CHAPTER XX.

By J. G. A. Meyer.

_LAYING OUT LINK-MOTION._

Fig. 19 is an outline of a link-motion such as is generally applied to the American locomotive. It can be adjusted to control the movement of the slide-valve in such a manner that equal portions of steam will be admitted alternately at each end of the cylinder.

In the following article we propose to explain how this can be accomplished.

Although we would not advise any person to be satisfied with approximate rules or constructions, yet cases do occur where the approximate constructions, being so very near correct, on account of their simplicity, are of greater practical value than the application of the rigid and more difficult theoretical rules.

By these remarks, we do not wish the reader to understand that the following constructions are all done according to the rules of thumb--not by any means; for all, with the exception of a few points, are theoretically correct. At the end of this article, we will point out those points which are, and which are not, approximately found; so that the reader may feel satisfied that our construction may always be relied upon as being correct for all practical purposes.

In what follows, the cylinder will always be regarded as lying on the right-hand side of the axle, the link being between cylinder and axle, and the axle located in the center of pedestal.

To avoid any misunderstanding, we will explain the meaning of some of the terms used.

The length of crank is the distance from center of axle to center of crank-pin.

For convenience, we shall call the total distance from center of eccentric-strap to the center of eccentric-pin in the link the length of the eccentric-rod.

The throw of eccentric is double the distance from center of axle to center of eccentric-wheel.

The length of the connecting rod is the distance from the center of crank-pin hole to center of cross-head pin-hole.

The length of link-hanger is the distance from center to center of holes.

CONDITIONS.

Since this article treats only on the adjustment of the link-motion, the following items are supposed to be known and established: The lap of valve, which in this case will be three-fourths of an inch; the throw of eccentrics, 5 inches; the stroke of the piston, 24 inches; the position of the rocker, as per Fig. 19; the length of the rocker-arms, which are in this case of equal length; length of link-hanger and all dimensions of link, complete, as shown in Fig. 19; and also the length of the connecting rod.

The adjustment of the link-motion may at first sight appear to be a difficult problem, as we must have a knowledge of the relative motions of the piston and slide-valve; but by reducing this problem to several elementary problems, so that the laws governing the relative motions may be discovered and clearly defined, a clear conception of our subject can be gained, and the solution of our original problem can be accomplished with comparative ease.

In order to find what kind of elementary problems are applicable, let us suppose that we are looking at a locomotive with a link-motion, as shown in Fig. 19, applied and correctly adjusted. Now let us examine it, commencing with the valve. We find that the valve receives its motion from the upper rocker-arm, and this receives its motion from the lower rocker-arm. According to our conditions, previously stated, both of the rocker-arms are of the same length; and, therefore, the arc described by the upper rocker-arm will be the same length as the arc described by the lower one. We also notice that the link which moves the lower rocker-arm is held in position by the lifting-shaft arm. The question, then, will naturally arise, Must this lifting-shaft arm have some particular length, and the center of lifting-shaft have some particular position? We answer, “Yes;” and this is one of our elementary problems to solve. Again, we notice that the saddle-pin is not in the center of the link; and we ask again, “Why?” To answer this will be another elementary problem. The next we notice are our eccentric-rods. These we find, on examination, to have some particular length; and to find this length is another elementary problem. Next we examine our eccentrics: these, we find, are fastened to the axle; and, since the crank is also fastened to the same axle, it follows that there are some relative positions between them; to find these positions is another elementary problem. Now let us look once more at the rocker, and we find that the two rocker-arms are not in the same straight line: hence, to find the amount of offset is another elementary problem. And, lastly, we must be able to find the position of crank-pin to correspond with the position of piston when at full stroke at either end of the cylinder, and also when at half stroke moving in either direction.

Here, then, we have all the elementary problems that are necessary to be understood for the solution of our original problem.

We will now explain all these problems, in an order the reverse to that in which we stated them; hence we have the following order:--

1st, To find position of crank at full and half stroke.

2d, To find center line of motion, and amount of offset in rocker-arms.

3d, To find relative positions of crank-pin and eccentrics when at full and half stroke.

4th, To determine the correct length of eccentric-rods.

5th, To find position of saddle-pin.

6th, To find the position of the center of lifting-shaft and length of arms.

PROBLEM 1, FIGS. 20 and 21.--_To find the position of crank when the piston is at full and half stroke._--Let the center of wheel and the axis of the cylinder be in the same straight line as _AD_, Fig. 20. With any point _C_ as a center, and a radius equal to the length of the crank, describe a circle _F_, ½_F_, _B_, ½_B_; and let us call this the crank-pin circle. The straight line _AD_ intersects the circumference of the circle in the points _F_ and _B_. The point _F_ will be the center of crank-pin when piston is at full stroke at the forward end of the cylinder, and point _B_ will be the center of crank-pin when the piston is at full stroke at the rear end of the cylinder. With the point _F_ as a center, and with a radius equal to the length of the connecting rod, describe an arc intersecting the line _AD_ in the point _F′_; with the point _B_ as a center, and with the same radius, describe an arc intersecting the straight line _AD_ in the point _B′_; and with the point _C_ as a center, and with the same radius, describe an arc intersecting the straight line _AD_ in the point _C′_. Point _F′_ will be the center of cross-head pin when the center of crank-pin is at _F_, and _B′_ the center of cross-head pin when the crank-pin is at _B_, and the point _C′_ will be the position of center of cross-head pin when piston is at half stroke. With point _C′_ as a center, and a radius equal to the length of the connecting rod, describe an arc passing through the point _C_, and intersecting the crank-pin circle in the points ½_F_ and ½_B_: these points will be the position of crank-pin when the piston is at half stroke, or when the center of cross-head pin is at _C′_.

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 ½_B_ and ½_F_: 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.

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.

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_, ½_F_, ½_B_, _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 ½_F_ (half stroke), draw the straight line ½_FC_ representing the center line of crank, and intersecting the eccentric-circle in the point ½_F″_. From the point ½_F″_, 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 ½_x_; also from ½_F″_ 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 ½_y_; then the point ½_x_ will be the position of forward eccentric, and the point ½_y_ will be the position of backward eccentric when the crank-pin is at ½_F_. 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 ½_a_ and ½_b_ when the crank-pin is at ½_B_. 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_.

Half stroke forward ½_F_, the forward eccentric will be at ½_x_.

Full stroke back end _B_, the forward eccentric will be at _a_.

Half stroke back end ½_B_, the forward eccentric will be at ½_a_.

Full Stroke forward _F_, the backward eccentric will be at _y_.

Half stroke forward ½_F_, the backward eccentric will be at ½_y_.

Full stroke back end _B_, the backward eccentric will be at _b_.

Half stroke back end ½_B_, the backward eccentric will be at ½_b_.

PROBLEM 4, FIG. 24.--_To determine the correct length of the eccentric-rods._--Let _c_ be the center of axle. Through this point draw the horizontal line _AD_, also a line _KL_ perpendicular to _AD_. The only purpose for which these two lines are drawn in this problem, as well as the others, is to have some lines from which we can locate other lines or points. Locate the center of rocker, and center lines of rocker-arms, as explained in Problem 2, and shown in Fig. 22; the lower arm standing perpendicular to the center line of motion, and the upper arm vertical. When the arms stand in this position, the rocker-pins will move through an equal distance on each side of these center lines during the time that the valve is making its full travel.

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. 6_a_ (p. 268). The arc _cᵥ{1}cᵥ{2}_ drawn through the center of opening of the link, we will call the link-arc; and the arc _dᵥ{1}dᵥ{2}_ 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. 6_b_ (link structure). This template is cut so that, if it is laid on the link, Fig. 6_a_, the arc of the template _cᵥ{3}cᵥ{4}_ will coincide with the link-arc _cᵥ{1}cᵥ{2}_, and _dᵥ{3}dᵥ{4}_ with the eccentric-pin arc _dᵥ{1}dᵥ{2}_, the end of template _dᵥ{3}cᵥ{3}_ with the line _dᵥ{1}cᵥ{1}_, and the end _dᵥ{4}cᵥ{4}_ with _dᵥ{2}cᵥ{2}_. On this template join the points _cᵥ{3}cᵥ{4}_ 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 _ffᵥ{1}_ parallel to _ee_, and on the other side draw _fᵥ{2}fᵥ{3}_ also parallel to _ee_; the distance from the point _f_ to the point _fᵥ{2}_ being equal to the distance between the centers of eccentric-rod pins, and _fe_ equal to _efᵥ{2}_. The points _f_ and _fᵥ{2}_ on the arc _dᵥ{3}dᵥ{4}_ 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 _v^1_ towards the axle, with a distance equal to _cᵥ{1}dᵥ{1}_, Fig. 6_a_; then with the point _x_ as a center, and _cv^1_ as a radius, describe the arc _xᵥ{1}xᵥ{2}_; 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 _cvᵥ{1}_ as a radius, describe the arc _yᵥ{1}yᵥ{2}_: 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 _cvᵥ{1}_ as a radius, describe an arc _aᵥ{1}aᵥ{2}_: 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 _cvᵥ{1}_ as a radius, describe the arc _bᵥ{1}bᵥ{2}_; 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 _xᵥ{1}xᵥ{2}_: point _fᵥ{2}_ in the arc _yᵥ{1}yᵥ{2}_ and the line _ee_ coincide with the center line of motion _cT_. Along the arc _cᵥ{3}cᵥ{4}_ of the template draw an arc on the paper. Next move the template so that the point _f_ will be in the arc _aᵥ{1}aᵥ{2}_, the point _fᵥ{2}_ in the arc _bᵥ{1}bᵥ{2}_, and the line _ee_ coincide with the center line of motion _cT_, and along the arc _cᵥ{3}cᵥ{4}_ 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 _cvᵥ{1}_ 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 _cvᵥ{1}_ of the eccentric-rod will be too short. In this case we must increase the length _cvᵥ{1}_ 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 _cvᵥ{1}_ 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 _cvᵥ{1}_ 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 _vvᵥ{2}_ (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 _vvᵥ{3}_ (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. 6_b_. 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 _cᵥ{3}cᵥ{4}_.

Since the inequality between the crank-angle _W_ and _Wᵥ{1}_, 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.

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 ½_x_ as a center, the arc ½_xᵥ{1}_ ½_xᵥ{2}_; also with the point ½_y_ as a center, and with the same radius, the arc ½_yᵥ{1}_ ½_yᵥ{2}_. Again, from the point ½_a_ as a center, describe the arc ½_aᵥ{1}_ ½_aᵥ{2}_; also with the point ½_b_ as a center, describe the arc ½_bᵥ{1}_ ½_bᵥ{2}_, using the length of eccentric-rods as a radius for all the arcs.

When the center of the forward eccentric is at ½_x_, the forward eccentric-rod pin will be located in the arc ½_xᵥ{1}_ ½_xᵥ{2}_. When the center of the forward eccentric is at ½_a_, the forward eccentric-rod pin will be located in the arc ½_aᵥ{1}_ ½_aᵥ{2}_. When the backward eccentric is at ½_y_, its eccentric-rod pin will be located in the arc ½_yᵥ{1}_ ½_yᵥ{2}_. When the center of the backward eccentric is at ½_b_, the eccentric-rod pin will be located in the arc ½_bᵥ{1}_ ½_bᵥ{2}_.

The next step is to find the relative position of the lower rocker-arm pin when steam is cut off at half stroke.

In Fig. 8 (p. 274) we have placed the slide-valve centrally over the ports, that is, it laps over each steam-port an equal amount, namely, ¾ of an inch, which is equal to the lap. In this position of the valve, the center line of the upper rocker-arm will stand perpendicular to the line drawn parallel to the valve-face, and the center line of the lower rocker-arm will stand perpendicular to the center line of motion _CT_: hence the center line of upper rocker-arm _MQ_ in Fig. 8 will coincide with the line _MQ_ in Fig. 25, and the center of lower arm _QU_ in Fig. 8 will coincide with the line _QU_ in Fig. 25.

Now let us follow the relative movement of the valve and piston. We find, that, when the piston commences its backward motion, the valve moves in the same direction, as shown by the arrow-points in Fig. 8_a_; and, during the time that the piston is completing the half stroke, the valve has finished its full travel backward, and commenced moving forward, as indicated by the arrow-points, Fig. 8_b_; and, at the time that the piston stands exactly at half stroke, the forward edge of the valve is just closing the forward steam-port, and consequently cutting off steam at half stroke when the piston is moving backward. From this we see, that, when the piston has completed the half stroke when moving backward, the center of the valve will be a little in the rear of the center of exhaust-port; the distance between the center of valve and the center of exhaust-port being ¾ of an inch, the amount of the lap: the upper rocker-pin will stand ¾ of an inch behind the line _MQ_, and the lower rocker-arm pin will be ¾ of an inch in front of the line _QU_, as shown in Fig. 8_b_. We therefore draw in Fig. 25 a straight line parallel to _QU_, and ¾ of an inch in front of it: this line will intersect the arc _RS_ in the point ½_Fᵥ{3}_; and this point is the position of the center of lower rocker-arm pin when the crank stands at ½_F_, and steam cut off at half stroke. Let the piston complete its backward stroke, and then commence moving forward towards half stroke, as shown by the arrow-point, Fig. 8_c_. During this time the valve has completed its full travel forward, and commenced traveling backward, as indicated by the arrow-point, Fig. 8_c_; and, when the piston stands exactly at half stroke, the rear edge of the valve is just closing the rear steam-port, and consequently cutting off steam at half stroke when the piston is moving forward. In this position the center line of the valve will be ¾ of an inch in front of the center of exhaust, the center of the upper rocker-arm pin will be ¾ of an inch in front of the line _MQ_, and lower rocker-pin ¾ of an inch in the rear of the line _QU_, as shown in Fig. 8_c_. We therefore draw in Fig. 25 a line parallel to _QU_, and ¾ of an inch behind it; this line will intersect the arc _RS_ in the point ½_Bᵥ{3}_; and this point will be the position of the center of lower rocker-arm pin when the crank stands at ½_B_, and steam cut off at half stroke. Now, remember, that when the crank stands at ½_F_, Fig. 25, the forward eccentric will be ½_x_, and the backward eccentric at ½_y_; and, if the link is raised or lowered while the eccentrics remain at ½_x_ and ½_y_, the forward eccentric-rod pin will move in the arc ½_xᵥ{1}_ ½_xᵥ{2}_, and the backward eccentric-rod pin will move in the arc ½_yᵥ{1}_ ½_yᵥ{2}_.

Let us now find the position of link when steam is cut off at half stroke at either end of the cylinder.

The points ½_Fᵥ{3}_ and ½_Bᵥ{3}_ in Fig. 25 being located, place the paper template on the drawing so that the point _f_ will lie in the arc _xᵥ{1}xᵥ{2}_, and the point _fᵥ{2}_ in the arc _yᵥ{1}yᵥ{2}_, and the link-arc _cᵥ{3}cᵥ{4}_ just touching the point ½_Fᵥ{3}_. While the template is in this position, draw on the paper along the edge _cᵥ{3}cᵥ{4}_ a portion of the link-arc, and mark the position that the line _ee_ occupied, so that, when the template is removed, the line _eᵥ{1}eᵥ{1}_ can be drawn on the paper to represent the line _ee_ of the template. Next place the template so that the point _f_ will lie in the arc ½_aᵥ{1}_ ½_aᵥ{2}_ the point _fᵥ{2}_ in the arc ½_bᵥ{1}_ ½_bᵥ{2}_ and the link-arc _cᵥ{3}cᵥ{4}_ just touching the point ½_Bᵥ{3}_ and, while in this position, draw part of the link-arc _cᵥ{3}cᵥ{4}_ on the paper, mark the position that the line _ee_ occupied, and, after the template is removed, draw the line _eᵥ{2}eᵥ{2}_ on the paper to represent the line _ee_ of the template. Now find by trial a point _xᵥ{1}_ on the line _cᵥ{1}cᵥ{1}_, and another point _xᵥ{2}_ on the line _eᵥ{2}eᵥ{2}_, so that the distances of these points from their link-arcs are equal, and that a straight line drawn through them will be parallel to the center line of motion.

The distance from _xᵥ{1}_ to the link-arc--or, which is the same thing, the distance from the point _xᵥ{2}_ to the link-arc--will be the correct distance between the center of saddle-pin and the link-arc _cᵥ{1}cᵥ{2}_, Fig. 6_a_. Or, in other words, the position of the point _xᵥ{1}_ or _xᵥ{2}_, Fig. 25, will indicate the proper position of the point of suspension on the link. For future reference, let us mark this point of suspension on the template, and indicate it by _X_, Fig. 6_b_.

PROBLEM 6, FIG. 26.--_To find the position of the center of lifting-shaft and the length of its arms._--In the last problem we found the point of suspension of the link, so that it will cause the valve to cut off equal portions of steam when the piston stands at half stroke. It now remains for us to find the position of the lifting-shaft and the length of the lifting-shaft arms, so that the greatest equal amounts of steam will be admitted alternately at each end of the cylinder. Here a little difficulty arises which needs explanation, so that our construction may not seem inconsistent to the reader. It would be an easy matter to place our lifting-shaft to accomplish the object just stated; but, if we do this, the lead will not be equal at each end of cylinder when the piston is at full stroke. Again, if we locate our lifting-shaft in such a manner that equal lead will be obtained, then the maximum cut-off will not be equal; but the difference will be comparatively so small that it will not injure the working of the engine. This small difference of the maximum cut-off is therefore considered among practical men of little or no importance, but it is always considered good practice to have an equal lead at full stroke. Let us therefore adjust the lifting-shaft to obtain an equal lead, and allow us to consider the maximum cut-off to be equal when the lead is equal at full stroke.

For this problem we have to combine all the foregoing problems. Through the center _C_ of axle draw the horizontal line _AD_, and the line _KL_ perpendicular to it. Find the positions of crank at full and half stroke, as per Problem 1. Locate the rocker, draw the center line of motion _CT_, and amount of offset in lower rocker-arm, according to Problem 2. Next, locate the relative positions of eccentrics when the crank stands at full and half stroke, as explained in Problem 3. Then with a radius equal to the correct length of eccentric-rods, as explained in Problem 4, draw

From the point _x_ as a center, the arc _xᵥ{1}_ _xᵥ{2}_ ” ” _y_ ” ” ” _yᵥ{1}_ _yᵥ{2}_ ” ” ½_x_ ” ” ” ½_xᵥ{1}_ ½_xᵥ{2}_ ” ” ½_y_ ” ” ” ½_yᵥ{1}_ ½_yᵥ{2}_ ” ” _a_ ” ” ” _aᵥ{1}_ _aᵥ{2}_ From the point _b_ as a center, the arc _bᵥ{1}_ _bᵥ{2}_ ” ” ½_a_ ” ” ” ½_aᵥ{1}_ ½_aᵥ{2}_ ” ” ½_b_ ” ” ” ½_bᵥ{1}_ ½_bᵥ{2}_ From the point _b_ as a center, the arc _bᵥ{1}_ _bᵥ{2}_ ” ” ½_a_ ” ” ” ½_aᵥ{1}_ ½_aᵥ{2}_ ” ” ½_b_ ” ” ” ½_bᵥ{1}_ ½_bᵥ{2}_

Locate the points ½_Bᵥ{3}_ and ½_Fᵥ{3}_, indicating the position of the center of lower rocker-pin when steam is cut off at half stroke; find the points _xᵥ{1}xᵥ{2}_ indicating the positions of the point of suspension when the link is lifted into the position to cut off at half stroke, as explained in Problem 5, and shown in Fig. 25.

Now, in order to find the position of lifting-shaft and length of arms, we must find four more additional points,--first, the position of the point of suspension of the link when the piston is at full stroke forward end of cylinder, and the crank-pin at _F_, the valve having 1/16 of an inch lead, and the engine moving forward, as indicated by the arrow-point 1; also the position of the point of suspension of the link when the piston is at full stroke at the opposite end of the cylinder, valve 1/16 inch lead, and engine going in the same direction. To find these two points, we must know the corresponding position of the center of lower rocker-pin. In Fig. 8_a_ we see, that when the piston is at full stroke forward, and valve with 1/16 inch lead, the center of valve is 13/16 of an inch in the rear of the center line of exhaust, and consequently the lower rocker-pin will be 13/16 of an inch in front of the line _QU_. In the same manner we can show that the center of lower rocker-pin will be 13/16 of an inch in the rear of the line _QU_ when the piston is at the opposite end of the cylinder.

Let us now locate the positions of the lower rocker-pin in Fig. 26, by drawing a line parallel to and in front of _QU_, with 13/16 of an inch between them: this line will intersect the arc _RS_ in the point _Fᵥ{3}_, and this point will be the center of lower rocker-pin when the piston is at full stroke forward. Draw another line 13/16 of an inch in the rear of _QU_ and parallel to it: this line will intersect arc _RS_ in the point _Bᵥ{3}_, and this point will be the center of rocker-pin when the piston is at full stroke in the rear end of the cylinder. Now place the paper template with the line _ee_ below the center line of motion _CT_, the point _fᵥ{1}_ on the arc _xᵥ{1}xᵥ{2}_, the point _fᵥ{2}_ on the arc _yᵥ{1}yᵥ{2}_, and the link-arc _cᵥ{3}cᵥ{4}_ just touching the point _Fᵥ{3}_, and, while in this position, mark the point _X_ of the template on the paper, which can be done with the aid of a needle, and indicate the point on the paper by _xᵥ{3}_. This point will be the position of the center of saddle-pin when the piston is at full stroke in the forward end of the cylinder, the valve having 1/16 inch lead. Again, slide the template along until the point _fᵥ{1}_ is on the arc _aᵥ{1}aᵥ{2}_, the point _fᵥ{2}_ on the arc _bᵥ{1}bᵥ{2}_, and the link-arc _cᵥ{3}cᵥ{4}_ in contact with the point _Bᵥ{3}_; mark the point _X_ of the template on the paper, and indicate this point by _xᵥ{4}_. This point will be the position of the center of saddle-pin when the piston is at full stroke in the rear end of the cylinder, the valve having 1/16 inch lead. Secondly, to find the position of the point of suspension of the link when the piston is at full stroke in the forward end of the cylinder, valve having 1/16 of an inch lead, and the engine moving backward, as indicated by the arrow-point 2; also the position of the point of suspension of the link when the piston is at full stroke at the opposite end of the cylinder, valve 1/16 of an inch lead, engine going in the same direction. For this purpose, slide the template along until the line _ee_ is above the line _CT_, and _fᵥ{1}_ in the arc _aᵥ{1}aᵥ{2}_, the point _fᵥ{2}_ in the arc _bᵥ{1}bᵥ{2}_, and the link-arc _cᵥ{3}cᵥ{4}_ in contact with the point _Bᵥ{3}_; mark the point _X_ on the paper, and indicate this point by _xᵥ{5}_. This point will be the position of the center of saddle-pin when the piston is at full stroke in the rear end of the cylinder, valve having 1/16 of an inch lead. Again, slide the template along until the point _fᵥ{1}_, will be in the arc _xᵥ{1}xᵥ{2}_, point _fᵥ{2}_ in the arc _yᵥ{1}yᵥ{2}_, and the link-arc _cᵥ{3}cᵥ{4}_ in contact with the point _Fᵥ{3}_; mark the point _X_ on the paper, and indicate this point by _xᵥ{6}_. This point will be the position of the center of saddle-pin (or the point of suspension) when the piston is at full stroke in the forward end of the cylinder, valve 1/16 of an inch lead, engine moving backward. Now, once more, with the point _xᵥ{3}_ as a center, and with the length of the link-hanger as a radius, describe an arc; and with the point _xᵥ{4}_ as a center, and the same radius, describe another arc. These two arcs will intersect each other in the point _g_. Again, with the length of the link-hanger as a radius, and the points _xᵥ{1}xᵥ{2}_ as centers, describe two arcs intersecting each other in the point _gᵥ{1}_, with the points _xᵥ{5}xᵥ{6}_ as centers; and, with the same radius, describe another two arcs intersecting each other in the point _gᵥ{2}_. Lastly, through the points _g_, _gᵥ{1}gᵥ{2}_, draw an arc. The center _h_, from which the arc has been described, will be the center of the lifting-shaft, and the radius _hg_ or _hgᵥ{2}_ will be the length of the lifting-shaft arms; that is, the length of the two arms to which the link-hangers are attached: the length of the other lifting-shaft arm, to which the reach-rod is attached, is made to suit the other details of the engine.

When the admittance of steam ceases at the same time that the piston has reached the half stroke, the practical man would say “that the valve is cutting off equal at half stroke.” When the greatest equal volume of steam is admitted alternately in each end of the cylinder, the valve is said to be cutting off equal when the link is in full gear.

It is always conceded among engineers, that when the link-motion is adjusted to cut off equal at half stroke, and also to cut off equal when the link is in full gear, equal volumes of steam will be admitted alternately when the link hangs at any intermediate point.

If, now, we examine Problem 5, we find, that, to obtain an equal cut-off at half stroke, it is necessary to find the proper position of saddle-pin.

Again, if we examine Problem 6, we find, that, in order to obtain an equal cut-off when the link is in full gear, also an equal cut-off for any point between full gear and half stroke, we have to determine the proper position of the center of lifting-shaft and the correct length of its arms.

Lastly, if we examine the first four problems, we find them simply to be preparatory problems.

According to promise, we will draw attention to those points which have been, and others which have not been, approximately found. Problems 1, 2, and 3 are theoretically correct. In Problems 4, 5, and 6, the use of the template will not be admitted for theoretical reasoning; but, if the construction is made with absolute accuracy, the result will be theoretically correct.

* * * * *

The following are a few dimensions of locomotives made by well-known makers:--

DIMENSIONS OF LOCOMOTIVES

============================+================================= | BALDWIN. +-----------+-------+------------- | Standard | Mogul |Consolidation | Passenger |Engine.| Engine. | Engine. | | ----------------------------+-----------+-------+------------- | | | | Inches. |Inches.| Inches. Dimensions of cylinders | 17 × 24 |18 × 24| 20 × 24 Length of steam-ports | 16 | 16 | 16 Width of steam-ports | 1¼ | 1¼ | 1¼ Width of exhaust-port | 2½ | 2½ | 2½ Throw of eccentrics | 5 | 5 | 5 Travel of valve | 5⅜ | 5⅜ | 5⅜ Outside lap of valve | ¾ to ⅞ | ¾ | ¾ Inside lap of valve |1/32 to ⅛ | 1/32 | 1/32 Distance between center of | | | axle and center of rocker | 56 | 46 | 81 Length of upper rocker-arm, | 10½ | 10½ | 10½ Length of lower rocker-arm, | 9½ | 9½ | 9½ Length of link | 55 | 46 | 81 Length of link-hanger | 13½ | 15 | 15 Length of tumbling-shaft | | | arm | 17 | 16 | 16 Length of connecting rod | 75 | 80 | 114½ Distance between centers of | | | eccentric-rod pins | 11-9/16| 10½ | 10¾ Suspension of link back of | | | link-center | ⅜ | 11/16 | 1 Lead of valve in full gear | 1/16 | 1/16 | 1/16 Lead of valve in center | ¼ | 1¼ | ¼ ============================+===========+=======+=============

DIMENSIONS OF LOCOMOTIVES--_Continued_. ============================+================================= | BROOKS. +-----------+-------+------------- | Standard | Mogul |Consolidation | Passenger |Engine.| Engine. | Engine. | | ----------------------------+-----------+-------+------------- | | | | Inches. |Inches.| Inches Dimensions of cylinders | 18 × 24 |18 × 24| 20 × 24 Length of steam-ports | 16 | 15 | 16 Width of steam-ports | 1⅛ | 1⅛ | 1¼ Width of exhaust-port | 2½ | 2½ | 2½ Throw of eccentrics | 5 | 4½ | 5 Travel of valve | 5 | 5 | 5½ Outside lap of valve | ¾ | ⅜ | 13/16 Inside lap of valve | 1/64 | 1/64| 1/64 Distance between center of | | | axle and center of rocker | 65 | 50 | 40 Length of upper rocker-arm, | 10½ | 9¾ | 10½ Length of lower rocker-arm, | 10½ | 8¾ | 9½ Length of link | 65 | 50 | 40 Length of link-hanger | 13 | 13½ | 15¼ Length of tumbling-shaft | | | arm | 17⅜ | 18 | 16 Length of connecting rod | 96 | 84 | 77 Distance between centers of | | | eccentric-rod pins | 11 |11-3/16| 11 Suspension of link back of | | | link-center | ⅜ | ⅜ | ⅜ Lead of valve in full gear | 1/10 | 1/10| 1/10 Lead of valve in center | ⅜ | ⅜ | ⅜ ============================+===========+=======+=============

DIMENSIONS OF LOCOMOTIVES--_Continued_. ============================+===========+=======+============= | GRANT. +-----------+-------+------------- | Standard | Mogul |Consolidation | Passenger |Engine.| Engine. | Engine. | | ----------------------------+-----------+-------+------------- | Inches. |Inches.| Inches. Dimensions of cylinders | 17 × 24 |18 × 24| 20 × 24 Length of steam-ports | 16 | 16 | 16 Width of steam-ports | 1¼ | 1¼ | 1¼ Width of exhaust-port | 2½ | 2½ | 2½ Throw of eccentrics | 5 | 5 | 5½ Travel of valve | 5 | 5¼ | 5 Outside lap of valve | ¾ | ⅞ | ⅞ Inside lap of valve | None. | 1/16| 1/16 Distance between center of | | | axle and center of rocker | 68-11/16| 40 | 80 Length of upper rocker-arm, | 10¼ | 9⅞ | 9⅞ Length of lower rocker-arm, | 10¼ | 8½ | 10½ Radius of link | 63 | 39⅜ | 80 Length of link-hanger | 12¾ | 14 | 12¾ Length of tumbling-shaft | | | arm | 17 | 16 | 14 Length of connecting rod | 85 | 69½ | 112 Distance between centers | | | of eccentric-rod pins | 12 | 12 | 12 Suspension of link back of | | | link-center | ⅜ | ⅜ | 1-3/16 Lead of valve in full gear | 1/32 | 1/32| 1/32 Lead of valve in center | 5/16 | ⅜ | ¼ ============================+===========+=======+=============

DIMENSIONS OF LOCOMOTIVES--_Continued_. ============================+===========+=======+============= | PITTSBURG. +-----------+-------+------------- | Standard | Mogul |Consolidation | Passenger |Engine.| Engine. | Engine. | | ----------------------------+-----------+-------+------------- | Inches. |Inches.| Inches. Dimensions of cylinders | 17 × 24 |18 × 24| 20 × 24 Length of steam-ports | 15 | 16 | 18 Width of steam-ports | 1¼ | 1¼ | 1¼ Width of exhaust-port | 2½ | 2½ | 2½ Throw of eccentrics | 5 | 5 | 5 Travel of valve | 5 | 5½ | 5-7/16 Outside lap of valve | ⅞ | ¾ | ¾ Inside lap of valve | ⅛ | 1/32| 1/32 Distance between center of | | | axle and center of rocker | 56 | 48 | 28 Length of upper rocker-arm, | 11 | 11¾ | 11¾ Length of lower rocker-arm, | 11 | 10¾ | 10¾ Radius of link | 56 | 48 | 80 Length of link-hanger | 14½ | 18¾ | 18½ Length of tumbling-shaft | | | arm | 18 | 15 | 15 Length of connecting rod | 88 | 85 | 114½ Distance between centers | | | of eccentric-rod pins | 12 | 12 | 12 Suspension of link back of | | | link-center | ⅜ | ⅜ | 9/16 Lead of valve in full gear | 1/10 | 1/10| 1/10 Lead of valve in center | 5/16 | 11/32| ¼ ============================+===========+=======+=============

DIMENSIONS OF LOCOMOTIVES--_Continued_. ============================+===========+=======+============= | SCHENECTADY. +-----------+-------+------------- | Standard | Mogul |Consolidation | Passenger |Engine.| Engine. | Engine. | | ----------------------------+-----------+-------+------------- | Inches. |Inches.| Inches. Dimensions of cylinders | 17 × 24 |19 × 24| 20 × 24 Length of steam-ports | 16 | 16 | 16 Width of steam-ports | 1¼ | 1¼ | 1¼ Width of exhaust-port | 2½ | 2¾ | 2¾ Throw of eccentrics | 5¼ | 5¼ | 5¼ Travel of valve | 5½ | 5½ | 5½ Outside lap of valve | ⅞ | ¾ | ¾ Inside lap of valve | 1/32 | 1/32| 1/32 Distance between center of | | | axle and center of rocker | 63 | 54 | 45 Length of upper rocker-arm, | 11 | 10¾ | 11 Length of lower rocker-arm, | 10 | 9¾ | 10 Radius of link | 63 | 54 | 45 Length of link-hanger | 13 | 16½ | 16½ Length of tumbling-shaft | | | arm | 16 | 16 | 16 Length of connecting rod | 92½ | 90½ | 79½ Distance between centers | | | of eccentric-rod pins | 12 | 12 | 12 Suspension of link back of | | | link-center | ⅞ | ⅞ | ⅜ Lead of valve in full gear | 1/16 | 1/16| 1/16 Lead of valve in center | 5/16 | 5/16| 5/16 ============================+===========+=======+=============

DIMENSIONS OF LOCOMOTIVES--_concluded_. ============================+===========+======= | MASON. +-----------+------- | Standard | Mogul | Passenger |Engine. | Engine. | ----------------------------+-----------+------- | Inches. |Inches. Dimensions of cylinders | 18 × 24 |19 × 24 Length of steam-ports | 17 | 18 Width of steam-ports | 1¼ | 1¼ Width of exhaust-port | 2½ | 2½ Throw of eccentrics | 5 | 5 Travel of valve | 5 | 5 Outside lap of valve | ¾ | ¾ Inside lap of valve | 1/64 | 1/32 Distance between center of | | axle and center of rocker | 38 | 49½ Length of upper rocker-arm, | 9 | 9 Length of lower rocker-arm, | 9 | 9 Radius of link | 58 | 49½ Length of link-hanger | 10 | 10 Length of tumbling-shaft | | arm | 20 | 20 Length of connecting rod | 91-3/16 | 84 Distance between centers | | of eccentric-rod pins | 12 | 12 Suspension of link back of | above link-center | 3⅜ | 3⅜ Lead of valve in full gear | 1/16 | 1/16 Lead of valve in center | 5/16 | 5/16 ============================+===========+=======