hereto- fore been determined directly from Q by the employment of friction circles. It follows from the above that the preceding construc- tions for the determination of journal friction can be entirely accurate only when the driving-force P lies in a plane perpendicular to the same axis as that to which the plane of the resistance Q is perpendicular, and when the diameters of the journals are equal. In reality the first condition is seldom fulfilled, and the journals also are seldom of the same size in a shaft. It therefore remains to bring this influence within the scope of cal- culation by force polygons. With this object in view let us suppose that the driving-force P (Fig. 26, plate III.) which is to over- JOURNAL FRICTION. 67 come the resistance Q is applied at P, a point outside the bearings A 1 and J. 2 , as is frequently the case in practice. It is further assumed that the resistance Q at C acts with a lever-arm A Cr = q, and the force P with a lever-arm AH = p at the point P. If no friction came into account we could assume P Q and Q as work- ing in the same plane, and get p = Ol p in the well-known way by uniting the intersection o t with centre A, and then resolving o x Q into this direction and that of P. Then imagine Q to be resolved into the forces Q x = Q~r~ 2 ~i~ ac ting at A v and Q 2 = Q * A 2 A 1 A^A 2 acting at ^4 2 . Then lay off these forces equal to o 1 Q 1 and o 1 Q 2 in the direction of Q. In the same way P can be resolved into two parallel forces, P x and P 2 , acting at the points A x and A 2 with the lever-arm AH — p. Their values are given by the equations Pi = p °f 2 f and p « = p »f a f- ^l 1 yi 2 ^1^2 Since P acts outside of the supporting points ^L x and J. 2 , P 1 and P 2 act in opposite directions o 1 P 1 and 0^2 along the line of P. We find the resultant of P 1 and Q x in the diagonal o 1 Z>, and this force acting in the plane through A x will call forth an equal and opposite re-action of the bearing A v This re-action R x does not act along the same line o 1 D as the resultant, how- ever, since among the forces acting at A x alone equilib- rium does not exist. The position of R x is found by 68 THE GRAPHICAL STATICS OF MECHANISM. drawing the line a x a x tangent to the friction circle at A x and parallel to o x D. Similarly we get the journal pressure at A 2 in the diagonal o x E, and in the line a. 2 a 2 parallel to o x E and tangent to thp friction circle at A 2 , the position of the re-action R 2 which the bearing A 2 exerts. We now see that the shaft is in equilibrium under the couple o x D and R v and the couple o x E and R 2 . Since we may suppose the couples to be slid along the axis until they lie in the same plane we can immedi- ately find P by uniting o x and the intersection o 2 of R x and R 2 , and resolving the resistance o^ Q parallel to the diagonal o x o 2 and the direction of P. We have therefore in o l P x the necessary turning-force P. In Fig. 27, plate III., the friction circles of A x and A 2 are drawn to a larger scale, and we see that the direction of R coincides nearly to the direction of a line drawn through o x tangent to a mean friction circle shown in dotted lines. We can therefore employ this simple con- struction with sufficient accuracy in the generality of cases, and especially in those where both P and Q fall within two bearings not far apart. But for exact deter- mination, and in cases where a force is applied outside of the bearing, the full construction is necessary. It will be noticed that the latter has still a slight inaccu- racy, since the component forces P x and P 2 are obtained from P instead of P. The error is quite inappreciable, however, and a correction unnecessary; though such cor- rection could be obtained by determining P x and P 2 anew from the value of P as deduced, and repeating the construction. We can now determine all the friction al resistances which occur in a screw. In Fig. 28 a , plate IV., S 2 S S is the direction of a helix at a mean distance from the JOURNAL FRICTION. 69 axis. S r s x and S v s 2 are the directions of re-action at two diametrically opposite points of the helix, so drawn as to make the angle BS l s 1 = BS x s 2 = $ -f- a with the axis of the screw, a being the pitch-angle of the screw. We then obtain the resistance q x = CB acting perpendicularly to the axis at each of these points by making BA X = Q, the load upon the screw, and draw- ing through A x and B the lines A X B and DB parallel respectively to S l s l and S x s 2 , and intersecting at B. The load Q also presses the nut M down upon the standard, and produces friction against the ring-shaped surface of contact A 2 A S . As pivot journal friction we can suppose this concentrated at a mean circumference A 2 A S , and acting at two diametrically opposite points. If, therefore, we draw the corresponding lines of re- action A l a 1 and A x a 2 inclined at the angle of friction to the axis A 1 B we have in CE = q 2 the amount of friction at each of the supporting points in the ring- shaped surface A 2 A S if the line BE is drawn parallel to A 1 a 2 . We can imagine these couples q x q x and q 2 q 2 as acting at the points bj) 2 (Fig. 28 6 , plate IV.) and c x c 2 respectively, and by compounding them get the resultant couple d l e 1 — q s and d 2 e 2 = q s . If the nut M is revolved by worm-gearing, we must represent the worm TFas acting upon it along the line w x w 2 with a force w applied at a mean helix of the worm. This one-sided working of the force w presses the nut M up against the sides of the standard K where the latter is bored out above A 2 A S , and calls forth a re-action parallel to w 2 w x and tangent to the 70 THE GRAPHICAL STATICS OF MECHANISM. friction circle of M in mm. To find w we join o x arid 2 , the points of intersection of the couple q s q s with m and w respectively, and making o 2 e = , a l being the pitch of the screw on the worm. The worm W in its turn is forced by the load to against its bearing i, and friction results along the mean circumference of the ring L 1 L r This friction can be supposed to act at two opposite points of this circumference, as at A 2 and A s in the case of the screw S. We have, therefore, a resisting couple TJ = p 2 (Fig. 28 c ) where TJ is obtained by drawing FJ and OJ through and F at the angle to OF. Finally, the load w upon the worm will also call forth frictional resistances at the bearings of the two neck- journals L and iV, since the one-sided action of w thrusts the journals against their bearings with a certain press- ure, the value of which, equal and opposite in the two cases, we have from the equation of moments w . W l w 1 = p s . iiV, JOURNAL FRICTION. 71 from which p s : w : : W x ii\ : LN : : LK X : K X K V From the last proportion we see that we can get p s by drawing through F (Fig. 28 c ) a parallel to LK 2 (Fig. 28 6 ), when OiT will represent the value of p s , the neck re-action exerted at L and N.* In order to determine the force P to be applied at a crank W s to the worm W, we must first unite the one- sided resistance p x and the two couples p 2 p% an d p^Pz- This can be done in the following way : First draw the two tangents l x and l 2 to the friction circle of the jour- * Throughout this discussion the author has assume! that the thrust w of the worm is applied along the line w x w, parallel to its axis. This is neither true in practice, nor does it correspond to that case shown in the figure. The thread of a rack is not square, but has the sides of its profile inclined at an angle of 75 degrees to the axis. The normal to the surfaces of contact would then be inclined at an angle of 15 degrees to the axis. The line of re-action would fall short of this inclination by an amount equal to — 5° 43' with a co-efficient ß = 0.10, so that its actual direction would be inclined 0° 17' to the axis. The re-action Oiin of the nut would be parallel to it, and the shape of the parallelogram o 2 ejf would be changed, and a different value for of = iv obtained. But since that component of this real value of to parallel to the axis WW would only differ from o 2 /by an inappreciable amount, which increases and decreases with the radius of the friction circle at A, and since the component perpendicular to the axis merely adds to the throat-friction at L an amount which it takes from that at N (its tendency being, of course, to thrust the worm bodily over to the right), the accuracy of the final result is practically unaffected. Moreover, as the gear may work in either direction, and as in the case of backward motion the condition of affairs would be exactly the reverse of that pointed out in the first part of this note, it is probable that the author assumed the average position, i.e., the one parallel to the axis, in order to have a general discussion applicable to all cases. — Trans. 72 THE GRAPHICAL STATICS OF MECHANISM. nal L representing the direction of re-action of p%, and then at a distance equal to the radius of the mean helix of the worm draw the vertical tangent tv s . Now lay- off from 6> 3 , the intersection of w s and Z x , the distance oJ: x = p x (= FH'm Fig. 28 c ) and o^K l = p 3 (= OBT in Fig. 28 c ) ; then the diagonal o 3 x gives the resultant of the forces p x and jp 3 , which latter acts along the tangent ? x . This resultant, when compounded with the other force 2h acting along l 2 in the opposite direction, gives a force passing through o 4 parallel and equal to o^i x = _p r We see from this that the influence of the two frictions p z produced by the neck-journal re-actions only causes the resistance to turning of the worm to act in the same direction and with equal force, but at a longer lever-arm, since it passes through o> 4 instead of o 3 . This corre- sponds to the well-known principle that the composition of a force and a couple merely effects a parallel shifting of the force unchanged in value. In the same way, by uniting the force p x going through # 4 with the couple P2P2 w hich corresponds to the friction offered by the ling-shapecl supporting surface L X L 2 (Fig. 28 6 ), we ob- tain merely a shifting of the force p x always parallel with itself from o 4 to o Q . It is done as follows : Draw the two forces p 2 p 2 (TJ in Fig. 29 c ) as two parallel tangents ? 3 / 4 to a circumference of the diameter L 1 L 2 . Then lay off from the point of intersection o~ of one of these tangents with the force p x now passing through o 4 the distance o b i = p 2 and o b h 2 = p v and the diag- onal o 5 y cuts the second tangent Z 4 in the point o 6 through with the force p x must pass. The further construction is familiar. Through the intersection o 1 of the resistance p x with the driving« force P the re-actions of the journal-bearings L and N must pass. This re- JOURNAL FRICTION. 73 action R is therefore acting along the line o 7 l 5 drawn tangent to the friction circle at L. If we then make o 7 ä 3 = p t = 0.1 for journal and thread friction, the construction gives, with a pitch n — tan a = Jg of the screw & for (> = 100, iv = 67, w = 11.8, and ri t —^ = 0.176, w where to is the force to be applied at the pitch-line of the worm. With a pitch n = tan a t = ^ f or the worm, for w = 67, P = 12.56, P = 4.3, and , 2 = 5q = 0.342. For the efficiency of the entire jack we have then t; :zz rj 1 rj 2 = 0.060. In the same way we can determine the efficiency for backward motion by finding the force (P), which must be exerted in the same " sense," or direction, as Q to produce a sinking cf the load.* * To fix the method to be followed in every case firmly in mind, it miy not be amiss to briefly sketch here the general aspect of the 74 THE GRAPHICAL STATICS OF MECHANISM. problem encountered in all these examples, giving the known quanti- ties in every case, and the way in which they are to be combined to determine the unknown, so that the student, in attempting to solve an outside problem, will know just what he has to work with, and just how to set about that work. First, by means of the friction angle and friction circle we can always draw the direction of the forces transmitted longitudinally through links joining two turning pairs, acting at sliding surfaces, or in links joining a sliding and a turning pair. Sometimes in the last case the exact position of the force is fixed, after its general direction has been determined by the angle of friction, by the requirement that it shall pass through the intersection of two others, instead of being tangent to a friction circle as in the simplest case. An example of this is the ordinary cross-head, Fig. 2, plate I. The line of the re-acting force at a lever or crank or bell- crank bearing is found by drawing a line from the intersection of the two other forces acting upon the lever, crank, or bell-crank tangent to the friction circle at its journal. The directions of P and Q are always given as the force of gravity, a piston-thrust, etc. We then have given, or can determine by these elementary methods, the direction of all the forces in any problem. We also have given the intensity of either P or 0. With these data the problem is solved as follows: Draw the force- polygon of all the forces acting upon the same piece as the known force, and dependent upon it. These can only be three in number if there is circular motion of the piece to which it is applied, and four if there is right-line motion. In the first case it is a question of drawing a triangle of forces, knowing the directions of all and the amount of one. In the second case combine the forces two and two, join the points of intersection thus obtained, getting in the line thus drawn the direction of the common resultant of the two pairs. Then resolve the known force in the direction of that force with which it is paired, and of the common resultant which here represents the com- bined effect of the other two forces. Having thus obtained the value of the resultant resolve it in the direction of the two forces making up the second pair, and all the forces are known in direction and amount. One of these becomes the known force acting on the next link in the mechanism, and by repeating the process all the forces acting throughout the machine may be determined. Whenever there are more than four forces acting on one piece they will be of such nature that they can be reduced to four; and generally where the limit is exceeded, as in the condensing beam-engine, and still further in the compound condensing beam-engine, it will be the JOURNAL FRICTION. 75 result of a compounding of several chains of mechanism, and in each simple chain either P' or Q' will be given, so that its resultant action on the common link can be determined, and combined with the P or Q of the main chain, their resultant action being regarded as one force. In the case of the condensing beam-engine the resistance of the air- pump (/would be known, and, as shown in plate VII., combined with the thrust of the piston to get the resultant force acting upon the beam. In the compound engine, P', the steam-pressure in the second cylinder, would be known. — Trans. 7G THE GRAPHICAL STATICS OF MECHANISM. §5.— ROLLING FRICTION, The resistance which is opposed to the rolling of a cylinder along a smooth path is of such small value that it may be left out of account in most cases in comparison with sliding and journal friction. We are accustomed, when it is taken into consideration, to assume it propor- tional to the pressure Q with which the roller is forced down upon the bearing surface, and inversely propor- tional to the radius r of the roller. For rollers and surfaces of iron and hard wood the formula P = 0.02^ will generally give the resistance, r being expressed in inches. If r is expressed in millimetres the formula becomes P = 0.5^. In order to get a graphic representation of this resist- ance let A (Fig. 29, plate IV.) be the centre of a cross- section of a cylinder, with radius AB ~ r, which is supported at B by a horizontal track. Let the load resulting from its own weight, and acting at the axis A, be represented by the vertical line AC — Q. To cause d rolling of the cylinder a horizontal force P = c . ~ must be applied at the axis A, AI) representing the ROLLING FRICTION. 77 intensity of this force. If we assume, as heretofore, the ' limiting condition of equilibrium for which the slightest increase of P will cause motion, the cylinder must be in equilibrium under the influence of the exterior forces P and Q, and of the re-action R offered by the surface GG. This is only possible if the re- action R is equal to the resultant of P and Q, and acts along the same line in the opposite direction. The surface GG then re-acts upon the cylinder with a force whose direction and value are given by the line PA. The cylinder, therefore, will remain at rest as long as the surface GG re-acts upon it along any line inclined to the normal at a less angle than that of PA as was the case in sliding friction. The plane surface then opposes the same character of resistance to the motion of the cylinder as in sliding friction, with this difference, how- ever, that while in the case of sliding friction the great- est possible deflection angle of the re-action depends only upon the nature of the material, and is constant for a given material, being, of course, the angle of friction for the same, in the case of rolling friction it depends both on the material and on the form, i.e., upon the size of the cylinder. From the expression given for the resistance to rolling r it will be readily seen that the co-efficient € is capable of geometric representation, since it follows that Q : r : : P : «, giving us directly in the figure BF = .e; which is, in 78 THE GRAPHICAL STATICS OF MECHANISM. other words, the greatest possible distance between the point of application F of the re-action and the theoretic point of contact B of roller and track. While in the case of sliding friction we have a constant angle for the friction angle <£, in the case of rolling friction we must deal with a linear value € which, for the same material, remains the same for rollers of all sizes. If it was of sufficient interest we could follow out the parallel still farther, showing that the friction cone in the case of sliding friction corresponds to the wedge-shaped space whose cross-section is FAF V whose edge is the axis A, and whose sides, shown in projection at FA and F X A, cut the supporting surface at the distance c to each side of the perpendicular AB* It follows that with P acting in the opposite direction the track would re-act from the other side of AB in the direction F 1 A. We can make this connection clear if we assume that in reality the roller is not supported on a line passing through B parallel to the axis, but upon a surface of a width e from each side of the normal AB, produced by a flattening of the roller and a corresponding indenta- tion of the supporting surface under the pressure of the load Q. This view corresponds also to the assumption of a fixed fulcrum for the roller at the constant distance € from the normal plane (that is, at the point F), and this should be kept in mind during the following discussion. The value of e, according to the above, is from 0.02 to 0.03 in inches, or from 0.5 to 0.75 in millimetres, for metals and hard wood. In the case of yielding materi- * For the connection between sliding and rolling friction see O. Reynolds, Philosophical Transactions, vol. 1C0, and Zeitschrift des Vereins deutscher Ingenieure, Jahrg., 1877, S., 417. ROLLING FRICTION. 79 als, as, for instance, carriage roads, the value of € is much greater, and in all such cases the correct value must be estimated. If a body KK (Fig. 30, plate IV.) rests upon a roller A which rolls along the horizontal track GG rolling friction occurs at both KK and GG. If Ave therefore draw through the centre A of the cross-section of the roller the normal DB to the two surfaces, we have, according to what precedes, the point of support of the fixed track in F at the distance BF = e from B, and also in E at the distance DE =z e from D the point at which the downward pressure of the moving body KK acts. If, therefore, EC := Q denotes the load upon the roller A, and P denotes the force acting in the horizontal plane KK necessary to move the body, these forces P and Q must be in equilibrium with the re-action R exerted by the track GG through the roller A upon the body KK, which re-action of courses takes the direction FE. We then have in the side EJoi the parallelogram ECHJ the necessary force P to produce motion. Equally well can be determined the force P (Fig. 31, plate IV.) necessary to move the load Q upon the wagon- wheel AB, by means of the value BF = e for rolling friction upon GG, and the friction circle of the journal AE upon which the load Q rests. The direction of the re-action B of the track GG against the axle-bearing of the wagon is along the line Fa drawn from F tan- gent to the friction circle of A, Therefore, by making AC = Q, and drawing through C a parallel CD to Fa, we get the driving-force P = AD. 80 THE GRAPHICAL STATICS OF MECHANISM. The investigation is practically the same when the track GG for the wheel has any desired inclination to the horizon, as in Fig. 32, plate IV. If we here draw the line AB through the centre A of the axis, and perpen- dicular to the track GG, make BF = e, and draw through .J 7 the tangent Fa to the friction circle of A, we have in this tangent the direction of re-action of the track GG against the axle-bearing. If, then, we make CA = Q, draw through C a parallel to the line of re- action R, and through A a parallel to the line of P, we get in AE the value of driving-force necessary, P = AE = DC. Without hurtful resistances the re-action of the track would lie along the normal BD , and we have in D C the theoretical driving-force P . With the grade or inclination of 1 in 3 for the track the drawing gives, for Q ad 100, P = 34.8, P = 33.3, and v = £j = 0.957. In the manner shown the resistance to running-gear of every description, upon both horizontal and inclined tracks, can be easily determined. As a further example we may take the roller-bearing for a swinging crane (Fig. 33, plate IV.). In this case the cylindrical sur- face of the stationary post or mast A serves as a track for the rollers B and C united to the brace L. If we connect the centres B and C with A, and make FI) z= GE = e, the lines Db and Sc drawn through B and E ROLLING FRICTION. bl tangent to the friction circles of B and G 7 , and intersect- ing one another in o v will give the directions of the re-actions R t and R 2 of the mast. If, now, a turning of the brace is produced by a force P acting in the direction HJ, and intersecting the line of pressure Q of the boom in o 2 , we have only to connect o x and o 2 , make Io 2 = Q, and draw through / a parallel to o x o 2 in the well-known way. We have thus determined in o 2 H the necessary turning-force P. In order to get the re- actions R 1 and R 2 of the mast against the rollers, resolve the resultant I //parallel to o x b and o x c^ and we have R x = III I and R 2 = II III A later example will show in what way the re-action in the bearing at the upper part of the mast is to be con- sidered. As already remarked rolling friction in most cases of mechanism is quite inappreciable as compared with other hinderances. At first thought it may not be clear how the geometric diagrams (Figs. 29 and 30, plate IV.) will always give the value FB constant and equal to e for any one material, whatever the load or size of roller, as stated on p. 78. The following analysis will render it clear that such a result does follow from the assumed relation r and the supposition that the roller is always a solid homogeneous cylinder. FAB being the angle made by the re-action R to the normal AB we have (1) tan FAB = - = — . From the second value we get (2) FB ^ r tan FAB. 82 THE GRAPHICAL STATICS OF MECHANISM. There are three ways in which the conditions may change : First, the size and consequent weight of the roller may remain constant, but a varying superimposed load acting upon a surface as KK (Fig. 30) may be applied. If in this case Q' represents the sum of the varying load and the constant weight of the roller we have, from the fundamental relation, r and, from equation (2), F'B = r tun F' AB, F' AB being the angle the new re-action R' makes with the normal. By supposition both e and r are constant, and therefore P' varies directly as Q' ; or, to put it in another form, F _. e P Q' r Q Evidently tan F 1 AB = — = - = tan FAB, Q' Q and therefore F'B = rtan FAB = rtan FAB = FB. That is, F'B, the distance of the intersection of R' with line GG measured from the foot of the normal, is equal to the value first obtained, FB or c. In the second case there is no superimposed load, but the roller varies in size and consequently in weight. The weight will vary as the cross-section of the cylinder, and r will vary as the square root of the cross-section or weight. If Q" is the varying weight we have O" P" P"= e.M-, tun F" AB = — , Q" 7 and F'B = r" tan F" AB. If m is the ratio of Q" to Q we have r" = \/mr y and Q" = mQ. ROLLING FRICTION. 83 Substituting these values in the three equations above, P" *=..J*SL=*-fa* to>F»AB = e sJTn r r — r^m 7)1 Q F"B - rSjm . ~= = e. r\m In this case also there is no variation from the original value e or FB. Thirdly, where both superimposed load and size of roller vary we have Q! = mQ, the former variable quantity, and Q" = m'Q, the latter ; while Q» = Q' + Q", their sum. As before r'" - r\[m\ and Q'" = mQ -f m'Q. Substituting in the three equations we have D ,„ Q'" wzQ + w?'Q Q (m + m') r'" r £ tan F'" AB = - Q (m + m') r ' \jm' £ Q(m + m') r\Jm! and F ,„ B _ rV /- £ the same result as in previous cases. This may seem like begging the question, since if £ is assumed to be a constant, and a certain line FB is found once to be its graphic equivalent, this intercept FB must always remain the same ; but the analysis may be of use in showing how the diagram adapts itself to this requirement under all conditions. — Trans. 84 THE GRAPHICAL STATICS OF MECHANISM. § 6. — CHAIN FRICTION. When a chain is wound on to or off of a drum or pulley there occur certain hurtful resistances on account of the change of direction in the links, which resistances may be determined in exactly the same manner as jour- nal friction. Let A (Fig. 34, plate IV.) be the journal of a chain-pulley whose radius AB = AD is represented by a. At the left side a weight Q is attached to the chain BC; and we are to ascertain the force P which must be applied to the other portion DE in order to cause a revolution of the pulley in the direction of the arrow, and a lifting of the weight Q. If we neglect friction of the journal A, it is evident that on account of the equality between the lever-arms AB and AD the forces P and Q must also be equal for the condition of equilibrium if there were no hurtful resistances at the points B and D where the chain winds on to and off of the drum, as would be the case if we suppose the chain replaced by an infinitely fine thread of perfect pliability. In this case there would be no slipping of the strands of the thread over one another, since the thread has no appreciable thickness ; and therefore the causes of fric- tion would be wanting, because the latter can only occur, as remarked in the introduction, where there is motion of two elements relative to one another. But as the links of the chain have a certain thickness, rela- tive motion will occur at the point of connection of two links at the instant of winding on or off, at the point B CHAIN FRICTION. 85 or D respectively, and this motion must be regarded as a turning. If we imagine the link B X B to be in motion from C to jB, it has evidently no relative motion toward the preceding link B 2 B S in which it hangs as long as the latter rises in the same straight line CB. At that instant when the preceding link B 2 B S , adjusting itself to the circumference of the pulley, begins to share in the latter's motion, there occurs relative motion between the links B 1 B and B. 2 B^ which is, as shown by the arrow in the figure, a left-handed turning of the link BB X about the link B 2 B S . In this turning the end of the link B X B serves as a journal, and the eye or loop of B 2 B S as a bearing. It is now clear, according to the laws of journal friction, that these links can only act upon one another along the tangents to the friction circles of these journals. Since the portion of chain i?G 7 is subjected to tensile strain we have the direction of this re-action in the tangent he which touches the friction circle at B on the opposite side from the centre J.. In other words, the lever-arm of the weight Q is increased through chain friction by an amount equal to the radius x of the friction circles at the chain joints. In the same way the link D 2 D Z about leaving the pulley at the point D on the other side undergoes left-handed revolution about the link DD X still upon the pulley, as shown by the arrow. Therefore the force P in the portion ED 2 of the chain will act along the tangent de to the friction circle at D. In other words, the lever- arm of the driving-force P is shortened through chain friction by an amount equal to x-> the radius of the friction circles. The two forces P and Q acting verti- cally downwards must be in equilibrium witli the re- action R offered by the bearing A i to the journal A. 86 THE GRAPHICAL STATICS OF MECHANISM. This re-action on account of journal friction can only act tangent to the friction circle of A, and on the same side as P. The investigation is now resolved, therefore, into the determination of two parallel forces P and Q, which have such relative values that their distances from the resultant lying between them shall be a — p — x anc l a + P +X respectively ; in which expressions p is the radius of the friction circle for the journal A, and x the radius of friction circles of the chain. We have, according to this, p = Q a + P + * This value can be readily constructed by drawing at any point the horizontal line (?5, making GrK — Q, drawing the horizontal line KJ, and then a line through J and L to M. We then have in MG = NH the necessary driving-force P, and in KM the re-action of the bearing R — P -j- Q. Since the links rub one another while in a dry condi- tion we assume a co-efficient of friction /x = 0.2. With this assumption, and that of /* = 0.1 at the journal, the figure gives, for Q = 100, P = 105; and since P = Q = 100 we have, for the fixed pulley, v = £) = 0.952. CHAIN FRICTION. 87 The investigation is the same if the directions of the chains are not parallel one to another. For the guide- pulley ABO (Fig. 35, plate IV.) we draw first the medial lines 00 and OB of the chain, and then the lines ob and oc along which the tension of the chain acts. We then have in the line ao drawn through the intersection c>, and tangent to the friction circle at A, the direction of the re-action R of the bearing. So that by making ol equal to Q, and drawing I II parallel to oc, we get III=P and oIIzzzB. For /x =: 0.1 and /x 2 = 0.2, the figure gives, for Q = P = 100, P = 104.4 and ^^^^ 0.958. In the same way may be determined the friction of an idler, or guide-pulley, which serves merely to support the chain and keep it from sagging (Fig. 36, plate IV.), and is often so employed in cranes. Draw the lines BE and BF (shown in dotted lines) along which the chain rolls off and on to the pulley, and draw parallel to them, at a distance equal to x-> the radius of friction for the chain, the lines of tension oc and od. Then the re-action of the journal A is given by the tangent oa to its friction circle passing through the point of intersec- tion 'o. By making ol equal to the resistance Z 1 acting in the portion BD of the chain, and drawing through I the parallel I II to oc, we have in I II the force Z 2 which is transmitted to the portion BO of the chain. In the case of the loose pulley (Fig. 37, plate IV.) upon whose journal the load Q hangs, and where one 88 THE GRAPHICAL STATICS OF MECHANISM. end of the chain is fastened at -Z), we have to determine the vertical force P acting at the other end of the chain CE. To do this we draw^ the directions of tension bd and ce parallel to the chain, and at the distance x of the friction radius of the chain, and also the vertical tangent ag to the friction circle at A Then draw through the centre of the journal, or at any other convenient place, the horizontal line BAC, make AG = Q, draw through Gi the line b^ parallel to BC, join B with c x , and draw through H the line JK parallel to BC We now have, as will be readily seen, in e x K = b x J the tension Z of the fast end of the chain, and in KG the force acting in the portion CE of the chain. By means of friction circles for journal and chain, whose radii will be denoted as heretofore by p and x respectively, the proportions of the various forces in all kinds of block and tackle and pulley gearing can be easily determined, as a few examples will show. Fig. 38, plate V., represents an ordinary block and tackle with two blocks, within each of which are three pulleys of equal size, and ranged side by side on the bolts A and B. When, by raising the load, the pulleys are turned in the direction shown by the arrows, and the chains wind on at E and D, and off at F and (7, it is evident that the pull of the load hanging on the hook üTacts upon the journal B of the lower block along the vertical tangent o 2 b to the friction circle at J5, while the re-action transmitted by the support Gr to the jour- nal of A acts along the tangent o x a to its friction circle. The forces of tension in the chains also will act at a distance x, the radius of friction of the chain, nearer to the centre of the pulleys at F and C, and at a distance increased by the same amount at E and D. Since the CHAIN FRICTION. 89 lower block B swings free, and opposes no resistance to side motion, it will shift its position a distance 2\ to the left when motion begins, so that the line of tension in the chain shall be vertical ; otherwise equilibrium could not exist. In the figure such a side movement of the block B is supposed to have taken place, so that the cen- tres A and B do not lie in the same vertical line, as would be the case when they are at rest. It is equally evident that with an opposite motion (that is, with a sinking of the load) a corresponding shifting of the block B to the opposite side must occur. We now denote the tension in the separate portions of chain by Z v Z 2 . . . Z v in such way that Z x is the tension of the first portion which hangs from the stationary block A and winds on to the first pulley of the block B at E, while Z 1 is the force which is to be applied to the free end of the rope to raise the load Q. It is then evident from foregoing principles that the relation existing between the tension of each portion of the chain and that of the next following is Z n {r + p -f x) = Z„ +1 (r - p - x) if r is the distance from the centre of the chain to the centre of the pulley. If, now, we draw at any convenient point a horizontal line which cuts the directions ce and fd of chain tension at J and K, and the directions of journal re-action at o l and o 2 , we know from the figure that JK — 2r, Jo { == o 2 K = r — p — x, and Jo 2 = o x K = r + p + x- 90 THE GRAPHICAL STATICS OF MECHANISM. From this follows immediately the construction given below for the determination of Z*. or P v Make J I equal the tension of the first portion Z x , draw through Zand o 1 the line cutting KD in II; K His then the ten- sion of the second portion Z 2 . From II draw through o 2 the line cutting off the distance J III, which is the tension in Z s . In the same way the lines IIIo^IV, IVo 2 V, Vo x VI, and VIo 2 VII give the tensions Z, = KIV, Z, = J V Z, = K VI, and z 1 = j vn. The load to be lifted is given by the equation Q = Z, + Z 2 + Z 3 + Z 4 + Z 5 + Z 6 , while Z 1 is the force P to be applied to the free end of the rope in order to lift it. In the figure the sum of the tensions from Z x to Z Q is shown by NL, and MO = Z 7 is the force P. By the construction here chosen we start with a value of Z v while in reality Z x is yet unknown, since only Q is given; but the method lends itself with equal ease to the solution under the latter condition. "We can assume Z x = J I oi any convenient length, and get, in the manner shown, NL = Z, + Z 2 + Z 3 + Z 4 + Z 5 + Z 6 and MO = Z r CHAIN FRICTION. 91 We then find from the given value of Q, and the pro- portion between NL and MO, the force v ivr which in reality only amounts to the assumption of a particular scale of force. On the contrary, a construc- tion direct from the value of Q would be unnecessarily tedious. This construction also holds for the backward motion of the tackle if only we regard the force P applied at the end of the rope to prevent any accelerated motion of the load Q as the tension in the first portion of the chain, and every following tension as increasing in the ratio — "*" p ~*~ . Then regarding J I as (P) or r — P — X Z v we have by the same construction the tension Z Q , Z h . . . Z v each one of which is greater than its prede- cessor, so that the tension Z x of the chain attached to the stationary block A is now the largest. Here also Q = Z, + Z 2 + Z s + Z 4 + Z 5 + Z fl , and Z, = (P). In the figure the sum of the series Z^ to Z 6 is shown by SB, and (i>) = Z 1 by TU. For forward motion, and Z x = 100, the figure gives § = 682.7, P = 134.6,- 92 THE GRAPHICAL STATICS OF MECHANISM. and since P = — = 113.8 v = 5d = 0.845. For backward motion, for Z 7 = (P) = 100, £ = 717.4; and since P = — = 119.6 (,) = iEi. = 0.837. ^0 If the pullej^s in the blocks are not of equal size, and therefore the ropes are not parallel, as when the pulleys are arranged on different centres in the block, the above determination for parallel ropes will in general have sufficient exactness on account of the slight divergence from parallelism in the case under consideration. An absolutely correct determination can be made, however, by taking into account the points of intersection of the ropes produced. An example in which this is done is shown in Fig. 40, plate V. We will next take up the differential pulley (Fig. 39, plate V.). Here also the load Q, hanging upon the hook ÜT, acts in the direction of the vertical tangent bit to the friction circle of B, while the re-action of the support G- to the journal A lies along the tangent ag. The direction of tension in the portion Z x of chain unwind- ing from the smaller pulley at 6\ and winding on to the loose pulley at L\ h again given in direction by ce, as is CHAIN FRICTION. 93 also the direction of tension in the portion Z 2 winding on and off at D and F respectively, by the linefd. The two tensions Z x and Z v which in the ordinary differ- ential pulley can be assumed as parallel, stand in the following relation one to another : ZJr - p - x) = Z^r + p + x) where r is the radius of the pulley JEF, and Q = Z x + Z 2 . If we then draw through any convenient point o of fd the line oM perpendicular to fd, and make MI = #, we have in the intersection b 2 of the line ol with the direction b x b of the load Q a point which will give us the proportion of Z l to Z 2 ; for, drawing through b 2 the horizontal b%N, we have the proportion bfa : Nil : ob x : b x M : : (r - p - X ) : (r + R + x) 5 from which we get b x b 2 = MN= Z x and N I = Z v These two forces 2\ and Z 2 must be in equilibrium with the tension Z z or force P applied to the free end of the chain JK along the line iK, and the re-action R given forth against the journal A along the line aj. We can most easily find the condition of equilibrium for these parallel forces through the drawing of an equi- 94 THE GRAPHICAL STATICS OF MECHANISM. librium polygon.* For this purpose let us regard o as the pole of the force polygon MNI, then draw through £ the line a£ parallel to oM, and ßt, parallel to oN\ then draw through ß the line ßti parallel to ol, and we have in the closing line a8 of the polygon the direction of the polar ray o II, which gives us in I II the driving-force P applied at i, and in II If the re-action of the journal A against the pulley. f The theoretic driving-force P can be determined by a similar construction, or more easily from the relation % p = Q^nA = q AD - al v 2E 1 V DJ In order to determine the proportion between the. forces for backward motion we have only to remember * For a more complete explanation of the principles here employed see Fart I. of The Elements of Graphic Statics by Karl Yon Ott. — Trans. t The figure &fö« is the funicular or equilibrium polygon, and it will be readily seen that the forces P, Z,, Z 2 , and the re-action at A acting upon the vertices r, C, /3, and a respectively of the polygon would keep it in equilibrium. — Trans. t This is, of course, derived as follows: Without friction *, = *, = $ so that the equation of moments about A becomes PoB, + { hi 2 = S ßl transposing P oRl = Q(^i ~ J* 2 ) or Po = qäi^zÄ2 t — Trans. . CHAIN FRICTION. 95 that the chain tensions and journal re-actions coincide with the broken lines in the figure. For backward motion also Q = (Z,) + (Z a ), but now (Z 2 )0 + P + x) = (Z x )(r - p - x) ; therefore the tension (Z{) of the chain CE is exactly equal to the tension Z 2 of the chain FD for forward motion, and also (Z 2 ) = z v If, then, we make MQN) = NI in the force polygon, and draw the polar ray o(N), we can construct the funicular polygon (a) (£)(/?) (8) for backward motion by drawing (")(£) parallel to' oM, (ß)(0 parallel to o(N), and 08)(8) parallel to oL Then (a)(8) is the closing line of the funicular polygon, and the polar ray o(II) drawn parallel to this gives us the force which must be applied at J during backward motion. Since this force is here acting upwards it is self-evident that the tackle cannot of itself commence backward motion, but that it is self-locking. The figure gives, for fi - 0.1 for journal friction, and fx x = 0.2 for chain friction, with a ratio - 1 = and for Q = 100, R l== AD _ 10 R'AL' 9 ' P = 12.8 and (P) = -2.7. 96 THE GRAPHICAL STATICS OF MECHANISM. Since P z= 5 we have — ^ = 0.391 and (y) = ^ = -0.54. F P As another example we wall investigate the arrange- ment of pulleys often employed in hydraulic lifting machinery.* In this case the chain upon which the load Q hangs is first led over guide-pulleys supported by the roof timbers at A and B (Fig. 40, plate V.), from which it hangs down in a loop containing the loose pulley (7, and then, after passing around the fixed pulley i>, comes back and is attached to the journal of C. Under the supposition of parallelism between the chains this arrangement would cause, for any distance trav- elled by (7, an elevation of the load Q through exactly three times that distance. The downward motion of is produced by aid of a second chain, one end of which is fastened to the journal of (7, and the other, after pass- ing around the loose pulley E and the fixed pulley F, returns again to the journal of E. The movement of the journal E is produced by the aid of the vertical piston-rod of a hydraulic cylinder, not shown in the drawing as not entering into the calculation. With the supposition again of parallelism between the ropes a downward motion of the piston w r ould cause the pulley C to traverse three times as great a distance, and the load Q to be raised by an amount equal to nine times the piston travel. Without hurtful resistances, therefore, the piston force F would equal 9$; and it is * See Weisbach, Ingenieur- und Maschinenmechanik, III., Theil; also, Rühlman, Allgemeine Maschinenlehre, IV., Bd. CHAIN FRICTION. 97 evident that this arrangement would find its only appli- cation where, as in hydraulic apparatus, the travel of the driving-force P is necessarily small, while the force itself is of great power. The proportion given above for travel of power and load under the assumption of parallelism would in reality vary but little from the actual result. On account of the non-existence of complete parallelism we will, however, follow out the investigation, taking this inclination of the chains one to another into account for the sake of showing the general method. The direction of the forces Z v Z 2 . . . Z 7 acting in the separate portions of the chain can easily be deter- mined by increasing the lever-arm by the radius x of chain friction at every point where the chain winds on to a pulley, as at A x , B x , C\, D v E^ and F v and by diminishing it by the same amount wherever the chain unwinds from a pulley, as at A 2 , B 2 . . . F 2 . The direction of re-action R 1 of the bearing of the fixed pulley A is given by the line o x a drawn through o x tangent to the friction circle at A. Its value is ascer- tained by making 01 = Q, and drawing through a parallel to o x a, and through I a parallel to o x o 2 . We have, then, III=Z V the tension in the chain A 2 B V In the same way the bearing re-acts against the journal B of the second guide-pulley along the line bo 2 ; therefore a resolution of the tension Z x = I II in the direction of bo 2 and o 2 o s will give in I III the tension Z 2 in the portion B 2 C\ of the chain, and in III II the re-action R 2 of the bearing at B. There are now acting upon the loose pulley C 98 THE GRAPHICAL STATICS OF MECHANISM. four forces ; namely, the tension Z 2 of the chain in the direction o 3 o 2 , that Z z of the portion G 2 D l in the direc- tion o 5 d v that Z± of the end of the chain in the direction o 5 d 2 , and finally the tension Z 5 of the second chain in the direction ce x . The lines of the two last, Z 4 and Z h , in ust evidently be tangent to the friction circle at C. Z and Zo intersect in o 3 , Z 4 and Z. in o 4 . The line OoO 3 V 4 joining the two is therefore the line of direction for the resultant of Z 2 and Z 3 as well as of Z± and Z 5 . If we then draw through I the line /.ZF" parallel to o s o±, and through ZZ7 the line III IV parallel to 2 d^ we have III IV = Z^ and in IV I the resultant of Z± and Z 5 . By resolving this resultant IV I into IV V parallel to d 2 o 5 , and / V parallel to ce v we get in VI V the tension Z 4 between C and D 2 , and in / Fthe force Z 5 with which the second chain pulls down upon the journal of the loose pulley C. We also have the re-action of the bearing against the journal D of the fixed pulley in V III, the resultant of Z s and Z 4 . We also know that the four forces Z 5 , Z 6 , Z 7 , and P acting upon the loose pulley E must be in equilibrium. The force Z b acts along the line ce v that of Z 6 along f 1 e 2 - Their intersection is at o Q . The force of tension in Z 7 and the piston-force P are tangential to the fric- tion circle of E. These last two forces intersect in o v The line o Q o 7 is then the direction of the resultant of Z 6 and Z & and that of Z 7 and P. We therefore draw through /the line I VI parallel to o Q o 7 , and V VI par- allel to f x e 2 , to get in VI V "the tension Z 6 of the portion E 2 F X of the chain; while VI I represents the resultant CHAIN FRICTION. 99 of Z x and P 8 . Revolving this force VI /parallel to f 2 o s or Z v and parallel to P, we have in VI VII the tension Z 7 , and in 2" VII the force P acting upon the piston-rod. The re-action of the bearing upon F is given by VII V, the resultant of Z 6 and Z v In apparatus of this kind the directions of the ten- sions in the chains vary but slightly one from another, so that their points of intersection o often fall without the limits of the drawing. This difficulty can be sur- mounted by the employment of the method used in Fig. 19, plate IL, in which the direction of the lines is obtained without having their points of intersection located. We must proceed, according to this method, when the points <9 3 , o 5 , o-, and ö 8 , fall beyond the limits of the drawing. When the position of the chains approaches parallelism there is the difficulty also that the point of intersection of two such lines cannot be determined with any degree of exactness. A sufficient degree of accuracy can, however, be obtained with the aid of the following construction : If the force Z 2 = I III is to be resolved into the directions III IV parallel to J? 3 , and I IV parallel to # 3 6> 4 , we can imagine this system of forces to be acted upon by two opposite and equal forces aß and a l ß l along the line C X C V which would not disturb the equilibrium. Let la represent a/3, and Ilia be the resultant of this force, and z 2 = mi Compound also the opposite force a 1 ß 1 with the yet unknown force Z 3 , and there will be another resultant whose direction my be determined. For when the resultant of Z 2 and aß is compounded with that of Z z 100 THE GRAPHICAL STATICS OF MECHANISM. and a 1 ^8 1 they will give a resultant which must coincide with o s o±. If, therefore, we draw through a a parallel to Ilia, to the point of intersection w with o s o^ the resultant of Z 3 and a 1 ß 1 must also pass through w, and is given by <*>a 1 . So that by resolving the force Ilia parallel to o s o^ and wa 1 , by drawing the triangle IIIm 1 ^ we get the point a 1 from which the point IVcslu be accurately determined by drawing through a x a line parallel to a 1 ß 1 intersecting a line from / parallel to o 3 6> 4 . The auxiliary forces aß and a^ may be chosen of any convenient value, but should be assumed so that the lines which are to determine the desired point by their intersection should be nearly at right angles one to another. In the same way the point of intersection VI can be determined by the application of two opposite and equal forces along the line E X E^ Instead of resolving the force Z 5 = V I directly parallel to o Q o 7 and Z 6 , the force F"S is substituted for Z b and resolved in the direc- tions o Q o 7 and wßy In this and similar ways we can employ upon every diagram constructions in which the lines will diverge sufficiently to determine accurately the points of inter- section. For Q ~ 100 the figure gives P = 1295; and since, with the assumption of parallelism, the theo- retic force I J = 900 ~ 9$, we have 7) = :5j} = 0.G95. CHAIN FRICTION. 101 In this value of the efficiency no account is taken of the resistances within the hydraulic cylinder. These latter must be determined in each case by a special investigation. It is evident, moreover, that the value of the piston and stuffing-box friction must be added to the force P already found to get the necessary pressure to be exerted upon the piston by the water from the accumulator. Any calculation for backward motion would have to take into consideration the weight of the chains, loose pulleys, and piston ; and we would get in the force which would have to be applied at the free end of the chain A x the counter-weight, which in this species of hoisting-gear is attached to the hook in order to render the backward motion automatic. In some cases the weight of the platform or cage is sufficient of itself to do this. 102 THE GRAPHICAL STATICS OE MECHANISM. § 7. — STIFFNESS OF ROPES. When pliable ropes and cords are wound on and off a pulley or drum there are certain hurtful resistances called forth, in part by friction between the strands of the rope produced by bending it, and in part by the resistance of these fibres to the expansion and contrac- tion which they are compelled to undergo. We can bring this resistance into the calculation in the same manner as chain friction by assuming that, in conse- quence of it, the lever-arm of the load is increased at the point where the rope winds on, and that lever-arm of the power where the rope runs off the pulley is decreased by a similar amount. This value has to be determined by a special investigation, and is generally expressed by an empirical formula. By such investi- gation it has been proven that this resistance is propor- tional to the tension in the rope, that it is inversely proportional to the radius of the pulley, and that it increases with the thickness d of the rope, not in the simple ratio, but as some higher power. For hemp ropes we assume that the resistance increases as the square of the thickness, and is given by the equation d 2 S = k-Z r where Z is the tension in the rope, and k is a constant co-efficient. STIFFNESS OF ROPES. 103 That the resistance due to stiffness of ropes really causes a lengthening and shortening of the lever-arms of the load Q and the force P respectively can be proven as follows : Let A (Fig. 41, plate V.) be the centre of a pulley, and AB = AC = r be the radius of the same extended to the centre of the rope. A load Q hanging on the rope at D produces a certain tension in the ele- ments of any section of the rope which w r e majr assume as equally distributed over the cross-section, so that the resultant of all these elementary tensions acts in a result- ant passing through the centre of the cross-section, and coinciding with the geometrical axis BD of any portion of the rope. In the condition of rest, therefore, as long as there is no turning of the pulley, the rope BD will arrange itself in such a position that the line of force Q will pass through the centre M of the section at B. If, however, we suppose an exterior force to be applied to the pulley which tends to cause a revolution of the same in the direction of the arrow, and a winding-up of the rope at B, there will be a bending of the rope at that point, and only the strands in the neutral plane M 1 31 2 will retain their original length, while those lying in the outer semicircle M X 3I 2 will be stretched, and those in the other semicircle 3I 1 31 2 J will be shortened by a force of compression. Each strand in the half 031 will now receive, beside the strain due to Q, a certain elastic tension o- which is proportional to the distance of that strand from the neutral plane 31 X 3I 2 . We may suppose all the elementary tensions in the half-section M 1 3I 2 to be combined in one resultant s which shall be applied at some point a. In the same way each strand of the inner half-section M 1 M 2 J will experience a certain compression which will also be proportional to 104 THE GRAPHICAL STATICS OF MECHANISM, its distance from the neutral plane M X M^ and if these are combined in one resultant we have a certain press- ure p acting at some point ß. In the bending of rigid bodies, as beams, it is known that the assumption s = p is made. It is not necessary here, however. We know from the preceding consideration that the rope BD upon which the exterior force Q acts at D is under the influ- ence of three forces at the section B: namely, the ten- sion Z acting upwards at the centre B, the force 6* also acting upwards at a, and that. p. acting downward at /?. These three forces must have a resultant which is equal and opposite to the force Q. It will be readily ^een that the resultant S of Z, *, and p which is to be equal to Q must be at a greater distance from A than that at which the centre 31 of the rope section is ; and this fact is proven when we combine Z and g, and then com- pound their resultant with p. From the relative posi- tions of B and a the resultant of Z and s must lie at a greater distance from the axis A than the radius r of the pulley, and the composition of this resultant with the opposite pressure p gives the point of application b of the final resultant S still farther off. For all pur- poses of the following demonstrations it is sufficient to represent the distance IB simply by o- which evidently corresponds to the value x used in chain friction. A demonstration similar to the preceding would show that friction shortens the lever-arm of the force acting in the portion CE of the rope. In that case also there would be acting in the section at C : First, a tension Z x opposite to P, and beside that a force s 1 acting upward at some point 8 of the inner semicircle X l N. 2 K, and the force p l acting downward at some point 8 of the outer semicircle. It follows, then, that by the unwinding of STIFFNESS OF ROPES. 105 the rope from the pulley at C the strands of the inner half-section N^^K are stretched, and those of the outer half N l N 2 L are pressed together. Therefore the resultant aS^ of Z v s v and p v equal to P, must have its point of application c between C and A, and the lever-arm of the acting-force is reduced by an amount Go = o-. If we suppose the ends of the rope BD and CE to swing free with weights suspended from them the weights will be shifted to one side or the other, accord- ing to the direction of revolution, by an amount a- as was shown for a similar case in the chain pulley. In every case of rope pulleys where p represents the radius of the friction circle at the journal the equation Q(r + P + cr) = P (r - P - cr) is true, which becomes the same as that for chain friction when x is substituted for o-. The investigation of all rope-gearing proceeds, there- fore, in the same way as for chain friction, and it only remains to get a graphic equivalent for the quantity o-. The way in which this value o- is to be applied can easily be determined in every case where the direction is known in which the forces act by remembering that the lever- arm of the unwinding rope is always shortened, and that of the winding-on rope always lengthened by this amount cr. If, for instance, a load Q (Fig. 42, plate V. ) hanging from the drum CITis to be raised by a revolu- tion of the pulley A we have the directions cle,fy, and hh for rope tensions at once. 106 THE GRAPHICAL STATICS OF MECHANISM. We have now to obtain a graphic representation for a- from some of the empiric formulae for the stiffness of ropes. Of the different formulae * the one most gener- ally used in practice is that of Eytelwein, which is simple in form and sufficiently accurate, at least in the case of hempen ropes and with large forces. This for- mula will then be assumed as the basis of our calcula- tions. It should be remarked, however, that in special cases, as with wire ropes, other formulae should be used, from which the value of o- can be determined in the same way as from Eytelwein's formula. According to this formula the entire resistance S due to stiffness of the rope at both winding-on and unwind- ing points is given by the expression d 2 S = 0.0186-0 where d is the thickness of the rope, and r the radius of the pulley, both in millimetres, and Q the tenison of the rope. S merely represents the force which is suffi- cient to overcome the stiffness of the rope at both sides, omitting journal friction. In order to produce motion, therefore, a force P = Q + S = Q(l + 0.0186^) must act upon the other end of the rope. Now, according to foregoing principles, in the absence * See Weisbach, Lehrbuch der Ingenieur- und Maschinenmechanik, Theil 1. STIFFNESS OF ROPES. 107 of journal friction there is the following relation between Panel Q: Q(r + a) = P (r - which the forces make one with another. If we imagine this pressure R of one wheel upon the other to be laid off on the lines CD and CE we have in the diagonal CF of the parallelogram the necessary driving-force P. Under the supposition again of equal pressures on the two pairs of teeth this resultant CF is parallel to the com- mon normal GH, and lies at a distance £ = \a x a 2 . tan 4> from the latter. If we now make the permissible assumption a x a 2 = £sina, a being the angle of the normal GH to the line of centres X 2 , we get, by substituting ^ = tan$, the desired distance fit . £ = -g sin a. The value of £ is therefore independent of the posi- tion of the two points of contact <( 1 b 1 and a 2 b 2 with reference to (9, it being supposed that they lie on oppo- site sides of 0, so that the lines of pressure may intersect at an angle 2$. According to these considerations we 114 THE GRAPHICAL STATICS OP MECHANISM. may conceive of tooth friction acting in the following way : While in the case of Motionless motion the force is transmitted from one wheel to the other in the direc- tion of a normal to the surfaces in contact, from which it results that the force P must be exactly equal to the resistance Q acting along the same line, in the case where friction is taken into account there is a parallel shifting to one side of the driving-force through a dis- tance £: so that the lever-arm of the driving-force with respect to the axis of the driven wheel B is shortened by an amount £, while with respect to the axis of the driving-wheel A it is increased by an equal amount. Since the line FC represents the re-action of the load Q for the wheel ^4, and OF = P is the driving-force with reference to the wheel B, we may say, as in chain and rope friction, that the arm of the power is diminished, and that of the resistance increased, by the amount £. In order to get a graphic representation of tooth friction Ave have only to determine the value £, and to then shift the line of force to one side of the theoretic line, so that it shall be parallel to the latter, and at the distance £ from it. It will not be difficult to show that the result of this shifting corresponds to that which we have been accus- tomed to find in practice by calculation, and which is given by the formula z = qM~ + -). Here Z is the force which" must be applied at the point of contact of the pitch circles to overcome tooth friction alone, and n 1 and n 2 are the number of teeth on TOOTH FRICTION. 115 the wheels respectively. To show this coincidence let r t be the radius of the pitch circle A, and r % that of B, and a = GrOO^ be again the inclination of the normal pressure to the line of centres. A resistance Q acting at along the line 0(7 to the wheel B has the lever-arm r 2 shia, to overcome which a force CF must be applied at G 7 , which, from the relation r 2 sin a and r 3 sin a — £ between the lever-arms, would be given by the equation _ n r 2 sina X =Q r Sill a This force X must be exerted by the driving-wheel A at the point C in the direction CF, with a lever-arm rjsina -f- £; therefore, there must be applied at the point of the wheel A a force in the direction Off given by the ratio of lever-arms in the equation V — Y r ! sin <* + £ _ q r 2 s ^ n a r i sin a -f- £ r x sin a r 2 sin a — £ r x sin a Substituting here — l, r 9 = -h% and £ = ^-< 2tt 2 2tt 2 we get, after reduction, y = £_^l_ . w i + p* = q( 1 + ff . ff Y >* 2 — /X7T w x \ w x ' w 2 / Since without friction § = P it follows that the 116 THE GRAPHICAL STATICS OF MECHANISM. value of the tooth friction, the force which must be applied at in the direction of Q, is z = r - p = qJ± + IX \n l n 2 / It may be remarked here that the latter formula, which is commonly used in calculating tooth friction, is deduced * under the supposition, employed in the pre- ceding investigation also, that the pressure is trans- mitted equally by two pairs of teeth. In the analytic determination of tooth friction it is customary to employ as Q the resistance which acts at the contact point of the pitch circles, normal to the line of centres, in order to avoid the value sin a. Even if such an approximation is permissible (sin a = sin 75° = 0.9659 approaching unity) it will be seen that it is unnecessary in the graphic method, as that loses nothing of its simplicity by drawing Q in its proper direction. It follows also that the construction remains the same when teeth of any other than the involute form are employed. If we only know the profiles of the teeth in contact the re- action is always inclined at the angle (f> to the normal to the surfaces where contact is taking place. In the case of profiles laid out by auxiliary circles the chord of the auxiliary circle passing the point of contact of the teeth, and that of the pitch circles at 0, is the desired normal. Of course in cycloidal and most other than involute profiles the normal will vary in its direction ; * Weisbach-Hermanu, Ingenieur- und Maschinenmechanik, Theil III., 1. TOOTH FRICTION. 117 for such cases we should assume that position of the teeth for the determination of £ at which the angle a of the normal with the line of centres has an average value. The variations of £ will only be small, how- ever, and can well be neglected as compared with the uncertainty which clings to all co-efficients of fric- tion. So far the investigation has been carried out on a basis of the equal transmission of power by two pairs of teeth. If there is contact between only one pair, which occurs when the arc within which the teeth mesh is less than f, the friction is somewhat less. In explanation let us suppose the teeth in Fig. 44, plate V., to be limited by the circles a 2 a s and b^^; then contact would exist between only one pair at a time, beginning in a 1 J 1 at the moment it ceased at a 2 b 2 . ^ that instant the tooth of the wheel A would act at the point a x with a force R in the direction a x C upon the wheel B. As long as motion continued the force R would remain parallel to a x C until the point of contact between the teeth came into the line of centres at 0, at which instant there would be no friction. In this case, as in all preceding ones, the arm of the power is diminished by an amount £, and that of the load increased by the same amount. Here we understand by £ the perpendicular distance of the point from the line a x G \ that is, £ z=z a ! 0. tan <£ ~ ^e x if the distance a x = e . Since this value of £ grows less and less, becoming equal to zero when the point of contact is at 0, we see that the friction has its maximum value when contact occurs at a x b v and its minimum 118 THE GRAPHICAL STATICS OF MECHANISM. when at 0. If we wish a mean value we have such an one at a point midway between and a 1 in £i = ¥i^ In the same way for motion from to a 2 we find the friction starting at zero, and reaching its maximum value at a 2 , where £ = Oa 2 tan ^ = /xe 2 if e 2 = 0a 2 . For this also we have the mean value Q 2 — 2/^2* There is this difference, however, in that after the point is passed the pressure acts along the line Ca 2 . If we suppose a x = 0a 2 , and consequently ^n^- |£sin a, we get for an average value of £ the equation This value of £ is only half as large as that £ = ±[d sin a obtained for the preceding case, where two pairs were in contact ; and, further, the arc of contact was twice as groat, =z 2f, in the first case discussed as here where it TOOTH FRICTION. 119 equals t. If, therefore, we denote the entire arc of con- tact on both sides of the line of centres by o> we have the general equation for both cases £ = l/xwsilla. This equation holds for all intermediate values of the arc of contact between t and 2t, when the transmission of force occurs part of the time through one pair and part through two pairs of teeth. We can then find the value of £ in every case by the following simple construction : Lay off on any straight line the distance OA — -J« (Fig. 45, plate V.) where w is the length of the arc of contact between the wheels ; draw the line OB at an angle a with the normal 00 x to OA, a being 75 degrees for involute teeth, and 80 degrees for cycioidal profiles ; then draw AB perpen- dicular to OB, and lay off the angle OBJD equal to the angle of friction $. We then have in the perpendicular OB to BO the value of £. We may assume ^ between 0.1 and 0.12. By making OA = ^w — ^t we get the same value for the friction that the formula /xr.( 1 ) \n t n 2 / gives. Having gotten the value of £ the tooth friction is readily determined. Suppose the load Q (Fig. 46, plate V.) to act upon the drum AB by means of a rope at B; it is then required to find the force P which must be applied in the direction EF to the crank EB in order to turn AB by means of the gear-wheel AC and the pinion CB, and lift the load. First find the direction in which Q acts, which is, of course, along the line o x b drawn at a distance o- from the mean circumference of 120 THE GRAPHICAL STATICS OF MECHANISM. the drum AB. The line of pressure Z between the teeth is given by the line o x c drawn at an angle of 75 degrees or 80 degrees to the line of centres AD, and intersecting the latter at a distance £ from the point of contact of the pitch circles. Then draw through o v the intersection of Z and Q, the line o 1 a tangent to the friction circle at A, and giving the direction of re-action at that journal. In a similar way we get the re-action do 2 of the journal D. The force polygon can now be drawn by making o x I = Q, and drawing I II parallel to Z, then resolving I II = Z in the direction of P and o 2 d. I III gives us the value of the force P which must be applied at the crank. To determine the theoretic force P we have only to draw the broken lines, as shown, through the centres of the journals, and perpendicular to AD through C. The drawing gives, for /x = 0.1, and Q = 100, P = 30.4, P = 28.1 ; and therefore ^^ 0.924. Tt will be readily seen that if in Fig. 44 the driving was done by the wheel B, but in such a manner that the points of contact remain at a 1 5 1 and a 2 b 2 (that is, if the motion was in an opposite direction to that shown by the arrows), the direction of pressure between the teeth would remain parallel to GH\ but the force Z would lie on the opposite side of this line, and would pass through C x between GrH and the axis of A, at a distance £ from the former. This case corresponds to the backward motion of the windlass ( Fig. 40) under the TOOTH FRICTION. 121 influence of Q. When, however, the wheel B (Fig. 44) is driven in the opposite direction to the arrows by the wheel A the investigation is similar in all respects, except that the direction of pressure is now along the line H'G'. 122 THE GRAPHICAL STATICS OF MECHANISM. § 9. — BELT GEARING. The principal source of loss in the transmission of rotary motion from one axis to another by means of belting is the friction which the axes suffer from being pressed against their bearings by the tension in the belt, since the resistance clue to stiffness where the belts wind on and off the pulleys is so small that it cannot be taken into consideration. But, on the con- trary, the journal friction is much greater for the trans- mission of a given force than by toothed gearing, since in belt gearing only the difference of tensions in the two portions of the belt represents force transmitted, while journal friction is caused by the sum of these two. The investigation of this resistance, with the determina- tion of the tension in the belt, is pursued in the follow- ing manner : — Let the shaft A (Fig. 47, plate V.) be driven from the shaft B by means of the belt and pullej^s CD and EF. We are to find the force P which must act on B with the lever-arm BK to overcome the resistance Q acting upon A with the lever-arm AL. If the belt surrounding the two pulleys is stretched to a certain tension this tension S is the same in both portions of the belt DE and CF while at rest. If, then, the pulley EF is acted upon by a force tending to revolve it in the direction shown by the arrow, the tension in the BELT GEARING. 123 belt OF increases to a value S v and at the same time that in DE diminishes to a value $ 2 , until #1 — ' ^2 is sufficient to overcome the resistance Q at the axis A, It being supposed that no slipping of the belt upon the pulleys occurs, the condition for such slipping is given by the formula S 1 = S 2 e» a , where S r is the tension in CF, S 2 that in DE, e is the base 2.71828 of the natural system of logarithms, jm is the co-efficient of friction between belt and pulley, and a the arc of contact between the same, the radius being unity. The greatest resistance W, therefore, which can be overcome with a tension 8% of the belt ED, when it (the resistance W) acts with a lever-arm AC from the axis A is W = S t -S 2 = 8 2 (e^- 1). The tension 8 in the belt when at rest must be deter- mined according to this relation, and is usually assumed 8 = l(S, + SJ, If we then suppose the two tensions S x and 8 2 to have a resultant Z we can draw the force polygon, as in pre- vious cases, by substituting this resultant for the tension in the two portions of the belt. The direction of the resultant is obtained by determining the proportion 124 THE GRAPHICAL STATICS OF MECHANISM. of the two tensions for the limiting condition of slip- ping —1 z= e^' S 2 so that by laying off from the intersection of the belts the distances OGr and OH of any convenient lengths, but in the ratio OH S x OG ~ S 2 ~ etX% we have in the diagonal OJof the completed parallelo- gram the resultant Z which may be substituted for the belt tensions themselves. To determine the value of this resultant, and of the belt tensions S x and S 2 , draw through o x and o. v the intersections of the resultant with P and Q, the tangents o x a and o 2 b to the friction circles of the journals A and B. Then make ol = (), and by drawing I II parallel to o x a we get the resultant Z = ojl of the belt tensions, and those tensions S 1 and # 2 , by resolving o x II in the directions o x IV parallel to CF, and II IV parallel to DE. From o x II = Z we get also the value of P by drawing o i III parallel to o 2 K, and II III parallel to o 2 b. In order to find the theoretic force P which would be sufficient to overcome Q in the absence of friction we can either employ the direction OJ of the resultant Z, and draw the re-actions o x A and <> 2 B through the centres of the journal, or take the intersections X and 2 of the forces Q and P with OF, BELT GEARING. 125 regarding the latter as a force acting at and F with- out friction. In Fig. 48, plate VI., the logarithmic spirals are drawn for the commonly occurring co-efficients of belt friction ,1 = 0.12, 0.18, 0.28, 0.38, and 0.47, in order to give a graphic representation of the ratio of belt tensions If we take the radius of the circle OA as unity, the radius vector OB drawn to the spiral, constructed with the particular co-efficient of friction /x, at the angle AOC = a, the angle of contact, gives the value e^ a . Therefore, if the tension S 2 of the slack side is repre- sented by OA, we have in OB the tension S 1 of the driving side, and in CB = aSj — >&2 the force transmitted. From the preceding it is easy to draw a comparison between the relative efficiencies of belt and toothed gearing. For this purpose we shall investigate the motion of a millstone, since these are as frequently driven by belts as by gearing. Let A (Fig. 49, plate VI.) be the mill-spindle, and BD the pulley on the same, which has the arc BMD in contact with the driving-belt. With the aid of the spiral for ^ = 0.28, as given in Fig. 48, plate VI., we find the ratio S 2 :S 1 126 THE GRAPHICAL STATICS OF MECHANISM. from that of OA : OB, having laid off the angle of con- tact BAD = a from the radius OA. Laying off these distances in KL and OK (Fig. 49) we have in OL the direction of the resultant Z of the two belt tensions which act upon the mill-spindle A as driving-force P. Since the resist- ance of the millstone upon its grinding-surface is exactly equal to this we must represent this resistance by a couple of forces Q and Q acting in the directions _ff/and FGr. This couple will not cause any side pressure of the mill-spindle against its bearing ; the re-action of the bearing will be that called forth by the resultant Z, and will therefore be equal to that resultant, and opposite in direction. We then draw R in the tangent ao 2 of the friction circle at A parallel to OL. For the existence of equilibrium between the four forces Q, Q, R, and Z, the resultant of any two must be coincident with and opposite to that of the remaining two. Uniting o x and o v making o x I = Q, and drawing I II parallel to o x o v we have in ^ 77 the resultant of the two tensions S x and aS y 2 ; and by resolving o x II into IIIo x , and II III parallel to the directions of the belt, we get IIIo x = S x and II III = S 2 . Without friction we should assume the direction of re- action at the bearing parallel to OL, and passing through the centre of A along the line A0 2 ; then drawing I II through /parallel to o x 2 , we should get P ^ 11,0,. BELT GEARING. 127 From the figure, with the assumed value /* = 0.28, and a journal friction of 0.1, we get, for Q = 100, P = 401.6, P = 379.2, and 77 = *~* = 0.944. If, on the other hand, we suppose the stone to be driven by gearing from the upright shaft B (Fig. 50, plate VI.), A again being the mill-spindle, Ave have the pressure transmitted by the gearing along the line o x c inclined at the angle of 75 degrees to the line of centres, at the distance £ from (7, and in the parallel line o 2 a the direction of re-action at the bearing of the spindle. Making Io 1 z=z Q, and drawing ///parallel to x ö 2 , we find P in o x II. To determine P we assume the thrust of the gearing along the perpendicular to AC passing through (7, and the re-action of the bearing parallel to this and passing through A. Then, if I O 1 = Q, we have in O 1 II the theoretic force P acting perpendicu- lar to AC through C In order to determine the effi- ciency we must compare, not the force P acting along o x c, but that component P f parallel to P , with the latter force. Let w be the point of intersection of the two directions, then we must resolve o x II = P in the direction of 1 C and wJ5. Draw o x III parallel with O x C, and II III parallel to uB, and we have in 128 THE GRAPHICAL STATICS OF MECHANISM. o x IH the force I" which must act at C perpendicular to the line of centres AB, thus giving the efficiency 17 = ^A 1 P' From the drawing we get, for Q = 100, P' = o x III = 226.4, P = 217.8, and v = 5; = 0.962. From these two examples we draw the conclusion that belt gearing is less economical than toothed gear- ing, as would have been expected from previous consid- erations. It is also evident that the result for belt transmission would be more unfavorable if the co- efficient of friction of belt upon pulley is less than 0.28, or if the arc of contact between the belt and the pulley of the mill-spindle subtends a less angle a, since the belt tension and resulting journal friction would then be much greater. In the foregoing comparison the journal friction of the main shaft is omitted in both cases, because in the general arrangement of several stones about one central shaft the opposing tensions or pressures would counteract one another, and the shaft would run freely in its bearings. If this were not the case, but only one stone was driven from the shaft, then. on account of the greater belt tension, the friction of the shaft in its bearing would be greater than in the case of toothed gearing. BELT GEARING. 129 The tensions in brake bands are to be estimated in the same way as with belt gearing. Here also there are two different tensions S ± and S 2 in the two ends of the band which have the relation one to the other, the greater tension S x being that which opposes the sliding of the brake pulley within the band. We can therefore introduce the resultant Z of the two tensions S 1 and S%, and regard it as a force preventing the motion which tends to occur. Let A (Fig. 51, plate VI.) be the axis of a drum on which is the brake pulley BC, whose brake band is fastened at one end to the stationary bolt E, and at the other to the bolt D of the brake-lever EDG which turns about E. We get the ratio of S x to S 2 from the spiral in Fig. 48, which corre- sponds to the co-efficient of friction for brake bands fi = 0.18, by making the angle a == CMB the arc of contact, and lay off the distances so determined along the direction of the ends of the band from their intersection in OUT and OJ. The diagonal OK of the completed par- allelogram gives the direction of the resultant Z. As regards the direction of tension of the brake band we see that the end CE fastened at E pulls in a line passing through the centre of E, since there is no relative turn- ing of the band about this point. But the line of tension in the end DB attached to the moving journal D would be tangential to the friction circle at D. If, now, a force Q acting with a lever-arm AF tends 130 THE GRAPHICAL STATICS OF MECHANISM. to turn the shaft A in the direction of the arrow the direction of journal pressure at A will be given by the line o x a drawn through the intersection o 1 of Q and Z tangent to the friction circle at A. Therefore, by making o x I = Q, and drawing through I a parallel to OK, we get in I II the necessary resultant Z of S x and >S f 2 . Resolving this parallel to HO and JO, we have III II = S 1 and IUI = S 2 . To determine the force P applied to the brake-lever G to produce the tension S 2 = IUI in the band BD we first draw from o 2 , the intersection of S 2 and P, the line o 2 e tangent to the friction circle at E to get the re-action of that bearing ; then a resolu- tion of I III = S 2 into I /{^parallel to o 2 e, and IV III parallel to o 2 G or P, gives us in IV III the force P which must be applied at Gr. The determination of the driving and brake forces in a whim or windlass, such as is shown in Fig. 52, plate VI., is of especial interest. Here two ropes FD and EC lead from the windlass-drum G- in such manner, that, if the drum is revolved in either direction (say, that indicated by the arrow), the rope EC is wound up, and that BD unwound ; and, in consequence, the weight Q consisting of useful load, and the weight Gr of cage, hooks, etc., hanging upon the rope ECB, is lifted, while the weight G of the empty cage hanging on the rope FDB sinks, and thus aids the revolution of the drum. If there were no hinderances to this motion we should at once assume that the working of BELT GEARING. 131 the entire apparatus amounted simply to lifting the useful load Q, since the cages balance one another and could be left out of consideration. Such an assumption is, however, not permissible on account of friction ; and Ave must regard the machine as a combination of two hoisting-gears, one of which burdened with the load Q -f- G- on the rope ECB is in forward motion, while the other is running backwards under the influence of the load G attached to the rope FDB. Accordingly the investigation would be pursued as follows : The line of tension in the ascending rope BCE must be assumed on account of stiffness in the rope along the lines b 1 o 1 and ce in such manner that the lever-arms at B and E shall be greater than the radii of the pulley and drum respectively by an amount o-, and at C less by an equal amount. On the other hand, the directions of tension for the rope FDB are given in fd and o 2 b 2 ; the lever- arms at D being increased, and those at F and B dimin- ished, by an amount o-. The direction of re-action at the bearing of the pulley A CB which revolves toward the left would be given in the line a x o x , while that of ABB revolving in the opposite direction would be along o 2 a 2 . Making o x I = Q -f- Gr, and o 2 III = Gr, and draw- ing through I a parallel i" II to ce, and through III a parallel III IV to df, we have in III = S t the tension in the rope EC, and in IV III = S 2 the tension in FD, The resultant Z of these tensions 132 THE GRAPHICAL STATICS OF MECHANISM. is given by V I if we draw II V equal and parallel to III IV, and complete the triangle. This resultant passes through the intersection o z of ec and fd, and Ave must therefore draw o 3 <9 4 parallel to / V If the drum is driven by a pinion HJ meshing with the gear-wheel GrJ we draw the line ii inclined at an angle of 75 degrees to GrH, and at a distance £ from J, for the direction of pressure between the teeth, and then draw from o 4 , the intersection of this line with that of Z, the tangent o±g to the friction circle at Gr. This gives us the direction of re-action at the bearings of Gr. Then resolve the force / V into I VI parallel to o±g, and VI V parallel to o 4 i, and we have in VI V the pressure on the teeth of the pinion HJ, or the necessary driving-force P. To determine the theoretic force P we have, as pre- viously remarked, only to lay off a distance equal to the useful load Q along the medial line of the rope EC from the intersection of that line with the common tangent JO to the pitch circles, so that OI = Q. Then draw through the radius 0(7, and through I the line I U parallel to the latter, and we have II = P . To determine the efficiency we must again compare P with that component of P obtained by resolving nv=p in the direction of P and W Ä This force P f is given by VI VII if VI VII is drawn parallel to OJ and V VII BELT GEARING. 133 parallel to wff. From the figure we get, for Q ~ 36 and G = 24, P f = VI VII = 47.3, P Q = 40.1, and therefore r, = £} = 0.848. For convenience, in order to use the same lines in the figure, we will suppose in determining the brake force that the loaded cage is now upon the rope FDB. Accordingly we lay off the broken and dotted lines o 2 (III) equal to (Q + Gr), and o { (I) equal to Gr; draw (ill) (IV) parallel to fd, and (I) (II) parallel to ce; we then have (JF)(1ZZ) = (£,) and {1-){I1) = (^). And by drawing tf 3 (JO equal to (III) (IV), and mak- ing (F)( VI) parallel and equal to (I)(II), we get in (VI)o s the resultant (Z) of (aS^) and (S 2 ). If the brake is applied by means of the brake-lever NLKM turning on the fixed point K, and attached to the ends of the brake band at iVand L, we first draw the lines of force W x n and W 2 l tangential to the brake jmlley T^IF 2 and to the friction circles of the bolts N and L. From the intersection o 5 lay off the distances o h TJ and o 5 T proportional to the tensions s 2 and s t in the brake band, which have been determined from Fig. 48. The diag- onal of the completed parallelogram then gives the resultant z of these tensions. If we now draw from o 6 , the intersection of z and (Z), the tangent o Q = 0.722. To determine the brake force p which is to be applied at W in the lever WU we must take the backward motion of the crane into consideration. In this case it is evident that the load Q = I II hanging upon the loose pulley causes, in sinking, a tension S { of the chain fastened at C equal to III I, and a tension S 2 = III II of the portion between A 2 and B x ; in other words, the strains in the two chains are reversed by backward motion. It will also be readily understood that the lines of tension in the chain at the pulleys B and D and at the drum will be those shown in broken and dotted lines ; that is, (o x ) (b^) and (o^) (6 2 ) at B, (fl 2 ) (^2) an d (°2 ) ( e \) at A an d ( e \) (^1) a * the drum, while the journal pressures at B and B take the direc- tion (oj) (7>) and (0 2 )OO- TJ ie direction in which the pressure (Z x ) of the teeth of the wheel ET now acts upon the pinion GF is given by the line (/) (/), which EXAMPLES. 145 cuts the line of centres EG at 75 degrees, and at a dis- tance £ from the point of contact F of the pitch circles toward the side of the driven axis G-, as was shown formerly in the discussion of tooth friction. The direc- tion of journal re-action at E corresponding to (/) (/) would vary but little from that, e£ X , found for forward motion ; it may therefore be assumed as the same. Taking these things into account the force polygon would be drawn as follows : Make (I)(III) = III II = (£,) ; draw (J) {IV) parallel to (oj) (ft), and (III) (IV) parallel to (o{) (6 2 ), also (IV)( V) parallel to (o 2 ) (, are supported by the pivot L in the head of the mast or post. This pivot has a side thrust upon it from the bearing L v while the two friction rollers W x and W 2 press upon the base T of the post. The pressures at L and T are determined from Fig. 57 as follows: Suppose S to be the centre of gravity of all the swinging portion of the crane, windlass, pulleys, chain, etc., and suppose G the weight of all these parts EXAMPLES. 147 acting downwards at S. This force may be combined with the weight Q hanging at A into one resultant force, acting at M. Now lay off on this line the distance in = Q + a, and draw the horizontal re-action R of the cylinder T through the middle of the rollers W. The re-action R x of the pivot L must then pass through the intersection of R and Q + (?, and must take the direction OL drawn through the centre of the bearing L v To determine R and R x we therefore resolve I II in the direction of OT and OL, and have IUI - R, the re-action against the rollers TT, and II III = R v the combined re-actions at L. This latter may be resolved back into horizontal and vertical components, /ZZ7 and II L The vertical re-action equal to G -f" Q produces end journal friction upon i, which may be regarded as the action of a couple, each force of which equals MQ + #)* and whose arm equals |c/, if d is the diameter of the journal. The horizontal component of the re-action R x 148 THE GRAPHICAL STATICS OF MECHANISM. is equal and opposite to the re-action R of the post against the rollers in the direction TO, so that these two forces form a couple in equilibrium with that formed by the vertical re-action Z, and the weight Q -\- Gi at M. To determine the turning-force p w r e draw, in Fig. 58, the horizontal pressure 1 2 equal to IUI in Fig. 57, and in such direction that it shall be tangent to the friction circle at L. Also lay off the two end journal frictions F = F X = l-^Q + G) at the distance \d on each side the centre of X, and in opposite directions, so as to bring this end journal fric- tion into the calculation. The compounding of this couple F, F l with the horizontal re-action 1 2 will simply result in a parallel shifting of that re-action undiminished in value to the position 4 5. To fix the amount of this shifting make 2 3 = jP, draw 3 1, and through o v the intersection of 1 2 and F, draw a parallel to 3 1 ; this will cut F x at a point o 2 , through which the resultant 4 5 of 1 2 and FF X must pass. The correct- ness of this construction is shown by the equilibrium of the four forces 1 2, F, F v and 5 4. Now draw the re- actions u l w 1 and u 2 w 2 of the post against the friction rollers in such way that they shall be tangent to the friction circles of W 1 and TF 2 , and shall pass to one side of the contact points L\ and U 2 of the rollers at the distance e determined previously for rolling friction. If the turning of the crane is produced by a pinion on the vertical shaft V, and working with a circular rack or internal gear Y on the base plate, we draw finally the direction vv of pressure between the teeth at 75 degrees EXAMPLES. 149 to the line of centres VL, and at the distance £ on the outer side of the contact point of the pitch circles. This line intersects the resultant of journal pressure 4 5 in o 3 , the two re-actions of the post against the rollers intersect in o 4 ; therefore o s o 4 must be the common resultant of these pairs of forces. Draw 5 6 parallel to vv, and through 4 a parallel to o 4 o s , and we have 5 6 = Z, the pressure which the teeth of the pinion on the shaft V must exert along the line vv. The method by which the force necessary to turn V by means of a crank would be determined is sufficiently well known from previous examples. In conclusion we will explain the diagram for the condensing beam-engine shown in plate VIII. Let I II = P, the force acting upon the piston (Fig. 59). This force acts along KC\ the geometric axis of the piston-rod, since the latter is rigidly fixed to both the piston and the cross-bar (7, and there can be no relative motion between those elements. If, then, we consider the piston, piston-rod, and cross-bar C to form one piece, we have at the point C three forces, P the piston force, and S x and S. 2 exerted by the link BC and the parallel rod EC respectively. In the present posi- tion of the engine these forces are in compression, and act along the tangents hc x and e^ shown on a larger scale in Fig. 60 a . The intersection of these forces is at o v and since P does not pass through this point the three forces P, S v and S 2 cannot be in equilibrium. There must therefore be a fourth force, which with P will give a resultant passing through o v and which must 150 THE GRAPHICAL STATICS OF MECHANISM. tend to produce right-handed revolution about o v since P alone would produce left-handed turning. This fourth force is the re-action R x exerted by the stuffing- box against the piston-rod ; it passes through the centre of the stuffing-box, is inclined at an angle cf> below the horizontal, and acts from the right toward the left. This re-action cuts the piston force in the point o 2 ; and therefore, as frequently shown before, the line o x o v not drawn in the figure, must be the common resultant of S x with aS' 2 , and P with R. So that, drawing through / a parallel to R v and through II a parallel to o 1 o 2 , we get in IUI = B 1 the re-action of the stuffing-box against the piston-rod. This indicates that in spite of the right-line motion there is a certain side thrust against the stuffing-box, as shown in Fig. 19, plate II., which is due to journal fric- tion on the cross-arm C\ and not to any inaccuracy in the parallel motion. This side thrust, which, as pre- viously shown, reverses its angle at every change in the direction of piston motion, is, however, so small that it is generally left out of account. Draw, next, through II and III the parallels II IV and III IV to e x e 2 and c x b, and we get the forces S 2 = II IV and S 1 = III IV. The last force S x = III IV is transmitted directly to the beam through the link CB. A second force, partly due to the resistance of EXAMPLES. 151 the air-pump L attached at F, and partly to the influ- ence of the radius-rod ED, acts upon the beam at G. This force is the next to be determined. Let II V represent the resistance L of the air-pump, which acts upon the link FG in the tangent to the friction circle of F passing through the point of intersection o s of the piston force L of the air-pump and the stuffing-box re- action R 2 . This small re-action R 2 , and its inappreci- able influence on the resistance L, will not be further taken into account. Looking at the link FGr we have acting upon it three forces : L the air-pump resistance in the direction fo s (see also Fig. 60^) ; the pressure aS y 2 of the parallel rod OF in the known direction c 2 e v and of the known value II IV; and finally a force aS y 3 exerted by the radius-rod ED upon the pin E, and which acts in the known direc- tion e 2 d v tangent to the friction circles at E and D. These three forces must be held in equilibrium by a force S 4 exerted by the journal G upon the link GE, and which remains to be determined. For this purpose we combine the two known forces of the air-pump, and II V S 2 = IV II of the parallel rod GE, in a resultant IV V. This resultant must pass through the intersection o 4 of its components (Fig. 60 6 ) ; and by drawing through o 4 a parallel to IVVwe get in o 5 , its intersection with the force # 3 acting along e 2 d l in the radius-rod DE, a point 152 THE GRAPHICAL STATICS OF MECHANISM. through which the desired re-action of the journal Gr must pass. If we draw the tangent o b g through o 5 to the friction circle of (2, the direction of # 4 , the resultant of S 2 , aS' 3 , and L is then found. In the force polygon we now resolve IV V, the resultant of L and # 2 , in the direction IV VI parallel to o 5 #, and VI V parallel to e % d x , thus getting V VI = s s , the tension in the radius-rod, and IV VI = s v the pressure of the link EG upon the journal Gr of the beam. There are now acting upon the beam from the two links the forces S\ = III IV in the direction c x b, and # 4 = IV VI in the direction go rQ . The resultant S of these two is given in value and direction by the line III VI of the force polygon. To determine the position of this re- action we must draw through the intersection of c x h and o h g a parallel to III VI Since this intersection lies beyond the limits of the drawing we can draw a funicular polygon derived from the force polygon in the well-known way to determine the position of S. Choose EXAMPLES. 153 any convenient point for the pole of the force poly- gon, and draw the polar rays IV, VI, and III, and construct the corresponding funicular polygon * a/38. Then the vertex 8 of the polygon is a point of the resultant $ which must be drawn through 8 parallel to VI III To check the construction a second funicular polygon ei 1 /5 1 8 1 has been drawn, the line 88 x giving the direction of the resultant S of the pressure S\ in the link OB, and the tension &\ in that GrE. If the construction is accurate 88 x will of course be parallel to Villi The other end of the beam acts upon the crank MJ with a force T in the connecting-rod HJ, the line of force T coinciding with hi, the common tangent to the friction circles at H and J, To determine this force T we must know the direction in which the bearing of the beam journal A re-acts against it. This re-action passes through the intersection of T and S, and is tangent to the friction circle at A. Since this point lies beyond the limits of the drawing we must employ the construction shown in Fig. 19, plate II. For this purpose draw a line through the centre of the journal A perpendicular to the supposed position of the desired re-action, cutting the directions of T and S in S and c, and the friction circle of A in v. Then draw a parallel to this line, cutting T and S in 8 X and e v and so locate the point v x that Oji/| * €^v^ • • o: V We then have in v x a point in the direction of the re- * The principles of the funicular polygon referred to here and in previous chapters are clearly set forth in Karl von Ott's little book on Graphic Statics. — Trans, 154 THE GRAPHICAL STATICS OF MECHANISM. action R which may be drawn through v 1 tangent to the friction circle at A. To locate the point v 1 so that the above proportion shall be true we have only to draw through 8 X and ^ any two parallel lines S 1 8 2 and e 1 € 2 , make 8-^2 = 81/ and E l€2 — C] /, and the intersection of 8 1 € 1 and S 2 *r 2 is the desired point v v Having thus obtained the direction of the re-action It we resolve the force S = IIIVI in the direction VI VII parallel to /«', and VII III parallel to v x a, thus getting T = VIIVI and B= VII III. If X is the point of contact of the two pitch circles of the wheels MX and NX by which the driving-force is transmitted from the engine shaft M to a line of shafting N we draw first the direction of pressure between the teeth in the line xx at 75 degrees to the line of centres MN, and at a distance £ from X. Taking the weight of the fly-wheel, which acts downward and increases the journal friction at M, into consideration we lay off in the force polygon the weight Gr equal to VI VIII, and have in VII VIII the resultant of con- necting-rod thrust T and weight Gr in direction and value. By drawing through o 6 , the intersection of the forces T and G, a line parallel to VII VIII, we get the intersection o 7 of their resultant with xx, the line of EXAMPLES. 155 pressure Z between the teeth of the gear-wheels. The tangent o 7 m drawn through o 7 to the friction circle of M gives the re-action of the fly-wheel bearing. We have finally to resolve the force VII VIII into VIII IX parallel to xx, and VII IX parallel to # 7 m, thus getting Z -IX VIII the force exerted by the teeth of the gear-wheel MX upon those of NX along the line xx. If r is the lever- arm NN 1 of this force Z with reference to the shaft iV, and if this shaft N encounters at the moment under consideration a resistance whose moment is Zr, the power of the steam-engine is sufficient to overcome that resistance. If, on the contrary, the moment of the resistance to be overcome has any other value Zr x there will be a moment Z(r — r x ) either expended in the acceleration of the moving masses when r — r t is positive, or given out by those masses if r — r x is negative. The discussion and investigation ill the case of the saw-grate (Fig. 53, plate VI.) apply also here. 156 THE GRAPHICAL STATICS OF MECHANISM. CONCLUDING REMARKS. The graphical determination of forces and of the resulting efficiency, as explained and applied in the foregoing chapters, will not present great difficulties in any form of mechanism, since in each case it is merely a question of determining the direction of the forces involved, and from them drawing the corresponding force polygon. This method possesses all the general advantages of graphical determinations over the ana- lytical methods until now alone employed. It rests upon the elementary laws of mechanics, does not re- quire a knowledge of analytics, leads to the desired end by a quicker and surer route, and has in nearly every case a sufficient degree of accuracy. To meet the last requirement it is only necessary to draw the diagrams to a sufficiently large scale, and in practice it would be well to employ a scale from six to ten times larger than that of the diagrams attached to this work. For the sake of clearness, and in consequence of the reduced scale of these plates, certain dimensions were chosen greater than they should be ; for instance, most of the friction circles will be found to be larger than the cor- responding co-efficient of friction would give. And therefore, as before remarked, the numerical results given for forces and efficiencies in the various examples were not taken from the figures in the plates, but from much larger drawings in which all dimensions were cor- rectly proportioned. That with large but manageable CONCLUDING REMARKS. 157 drawings sufficient accuracy for practical purposes is obtained is evident when we consider the uncertainty there is about the co-efficient of friction itself which is determined empirically. A co-efficient of friction is never given with certainty beyond two decimal places, as a glance over the tables of these co-efficients shows, and it is safe to assume that in the average case there is an uncertainty of several per cent. In the light of these facts how worthless is the determination of forces carried out to many decimal places, to hundred-thou- sandths even, as is the case in many analytical deduc- tions ! At any rate it follows from what has been said, that with moderately large drawings and with fine lines the attainable accuracy is all-sufficient. Supposing, for instance, that an untrained e} r e would permit an error of one millimetre in laying off distances, the error would be at the most only one per cent if the shortest line of force was one decimetre. In reference to the accuracy of results obtained by graphic methods we may also remark that in such determinations there is no temptation to employ certain approximations which are generally introduced in ana- lytic calculations to render the formulae less unwieldy, and, in fact, often have to be employed to render further progress possible. Many of the preceding examples correspond to just such cases. Thus in pulleys where the ropes are not parallel such parallelism has to be assumed in determining the journal pressure ; in slider- crank gearing the inclination of the connecting-rod is neglected, etc. All such assumptions are unnecessary in graphic methods. In the determination of the force acting upon the lever of a brake-band the friction of the journal by which the band is connected to the lever 158 THE GRAPHICAL STATICS OF MECHANISM. is neglected in analysis, while in the graphic method it simply amounts to drawing the friction circle of that journal, and the determination is in no way complicated by thus taking it into consideration. Even though, as it may be urged, such small resistances may be safely neglected as inappreciable, it goes to strengthen the assertion that the graphic method is more accurate in such cases than the analytical. A special advantage of the graphical method is its great clearness. This is of the highest importance to the designer or engineer. Take, for instance, the graphical analysis of a hoisting-gear. It presents to the eye the position, direction, and intensity of all the forces at one glance ; and having these it is not difficult to determine by the laws of graphical statics the mo- ments at any section of any piece, and from that the necessary dimensions. All such determinations are without the scope of the present work, however ; they will be found in the standard works on machine design and graphical statics. Plate I. Plate Plate Ml W~^f Plate IV. _D_ Fig. 30 TTi K . 31 Plate V. Plate VII. Plate VIII. * 42 43 45 44 46 47 49 48 50 51