It often happens that the axis of the cylinder is above the center of axle. When such is the case, we must follow the construction as shown in [Fig. 21]. Let two inches be the distance that the center of axis of cylinder is above the center of axle.
Draw a straight line AD through the center of axle C; two inches above this draw a straight line GH parallel to AD; this line will then pass through the axis of cylinder. With the center of axle C on the straight line AD as a center, and a radius equal to the length of the crank, describe a circle F, ½ F, B, ½ B: this circle will be the crank-pin circle. With the point C as a center, and a radius equal to the length of the connecting rod plus the length of the crank, describe an arc intersecting the straight line GH in the point F′: this point will be the position of the cross-head pin when the piston is at full stroke forward. Through the points F′ and C draw a straight line, intersecting the crank-pin circle in the point F: this point will be the position of the center of the crank-pin when the piston is at full stroke forward. Again, with the point C as a center, and a radius equal to the length of the connecting rod minus the length of the crank, describe an arc intersecting the straight line GH in the point B′: this point will be the position of the center of the cross-head pin when the piston is at full stroke in the rear end of the cylinder. Through the points B and C draw a straight line, intersecting the crank-pin circle in the point B: this point will be the position of the center of crank-pin when the piston is at full stroke in the rear end of the cylinder. Find a point C′ exactly central between the points B′ and F′ on the line GH: in other words, bisect the distance B′ F′ by the point C′. With the point C′ as a center, and a radius equal to the length of the connecting rod, describe an arc intersecting the crank-pin circle in the points ½B and ½F: these two points will be the center of crank-pin when the piston stands at half stroke. In the link-motion, as shown in [Fig. 19], the axis of the cylinder is supposed to be 2 inches higher than the center of axle. For this reason the construction shown in [Fig. 21] will hereafter be used.
Problem 2, [Fig. 22].—To find the center line of motion and the amount of offset in the lower rocker-arm.—Let C be the center of axle: through C draw the straight lines AD and KL perpendicular to AD. The center of rocker we find in [Fig. 19] to be 37½ inches in front of the center of axle, and 7½ inches above. We therefore continue our construction in [Fig. 22] by drawing a straight line MN 37½ inches in front of, and parallel to, the straight line KL, and another straight line OP parallel to AD, and 7½ inches above it. These two lines intersect in the point Q, and this point is the center of rocker. With Q as a center, and a radius equal to the length of the lower rocker-arm, describe the arc RS: through the point C draw a straight line CT tangent to the arc RS, then CT will be the center line of motion.
Fig. 22.
To find the amount of offset in the lower rocker-arm, let us place the center line of the upper rocker-arm perpendicular to a line drawn parallel to the valve surface: but in our case this valve surface is parallel to the line AD; hence our line drawn parallel to the valve surface will also be parallel to the line AD, and the center line of upper rocker-arm will be perpendicular to AD, and coincide with the line MN. Through the point Q draw a straight line perpendicular to the line CT, and intersecting the arc RS in the point U: then the distance from the point U to the line MN will be the amount of the offset in the lower rocker-arm.
Fig. 23.
Problem 3, [Fig. 23].—To find the relative positions of crank-pin and eccentrics when the piston is at full and half stroke.—Let C be the center of axle. Through C draw the horizontal line AD, and find the positions of center of crank-pin at full and half stroke; namely, the points F, ½F, ½B, B, as explained in Problem 1, and shown in [Fig. 21]. Next draw the center line of motion as explained in Problem 2 and [Fig. 22]. With the point C as a center, and a radius equal to ½ the throw of the eccentric (2½ inches), draw a circle; and let us call this circle the “eccentric-circle.” On the line of motion CT, lay off a point towards the rocker 13/16 of an inch from C (this being the sum of the lap and lead,—¾ of an inch for the lap, and 1/16 of an inch for the lead): through this point draw a straight line perpendicular to the line of motion CT, and intersecting the eccentric-circle in the points x and y. The point x will be the center of the forward eccentric; and the point y will be the center of backward eccentric when the center of crank-pin is at F, full stroke forward. Through the points F and C draw a straight line, intersecting the eccentric-circle in the point F″. The line FC will represent the center line of crank; and the distance between the points F″ and x, measured on the eccentric-circle, is the amount that the center of forward eccentric is set back of the center line of crank; and the distance between the points F″ and y is the amount that the backward eccentric is set ahead of the center line of crank. Since both the crank and eccentrics are fastened to the same axle, it follows, that, whatever position the center line of crank may be in, the distances between center line of crank and eccentrics—that is, the distances between F″ and x, also F″ and y, measured on the eccentric-circle—remain constant. Therefore, to find the position of eccentrics when the crank stands at ½F (half stroke), draw the straight line ½FC representing the center line of crank, and intersecting the eccentric-circle in the point ½F″. From the point ½F″, lay off on the eccentric-circle a point with a distance equal to F″x, back of the center line of crank, and indicate this point by ½x; also from ½F″ measured on the same circle, lay off a point in the front of the center line of crank, and with a distance equal to F″y, and mark this point ½y; then the point ½x will be the position of forward eccentric, and the point ½y will be the position of backward eccentric when the crank-pin is at ½F. In precisely the same manner we find the position of eccentrics when the center of crank-pin is at B (full stroke back). Through the points C and B draw a straight line, intersecting the eccentric-circle in the point B″. From the point B″, and with a distance equal to F″x, lay off a point on the eccentric-circle back of crank; this point will be the position of forward eccentric when crank is at full stroke back; and, in order to distinguish this from the other position of eccentric, call this point a: also from B″, lay off in front of the crank the position of backward eccentric at a distance equal to F″y, and call this point b. In the same manner find points ½a and ½b when the crank-pin is at ½B. We have now found the position of eccentrics when the crank-pin stands in the following positions:—
Full stroke forward F, the forward eccentric will be at x.