Next find centers of eccentrics x and y when the crank is at full stroke forward; also a and b when the crank is at full stroke back, as explained in Problem 3, and shown in [Fig. 23]. Before we proceed, let us give names to some of the lines, as shown in [Fig. 6a] ([p. 268]). The arc c1c2 drawn through the center of opening of the link, we will call the link-arc; and the arc d1d2 drawn through the center of eccentric-rod pin-holes, we will call the eccentric-rod pin-arc. Both of these arcs are drawn from the same center; that is, the center from which the link is drawn. Let us now cut a paper template, as shown in [Fig. 6b] (link structure). This template is cut so that, if it is laid on the link, [Fig. 6a], the arc of the template c3c4 will coincide with the link-arc c1c2, and d3d4 with the eccentric-pin arc d1d2, the end of template d3c3 with the line d1c1, and the end d4c4 with d2c2. On this template join the points c3c4 by a straight line, and bisect this line by the perpendicular line ee: on this line the center of the saddle-pin will be located. On one side of this line draw the line ff1 parallel to ee, and on the other side draw f2f3 also parallel to ee; the distance from the point f to the point f2 being equal to the distance between the centers of eccentric-rod pins, and fe equal to ef2. The points f and f2 on the arc d3d4 indicate the position on the template of the centers of eccentric-rod pins. On the center line of motion cT, lay off from v a point v1 towards the axle, with a distance equal to c1d1, [Fig. 6a]; then with the point x as a center, and cv1 as a radius, describe the arc x1x2; in this arc the upper eccentric-rod will be located as long as the center of forward eccentric remains at x. With the point y as a center, and cv1 as a radius, describe the arc y1y2: in this arc the center of lower eccentric-rod will be located as long as the backward eccentric remains at y. With the point a as a center, and cv1 as a radius, describe an arc a1a2: in this arc the upper eccentric-rod pin will be located while the forward eccentric is at a. With the point b as a center, and cv1 as a radius, describe the arc b1b2; and in this arc the center of lower eccentric-rod pin will be located when the backward eccentric is at b. Now adjust the template on the drawing so that the point f will be in the arc x1x2: point f2 in the arc y1y2 and the line ee coincide with the center line of motion cT. Along the arc c3c4 of the template draw an arc on the paper. Next move the template so that the point f will be in the arc a1a2, the point f2 in the arc b1b2, and the line ee coincide with the center line of motion cT, and along the arc c3c4 of the template draw the second arc on the paper. Now, if the distance measured on the arc RS from the point v (the center of the lower rocker-arm pin) to the first arc drawn, is equal to the distance measured on the arc RS from v to the second arc, the radius cv1 will be the correct length of the eccentric-rods. But, if the distance from v to the first arc is less than the distance from v to the second arc, the length cv1 of the eccentric-rod will be too short. In this case we must increase the length cv1 by adding an amount equal to one-half the difference of the distances from v to the first arc, and from v to the second arc previously drawn; and this last length so found will be the correct length of eccentric-rods.
It will be proper to remark here, that the radius cv1 was assumed to be the correct length of eccentric-rods; but since the rods cross each other when the eccentrics are at a and b, and do not cross each other when at x and y, the radius cv1 will always be a trifle short. It is therefore necessary to make the correction as explained.
In every case, the length of eccentric-rods must be so adjusted, that, when the line ee coincides with the center line of motion cT, the arc vv2 (which is the amount that the rocker-pin is drawn towards the axle from the line Qv when the eccentrics are at a and b) must be equal to the arc vv3 (which is the amount that the rocker-pin is moved towards the cylinders from the line Qv when the eccentrics are at x and y); the straight line Qv being perpendicular to the center line of motion cT.
Problem 5, [Fig. 25].—To find the position of the center of saddle-pin.—For this problem we again call to our aid the paper template shown in [Fig. 6b]. We have already seen in Problem 4 that the center of saddle-pin will be located on the line ee drawn on this template: it now only remains to determine the distance of this point from the link-arc c3c4.
Since the inequality between the crank-angle W and W1, [Fig. 23], becomes the greatest when the crank stands at half stroke, it is of the utmost importance to find such a position for the center of saddle-pin that equal portions of steam will be admitted alternately when the crank stands at half stroke. Or, in other words, the admittance of steam must cease at the moment that the piston has completed one-half stroke.
Fig. 25.
Let us commence this problem as we began the others; namely, Through the center of axle C, [Fig. 25], draw the horizontal line AD, also the vertical line LK. Find the position of crank at half stroke, as shown in [Fig. 21]. Next find the position of center line of motion CT, and position of rocker, as shown in [Fig. 22]. Find the relative position of eccentrics and crank when at half stroke, as shown in [Fig. 23]. Now, with a radius equal to the correct length of eccentric-rods, previously determined (shown in [Fig. 24]), describe from the point ½x as a center, the arc ½x1 ½x2; also with the point ½y as a center, and with the same radius, the arc ½y1 ½y2. Again, from the point ½a as a center, describe the arc ½a1 ½a2; also with the point ½b as a center, describe the arc ½b1 ½b2, using the length of eccentric-rods as a radius for all the arcs.
When the center of the forward eccentric is at ½x, the forward eccentric-rod pin will be located in the arc ½x1 ½x2. When the center of the forward eccentric is at ½a, the forward eccentric-rod pin will be located in the arc ½a1 ½a2. When the backward eccentric is at ½y, its eccentric-rod pin will be located in the arc ½y1 ½y2. When the center of the backward eccentric is at ½b, the eccentric-rod pin will be located in the arc ½b1 ½b2.
The next step is to find the relative position of the lower rocker-arm pin when steam is cut off at half stroke.