If the limit of rim velocity for all wheels be assumed to be 88 ft. per second, equal to 1 mile per minute, F= 6000 lbs., the formula becomes

When wheels are made in halves or in sections, the bending strain may be such as to make t greater than that given above. Thus, when the joint comes half way between the arms, the bending action is similar to a beam supported simply at the ends, uniformly loaded, and t is 50% greater. Then the formula becomes

or for a fixed maximum rim velocity of 88 ft. per second and F = 6000 lbs., 1.05 t = In segmental wheels it is preferable to have the joints opposite N2 the arms. Wheels in halves, if very thin rims are to be employed, should have double arms along the line of separation,

Attention should be given to the proportions of large receiving and tightening pulleys. The thickness of rim for a 48-in. wheel (shown in table) with a rim velocity of 88 ft. per second, is 15/16 in. Many wrecks have been caused by the failure of receiving or tightening pulleys whose rims have beer too thin. Fly-wheels calculated for a given coefficient of steadiness are frequently lighter than the minimum safe weight. This is true especially of large wheels. A rough guide to the minimum weight of wheels can be deduced from our formulæ. The arms, hub, lugs, etc., usually form from one quarter to one third the entire weight of the wheel. If b represents the face of a wheel in inches, the weight of the rim (considered as a simple annular ring) will be w = .82dtb lbs. If the limit of speed is 88 ft. per second, then for solid wheels t=0.7dN2. For sectional wheels (joint between arms) t = 1.05d÷ N2. Weight of rim for solid wheels, w = .57d2b+N2 in pounds. Weight of rim in sectional wheels with joints between arms, w = .86d2b÷ N2 in pounds. Total weight of wheel: for solid wheel, W = .76d2b ÷ N2 to .86d2b N2, in pounds. For segmental wheels with joint between arms, W = 1.05d2b÷Ñ2 to 1.3d2b N2, in pounds.

(This subject is further discussed by Mr. Stanwood, in vol. xv., and by Prof. Gaetano Lanza, in vol. xvi., Trans. A. S. M. E.)

A Wooden Rim Fly-wheel, built in 1891 for a pair of Corliss engines at the Amoskeag Mfg. Co.'s mill, Manchester, N. H., is described by C. H. Manning in Trans. A. S. M. E., xiii. 618. It is 30 ft. diam, and 108 in. face. The rim is 12 inches thick, and is built up of 44 courses of ash plank, 2, 3, and 4 inches thick, reduced about 1⁄2 inch in dressing, set edgewise, so as to break joints, and glued and bolted together. There are two hubs and two 'sets of arms, 12 in each, all of cast iron. The weights are as follows:

The wheel was tested at 76 revs. per min., being a surface speed of nearly

7200 feet per minute.

Mr. Manning discusses the relative safety of cast iron and of wooden wheels as follows: As for safety, the speeds being the same in both cases, the hoop tension in the rim per unit of cross-section would be directly as the weight per cubic unit; and its capacity to stand the strain directly as the tensile strength per square unit; therefore the tensile strengths divided by the weights will give relative values of different materials. Cast iron weighing 450 lbs. per cubic foot and with a tensile strength of 1,440,000 lbs. per square foot would give a value of 1,440,000÷450= 3200, whilst ash, of which the rim was made, weghing 34 lbs. per cubic foot, and with 1,152,000 lbs. tensile strength per square foot, gives a result 1,152,000 3433,882, and 33,882 3200 10.58, or the wood-rimmed pulley is ten times safer than the cast-iron when the castings are good. This would allow the woodrimmed pulley to increase its speed to /10.58 3.25 times that of a sound cast-iron one with equal safety.

=

Wooden Fly-wheel of the Willimantic Linen Co. (Illustrated in Power, March, 1893.)-Rim 28 ft. diam., 110 in. face. The rim is carried upon three sets of arms, one under the centre of each belt, with 12 arms in each set.

The material of the rim is ordinary whitewood, % in. in thickness, cut into segments not exceeding 4 feet in length, and either 5 or 8 inches in width. These were assembled by building a complete circle 13 inches in width, first with the 8 inch inside and the 5-inch outside, and then beside it another circle with the widths reversed, so as to break joints. Each piece as it was added was brushed over with glue and nailed with three-inch wire nails to the pieces already in position. The nails pass through three and into the fourth thickness. At the end of each arm four 14-inch bolts secure the rim, the ends being covered by wooden plugs glued and driven into the face of the wheel.

Wire-wound Fly-wheels for Extreme Speeds. (Eng'g News, August 2, 1890.)-The power required to produce the Mannesmann tubes is very large, varying from 2000 to 10,000 H.P., according to the dimensions of the tube. Since this power is only needed for a short time (it takes only 30 to 45 seconds to convert a bar 10 to 12 ft. long and 4 in. in diameter into a tube), and then some time elapses before the next bar is ready, an engine of 1200 H.P. provided with a large fly-wheel for storing the energy will supply power enough for one set of rolls. These fly-wheels are so large and run at such great speeds that the ordinary method of constructing them cannot be followed. A wheel at the Mannesmann Works, made in Komotau, Hungary, in the usual manner, broke at a tangential velocity of 125 ft. per second. The fly-wheels designed to hold at more than double this speed consist of a cast-iron hub to which two steel disks, 20 ft. in diameter, are bolted; around the circumference of the wheel thus formed 70 tons of No. 5 wire are wound under a tension of 50 lbs. In the Mannesmann Works at Landore, Wales, such a wheel makes 240 revolutions a minute, corresponding to a tangential velocity of 15,080 ft. or 2.85 miles per minute.

THE SLIDE-VALVE.

Definitions.-Travel = total distance moved by the valve.

Throw of the Eccentric eccentricity of the eccentric distance from the centre of the shaft to the centre of the eccentric disk = 1⁄2 the travel of the valve. (Some writers use the term "throw " to mean the whole travel of the valve.)

Lap of the valve, also called outside lap or steam-lap distance the outer Dr steam edge of the valve extends beyond or laps over the steam edge of the port when the valve is in its central position.

Inside lap, or exhaust-lap distance the inner or exhaust edge of the valve extends beyond or laps over the exhaust edge of the port when the valve is in its central position. The inside lap is sometimes made zero, or even negative, in which latter case the distance between the edge of the valve and the edge of the port is sometimes called exhaust clearance, or inside clearance.

Lead of the valve the distance the steam-port is opened when the engine is on its centre and the piston is at the beginning of the stroke.

Lead-angle the angle between the position of the crank when the valve begins to be opened and its position when the piston is at the beginning of

the stroke.

The valve is said to have lead when the steam-port opens before the piston

begins its stroke. If the piston begins its stroke before the admission of steam begins the valve is said to have negative lead, and its amount is the lap of the edge of the valve over the edge of the port at the instant when the piston stroke begins.

Lap-angle the angle through which the eccentric must be rotated to cause the steam edge to travel from its central position the distance of the lap.

=

Angular advance of the eccentric lap-angle + lead angle.
Linear advance = lap + lead.

Effect of Lap, Lead, etc., upon the Steam Distribution.Given valve-travel 234 in., lap 34 in., lead 1/16 in., exhaust-lap in., required crank position for admission, cut-off, release and compression. and greatest port-opening. (Halsey on Slide-valve Gears.) Draw a circle of diameter fh travel of valve. From O the centre set off Ou lap and ab lead, erect perpendiculars Oe, ac, bd; then ec is the lap-angle and cd the lead-angle, measured as arcs. Set off fg cd, the lead-angle, then Og is the position of the crank for steam admission. Set off 2ec+cd from h to i; then Oi is the crank-angle for cut-off, and fk+fh is the fraction of stroke completed at cut-off. Set off Ol exhaust-lap and draw Im; em is the exhaust-lap angle. Set off hn ec + cd -em, and On is the position of crank at release. Set off fp =ec+ca + em, and Op is the position of crank for compression, fo÷fh is the fraction of stroke completed at release, and hqhf is the fraction of the return stroke completed when compression begins; Oh, the throw of the eccentric, minus Oa the lap, equals ah the maximum port-opening,

If a valve has neither lap nor lead, the line joining the centre of the eccen

tric disk and the centre of the shaft being at right angles to the line of the crank, the engine would follow full stroke, admission of steam beginning at the beginning of the stroke and ending at the end of the stroke.

Adding lap to the valve enables us to cut off steam before the end of the stroke; the eccentric being advanced on the shaft an amount equal to the lap-angle enables steam to be admitted at the beginning of the stroke, as

before lap was added, and advancing it a further amount equal to the lead angle causes steam to be admitted before the beginning of the stroke. Having given lap to the valve, and having advanced the eccentric on the shaft from its central position at right angles to the crank, through the angular advance lap-angle and lead-angle, the four events, admission, cut-off, release or exhaust-opening, and compression or exhaust-closure, take place as follows: Admission, when the crank lacks the lead-angle of having reached the centre; cut-off, when the crank lacks two lap-angles and one lead-angle of having reached the centre. During the admission of steam the crank turns through a semicircle less twice the lap-angle. The greatest port-opening is equal to half the travel of the valve less the lap. Therefore for a given port-opening the travel of the valve must be increased if the lap is increased. When exhaust-lap is added to the valve it delays the opening of the exhaust and hastens its closing by an angle of rotation equal to the exhaust-lap angle, which is the angle through which the eccentric rotates from its middle position while the exhaust edge of the valve uncovers its lap. Release then takes place when the crank lacks one lap-angle and one lead-angle minus one exhaust-lap angle of having reached the centre, and compression when the crank lacks lap-angle + lead-angle+ exhaust-lap angle of having reached the centre.

The above discussion of the relative position of the crank, piston, and valve for the different points of the stroke is accurate only with a connecting-rod of infinite length.

For actual connecting-rods the angular position of the rod causes a distortion of the position of the valve, causing the events to take place too late in the forward stroke and too early in the return. The correction of this distortion may be accomplished to some extent by setting the valve so as to give equal lead on both forward and return stroke, and by altering the exhaust-lap on one end so as to equalize the release and compression. F. A. Halsey, in his Slide-valve Gears, describes a method of equalizing the cut-off without at the same time affecting the equality of the lead. In designing slide-valves the effect of angularity of the connecting-rod should be studied on the drawing-board, and preferably by the use of a model.

Sweet's Valve-diagram.-To find outside and inside lap of valve for different cut-offs and compressions (see Fig. 147): Draw a circle whose

diameter equals travel of valve. Draw diameter BA and continue to A1, so that the length AA1 bears the same ratio to XA as the length of connecting-rod does to length of engine-crank. Draw small circle K with a radius equal to lead. Lay off AC so that ratio of AC to AB = cut-off in parts of the stroke. Erect perpendicular CD. Draw DL tangent to K; draw XS perpendicular to DL; XS is then outside lap of valve.

To find release and compression: If there is no inside lap, draw FE through X parallel to DL. F and E will be position of crank for release and compression. If there is an inside lap, draw a circle about X, in which radius XY equals inside lap. Draw HG tangent to this circle and parallel to DL; then H and G are crank position for release and compression. Draw HN and MG, then AN is piston position at release and AM piston position at compression, AB being considered stroke of engine.

To make compression alike on each stroke it is necessary to increase the inside lap on crank end of valve, and to decrease by the same amount the

inside lap on back end of valve. To determine this amount, through M with a radius MM AA', draw arc M P, from P draw PT perpendicular to AB, then TM is the amount to be added to inside lap on crank end, and to be deducted from inside lap on back end of valve, inside lap being XY.

For the Bilgram Valve Diagram, see Halsey on Slide-valve Gears. The Zeuner Valve-diagram is given in most of the works on the steam-engine, and in treatises on valve-gears, as Zeuner's, Peabody's, and

Spangler's. The following is condensed from Holmes on the Steam-engine: Describe a circle, with radius OA equal to the half travel of the valve. From O measure off OB equal to the outside lap, and BC equal to the lead. When the crank-pin occupies the dead centre 4, the valve has already moved to the right of its central position by the space OB+BC. From C erect the perpendicular CE and join OE. Then will OE be the position occupied by the line joining the centre of the eccentric with the centre of the crank-shaft at the commencement of the stroke. On the line OE as diameter describe the circle OCE; then any chords, as Oe, OE, Oe', will represent the spaces travelled by the valve from its central position when the crank-pin occupies respectively the positions opposite to D, E, and F. Before the port is opened at all the valve must have moved from its central position by an amount equal to the lap OB. Hence, to obtain the space by which the port is opened, subtract from each of the arcs Oe, OE, etc., a length equal to OB. This is represented graphically by describing from centre 0 a circle with radius equal to the lap OB; then the spaces fe, gE, etc., intercepted between the circumferences of the lap-circle Bfe' and the valve-circle OCE, will give the extent to which the steam-port is opened. At the point k, at which the chorl Ok is common to both valve and lap circles, it is evident that the valve has moved to the right by the amount of the lap, and is consequently just on the point of opening the steam-port. Hence the steam is admitted before the commencement of the stroke, when the crank occupies the position OH, and while the portion HA of the revo