is fitted with gun-metal bushes adjustable by a cotter attached to a screw on the end of the gib.

Thickness of strap at the end = diameter of bearing multiplied by 33. Thickness of strap at the side diameter of bearing multiplied by 24. Thickness of strap at cotter-hole diameter of bearing multiplied by 4. Width of strap = length of bearing multiplied by 7.

Length of strap beyond the cotter-hole = diameter of bearing multiplied

by '54.

Distance from end of gun-metal bush to edge of cotter diameter of bearing multiplied by '54.

Thickness of gun-metal bush at the end = diameter of bearing multiplied by 25.

Thickness of gun-metal bush at the side thickness of gun-metal bush at the end multiplied by '75.

The bushes should be carefully adjusted to the crank-pin. Play in the bushes tends to wear the crank-pin oval, and excessive play causes a knock at each end of the stroke.

Width of gib and cotter at the centre the diameter of the bearing. Thickness of gib and cotter diameter of bearing multiplied by 22.

Taper of cotter, inch per foot.

Depth and width of the clip of the gib, each the thickness of the gib. Diameter of the connecting-rod at the small end the diameter of the piston-rod.

Diameter of connecting-rod at the end next to the crank, or the large end the diameter of the piston-rod multiplied by 125.

Diameter of connecting-rod at the centre = the diameter of large end plus of an inch per foot of length of the rod.

The diameter of the smallest part of the rod should not be less, when of wrought-iron, than = diameter of cylinder in inches x 018 x square root of the initial pressure of the steam in lbs. per square inch.

Fig. 57.-Connecting-Rod with Cap-End.

=

Length of connecting-rod twice the length of the stroke.

The bending action is considered to be greatest at a point equal 6 of the length of the rod, measured from the cross-head-end of the rod.

The Connecting-rod with Capend, shown in Fig. 57, has bushes adjustable by bolts.

Cap-bolts.-The sectional area of each bolt should be equal to one-half the sectional area of the piston-rod.

Thickness of cap diameter of bearing multiplied by 5.

Width of cap and rod-end = length of bearing multiplied by 7.

Thickness of gun-metal bush = diameter of bearing multiplied by '2.

OF

CALIFI

The Connecting-rod with Cap-end, shown in Fig. 58, is of very strong construction. The cap hooks over projections on the jaws, turned concentrically with the rod: the bushes are flanged on both sides; the wear of the bushes is taken up by a wedge, adjustable by screws. the connecting-rod is shown in Fig. 59.

An end view of

Cap-Bolts. The sectional area of each cap bolt to be at least equal to one-fourth the sectional area of the piston rod.

Thickness of gun-metal bush at the end-diameter of bearing multiplied by 25.

Fig. 58.

Fig. 59.

Connecting Rod with Cap-end.

Fig. 60.-Lubricator for Connecting-Rod.

Thickness of gun-metal bush at the side diameter of bearing multiplied by 2.

All the above proportions are for connecting-rods of wrought-iron. The proportions of connecting-rods of steel may, in a general way, be 20 per cent. less than those for wrought-iron.

A Lubricator for the large end of a Connecting-Rod is shown in Fig. 60. It is fitted with a valve having a spindle resting on the crank-pin. The motion of the crank causes the valve to rise and fall, and the oil is delivered on the journal drop by drop. The lift of the valve is regulated by a screw. Engine-bed when made Box-pattern,

like the section of bed shown in Fig. 61.

Full width across the top diameter of

cylinder multiplied by 2.

Width of each side frame or box= diameter

of cylinder multiplied by 5.

Width inside the two frames = diameter of the cylinder.

Thickness of metal thickness of metal of the cylinder multiplied by 7.

Fig. 61.-Section of Engine-Bed.

Depth of bed = diameter of cylinder multiplied by 5 to 6.

To Adjust an Engine-Bed, or to steady a slack corner of the bed :Wedge the bed level, lute the bottom of the bed with clay at the defective part, and run in expansive-metal between the bed and the foundationstone. A good expansive-metal for this purpose consists of lead, 9 parts; antimony, 2 parts; bismuth, I part.

E

Weight of Foundation for an Engine.-The weight of a foundation in stone or brick may be one ton per nominal horse-power of the engine. Wall-Engines. A steam-engine when arranged as a wall-engine, as shown in Fig. 62, dispenses with a foundation, and is suitable for the storey of a factory, or a confined space. It is self-contained on a strong frame, and may be either bolted vertically or diagonally to a wall. It is essential to balance the momentum of the moving parts, to secure a steady motion and prevent vibration and injury to the walls of the building.

Fly-wheel. A fly-wheel temporarily absorbs, keeps, and equalizes power, and conduces to uniformity of motion. It stores up surplus energy of motion during the temporary development of excess of power by the engine, and gives it up again during a decrease of the driving-power. It therefore tends to neutralize variations in the motion of the crank-shaft, due to increase and decrease of power, and it consequently steadies the speed of the engine. The weight of a fly-wheel should be sufficient to permit the crank to pass its dead-centres without sensible diminution of speed.

=

The diameter of a fly-wheel is generally the length of the stroke of the engine multiplied by 3 to 4.

The arms of fly-wheels present the least resistance to the air when of oval-section.

The weight of a fly-wheel, in cwts., is generally the nominal horsepower of the engine multiplied by 3.

The weight of the Fly-wheel of a Steam-engine of a given indicated Horse-power may be found by this rule :

:

Multiply the constant number 7,000,000 by the indicated horse-power to be developed by the engine, and divide by the product of the square of the radius of the fly-wheel in feet by the cube of the number of revolutions per minute, the quotient will be the weight of the fly-wheel in cwts.

Example. A fly-wheel of 10 feet diameter making 56 revolutions per minute is required for a steam engine of 40 indicated horse-power. Required the weight of the wheel in cwts. ?

Then

7000000 × 40 indicated horse-power (5 ft. radius of wheel)2 × (56 revolutons) the weight of fly-wheel required for that engine.

This rule gives the weight of the fly-wheel complete, that is, including the boss, arms, and rim. The weight of the arms and boss together is generally equal to about one-third the weight of the wheel, and the weight of the rim equals about two-thirds the weight of the wheel.

Maximum velocity of Fly-wheels.-The maximum safe velocity for good cast-iron is 80 feet per second. The wheel in the previous example is 10 feet × 31416 = 31°416 feet circumference, and as it makes 56 revolutions per minute, the velocity of the circumference of the rim is

=

31416 feet x 56 revolutions

60 seconds

= 29°32, or say 30 feet per second, or a

little more than one-third the velocity at which it is safe to run a fly-wheel.

Strength of Fly-wheels.-Centrifugal force is the inertia of matter, or the resistance which matter always opposes to a force exerted to put it in motion from a state of rest. Unhindered motion is always in a straight or

direct line. A moving body must move in a direct line, unless some force is exerted upon it to change the direction of its motion. Each particle of matter in the rim of a wheel in motion, if the bond that holds it to the centre were broken, would move on in a straight line, in the direction it was

on a Wall.

moving at that instant, and all these straight lines would be tangents to the circle, as the water flying from the face of a grindstone perfectly illustrates. But while this bond holds, each particle is being continually deflected from a direct line of motion, and compelled to travel in a circle. At every point in its path it is moved directly towards the centre around which it is revolving. This second motion is exactly at right angles with the direction in which its momentum would carry it if it were free. The latter is tangential, and the former is radial. The force drawing the body towards the centre acts upon it precisely as if it were at rest, because it is exerted at right angles with its line of motion at that instant. The body resists this deflection precisely as a body at rest resists being put in motion, and that resistance is what is called centrifugal force. It varies with different rates of deflection, according to the law of inertia, namely, directly as the square of the velocity imparted to the body in moving through a given distance towards the centre. The application of this law causes the force required to produce the deflection to vary directly as the diameter of the circle at a given rate of revolution, and as the square of the speed in any given circle. Centrifugal force may then be properly defined in either of two ways, namely, as the tendency of a body at rest to continue in that state, or as the tendency of a moving body to move in a direct line. The mode of computing the force required to produce this deflection from a direct line of motion in revolving bodies is as follows:

The resistance to deflection from a direct line of motion that is offered by a body making one revolution in a minute, in a circle of 1 ft. radius, is 000341 of its own weight. And this resistance varies directly as the radius of the circle, and as the square of the number of revolutions per minute. Thus let

then A x R x C = the centrifugal force of this wheel, in terms of its weight.

Example: Required the centrifugal force of a wheel 20ft. in diameter, the rim of which weighs 10,000 lbs. at 100 revolutions per minute? Then, 10x 1002 × 000341 × 10,000 lbs. = 341,000lbs. This is the sum of all the centrifugal strains in the rim of this wheel. In these computations the arms and boss are omitted, as their strength is always in excess of their centrifugal strains. The only factors considered are the centrifugal force of the rim, and its cohesive strength. The radius of the wheel should be measured to the centre of the cross-section of the rim. This force is exerted to burst a wheel of uniform section, in the same manner precisely in which the force of steam is exerted to burst a cylindrical boiler, that is, equally in all directions, and it is resisted in the rim of the wheel precisely as the latter is resisted in the shell of the boiler, therefore the computation for the strength required in the rim or in the shell is made on the same principle.