another reason which tends to render the fitting of a flywheel to this type of engine unnecessary, and this we will now consider.

Effect of Reciprocating Parts.—We have so far considered these parts as weightless in the present chapter. But they are actually heavy, and will accordingly affect the motion, and we will now examine their effect.

In the case of the Bull engine (page 206) the reciprocating parts were very heavy, and their motion was determined simply by the relation between their weight, the varying effort, and the resistance. In our present case they are not nearly so heavy comparatively, and, moreover, their motion is determined to a very great extent simply by the connection between them and the rotating parts.

Let us, for simplicity, neglect the effect of obliquity. Then the piston, rod, and connecting rod of the leading crank, OP in the figure, all move forward at the instant with velocity Vp sin 0.

Fig. 153.

P

Vsino

since necessarily Vo=Vp.

The following crank will be at OQ, and the velocity of its reciprocating parts will be

Vo cos 0, or Vp cos 0,

The total K. E. then of the two sets of reciprocating

pieces is

W/Vp2 sin20 W'V2 cos2 0

2g

+

2g

or

W/Vp2

2g

(W' being the weight of one set).

The two sets together then have always the same K. E. as a fly-wheel of weight W', mean radius of rim a,

We

would have, if such a wheel were fixed to the shaft. say always, because OP is any crank position whatever. Thus then, the motion of the shaft will be identical with what it would be if the reciprocating parts were weightless, and such a heavy fly-wheel as we have described were fixed to the shaft.

Hence the double crank engine is preferable to the single, not only by reason of the greater regularity of effort, but also by containing in itself an equivalent to a fly-wheel.

It does not follow, however, from what we have just said that it is necessarily advantageous to have heavy reciprocating parts. We have shown that such will be advantageous as regards regularity of motion of the engine as a whole; but it may very likely be that such regularity is obtained at the expense of great irregularity of force in the cylinders separately. We must therefore examine the effect on one cylinder separately.

Inertia of Reciprocating Parts.-To effect this we shall have, for simplicity, to make the assumption that the crank revolves uniformly.

[This assumption would of course be unwarranted if we were attacking the question for the first time, since the motion of the crank would be one of the results to be obtained. But we know already that the crank does rotate very nearly uniformly, and hence we make the assumption, knowing the error caused will be only slight.]

We will also neglect obliquity.

Suppose now that the crank is rotating uniformly ; then before the steam pressure on the piston can produce any effort on the crank, it must first keep the reciprocating parts up to the crank pin. For example, let us start the stroke with the crank pin accurately centred in its bearing (Fig. 154), and therefore, since there is usually some clearance, not touching it anywhere.

[We must of course suppose the mere dead weight of the rod kept off the bearings in some way.]

Now as the crank revolves uniformly, a certain steam

pressure is required to keep the rod up to and centred on the pin, quite irrespective of any effort being exerted

Fig. 154.

Fig. 155.

on the pin; and if this pressure be not supplied, the pin will drag the rods with it, the pin bearing as in Fig. 155. If just the necessary pressure be applied the pin keeps centred, as in Fig. 156, while if more than this be applied, the rod drives the pin (Fig. 157), but the effort

it exerts is only due to the excess of the actual pressure over that required to keep the bearing centred relative to the pin.

[In the figures the clearance is much exaggerated for clearness. On the scale of the figures it would not actually be visible, but can seldom be entirely absent.

We propose now to find what pressure is necessary to keep the rods up to the crank.

There is no work done because the pin is not touched by the rod, or because that is one of our conditions,

.. Energy exerted by steam = Final K. E. - Initial K. E., W.V.2 QN2 W'V.2 PM2

=

mean pressure during the motion. Then

Energy exerted = P(x1 − x2),

P is the mean value during the piston movement MN but if we make MN very small, that is, make x2 equal to x, P becomes the actual value at M, and its value is given by

P then varies as x; and it follows that if we draw a curve of effort by setting up, at each point M (Fig. 159), the value of P at that point, then the curve will be a straight line passing through O. For then at any point, Pr tan a, or P varies as x.

=

a

Fig. 159.

Since the curve passes through O, the part to the right of O lies below the base line, which in the ordinary way would indicate that P was negative. It can be easily seen that such is the case by considering a crank position to the right of the upright. Since x when measured to the left is taken as plus, it must when to the right be negative, and hence P becomes negative. Or, we can see that the velocity of the parts must be retarded, which requires a pull to the left. Actually,

of course, the retardation is effected by the resistance of the crank pin; and what our figure shows is, that during the latter half of the stroke the reciprocating parts can exert, independent of any steam pressure, an effort equivalent to a steam pressure on the piston represented by the ordinate of OD.

There is, of course, on the whole no total effect; for during the first half of the stroke the parts abstract from the steam energy represented by OAC, while during the latter half they give out energy represented by OA'D, and these are of course equal. Plainly this must always be the case, since we start with no velocity, and end with the same.

Correction of Indicator Diagram.—The pressure we have so far spoken of is a total pressure; but it will be convenient to express our result as a pressure per sq. in., and this we proceed to do.

Let be the pressure per sq. inch on the piston equivalent to P. Then

But W'/A is a pressure per sq. inch, being lbs. divided by sq. inches, and it is the pressure which if applied to the piston would produce a total pressure W'. Hence representing it by po, we call the pressure equivalent to the weight of reciprocating parts.

We have then

To find its initial value put x = a, and then

V2 p=po ga

and this pressure is required simply to start the piston.

Now the actual pressure per square inch on the piston