CE will be one force acting tangentially to crank, and tending to turn it round, and 1 E will be the other force acting through the crank, resulting in pressure on bearings. We find the tangential forces for the remaining positions of crank, and a curve drawn through these points will give a circumferential diagram showing the turning efforts on the crank pin exerted by pressure on the piston. The same process is repeated for fig. 63, showing the negative and positive pressures on the exhausting stroke, fig. 64 on the charging stroke, and fig. 65 on the compressing stroke. Fig. 66 shows in a linear diagram the pressures at the various points of the stroke, G H representing the path of the crank pin, the positive pressures being above and the negative ones below the line, showing at a glance how the pressure on the piston is distributed to the crank pin, being diminished or increased as the inertia of the weight of the moving parts is acting with or against it. Thus, commencing with the charging stroke, it is assumed that the charge enters at atmospheric pressure (although in practice this is found to be 1 or 2 lbs. below), so that the pressure on the crank pin is governed altogether by the inertia of the moving parts, as through the first three divisions the energy stored in the flywheel has to be utilised in drawing the piston forward, imparting to the moving parts their required amount of inertia. From the middle of the stroke to the end the work is done by the pressure required to bring the moving parts to a state of rest, and during the compression stroke the effect of the inertia is to make the work done by the flywheel equal at each end of the stroke, and greatest in the middle, whilst on the firing stroke the inertia has the effect of reducing the pressure of explosion on the crank pin at its commencement, and does not reach its maximum until about a quarter of the stroke has been completed, when the pressure gradually decreases until the exhausting stroke commences, when the energy stored by the flywheel drives the piston back, expelling the products of combustion; and on reaching the middle of the stroke the energy stored in the moving parts again comes into play, so that no power is lost by this inertia, as the amount absorbed at one part of the stroke is given out at the other. The mean of the pressures contained in the six divisions of the compression stroke (fig. 60) is 15 lbs. pressure on the crank pin per square inch of piston area, which work has to be done by the stored-up energy in the flywheel. In the same way the 420 pressure for the firing stroke will be = 70 lbs. per square 6 inch of piston area on crank pin; subtracting the negative mean tangential pressure during the compression stroke from the positive mean tangential pressure during the firing stroke, = 70–15 = 55 lbs. which mean pressure, having to be distributed between the number of strokes in a complete cycle, 55 13.75 lbs., and this deducted from that of the firing 4 gives = = W = stroke, viz. 70-13.75 56.25 lbs. the excess tangential pressure obtained during the firing stroke. The work supplied by a flywheel during its retardation = (V2 max. - V2 min.), where 2g W = weight in lbs., g = acceleration of gravity, V maximum= the highest, and V minimum = the reduced velocity, in feet per second. The variation in speed should not exceed 5 per cent. for commercial engines, and where extreme steadiness of running is required the variation should not exceed 2 per cent. With 5 per cent. variation (that is, from 1.025 to 975 of the mean or reputed speed) V = (V max.2 — V min.2) = 1 V2, hence the •1 W V2 work of retardation of the wheel = = 2 g foot-lb., and the x work to be stored up in the flywheel 56-25 × 3.9 × 283.5 (square inches) 62,171 foot-lbs., where 56.25 equals the excess of tangential pressure during the firing stroke in pounds per square inch of piston area, 3.9 equals half the circumference of the crank pin circle in feet, and 283-5 square inches equals the piston area. Assuming the diameter of the flywheel to be 10 feet, the 10 x 31416 x 120 revolutions mean velocity of rim feet per second = V. We can now resolve the equation into = 62.8 the weight of flywheel required for the above engine. For approximate weight of flywheels the above diagrams and formulæ may be taken as a standard, as all gas engine indicator cards are much the same. The different methods by which the flywheels are keyed to the crank shaft are illustrated in figs. 67 and 68. Fig. 68 is the method used for small and medium sized engines. But for larger sized engines two keys are generally fitted, as shown in fig. 67, one of which is a saddle key. Fig. 69 shows a special method of keying flywheels on large engines which was designed by Mr. S. Quipp, one of the staff of Robey & Co., and differs from fig. 67 in that, instead of a saddle key, a special form is used, sunk deep in the flywheel at one side, and deep into the shaft at the opposite side, the direction of running determining the deep side, the shallow side in the shaft always leading in the direction in which the engine is running. This key drives as a strut in the manner indicated by the arrows, the bearing surface in proportion to cross section being considerably increased, taking all the driving strain due to the pressure of the explosion, and effectually resisting the shock due to the starting impulse, which in all pressure starters is very great, the heavy flywheel having to get up the required velocity due to the force of this impulse almost instantaneously, and the inertia of the flywheel acting against this impulse causes a very severe strain on the sides of the key beds in both shaft and flywheel, the result being that it soon becomes loose when a single key is used. Flywheels over 7 feet in diameter should be cast in halves, as the greater frictional grip obtained by the bolts assists the keys in a very efficient manner. Until recently the practice of gas engine makers was to fit |