In the double-acting engine, the fly-wheel is calculated in a similar manner. The work done by the power during one revolution is 4r. F; hence, in this case, we have

And by reasoning in the same manner as before, we find

AD 6366.r;

:. BD=√AB2 — AD2 = √‚2 — (·6366 . r)2

= √59474044 r2 = ·77119 r.

The arc whose sine is 77119, is rather more than 50° 27';

.. arc BB' = 100° 54′,

180° 100° 54′ :: 3.1416: x

x=1·761 r = the length of the arc BB'.

Now, the work of the resistance is as before, represented by

The work of the power in the same interval is: F multiplied by the chord

BB' = F × ·77119 .r × 2 = 1·5424.r.F.

Hence the accumulated work in this interval is

P

} . 2 (V2 — v2) = (1·5424

g

(V2 — v2) = (1·5424 — 1∙1211) r. F = 4213 . r . F.

or,

2P. V/2
gn

·4213.r. F.

(18.)

Now, 2r. F = the work done in a single stroke, hence, work done in one revolution of the fly-wheel;

4r. F

Substituting this value of r. F, in equation (18),

Introducing for g its value 321 feet per second,

1737-8625 × 900 × 321. H.n

(3.1416)2R, 2N3

P=

5097558. H.n

in lbs.,

R2N3

.....

.(20.)

2275.H.n in tons. (Rule 1, page 27.)

R2N3

Substituting the value from equation (11) in equation

(20) we have

=

N3 x 2π.A. w N3 x 2 x 3.1416.A.w'

but w 450 lbs., the weight of a cubic foot of iron,

If we had put N for the number of single strokes, and

Also, the area of the section of the rim,

Equation (21) gives the weight in tons of a fly-wheel of a given mean radius R', so that, being applied to an engine whose horse-power is H, making a given number N, of revolutions per minute, it shall cause the angular velocity

1

of the wheel not to vary more thanth from the mean

n

angular velocity. The same may be said of equation (24),

N

taking for the revolutions per minute, instead of N.

2

Similar observations apply to R' and A. It is evident that the weight of the wheel varies inversely as the cube of the number of strokes per minute, so that an engine making twice as many strokes as another of equal horsepower, would receive an equal steadiness of motion from a fly-wheel of one-eighth the weight, the radius being the same.

On the Friction of a Fly-wheel.

Let limiting angle of resistance, then tan is the coefficient of friction, and P. tan is the resistance from friction at the surface of the shaft; but in the time that the fly-wheel makes one revolution, this resistance acts through the whole circumference of the shaft 2πr, if r be the radius of the shaft; hence, the work expended on friction is

2πr. Ptan ;

and if N be the number of single strokes per minute,

and, therefore, = number of revolutions of the fly

N
2

wheel per minute, the work consumed by friction per minute is

and in horses' power

NP.πr.tan.

NP.πr.tan.

33000

ON THE ECCENTRIC WHEEL.

Let S represent the space the end A is moved through by the eccentric wheel, and s the space the slide moves. AB × = BC × S.

Then