REGULATION OF STEAM.

119

the first place, be found due to the length of the crank and connectingrod, as likewise the point of cut-off; then when the versed sine ofthe eccentric is likewise known, which equals the opening of the portby valve, minus one-half of the lead, then the eccentric circle can befound by the rule of three (based on the known property that theversed sines of circles of similar Segments are as the diameters ofthe respective circles). Thus, supposing the crank circle was18 inches in diameter, the versed sine E F being 4 inches, while theversed sine GH is iy inch: we have

4:18:: i'25 = 5'Ö2 inches diameter of the eccentric circle.

THE CRANK AND ECCENTRIC PATHS DELINEATED AS REGARDSTHE COVER, LEAD, AND CUT-OFF.

The length of the eccentric rod being not less than six times thethrow of the eccentric, the versed sine of the chord described by thearc on the eccentric path equals the opening of the port by valve,minus one-half of the lead nearly. Thus, supposing the versed sineof the chord of the arc of supply on the crank path was 4 inches,the opening of the port by valve 1 y 2 inch, and, for the sake ofillustration, the lead or opening of the port at the commencementof the stroke of the piston is x / 2 inch, we have 1 y 2 — y~iy inch,the versed sine of the chord of the arc of supply of the eccentric.

C, o*- jujtnt, of Cut off

OUT Centre Lvnc

Fig. 65.—The Crank and Eccentric Paths delineated as regards the cover, lead, and cut-off.

Let A B represent the line of the crank at the commencement ofthe IN stroke; as the point A or crank pin travels from A to C, thepoint of cut-off, it is evident that the valve must open and shutwhile the crank-pin centre describes the arc A C. It is thereforenecessary to lay off the eccentric centre on the opposite side of thepath, as at F. Set off F E, which equals the full opening of the port