axes intersecting at a distance wo behind the front axle, given by tan a= p from which wo=147 inches. (1 + cos 0°), Taking the maximum angle of lock for this car to be 40°, for this deflexion of the inner steering wheel the front stub axles intersect at a distance wm behind the front axle given by tana= Р from which wm=100 inches. (1+cos 1·63 × 40°), The gear thus gives the steering-wheel axes intersecting 27 inches behind the back axle at the beginning of the steering movement, intersecting on the back axle when the deflexion of the inner wheel is 30°, and intersecting 20 inches in front of the rear axle at the end of the steering movement. To design a gear to give the point of intersection of the steering-wheel axes as much behind the back axle at the beginning of the steering movement (deflexion 0°) as it is in front of the rear axle at the end of the steering movement (deflexion ), we have from (6) w。+2r=p(1+ cos 0°)/2 tan a, um+2r=p(1+cos 1·634)/2 tan ɑ, and since we require w=(w+wm) we get w+2r=p(3+cos 1.634)/4 tan a or tan α= P [3+cos 1·634].. . . (7) With the above dimensions this gives a setting of 17° for the steering arm angle a. The design formula (5) may be used to calculate the performance of an existing gear. Take, for example, a car having p=52 ins., w=100 ins., t=6 ins., s=2 ins. To find the angle at which correct steering obtains, we have tan a=s/t=2/6, so that substituting in (5) 1/3=52(1+cos 1·630)/2(100 + 12), whence 0 39°. = Historically, the gear with steering arms converging to the centre of the back axle is of interest. For a general notion of the performance of this gear, for which tan a=p/2w, we may neglect t in (5). We then obtain 1=1+ cos 1.630, from which 0=55°, an angle much outside the working range usual in motor-car practice. The inclinations sometimes given to steering pivots and to stub-axles do not appreciably affect the design of the linkage; the actual angle of deflexion of the stub-axle is sensibly the same as the angle in plan, while in plan the path of the steering-arm end is an ellipse which differs but little from the actual circular path. 6. Graphical Solution of Design Problem. It may not be without interest to give a graphical solution of (3) when the known magnitudes are 0, 4, and t. The given data enable us to draw the lines AB, AM, BN, and the perpendiculars MDL, NCQ at M and N. figs. 3 and 4. The problem evidently now is to fit in CD between ML and NQ so that NC= MD and NC+CD+DM=AB. From N draw NR parallel to ML and make NR equal to AB, the pitch of the steering pivots. Draw NP bisecting the angle QNR and from R drop RP perpendicular to NP. Join PM. With centre N and radius NR describe a circle cutting PM in E. Join EN and through M draw MF parallel to EN cutting NP in F. Through F draw FC parallel to MD cutting NQ in C. Through C draw CD parallel to MF cutting ML in D. Join BC, AD. Then BCDA is the required linkage. The construction evidently makes NC=CF=MD. Draw NK parallel to RP and let CF be produced to meet RP in J and NK in K. The triangles PNE, PFM are similar, therefore NE: FM:: NP: FP. Again, since the triangles PNR, PFJ are similar, NP: FP :: NR: FJ. Hence NE: FM:: NR: FJ. And since NE NR we therefore have FM = FJ. But FM CD, hence CD=FJ. Since NC CF and the angle FNK is a right angle, the circle centre C which passes through F and N will pass through K so that KC=NC. Hence finally we have NC+CD+DM=KC+CF+FJ=KJ=NR=AB as required. A LXXV. On the Decay of Vortices in a Viscous Fluid. NUMBER of problems have been solved in which the rate of decay of small oscillations or waves in a viscous fluid has been found, but the simplification brought about by considering only small motions excludes many of the most important problems of fluid motion. On the other hand, when the complete equations of motion involving terms containing the square of the velocity have been used very few solutions have been obtained. Certain problems in steady motion †, problems concerning the two-dimensional motion. of a viscous liquid when it is symmetrical about an axis ‡, and problems concerning laminar motion parallel to a plane § probably complete the list. The object of the present paper is to draw attention to at class of cases in which solutions of the complete equations * Communicated by the Author. Flow through a pipe or between concentric cylinders. See Taylor,' Reports and Memoranda of the Advisory Committee for Aeronautics, 1918-19.' Also Terazawa, 'Report of Aeronautical Research Institute,' Tokio, 1922. § See Lamb, Hydrodynamics,' chap. xi. of motion, including the "inertia " terms, may be obtained for two-dimensional viscous flow. where v is the coefficient of kinematic viscosity. Now consider functions which satisfy the equation ▼2¥ = K¥, where K is a constant. These functions will also satisfy (1) if ду the "inertia " terms (2) vanishing because the stream lines are also lines of constant vorticity. Equation (2) is satisfied if where is a function of x and y only. Hence, if ; is a solution of Analogy with the theory of vibrating membranes. The equation of motion for a vibrating membrane is (3) (4) where is the displacement at any point from the position of equilibrium and e depends only on the tension and mass of the membrane. If the membrane vibrates in simple harmonic motion of period T, (4) becomes This equation is of the same form as (3), and the amplitude of vibration of the membrane may be taken to represent it It appears therefore that if a solution of the problem of a vibrating membrane has been obtained so that the period and contours, or curves of equal displacement of the membrane, have been determined, a problem of viscous motion has also been solved in which the stream lines are the same as the contours of the vibrating membrane. The velocity of the flow dies down exponentially so that it is reduced in the c2T2 4π2 ratio e: 1 in time . Though the analogy is exact, so far as it goes, it should be borne in mind that in general the boundary conditions in the two cases are different. In the case of a membrane held round its edges, for instance, the boundary condition is =0, while for the viscous fluid both and ǝyǝn must be zero at a solid wall. Another point of difference is that if two different solutions have been obtained in the membrane problem, they can be superposed. This is not true in the case of the viscous motion problem because the “inertia " terms in (1), namely, but they do not vanish when is a sum of solutions of (6) and (7). Phil. Mag. S. 6. Vol. 46. No. 274. Oct. 1923. 2 X |