(a)The plane through O to which the shaft is perpendicular is called thereference plane, because all the transferred forces act in that plane at the point O. The plane through the radius of the weight containing the axis OX is called theaxial planebecause it contains the forces forming the couple due to the transference of F to the reference plane. Substituting the values of F in (a) the two conditions become(1) (W1r1+ W2r2+ W3r3+ ...)α2= 0g(2) (W1a1r1+ W2a2r2+ ... )α2= 0g(b)In order that these conditions may obtain, the quantities in the brackets must be zero, since the factor α2/g is not zero. Hence finally the conditions which must be satisfied by the system of weights in order that the axis of rotation may be a permanent axis is(1) (W1r1+ W2r2+ W3r3) = 0(2) (W1a1r1+ W2a2r2+ W3a3r3) = 0(c)It must be remembered that these are all directed quantities, and that their respective sums are to be taken by drawing vector polygons. In drawing these polygons the magnitude of the vector of the type Wr is the product Wr, and the direction of the vector is from the shaft outwards towards the weight W, parallel to the radius r. For the vector representing a couple of the type War, if the masses are all on the same side of the reference plane, the direction of drawing is from the axis outwards; if the masses are some on one side of the reference plane and some on the other side, the direction of drawing is from the axis outwards towards the weight for all masses on the one side, and from the mass inwards towards the axis for all weights on the other side, drawing always parallel to the direction defined by the radius r. The magnitude of the vector is the product War. The conditions (c) may thus be expressed: first, that the sum of the vectors Wr must form a closed polygon, and, second, that the sum of the vectors War must form a closed polygon. The general problem in practice is, given a system of weights attached to a shaft, to find the respective weights and positions of two balance weights or counterpoises which must be added to the system in order to make the shaft a permanent axis, the planes in which the balance weights are to revolve also being given. To solve this the reference plane must be chosen so that it coincides with the plane of revolution of one of the as yet unknown balance weights. The balance weight in this plane has therefore no couple corresponding to it. Hence by drawing a couple polygon for the given weights the vector which is required to close the polygon is at once found and from it the magnitude and position of the balance weight which must be added to the system to balance the couples follow at once. Then, transferring the product Wr corresponding with this balance weight to the reference plane, proceed to draw the force polygon. The vector required to close it will determine the second balance weight, the work may be checked by taking the reference plane to coincide with the plane of revolution of the second balance weight and then re-determining them, or by taking a reference plane anywhere and including the two balance weights trying if condition (c) is satisfied.When a weight is reciprocated, the equal and opposite force required for its acceleration at any instant appears as an unbalanced force on the frame of the machine to which the weight belongs. In the particular case, where the motion is of the kind known as “simple harmonic” the disturbing force on the frame due to the reciprocation of the weight is equal to the component of the centrifugal force in the line of stroke due to a weight equal to the reciprocated weight supposed concentrated at the crank pin. Using this principle the method of finding the balance weights to be added to a given system of reciprocating weights in order to produce a system of forces on the frame continuously in equilibrium is exactly the same as that just explained for a system of revolving weights, because for the purpose of finding the balance weights each reciprocating weight may be supposed attached to the crank pin which operates it, thus forming an equivalent revolving system. The balance weights found as part of the equivalent revolving system when reciprocated by their respective crank pins form the balance weights for the given reciprocating system. These conditions may be exactly realized by a system of weights reciprocated by slotted bars, the crank shaft driving the slotted bars rotating uniformly. In practice reciprocation is usually effected through a connecting rod, as in the case of steam engines. In balancing the mechanism of a steam engine it is often sufficiently accurate to consider the motion of the pistons as simple harmonic, and the effect on the framework of the acceleration of the connecting rod may be approximately allowed for by distributing the weight of the rod between the crank pin and the piston inversely as the centre of gravity of the rod divides the distance between the centre of the cross head pin and the centre of the crank pin. The moving parts of the engine are then divided into two complete and independent systems, namely, one system of revolving weights consisting of crank pins, crank arms, &c., attached to and revolving with the crank shaft, and a second system of reciprocating weights consisting of the pistons, cross-heads, &c., supposed to be moving each in its line of stroke with simple harmonic motion. The balance weights are to be separately calculated for each system, the one set being added to the crank shaft as revolving weights, and the second set being included with the reciprocating weights and operated by a properly placed crank on the crank shaft. Balance weights added in this way to a set of reciprocating weights are sometimes called bob-weights. In the case of locomotives the balance weights required to balance the pistons are added as revolving weights to the crank shaft system, and in fact are generally combined with the weights required to balance the revolving system so as to form one weight, the counterpoise referred to in the preceding section, which is seen between the spokes of the wheels of a locomotive. Although this method balances the pistons in the horizontal plane, and thus allows the pull of the engine on the train to be exerted without the variation due to the reciprocation of the pistons, yet the force balanced horizontally is introduced vertically and appears as a variation of pressure on the rail. In practice about two-thirds of the reciprocating weight is balanced in order to keep this variation of rail pressure within safe limits. The assumption that the pistons of an engine move with simple harmonic motion is increasingly erroneous as the ratio of the length of the crank r, to the length of the connecting rod l increases. A more accurate though still approximate expression for the force on the frame due to the acceleration of the piston whose weight is W is given byWω2r{cos θ +rcos 2θ}glThe conditions regulating the balancing of a system of weights reciprocating under the action of accelerating forces given by the above expression are investigated in a paper by Otto Schlick, “On Balancing of Steam Engines,”Trans, Inst. Nav. Arch.(1900), and in a paper by W. E. Dalby, “On the Balancing of the Reciprocating Parts of Engines, including the Effect of the Connecting Rod” (ibid., 1901). A still more accurate expression than the above is obtained by expansion in a Fourier series, regarding which and its bearing on balancing engines see a paper by J. H. Macalpine, “A Solution of the Vibration Problem” (ibid., 1901). The whole subject is dealt with in a treatise,The Balancing of Engines, by W. E. Dalby (London, 1906). Most of the original papers on this subject of engine balancing are to be found in theTransactionsof the Institution of Naval Architects.§ 113.*Centrifugal Whirling of Shafts.—When a system of revolving masses is balanced so that the conditions of the preceding section are fulfilled, the centre of gravity of the system lies on the axis of revolution. If there is the slightest displacement of the centre of gravity of the system from the axis of revolution a force acts on the shaft tending to deflect it, and varies as the deflexion and as the square of the speed. If the shaft is therefore to revolve stably, this force must be balanced at any instant by the elastic resistance of the shaft to deflexion. To take a simple case, suppose a shaft,supported on two bearings to carry a disk of weight W at its centre, and let the centre of gravity of the disk be at a distance e from the axis of rotation, this small distance being due to imperfections of material or faulty construction. Neglecting the mass of the shaft itself, when the shaft rotates with an angular velocity a, the centrifugal force Wa2e/g will act upon the shaft and cause its axis to deflect from the axis of rotation a distance, y say. The elastic resistance evoked by this deflexion is proportional to the deflexion, so that if c is a constant depending upon the form, material and method of support of the shaft, the following equality must hold if the shaft is to rotate stably at the stated speed—W(y + e) a2= cy,gfrom which y = Wa2e / (gc − Wa2).This expression shows that as a increases y increases until when Wa2= gc, y becomes infinitely large. The corresponding value of a, namely √(gc/W), is called thecritical velocityof the shaft, and is the speed at which the shaft ceases to rotate stably and at which centrifugal whirling begins. The general problem is to find the value of a corresponding to all kinds of loadings on shafts supported in any manner. The question was investigated by Rankine in an article in theEngineer(April 9, 1869). Professor A. G. Greenhill treated the problem of the centrifugal whirling of an unloaded shaft with different supporting conditions in a paper “On the Strength of Shafting exposed both to torsion and to end thrust,”Proc. Inst. Mech. Eng.(1883). Professor S. Dunkerley (“On the Whirling and Vibration of Shafts,”Phil. Trans., 1894) investigated the question for the cases of loaded and unloaded shafts, and, owing to the complication arising from the application of the general theory to the cases of loaded shafts, devised empirical formulae for the critical speeds of shafts loaded with heavy pulleys, based generally upon the following assumption, which is stated for the case of a shaft carrying one pulley: If N1, N2be the separate speeds of whirl of the shaft and pulley on the assumption that the effect of one is neglected when that of the other is under consideration, then the resulting speed of whirl due to both causes combined may be taken to be of the form N1N2√(N21+ N12) where N means revolutions per minute. This form is extended to include the cases of several pulleys on the same shaft. The interesting and important part of the investigation is that a number of experiments were made on small shafts arranged in different ways and loaded in different ways, and the speed at which whirling actually occurred was compared with the speed calculated from formulae of the general type indicated above. The agreement between the observed and calculated values of the critical speeds was in most cases quite remarkable. In a paper by Dr C. Chree, “The Whirling and Transverse Vibrations of Rotating Shafts,”Proc. Phys. Soc. Lon., vol. 19 (1904); alsoPhil. Mag., vol. 7 (1904), the question is investigated from a new mathematical point of view, and expressions for the whirling of loaded shafts are obtained without the necessity of any assumption of the kind stated above. An elementary presentation of the problem from a practical point of view will be found inSteam Turbines, by Dr A. Stodola (London, 1905).Fig. 131.§ 114.Revolving Pendulum. Governors.—In fig. 131 AO represents an upright axis or spindle; B a weight called abob, suspended by rod OB from a horizontal axis at O, carried by the vertical axis. When the spindle is at rest the bob hangs close to it; when the spindle rotates, the bob, being made to revolve round it, diverges until the resultant of the centrifugal force and the weight of the bob is a force acting at O in the direction OB, and then it revolves steadily in a circle. This combination is called arevolving,centrifugal, orconical pendulum. Revolving pendulums are usually constructed withpairsof rods and bobs, as OB, Ob, hung at opposite sides of the spindle, that the centrifugal forces exerted at the point O may balance each other.In finding the position in which the bob will revolve with a given angular velocity, a, for most practical cases connected with machinery the mass of the rod may be considered as insensible compared with that of the bob. Let the bob be a sphere, and from the centre of that sphere draw BH = y perpendicular to OA. Let OH = z; let W be the weight of the bob, F its centrifugal force. Then the condition of its steady revolution is W : F :: z : y; that is to say, y/z = F/W = yα2/g; consequentlyz = g/α2(69)Or, if n = α 2π = α/6.2832 be the number of turns or fractions of a turn in a second,z =g=0.8165 ft.=9.79771 in.4π2n2n2n2(70)z is called thealtitude of the pendulum.Fig. 132.If the rod of a revolving pendulum be jointed, as in fig. 132, not to a point in the vertical axis, but to the end of a projecting arm C, the position in which the bob will revolve will be the same as if the rod were jointed to the point O, where its prolongation cuts the vertical axis.A revolving pendulum is an essential part of most of the contrivances calledgovernors, for regulating the speed of prime movers, for further particulars of which seeSteam Engine.Division 3. Working of Machines of Varying Velocity.§ 115.General Principles.—In order that the velocity of every piece of a machine may be uniform, it is necessary that the forces acting on each piece should be always exactly balanced. Also, in order that the forces acting on each piece of a machine may be always exactly balanced, it is necessary that the velocity of that piece should be uniform.An excess of the effort exerted on any piece, above that which is necessary to balance the resistance, is accompanied with acceleration; a deficiency of the effort, with retardation.When a machine is being started from a state of rest, and brought by degrees up to its proper speed, the effort must be in excess; when it is being retarded for the purpose of stopping it, the resistance must be in excess.An excess of effort above resistance involves an excess of energy exerted above work performed; that excess of energy is employed in producing acceleration.An excess of resistance above effort involves an excess of work performed above energy expended; that excess of work is performed by means of the retardation of the machinery.When a machine undergoes alternate acceleration and retardation, so that at certain instants of time, occurring at the end of intervals calledperiodsorcycles, it returns to its original speed, then in each of those periods or cycles the alternate excesses of energy and of work neutralize each other; and at the end of each cycle the principle of the equality of energy and work stated in § 87, with all its consequences, is verified exactly as in the case of machines of uniform speed.At intermediate instants, however, other principles have also to be taken into account, which are deduced from the second law of motion, as applied todirect deviation, or acceleration and retardation.§ 116.Energy of Acceleration and Work of Retardation for a Shifting Body.—Let w be the weight of a body which has a motion of translation in any path, and in the course of the interval of time Δt let its velocity be increased at a uniform rate of acceleration from v1to v2. The rate of acceleration will bedv/dt = const. = (v2− v1) Δt;and to produce this acceleration a uniform effort will be required, expressed byP = w (v2− v1) gΔt(71)(The product wv/g of the mass of a body by its velocity is called itsmomentum; so that the effort required is found by dividing the increase of momentum by the time in which it is produced.)To find theenergywhich has to be exerted to produce the acceleration from v1to v2, it is to be observed that thedistancethrough which the effort P acts during the acceleration isΔs = (v2+ v1) Δt/2;consequently, theenergy of accelerationisPΔs = w (v2− v1) (v2+ v1) / 2g = w (v22− v12) 2g,(72)being proportional to the increase in the square of the velocity, andindependent of the time.In order to produce aretardationfrom the greater velocity v2to the less velocity v1, it is necessary to apply to the body aresistanceconnected with the retardation and the time by an equation identical in every respect with equation (71), except by the substitution of a resistance for an effort; and in overcoming that resistance the bodyperforms workto an amount determined by equation (72), putting Rds for Pas.§ 117.Energy Stored and Restored by Deviations of Velocity.—Thus a body alternately accelerated and retarded, so as to be brought back to its original speed, performs work during its retardation exactly equal in amount to the energy exerted upon it during its acceleration; so that that energy may be considered asstoredduring the acceleration, andrestoredduring the retardation, in a manner analogous to the operation of a reciprocating force (§ 108).Let there be given the mean velocity V =1⁄2(v2+ v1) of a body whose weight is w, and let it be required to determine the fluctuation of velocity v2− v1, and the extreme velocities v1, v2, which that body must have, in order alternately to store and restore an amount of energy E. By equation (72) we haveE = w (v22− v12) / 2gwhich, being divided by V =1⁄2(v2+ v1), givesE/V = w (v2− v1) / g;and consequentlyv2− v1= gE / Vw(73)The ratio of this fluctuation to the mean velocity, sometimes called the unsteadiness of the motion of the body, is(v2− v1) V = gE / V2w.(74)§ 118.Actual Energy of a Shifting Body.—The energy which must be exerted on a body of the weight w, to accelerate it from a state of rest up to a given velocity of translation v, and the equal amount of work which that body is capable of performing by overcoming resistance while being retarded from the same velocity of translation v to a state of rest, iswv2/ 2g.(75)This is called theactual energyof the motion of the body, and is half the quantity which in some treatises is called vis viva.The energy stored or restored, as the case may be, by the deviations of velocity of a body or a system of bodies, is the amount by which the actual energy is increased or diminished.§ 119.Principle of the Conservation of Energy in Machines.—The following principle, expressing the general law of the action of machines with a velocity uniform or varying, includes the law of the equality of energy and work stated in § 89 for machines of uniform speed.In any given interval during the working of a machine, the energy exerted added to the energy restored is equal to the energy stored added to the work performed.§ 120.Actual Energy of Circular Translation—Moment of Inertia.—Let a small body of the weight w undergo translation in a circular path of the radius ρ, with the angular velocity of deflexion α, so that the common linear velocity of all its particles is v = αρ. Then the actual energy of that body iswv2/ 2g = wa2ρ2/ 2g.(76)By comparing this with the expression for the centrifugal force (wa2ρ/g), it appears that the actual energy of a revolving body is equal to the potential energy Fρ/2 due to the action of the deflecting force along one-half of the radius of curvature of the path of the body.The product wρ2/g, by which the half-square of the angular velocity is multiplied, is called themoment of inertiaof the revolving body.§ 121.Flywheels.—A flywheel is a rotating piece in a machine, generally shaped like a wheel (that is to say, consisting of a rim with spokes), and suited to store and restore energy by the periodical variations in its angular velocity.The principles according to which variations of angular velocity store and restore energy are the same as those of § 117, only substitutingmoment of inertiaformass, andangularforlinearvelocity.Let W be the weight of a flywheel, R its radius of gyration, a2its maximum, a1its minimum, and A =1⁄2(α2+ α1) its mean angular velocity. LetI/S = (α2− α2) / Adenote theunsteadinessof the motion of the flywheel; the denominator S of this fraction is called thesteadiness. Let e denote the quantity by which the energy exerted in each cycle of the working of the machine alternately exceeds and falls short of the work performed, and which has consequently to be alternately stored by acceleration and restored by retardation of the flywheel. The value of thisperiodical excessis—e = R2W (α22− α12), 2g,(77)from which, dividing both sides by A2, we obtain the following equations:—e / A2= R2W / gSR2WA2/ 2g = Se / 2.(78)The latter of these equations may be thus expressed in words:The actual energy due to the rotation of the fly, with its mean angular velocity, is equal to one-half of the periodical excess of energy multiplied by the steadiness.In ordinary machinery S = about 32; in machinery for fine purposes S = from 50 to 60; and when great steadiness is required S = from 100 to 150.The periodical excess e may arise either from variations in the effort exerted by the prime mover, or from variations in the resistance of the work, or from both these causes combined. When but one flywheel is used, it should be placed in as direct connexion as possible with that part of the mechanism where the greatest amount of the periodical excess originates; but when it originates at two or more points, it is best to have a flywheel in connexion with each of these points. For example, in a machine-work, the steam-engine, which is the prime mover of the various tools, has a flywheel on the crank-shaft to store and restore the periodical excess of energy arising from the variations in the effort exerted by the connecting-rod upon the crank; and each of the slotting machines, punching machines, riveting machines, and other tools has a flywheel of its own to store and restore energy, so as to enable the very different resistances opposed to those tools at different times to be overcome without too great unsteadiness of motion. For tools performing useful work at intervals, and having only their own friction to overcome during the intermediate intervals, e should be assumed equal to the whole work performed at each separate operation.§ 122.Brakes.—A brake is an apparatus for stopping and diminishing the velocity of a machine by friction, such as the friction-strap already referred to in § 103. To find the distance s through which a brake, exerting the friction F, must rub in order to stop a machine having the total actual energy E at the moment when the brake begins to act, reduce, by the principles of § 96, the various efforts and other resistances of the machine which act at the same time with the friction of the brake to the rubbing surface of the brake, and let R be their resultant—positive ifresistance,negativeif effort preponderates. Thens = E / (F + R).(79)§ 123.Energy distributed between two Bodies: Projection and Propulsion.—Hitherto the effort by which a machine is moved has been treated as a force exerted between a movable body and a fixed body, so that the whole energy exerted by it is employed upon the movable body, and none upon the fixed body. This conception is sensibly realized in practice when one of the two bodies between which the effort acts is either so heavy as compared with the other, or has so great a resistance opposed to its motion, that it may, without sensible error, be treated as fixed. But there are cases in which the motions of both bodies are appreciable, and must be taken into account—such as the projection of projectiles, where the velocity of therecoilor backward motion of the gun bears an appreciable proportion to the forward motion of the projectile; and such as the propulsion of vessels, where the velocity of the water thrown backward by the paddle, screw or other propeller bears a very considerable proportion to the velocity of the water moved forwards and sideways by the ship. In cases of this kind the energy exerted by the effort isdistributedbetween the two bodies between which the effort is exerted in shares proportional to the velocities of the two bodies during the action of the effort; and those velocities are to each other directly as the portions of the effort unbalanced by resistance on the respective bodies, and inversely as the weights of the bodies.To express this symbolically, let W1, W2be the weights of the bodies; P the effort exerted between them; S the distance through which it acts; R1, R2the resistances opposed to the effort overcome by W1, W2respectively; E1, E2the shares of the whole energy E exerted upon W1, W2respectively. ThenE:E1:E2::W2(P − R1) + W1(P − R2):P − R1:P − R2.W1W2W1W2(80)If R1= R2, which is the case when the resistance, as well as the effort, arises from the mutual actions of the two bodies, the above becomes,E : E1: E2:: W1+ W2: W2: W1,(81)that is to say, the energy is exerted on the bodies in shares inversely proportional to their weights; and they receive accelerations inversely proportional to their weights, according to the principle of dynamics, already quoted in a note to § 110, that the mutual actions of a system of bodies do not affect the motion of their common centre of gravity.For example, if the weight of a gun be 160 times that of its ball160⁄161of the energy exerted by the powder in exploding will be employed in propelling the ball, and1⁄161in producing the recoil of the gun, provided the gun up to the instant of the ball’s quitting the muzzle meets with no resistance to its recoil except the friction of the ball.§ 124.Centre of Percussion.—It is obviously desirable that the deviations or changes of motion of oscillating pieces in machinery should, as far as possible, be effected by forces applied at their centres of percussion.If the deviation be atranslation—that is, an equal change of motion of all the particles of the body—the centre of percussion is obviously the centre of gravity itself; and, according to the second law of motion, if dv be the deviation of velocity to be produced in the interval dt, and W the weight of the body, thenP =W·dvgdt(82)is the unbalanced effort required.If the deviation be a rotation about an axis traversing the centre of gravity, there is no centre of percussion; for such a deviation can only be produced by acoupleof forces, and not by any single force. Let dα be the deviation of angular velocity to be produced in the interval dt, and I the moment of the inertia of the body about an axis through its centre of gravity; then1⁄2Id(α2) = Iα dα is the variation of the body’s actual energy. Let M be the moment of the unbalanced couple required to produce the deviation; then by equation 57, § 104, the energy exerted by this couple in the interval dt is Mα dt, which, being equated to the variation of energy, givesM = Idα=R2W·dα.dtgdt(83)R is called the radius of gyration of the body with regard to an axis through its centre of gravity.Fig. 133.Now (fig. 133) let the required deviation be a rotation of the body BB about an axis O, not traversing the centre of gravity G, dαbeing, as before, the deviation of angular velocity to be produced in the interval dt. A rotation with the angular velocity α about an axis O may be considered as compounded of a rotation with the same angular velocity about an axis drawn through G parallel to O and a translation with the velocity α. OG, OG being the perpendicular distance between the two axes. Hence the required deviation may be regarded as compounded of a deviation of translation dv = OG · dα, to produce which there would be required, according to equation (82), a force applied at G perpendicular to the plane OG—P =W· OG ·dαgdt(84)and a deviation dα of rotation about an axis drawn through G parallel to O, to produce which there would be required a couple of the moment M given by equation (83). According to the principles of statics, the resultant of the force P, applied at G perpendicular to the plane OG, and the couple M is a force equal and parallel to P, but applied at a distance GC from G, in the prolongation of the perpendicular OG, whose value isGC = M / P = R2/ OG.(85)Thus is determined the position of the centre of percussion C, corresponding to the axis of rotation O. It is obvious from this equation that, for an axis of rotation parallel to O traversing C, the centre of percussion is at the point where the perpendicular OG meets O.§ 125.*To find the moment of inertia of a body about an axis through its centre of gravity experimentally.—Suspend the body from any conveniently selected axis O (fig. 48) and hang near it a small plumb bob. Adjust the length of the plumb-line until it and the body oscillate together in unison. The length of the plumb-line, measured from its point of suspension to the centre of the bob, is for all practical purposes equal to the length OC, C being therefore the centre of percussion corresponding to the selected axis O. From equation (85)R2= CG × OG = (OC − OG) OG.The position of G can be found experimentally; hence OG is known, and the quantity R2can be calculated, from which and the ascertained weight W of the body the moment of inertia about an axis through G, namely, W/g × R2, can be computed.Fig. 134.§ 126.*To find the force competent to produce the instantaneous acceleration of any link of a mechanism.—In many practical problems it is necessary to know the magnitude and position of the forces acting to produce the accelerations of the several links of a mechanism. For a given link, this force is the resultant of all the accelerating forces distributed through the substance of the material of the link required to produce the requisite acceleration of each particle, and the determination of this force depends upon the principles of the two preceding sections. The investigation of the distribution of the forces through the material and the stress consequently produced belongs to the subject of theStrength of Materials(q.v.). Let BK (fig. 134) be any link moving in any manner in a plane, and let G be its centre of gravity. Then its motion may be analysed into (1) a translation of its centre of gravity; and (2) a rotation about an axis through its centre of gravity perpendicular to its plane of motion. Let α be the acceleration of the centre of gravity and let A be the angular acceleration about the axis through the centre of gravity; then the force required to produce the translation of the centre of gravity is F = Wα/g, and the couple required to produce the angular acceleration about the centre of gravity is M = IA/g, W and I being respectively the weight and the moment of inertia of the link about the axis through the centre of gravity. The couple M may be produced by shifting the force F parallel to itself through a distance x. such that Fx = M. When the link forms part of a mechanism the respective accelerations of two points in the link can be determined by means of the velocity and acceleration diagrams described in § 82, it being understood that the motion of one link in the mechanism is prescribed, for instance, in the steam-engine’s mechanism that the crank shall revolve uniformly. Let the acceleration of the two points B and K therefore be supposed known. The problem is now to find the acceleration α and A. Take any pole O (fig. 49), and set out Ob equal to the acceleration of B and Ok equal to the acceleration of K. Join bk and take the point g so that KG: GB = kg : gb. Og is then the acceleration of the centre of gravity and the force F can therefore be immediately calculated. To find the angular acceleration A, draw kt, bt respectively parallel to and at right angles to the link KB. Then tb represents the angular acceleration of the point B relatively to the point K and hence tb/KB is the value of A, the angular acceleration of the link. Its moment of inertia about G can be found experimentally by the method explained in § 125, and then the value of the couple M can be computed. The value of x is found immediately from the quotient M/F. Hence the magnitude F and the position of F relatively to the centre of gravity of the link, necessary to give rise to the couple M, are known, and this force is therefore the resultant force required.Fig. 135.§ 127.*Alternative construction for finding the position of F relatively to the centre of gravity of the link.—Let B and K be any two points in the link which for greater generality are taken in fig. 135, so that the centre of gravity G is not in the line joining them. First find the value of R experimentally. Then produce the given directions of acceleration of B and K to meet in O; draw a circle through the three points B, K and O; produce the line joining O and G to cut the circle in Y; and take a point Z on the line OY so that YG × GZ = R2. Then Z is a point in the line of action of the force F. This useful theorem is due to G. T. Bennett, of Emmanuel College, Cambridge. A proof of it and three corollaries are given in appendix 4 of the second edition of Dalby’sBalancing of Engines(London, 1906). It is to be noticed that only the directions of the accelerations of two points are required to find the point Z.For an example of the application of the principles of the two preceding sections to a practical problem seeValve and Valve Gear Mechanisms, by W. E. Dalby (London, 1906), where the inertia stresses brought upon the several links of a Joy valve gear, belonging to an express passenger engine of the Lancashire & Yorkshire railway, are investigated for an engine-speed of 68 m. an hour.Fig. 136.§ 128.*The Connecting Rod Problem.—A particular problem of practical importance is the determination of the force producing the motion of the connecting rod of a steam-engine mechanism of the usual type. The methods of the two preceding sections may be used when the acceleration of two points in the rod are known. In this problem it is usually assumed that the crank pin K (fig. 136) moves with uniform velocity, so that if α is its angular velocity and r its radius, the acceleration is α2r in a direction along the crank arm from the crank pin to the centre of the shaft. Thus the acceleration of one point K is known completely. The acceleration of a second point, usually taken at the centre of the crosshead pin, can be found by the principles of § 82, but several special geometrical constructions have been devised for this purpose, notably the construction of Klein,4discovered also independently by Kirsch.5But probably the most convenient is the construction due to G. T. Bennett6which is as follows: Let OK be the crank and KB the connecting rod. On the connecting rod take a point L such that KL × KB = KO2. Then, the crank standing at any angle with the line of stroke, draw LP at right angles to the connecting rod, PN at right angles to the line of stroke OB and NA at right angles to the connecting rod; then AO is the acceleration of the point B to the scale on which KO represents the acceleration of the point K. The proof of this construction is given inThe Balancing of Engines.The finding of F may be continued thus: join AK, then AK is the acceleration image of the rod, OKA being the acceleration diagram. Through G, the centre of gravity of the rod, draw Gg parallel to the line of stroke, thus dividing the image at g in the proportion that the connecting rod is divided by G. Hence Og represents the acceleration of the centre of gravity and, the weight of the connectingrod being ascertained, F can be immediately calculated. To find a point in its line of action, take a point Q on the rod such that KG × GQ = R2, R having been determined experimentally by the method of § 125; join G with O and through Q draw a line parallel to BO to cut GO in Z. Z is a point in the line of action of the resultant force F; hence through Z draw a line parallel to Og. The force F acts in this line, and thus the problem is completely solved. The above construction for Z is a corollary of the general theorem given in § 127.§ 129.Impact.Impact or collision is a pressure of short duration exerted between two bodies.The effects of impact are sometimes an alteration of the distribution of actual energy between the two bodies, and always a loss of a portion of that energy, depending on the imperfection of the elasticity of the bodies, in permanently altering their figures, and producing heat. The determination of the distribution of the actual energy after collision and of the loss of energy is effected by means of the following principles:—I. The motion of the common centre of gravity of the two bodies is unchanged by the collision.II. The loss of energy consists of a certain proportion of that part of the actual energy of the bodies which is due to their motion relatively to their common centre of gravity.Unless there is some special reason for using impact in machines, it ought to be avoided, on account not only of the waste of energy which it causes, but from the damage which it occasions to the frame and mechanism.
(a)
The plane through O to which the shaft is perpendicular is called thereference plane, because all the transferred forces act in that plane at the point O. The plane through the radius of the weight containing the axis OX is called theaxial planebecause it contains the forces forming the couple due to the transference of F to the reference plane. Substituting the values of F in (a) the two conditions become
(b)
In order that these conditions may obtain, the quantities in the brackets must be zero, since the factor α2/g is not zero. Hence finally the conditions which must be satisfied by the system of weights in order that the axis of rotation may be a permanent axis is
(c)
It must be remembered that these are all directed quantities, and that their respective sums are to be taken by drawing vector polygons. In drawing these polygons the magnitude of the vector of the type Wr is the product Wr, and the direction of the vector is from the shaft outwards towards the weight W, parallel to the radius r. For the vector representing a couple of the type War, if the masses are all on the same side of the reference plane, the direction of drawing is from the axis outwards; if the masses are some on one side of the reference plane and some on the other side, the direction of drawing is from the axis outwards towards the weight for all masses on the one side, and from the mass inwards towards the axis for all weights on the other side, drawing always parallel to the direction defined by the radius r. The magnitude of the vector is the product War. The conditions (c) may thus be expressed: first, that the sum of the vectors Wr must form a closed polygon, and, second, that the sum of the vectors War must form a closed polygon. The general problem in practice is, given a system of weights attached to a shaft, to find the respective weights and positions of two balance weights or counterpoises which must be added to the system in order to make the shaft a permanent axis, the planes in which the balance weights are to revolve also being given. To solve this the reference plane must be chosen so that it coincides with the plane of revolution of one of the as yet unknown balance weights. The balance weight in this plane has therefore no couple corresponding to it. Hence by drawing a couple polygon for the given weights the vector which is required to close the polygon is at once found and from it the magnitude and position of the balance weight which must be added to the system to balance the couples follow at once. Then, transferring the product Wr corresponding with this balance weight to the reference plane, proceed to draw the force polygon. The vector required to close it will determine the second balance weight, the work may be checked by taking the reference plane to coincide with the plane of revolution of the second balance weight and then re-determining them, or by taking a reference plane anywhere and including the two balance weights trying if condition (c) is satisfied.
When a weight is reciprocated, the equal and opposite force required for its acceleration at any instant appears as an unbalanced force on the frame of the machine to which the weight belongs. In the particular case, where the motion is of the kind known as “simple harmonic” the disturbing force on the frame due to the reciprocation of the weight is equal to the component of the centrifugal force in the line of stroke due to a weight equal to the reciprocated weight supposed concentrated at the crank pin. Using this principle the method of finding the balance weights to be added to a given system of reciprocating weights in order to produce a system of forces on the frame continuously in equilibrium is exactly the same as that just explained for a system of revolving weights, because for the purpose of finding the balance weights each reciprocating weight may be supposed attached to the crank pin which operates it, thus forming an equivalent revolving system. The balance weights found as part of the equivalent revolving system when reciprocated by their respective crank pins form the balance weights for the given reciprocating system. These conditions may be exactly realized by a system of weights reciprocated by slotted bars, the crank shaft driving the slotted bars rotating uniformly. In practice reciprocation is usually effected through a connecting rod, as in the case of steam engines. In balancing the mechanism of a steam engine it is often sufficiently accurate to consider the motion of the pistons as simple harmonic, and the effect on the framework of the acceleration of the connecting rod may be approximately allowed for by distributing the weight of the rod between the crank pin and the piston inversely as the centre of gravity of the rod divides the distance between the centre of the cross head pin and the centre of the crank pin. The moving parts of the engine are then divided into two complete and independent systems, namely, one system of revolving weights consisting of crank pins, crank arms, &c., attached to and revolving with the crank shaft, and a second system of reciprocating weights consisting of the pistons, cross-heads, &c., supposed to be moving each in its line of stroke with simple harmonic motion. The balance weights are to be separately calculated for each system, the one set being added to the crank shaft as revolving weights, and the second set being included with the reciprocating weights and operated by a properly placed crank on the crank shaft. Balance weights added in this way to a set of reciprocating weights are sometimes called bob-weights. In the case of locomotives the balance weights required to balance the pistons are added as revolving weights to the crank shaft system, and in fact are generally combined with the weights required to balance the revolving system so as to form one weight, the counterpoise referred to in the preceding section, which is seen between the spokes of the wheels of a locomotive. Although this method balances the pistons in the horizontal plane, and thus allows the pull of the engine on the train to be exerted without the variation due to the reciprocation of the pistons, yet the force balanced horizontally is introduced vertically and appears as a variation of pressure on the rail. In practice about two-thirds of the reciprocating weight is balanced in order to keep this variation of rail pressure within safe limits. The assumption that the pistons of an engine move with simple harmonic motion is increasingly erroneous as the ratio of the length of the crank r, to the length of the connecting rod l increases. A more accurate though still approximate expression for the force on the frame due to the acceleration of the piston whose weight is W is given by
The conditions regulating the balancing of a system of weights reciprocating under the action of accelerating forces given by the above expression are investigated in a paper by Otto Schlick, “On Balancing of Steam Engines,”Trans, Inst. Nav. Arch.(1900), and in a paper by W. E. Dalby, “On the Balancing of the Reciprocating Parts of Engines, including the Effect of the Connecting Rod” (ibid., 1901). A still more accurate expression than the above is obtained by expansion in a Fourier series, regarding which and its bearing on balancing engines see a paper by J. H. Macalpine, “A Solution of the Vibration Problem” (ibid., 1901). The whole subject is dealt with in a treatise,The Balancing of Engines, by W. E. Dalby (London, 1906). Most of the original papers on this subject of engine balancing are to be found in theTransactionsof the Institution of Naval Architects.
§ 113.*Centrifugal Whirling of Shafts.—When a system of revolving masses is balanced so that the conditions of the preceding section are fulfilled, the centre of gravity of the system lies on the axis of revolution. If there is the slightest displacement of the centre of gravity of the system from the axis of revolution a force acts on the shaft tending to deflect it, and varies as the deflexion and as the square of the speed. If the shaft is therefore to revolve stably, this force must be balanced at any instant by the elastic resistance of the shaft to deflexion. To take a simple case, suppose a shaft,supported on two bearings to carry a disk of weight W at its centre, and let the centre of gravity of the disk be at a distance e from the axis of rotation, this small distance being due to imperfections of material or faulty construction. Neglecting the mass of the shaft itself, when the shaft rotates with an angular velocity a, the centrifugal force Wa2e/g will act upon the shaft and cause its axis to deflect from the axis of rotation a distance, y say. The elastic resistance evoked by this deflexion is proportional to the deflexion, so that if c is a constant depending upon the form, material and method of support of the shaft, the following equality must hold if the shaft is to rotate stably at the stated speed—
from which y = Wa2e / (gc − Wa2).
This expression shows that as a increases y increases until when Wa2= gc, y becomes infinitely large. The corresponding value of a, namely √(gc/W), is called thecritical velocityof the shaft, and is the speed at which the shaft ceases to rotate stably and at which centrifugal whirling begins. The general problem is to find the value of a corresponding to all kinds of loadings on shafts supported in any manner. The question was investigated by Rankine in an article in theEngineer(April 9, 1869). Professor A. G. Greenhill treated the problem of the centrifugal whirling of an unloaded shaft with different supporting conditions in a paper “On the Strength of Shafting exposed both to torsion and to end thrust,”Proc. Inst. Mech. Eng.(1883). Professor S. Dunkerley (“On the Whirling and Vibration of Shafts,”Phil. Trans., 1894) investigated the question for the cases of loaded and unloaded shafts, and, owing to the complication arising from the application of the general theory to the cases of loaded shafts, devised empirical formulae for the critical speeds of shafts loaded with heavy pulleys, based generally upon the following assumption, which is stated for the case of a shaft carrying one pulley: If N1, N2be the separate speeds of whirl of the shaft and pulley on the assumption that the effect of one is neglected when that of the other is under consideration, then the resulting speed of whirl due to both causes combined may be taken to be of the form N1N2√(N21+ N12) where N means revolutions per minute. This form is extended to include the cases of several pulleys on the same shaft. The interesting and important part of the investigation is that a number of experiments were made on small shafts arranged in different ways and loaded in different ways, and the speed at which whirling actually occurred was compared with the speed calculated from formulae of the general type indicated above. The agreement between the observed and calculated values of the critical speeds was in most cases quite remarkable. In a paper by Dr C. Chree, “The Whirling and Transverse Vibrations of Rotating Shafts,”Proc. Phys. Soc. Lon., vol. 19 (1904); alsoPhil. Mag., vol. 7 (1904), the question is investigated from a new mathematical point of view, and expressions for the whirling of loaded shafts are obtained without the necessity of any assumption of the kind stated above. An elementary presentation of the problem from a practical point of view will be found inSteam Turbines, by Dr A. Stodola (London, 1905).
§ 114.Revolving Pendulum. Governors.—In fig. 131 AO represents an upright axis or spindle; B a weight called abob, suspended by rod OB from a horizontal axis at O, carried by the vertical axis. When the spindle is at rest the bob hangs close to it; when the spindle rotates, the bob, being made to revolve round it, diverges until the resultant of the centrifugal force and the weight of the bob is a force acting at O in the direction OB, and then it revolves steadily in a circle. This combination is called arevolving,centrifugal, orconical pendulum. Revolving pendulums are usually constructed withpairsof rods and bobs, as OB, Ob, hung at opposite sides of the spindle, that the centrifugal forces exerted at the point O may balance each other.
In finding the position in which the bob will revolve with a given angular velocity, a, for most practical cases connected with machinery the mass of the rod may be considered as insensible compared with that of the bob. Let the bob be a sphere, and from the centre of that sphere draw BH = y perpendicular to OA. Let OH = z; let W be the weight of the bob, F its centrifugal force. Then the condition of its steady revolution is W : F :: z : y; that is to say, y/z = F/W = yα2/g; consequently
z = g/α2
(69)
Or, if n = α 2π = α/6.2832 be the number of turns or fractions of a turn in a second,
(70)
z is called thealtitude of the pendulum.
If the rod of a revolving pendulum be jointed, as in fig. 132, not to a point in the vertical axis, but to the end of a projecting arm C, the position in which the bob will revolve will be the same as if the rod were jointed to the point O, where its prolongation cuts the vertical axis.
A revolving pendulum is an essential part of most of the contrivances calledgovernors, for regulating the speed of prime movers, for further particulars of which seeSteam Engine.
Division 3. Working of Machines of Varying Velocity.
§ 115.General Principles.—In order that the velocity of every piece of a machine may be uniform, it is necessary that the forces acting on each piece should be always exactly balanced. Also, in order that the forces acting on each piece of a machine may be always exactly balanced, it is necessary that the velocity of that piece should be uniform.
An excess of the effort exerted on any piece, above that which is necessary to balance the resistance, is accompanied with acceleration; a deficiency of the effort, with retardation.
When a machine is being started from a state of rest, and brought by degrees up to its proper speed, the effort must be in excess; when it is being retarded for the purpose of stopping it, the resistance must be in excess.
An excess of effort above resistance involves an excess of energy exerted above work performed; that excess of energy is employed in producing acceleration.
An excess of resistance above effort involves an excess of work performed above energy expended; that excess of work is performed by means of the retardation of the machinery.
When a machine undergoes alternate acceleration and retardation, so that at certain instants of time, occurring at the end of intervals calledperiodsorcycles, it returns to its original speed, then in each of those periods or cycles the alternate excesses of energy and of work neutralize each other; and at the end of each cycle the principle of the equality of energy and work stated in § 87, with all its consequences, is verified exactly as in the case of machines of uniform speed.
At intermediate instants, however, other principles have also to be taken into account, which are deduced from the second law of motion, as applied todirect deviation, or acceleration and retardation.
§ 116.Energy of Acceleration and Work of Retardation for a Shifting Body.—Let w be the weight of a body which has a motion of translation in any path, and in the course of the interval of time Δt let its velocity be increased at a uniform rate of acceleration from v1to v2. The rate of acceleration will be
dv/dt = const. = (v2− v1) Δt;
and to produce this acceleration a uniform effort will be required, expressed by
P = w (v2− v1) gΔt
(71)
(The product wv/g of the mass of a body by its velocity is called itsmomentum; so that the effort required is found by dividing the increase of momentum by the time in which it is produced.)
To find theenergywhich has to be exerted to produce the acceleration from v1to v2, it is to be observed that thedistancethrough which the effort P acts during the acceleration is
Δs = (v2+ v1) Δt/2;
consequently, theenergy of accelerationis
PΔs = w (v2− v1) (v2+ v1) / 2g = w (v22− v12) 2g,
(72)
being proportional to the increase in the square of the velocity, andindependent of the time.
In order to produce aretardationfrom the greater velocity v2to the less velocity v1, it is necessary to apply to the body aresistanceconnected with the retardation and the time by an equation identical in every respect with equation (71), except by the substitution of a resistance for an effort; and in overcoming that resistance the bodyperforms workto an amount determined by equation (72), putting Rds for Pas.
§ 117.Energy Stored and Restored by Deviations of Velocity.—Thus a body alternately accelerated and retarded, so as to be brought back to its original speed, performs work during its retardation exactly equal in amount to the energy exerted upon it during its acceleration; so that that energy may be considered asstoredduring the acceleration, andrestoredduring the retardation, in a manner analogous to the operation of a reciprocating force (§ 108).
Let there be given the mean velocity V =1⁄2(v2+ v1) of a body whose weight is w, and let it be required to determine the fluctuation of velocity v2− v1, and the extreme velocities v1, v2, which that body must have, in order alternately to store and restore an amount of energy E. By equation (72) we have
E = w (v22− v12) / 2g
which, being divided by V =1⁄2(v2+ v1), gives
E/V = w (v2− v1) / g;
and consequently
v2− v1= gE / Vw
(73)
The ratio of this fluctuation to the mean velocity, sometimes called the unsteadiness of the motion of the body, is
(v2− v1) V = gE / V2w.
(74)
§ 118.Actual Energy of a Shifting Body.—The energy which must be exerted on a body of the weight w, to accelerate it from a state of rest up to a given velocity of translation v, and the equal amount of work which that body is capable of performing by overcoming resistance while being retarded from the same velocity of translation v to a state of rest, is
wv2/ 2g.
(75)
This is called theactual energyof the motion of the body, and is half the quantity which in some treatises is called vis viva.
The energy stored or restored, as the case may be, by the deviations of velocity of a body or a system of bodies, is the amount by which the actual energy is increased or diminished.
§ 119.Principle of the Conservation of Energy in Machines.—The following principle, expressing the general law of the action of machines with a velocity uniform or varying, includes the law of the equality of energy and work stated in § 89 for machines of uniform speed.
In any given interval during the working of a machine, the energy exerted added to the energy restored is equal to the energy stored added to the work performed.
§ 120.Actual Energy of Circular Translation—Moment of Inertia.—Let a small body of the weight w undergo translation in a circular path of the radius ρ, with the angular velocity of deflexion α, so that the common linear velocity of all its particles is v = αρ. Then the actual energy of that body is
wv2/ 2g = wa2ρ2/ 2g.
(76)
By comparing this with the expression for the centrifugal force (wa2ρ/g), it appears that the actual energy of a revolving body is equal to the potential energy Fρ/2 due to the action of the deflecting force along one-half of the radius of curvature of the path of the body.
The product wρ2/g, by which the half-square of the angular velocity is multiplied, is called themoment of inertiaof the revolving body.
§ 121.Flywheels.—A flywheel is a rotating piece in a machine, generally shaped like a wheel (that is to say, consisting of a rim with spokes), and suited to store and restore energy by the periodical variations in its angular velocity.
The principles according to which variations of angular velocity store and restore energy are the same as those of § 117, only substitutingmoment of inertiaformass, andangularforlinearvelocity.
Let W be the weight of a flywheel, R its radius of gyration, a2its maximum, a1its minimum, and A =1⁄2(α2+ α1) its mean angular velocity. Let
I/S = (α2− α2) / A
denote theunsteadinessof the motion of the flywheel; the denominator S of this fraction is called thesteadiness. Let e denote the quantity by which the energy exerted in each cycle of the working of the machine alternately exceeds and falls short of the work performed, and which has consequently to be alternately stored by acceleration and restored by retardation of the flywheel. The value of thisperiodical excessis—
e = R2W (α22− α12), 2g,
(77)
from which, dividing both sides by A2, we obtain the following equations:—
e / A2= R2W / gSR2WA2/ 2g = Se / 2.
(78)
The latter of these equations may be thus expressed in words:The actual energy due to the rotation of the fly, with its mean angular velocity, is equal to one-half of the periodical excess of energy multiplied by the steadiness.
In ordinary machinery S = about 32; in machinery for fine purposes S = from 50 to 60; and when great steadiness is required S = from 100 to 150.
The periodical excess e may arise either from variations in the effort exerted by the prime mover, or from variations in the resistance of the work, or from both these causes combined. When but one flywheel is used, it should be placed in as direct connexion as possible with that part of the mechanism where the greatest amount of the periodical excess originates; but when it originates at two or more points, it is best to have a flywheel in connexion with each of these points. For example, in a machine-work, the steam-engine, which is the prime mover of the various tools, has a flywheel on the crank-shaft to store and restore the periodical excess of energy arising from the variations in the effort exerted by the connecting-rod upon the crank; and each of the slotting machines, punching machines, riveting machines, and other tools has a flywheel of its own to store and restore energy, so as to enable the very different resistances opposed to those tools at different times to be overcome without too great unsteadiness of motion. For tools performing useful work at intervals, and having only their own friction to overcome during the intermediate intervals, e should be assumed equal to the whole work performed at each separate operation.
§ 122.Brakes.—A brake is an apparatus for stopping and diminishing the velocity of a machine by friction, such as the friction-strap already referred to in § 103. To find the distance s through which a brake, exerting the friction F, must rub in order to stop a machine having the total actual energy E at the moment when the brake begins to act, reduce, by the principles of § 96, the various efforts and other resistances of the machine which act at the same time with the friction of the brake to the rubbing surface of the brake, and let R be their resultant—positive ifresistance,negativeif effort preponderates. Then
s = E / (F + R).
(79)
§ 123.Energy distributed between two Bodies: Projection and Propulsion.—Hitherto the effort by which a machine is moved has been treated as a force exerted between a movable body and a fixed body, so that the whole energy exerted by it is employed upon the movable body, and none upon the fixed body. This conception is sensibly realized in practice when one of the two bodies between which the effort acts is either so heavy as compared with the other, or has so great a resistance opposed to its motion, that it may, without sensible error, be treated as fixed. But there are cases in which the motions of both bodies are appreciable, and must be taken into account—such as the projection of projectiles, where the velocity of therecoilor backward motion of the gun bears an appreciable proportion to the forward motion of the projectile; and such as the propulsion of vessels, where the velocity of the water thrown backward by the paddle, screw or other propeller bears a very considerable proportion to the velocity of the water moved forwards and sideways by the ship. In cases of this kind the energy exerted by the effort isdistributedbetween the two bodies between which the effort is exerted in shares proportional to the velocities of the two bodies during the action of the effort; and those velocities are to each other directly as the portions of the effort unbalanced by resistance on the respective bodies, and inversely as the weights of the bodies.
To express this symbolically, let W1, W2be the weights of the bodies; P the effort exerted between them; S the distance through which it acts; R1, R2the resistances opposed to the effort overcome by W1, W2respectively; E1, E2the shares of the whole energy E exerted upon W1, W2respectively. Then
(80)
If R1= R2, which is the case when the resistance, as well as the effort, arises from the mutual actions of the two bodies, the above becomes,
E : E1: E2:: W1+ W2: W2: W1,
(81)
that is to say, the energy is exerted on the bodies in shares inversely proportional to their weights; and they receive accelerations inversely proportional to their weights, according to the principle of dynamics, already quoted in a note to § 110, that the mutual actions of a system of bodies do not affect the motion of their common centre of gravity.
For example, if the weight of a gun be 160 times that of its ball160⁄161of the energy exerted by the powder in exploding will be employed in propelling the ball, and1⁄161in producing the recoil of the gun, provided the gun up to the instant of the ball’s quitting the muzzle meets with no resistance to its recoil except the friction of the ball.
§ 124.Centre of Percussion.—It is obviously desirable that the deviations or changes of motion of oscillating pieces in machinery should, as far as possible, be effected by forces applied at their centres of percussion.
If the deviation be atranslation—that is, an equal change of motion of all the particles of the body—the centre of percussion is obviously the centre of gravity itself; and, according to the second law of motion, if dv be the deviation of velocity to be produced in the interval dt, and W the weight of the body, then
(82)
is the unbalanced effort required.
If the deviation be a rotation about an axis traversing the centre of gravity, there is no centre of percussion; for such a deviation can only be produced by acoupleof forces, and not by any single force. Let dα be the deviation of angular velocity to be produced in the interval dt, and I the moment of the inertia of the body about an axis through its centre of gravity; then1⁄2Id(α2) = Iα dα is the variation of the body’s actual energy. Let M be the moment of the unbalanced couple required to produce the deviation; then by equation 57, § 104, the energy exerted by this couple in the interval dt is Mα dt, which, being equated to the variation of energy, gives
(83)
R is called the radius of gyration of the body with regard to an axis through its centre of gravity.
Now (fig. 133) let the required deviation be a rotation of the body BB about an axis O, not traversing the centre of gravity G, dαbeing, as before, the deviation of angular velocity to be produced in the interval dt. A rotation with the angular velocity α about an axis O may be considered as compounded of a rotation with the same angular velocity about an axis drawn through G parallel to O and a translation with the velocity α. OG, OG being the perpendicular distance between the two axes. Hence the required deviation may be regarded as compounded of a deviation of translation dv = OG · dα, to produce which there would be required, according to equation (82), a force applied at G perpendicular to the plane OG—
(84)
and a deviation dα of rotation about an axis drawn through G parallel to O, to produce which there would be required a couple of the moment M given by equation (83). According to the principles of statics, the resultant of the force P, applied at G perpendicular to the plane OG, and the couple M is a force equal and parallel to P, but applied at a distance GC from G, in the prolongation of the perpendicular OG, whose value is
GC = M / P = R2/ OG.
(85)
Thus is determined the position of the centre of percussion C, corresponding to the axis of rotation O. It is obvious from this equation that, for an axis of rotation parallel to O traversing C, the centre of percussion is at the point where the perpendicular OG meets O.
§ 125.*To find the moment of inertia of a body about an axis through its centre of gravity experimentally.—Suspend the body from any conveniently selected axis O (fig. 48) and hang near it a small plumb bob. Adjust the length of the plumb-line until it and the body oscillate together in unison. The length of the plumb-line, measured from its point of suspension to the centre of the bob, is for all practical purposes equal to the length OC, C being therefore the centre of percussion corresponding to the selected axis O. From equation (85)
R2= CG × OG = (OC − OG) OG.
The position of G can be found experimentally; hence OG is known, and the quantity R2can be calculated, from which and the ascertained weight W of the body the moment of inertia about an axis through G, namely, W/g × R2, can be computed.
§ 126.*To find the force competent to produce the instantaneous acceleration of any link of a mechanism.—In many practical problems it is necessary to know the magnitude and position of the forces acting to produce the accelerations of the several links of a mechanism. For a given link, this force is the resultant of all the accelerating forces distributed through the substance of the material of the link required to produce the requisite acceleration of each particle, and the determination of this force depends upon the principles of the two preceding sections. The investigation of the distribution of the forces through the material and the stress consequently produced belongs to the subject of theStrength of Materials(q.v.). Let BK (fig. 134) be any link moving in any manner in a plane, and let G be its centre of gravity. Then its motion may be analysed into (1) a translation of its centre of gravity; and (2) a rotation about an axis through its centre of gravity perpendicular to its plane of motion. Let α be the acceleration of the centre of gravity and let A be the angular acceleration about the axis through the centre of gravity; then the force required to produce the translation of the centre of gravity is F = Wα/g, and the couple required to produce the angular acceleration about the centre of gravity is M = IA/g, W and I being respectively the weight and the moment of inertia of the link about the axis through the centre of gravity. The couple M may be produced by shifting the force F parallel to itself through a distance x. such that Fx = M. When the link forms part of a mechanism the respective accelerations of two points in the link can be determined by means of the velocity and acceleration diagrams described in § 82, it being understood that the motion of one link in the mechanism is prescribed, for instance, in the steam-engine’s mechanism that the crank shall revolve uniformly. Let the acceleration of the two points B and K therefore be supposed known. The problem is now to find the acceleration α and A. Take any pole O (fig. 49), and set out Ob equal to the acceleration of B and Ok equal to the acceleration of K. Join bk and take the point g so that KG: GB = kg : gb. Og is then the acceleration of the centre of gravity and the force F can therefore be immediately calculated. To find the angular acceleration A, draw kt, bt respectively parallel to and at right angles to the link KB. Then tb represents the angular acceleration of the point B relatively to the point K and hence tb/KB is the value of A, the angular acceleration of the link. Its moment of inertia about G can be found experimentally by the method explained in § 125, and then the value of the couple M can be computed. The value of x is found immediately from the quotient M/F. Hence the magnitude F and the position of F relatively to the centre of gravity of the link, necessary to give rise to the couple M, are known, and this force is therefore the resultant force required.
§ 127.*Alternative construction for finding the position of F relatively to the centre of gravity of the link.—Let B and K be any two points in the link which for greater generality are taken in fig. 135, so that the centre of gravity G is not in the line joining them. First find the value of R experimentally. Then produce the given directions of acceleration of B and K to meet in O; draw a circle through the three points B, K and O; produce the line joining O and G to cut the circle in Y; and take a point Z on the line OY so that YG × GZ = R2. Then Z is a point in the line of action of the force F. This useful theorem is due to G. T. Bennett, of Emmanuel College, Cambridge. A proof of it and three corollaries are given in appendix 4 of the second edition of Dalby’sBalancing of Engines(London, 1906). It is to be noticed that only the directions of the accelerations of two points are required to find the point Z.
For an example of the application of the principles of the two preceding sections to a practical problem seeValve and Valve Gear Mechanisms, by W. E. Dalby (London, 1906), where the inertia stresses brought upon the several links of a Joy valve gear, belonging to an express passenger engine of the Lancashire & Yorkshire railway, are investigated for an engine-speed of 68 m. an hour.
§ 128.*The Connecting Rod Problem.—A particular problem of practical importance is the determination of the force producing the motion of the connecting rod of a steam-engine mechanism of the usual type. The methods of the two preceding sections may be used when the acceleration of two points in the rod are known. In this problem it is usually assumed that the crank pin K (fig. 136) moves with uniform velocity, so that if α is its angular velocity and r its radius, the acceleration is α2r in a direction along the crank arm from the crank pin to the centre of the shaft. Thus the acceleration of one point K is known completely. The acceleration of a second point, usually taken at the centre of the crosshead pin, can be found by the principles of § 82, but several special geometrical constructions have been devised for this purpose, notably the construction of Klein,4discovered also independently by Kirsch.5But probably the most convenient is the construction due to G. T. Bennett6which is as follows: Let OK be the crank and KB the connecting rod. On the connecting rod take a point L such that KL × KB = KO2. Then, the crank standing at any angle with the line of stroke, draw LP at right angles to the connecting rod, PN at right angles to the line of stroke OB and NA at right angles to the connecting rod; then AO is the acceleration of the point B to the scale on which KO represents the acceleration of the point K. The proof of this construction is given inThe Balancing of Engines.
The finding of F may be continued thus: join AK, then AK is the acceleration image of the rod, OKA being the acceleration diagram. Through G, the centre of gravity of the rod, draw Gg parallel to the line of stroke, thus dividing the image at g in the proportion that the connecting rod is divided by G. Hence Og represents the acceleration of the centre of gravity and, the weight of the connectingrod being ascertained, F can be immediately calculated. To find a point in its line of action, take a point Q on the rod such that KG × GQ = R2, R having been determined experimentally by the method of § 125; join G with O and through Q draw a line parallel to BO to cut GO in Z. Z is a point in the line of action of the resultant force F; hence through Z draw a line parallel to Og. The force F acts in this line, and thus the problem is completely solved. The above construction for Z is a corollary of the general theorem given in § 127.
§ 129.Impact.Impact or collision is a pressure of short duration exerted between two bodies.
The effects of impact are sometimes an alteration of the distribution of actual energy between the two bodies, and always a loss of a portion of that energy, depending on the imperfection of the elasticity of the bodies, in permanently altering their figures, and producing heat. The determination of the distribution of the actual energy after collision and of the loss of energy is effected by means of the following principles:—
I. The motion of the common centre of gravity of the two bodies is unchanged by the collision.
II. The loss of energy consists of a certain proportion of that part of the actual energy of the bodies which is due to their motion relatively to their common centre of gravity.
Unless there is some special reason for using impact in machines, it ought to be avoided, on account not only of the waste of energy which it causes, but from the damage which it occasions to the frame and mechanism.