of A and B have the same frequency. The advance angles of the cranks oa and ob are a and ẞ respectively. C is a point in AB or in AB produced either way. The vertical component of the motion of C is obviously a simple harmonic motion and it is required to find the representative crank for this motion. From a point o' draw o'a' parallel and equal to oa. Draw o'b' parallel and equal to ob. Join a'b' and divide it at c' so that a'c': c'b' :: AC: CB. Then o'c' is the crank required and y is its advance angle. The proof is as follows. Draw e'r parallel to b'o' to meet o'a' at r. Draw c's parallel to a'o' to meet o'b' at 8. Let the vertical motion of B be destroyed, then the vertical motion of C will be to that of A as CB: AB or as cb': a'b' or as o'r o'a'. o'r is therefore the representative crank for the vertical harmonic motion of C when B has no vertical motion. In like manner when the vertical motion of A is destroyed and B is driven, the representative crank for the vertical motion of C will be o's. Hence, when A and B are both driven the representative crank for the vertical motion of C will be o'c' which is the diagonal of the parallelogram o'rc's as was proved in Art. 129. Denoting the lengths of the cranks oa, ob, and oc by a, b, and c respectively the equations for the vertical displacements of A, B, and C from their mean positions are y1 = a sin (0+ a), y = b sin (0 + B), and y1 = c sin (0 + y) where is measured from the initial positions oa, ob, and oc. The above problem occurs in connection with reciprocating steam engine valve gears. oa and ob are eccentrics which drive the link AB. The valve is driven from a point C in AB. o'c' is the equivalent eccentric, that is, o'c' is an eccentric which would give the same motion to the valve, driving it directly, as the two eccentrics and the link AB give it. In a vertical engine if the crank is in the vertical position oY, when oa and ob are in the positions shown, then 180 a is the angle of advance of the eccentric oa, and ẞ is the angle of advance of the eccentric ob. In the actual engine the eccentrics oa and ob are on the same shaft. 132. Velocity and Acceleration in Harmonic Motion.-CP (Fig. 279) is the representative crank for the simple harmonic motions of the points M and N which reciprocate along the vertical and horizontal diameters respectively of the auxiliary circle. Vy Let CP r and let the position of CP be defined by the angle which it makes with CX. Let be the angular velocity of CP and let V be the linear velocity of P. Then V = wr. Let the velocities of M and N when in the positions shown be V, and V, respectively. These velocities are the vertical and horizontal components of V respectively. Hence V = V cos = wr cos = or cos wt, and FIG. 279. I V1 = V sin @ = wr sin 0 = wr sin wt, where t is the time taken by P to travel from X to P. = V If a scale for velocity be chosen such that CP = V, then PM and PN = V. The auxiliary circle therefore becomes a velocity diagram on a displacement base for the simple harmonic motion of the point M or the point N. = If the angular position of CP be measured from a radius CA which makes an angle a with CX and if the angle ACP $ then V, V cos (+ a) = or cos (+ a) = wr cos (wt + a) and V1 = V sin ( + a) = wr sin (+ a) = wr cos (wt + a) where t is now the time taken by P to travel from A to P. = The velocities V, and V, may be plotted on a time t or angle or angle base in exactly the same way as displacements were plotted in Arts. 127 and 128. = V2 = = w2r. Let the The radial acceleration of the point P is f: acceleration of the points M and N be f, and f, respectively. These accelerations are the vertical and horizontal components of ƒ respectively. Hence (Fig. 280) x Since sin 6 = 2, and cos 0 ==- the above equation may be written, g If a scale for acceleration be chosen such that CP = and f = x. = Y = w2y, V2 x = w2x. The acceleration of the point M is shown plotted on the vertical diameter YY, as a base. If the angular position of CP be measured from a radius CA making an angle a with CX and if the angle ACP then for 0 in the above equations substitute + a. The accelerations f, and f, may be plotted on a time t or angle or angle base in exactly the same way as displacements were plotted in Arts. 127 and 128. When a point has a motion which is compounded of two or more simple har r FIG. 280. monic motions the resultant velocity and resultant acceleration at any instant are determined from the component velocities and component accelerations exactly as for displacements. 133. Harmonic Analysis.-It has been shown that the displacement of a point from its mean position, when it has simple harmonic motion in a straight line, is given by the equation y = r sin (0 + a) or by xr cos (0 + a). Again, if a point has periodic motion in a straight line, and if its motion is compounded of a number of simple harmonic motions in which the separate displacements from the mean position are given by the equations ï1 = r1 cos (0 + a1), x2 = r2 cos (20 + a2), x3 = r3 cos (30 + a), and so on, then the resultant displacement is given by the equation— x = r1 cos (0 + a1) + r2 cos (20 + α) + r, cos (30 + a) + and it has been shown how such a resultant displacement may be found graphically and plotted to obtain a curve of resultant displacement on a time t or angle 0 base. If instead of measuring the resultant displacement from the mean position it is measured from a point at a distance c from it, then, x = c + r1 cos (0 + a1) + r2 cos (20 + α2) + r ̧ cos (30 + a3) + The right hand side of this equation is known as a Fourier series. Any periodic curve of displacement being given it may be resolved into a number of simple harmonic components. The number of these components is greater the greater the complexity of the given periodic curve. The process of breaking up a given periodic curve into a number of simple harmonic curves, or the process of finding the equation to the curve in the form x = c + r1 cos (0 + a1) + 11⁄2 cos (20 + a) + r ̧ cos (30 + as) + is called harmonic analysis. The process of harmonic analysis is one of great importance in connection with the study of the periodic motions of machines and of alternating electric currents. A graphic method of harmonic analysis which is comparatively simple and easily applied will now be described. This method is due to Mr. Joseph Harrison and was described by him in Engineering of Feb. 16th, 1906. The procedure will be illustrated by reference to a definite example. The full curve EFGH (Fig. 281) has been plotted on the base line ŎL. The base represents one complete revolution of a shaft upon which there is an eccentric which drives a slide valve through certain intermediate link-work. The ordinates of the curve represent the displacements of the valve from a certain fixed position. In order that the student may transfer this curve accurately and of full size to his drawing paper, and work out the example for himself, the ordinates have been dimensioned. Twelve ordinates at equal intervals will be used and these are numbered 0 to 11. In most engineering problems in harmonic analysis it will be found that not more than three of the harmonic terms are required and in many cases two terms are sufficient. In the example to be worked three terms in addition to the constant will be found. x=c+r, cos (0+ a1) + r2 cos (20+ a2) + r, cos (30+ a3) and 9 a21 az. and it is required to find c, r1, T2, and r3, also a1 tan a1 = b1 a1 Also, let-r, sin a = b1, -r2 sin a2 = b2, and -rs Then x = c + a1 cos 0+ a cos 20+ a, cos 30 +b, sin b2 sin 20 + b, sin 30. Since r1 cos a1 = α19 and -Ti sin α1 = Also r1 = a + b2 and r1 = sin az = b3. b, it follows by division that similar relations for the other corresponding constants. The constant c is the mean of all the given ordinates and in this case it is 163. From an origin O, beginning with the zero direction draw 12 radial lines at equal angular intervals. Along these radial lines, beginning with the zero line, mark off in succession from O, the ordinates 0 to 11. Find the horizontal and vertical projections of these radiating vectors as shown. Find the sum of the horizontal projections having regard to their signs. This sum is 4.00 and is equal to 6a,, the coefficient 6 of a being half the number of ordinates used. Therefore a1 = 0·67. Find the sum of the vertical projections, having regard to their signs. This sum is 7.26 and is equal to 6b. Therefore b1 = 1·21. b1 = = +1.38. and a1 +1.38. Since r1 cos a1 = α1 and cos are here both positive it follows that r1 The first harmonic term or the fundamental term is therefore +1.38 cos (0 – 61°). Next, from an origin O, beginning with the zero direction, draw, at equal angular intervals, half the number of radial lines that were drawn from O1. On these radial lines, beginning with the zero line, mark off in succession from O, the ordinates 0 to 11, going twice round. Find the sums of the horizontal and vertical projections of these radiating vectors. In this case these sums are -1.09 and -0.36 - 0·333 and a2 = -18° to the nearest degree. 019. Since r2 cos α = α, and since cos az is positive and a, is negative r, must be negative. Hence r = -0.19. The second harmonic term or the octave term is therefore -0.19 cos (20 — 18°). Lastly, from an origin O, beginning with the zero direction, draw, at equal angular intervals, one third of the number of radial lines that were drawn from O1. On these radial lines, which in this case are at right angle intervals, mark off in succession from O, the ordinates 0 to 11, beginning with the zero line, and going three times round as shown. The sums of the horizontal and vertical projections of these radiating vectors in this case are +0.24 and +004 respectively. 3 |