Page:A Treatise on the Steam Engine (1847).djvu/111

It will be observed that the greater the difference between the lengths of the radius bar and beam, the greater is the amount of the correction to be made. As an example of the use of the table, suppose we have the length of the beam C G = 38 inches, and that the radius bar has been found by proportion (A) to be 15½ inches; 15½ ÷ 38 is nearly 0.4: we shall therefore get for the corrected length of the radius bar 15½ + 0.0817 x 38 = 16.76 inches. The application of this correction to the length of the radius bar, will in all cases diminish the amount of the deviation from the perpendicular by one half; but it should be borne in mind that the parallel motion will work most accurately when the radius bar and beam C G are of the same length; and they should therefore be kept as nearly equal as circumstances will permit.

The simple arrangement of two levers connected by a link, which we have now described, is sometimes used as a parallel motion; but additional rods and connecting links are generally superadded, which render the application of the principle more convenient in practice. Fig. 3 represents the parallel motion in most general use for land engines; and fig. 4 a similar one adapted for marine side-lever engines. The general arrangement in both of these is the same, except that the marine engine motion has the reverse side uppermost. We have marked all the corresponding parts in the two figures with the same letters, so that what we have to say of the one will refer equally to the other. In land engines, the radius bar H F is generally either equal to or longer than G C, the part of the beam which works it; while, in the marine engine motion, the radius bar is usually shorter than the part of the beam that gives it motion. This difference is shown in the figures. Referring now to figs. 3 and 4., we shall explain the method of applying the principles we have already described to the calculation of the proportions of the several parts. D C is the beam or side lever moving on the centre C ; H F, the radius bar moving on the centre H, and connected with the beam by the rod F G. D P is another rod equal in length to G F; and P F is a rod equal in length to D G. The three rods, therefore, D P, P F, and F G, with the part D G of the beam, form a parallelogram; and the piston-rod is attached to the point P, which, at half stroke, is in the perpendicular line R S, bisecting the versed sine D K. When the engine is put in motion, this point P will move nearly in the perpendicular line R S. For, suppose the beam to be moved upwards into the position D’ C, the point will then be in the position P’, and the positions of the various other points are denoted by corresponding accented letters. Join C P and C P’, and also E E’, and we will have, by similar triangles, PE: EC :: DG : GC :: D’G’ : G’ C, and P’E’: E’ C :: D’ G. G C; hence, PE : EC :: P’E’ : E’ C.

Therefore (Euclid vi. 2.) P P’ is parallel to E E’. Let the length of F H be made such, that $HF={\tfrac {GC+EG}{FE}}$, and by equation A, the point E will move in a perpendicular line; but P P’ having been proved to be parallel to the line E E’ in which E moves, the point P must also move in a perpendicular line. Again, referring to equation (A), we have
E F : EG :: G C: H F, but by similar triangles
E F : E G :: F P : G C :: D G : G C
hence, D G : G C :: G C : H F = ${\tfrac {GC^{2}}{DG}}$ ... (B).

This equation will give us the length of the radius bar H F, when the length of C G is determined. In land engines the air pump is generally wrought from the point E, and the centre G is fixed so as to be most convenient for that purpose. Formula (B), is therefore very useful for determining the proper length of the radius rod in such cases. In marine engines, however, the position of the centre H of the radius bar is generally fixed from other considerations, and it becomes necessary to find at what point the centre G should be placed, so as to make P move in a perpendicular line.

To enable us to determine this,
Let H F + G C = a = the horizontal distance of the centre^[1] H, from the main centre, plus half the versed sine of the arc, described by the extremity D of the beam
$DG+GC$, or $DC=b$
then, $GC=a-HF\therefore DG=b+HF-a$

hence, (B) ... $HF={\tfrac {(a-HF)^{2}}{b+HF-a}}={\tfrac {a^{2}+HF^{2}-2a\cdot HF}{b^{2}+HF-a}}$
$\therefore HF^{2}+HF(b-a)=a^{2}+HF^{2}-2a\cdot HF$
$\therefore HF={\tfrac {a^{2}}{b+a}}$
or, $HF={\tfrac {(GF+GC)^{2}}{GF+GC+DC}}$ .... (C)

This equation will give us the length of H F ; and to get G C, we have $GC=a-HF$ ... (C’)

These three equations therefore, (B), (C), and (C’) enable us to calculate the proportions of this form of parallel motion in all circumstances.

In figs. 5 and 6, two other forms of parallel motion in common use for marine engines are represented. In fig. 5 the radius bar, instead of being connected as formerly to the crosshead, is connected by the rod F Q to the side rod by a pin at Q. F Q is made equal, and parallel to G D ; and F G equal, and parallel to Q D. In fig. 6 the centre H of the radius bar is placed much farther from the main centre. It may be on either side of the piston rod, as is found most convenient. In this motion, too, the end F of the radius bar is connected by a pin to the upright rod G V. The rod G V is made equal and parallel to the side rod D P, and the rod P V is made equal and parallel to D G.

To enable us to investigate the method of calculating the lengths of the parts in these motions, join as before C P, and C P’, and from the point T where C P cuts F Q in fig. 5 or H F in fig. 6 draw T T’ parallel to DP. These lines being drawn in both figures, the following investigation will apply to either: —

Let $DC=b=CG+GT'+T'D$

$PQ=e$

$FH=r$

$PD=s$

$GC+FH=a=\mathrm {the\ horizontal\ distance\ of\ H\ from\ the\ main\ centre} \ +{\tfrac {1}{2}}\mathrm {versine\ arc} \ D'DD''\times {\tfrac {s-c}{s}}$^[2]

It may be proved in the same way, as for fig. 4 that the point P moves in a line parallel to that described by the point E, so that the inquiry resolves itself as before into an investigation of the relative lengths of C G, and F H, that will cause the point E to move in a perpendicular line. Let us suppose, first, that the height of the centre H of the radius bar, and consequently the height also of the pin Q, is fixed, and that the point G, or the length of C G, is also determined. It is required to find from these data the proper length of H F. By equation (A) we have

C G : F H :: F E : E G, but by similar triangles F T : G C :: F E : EG;
hence, TF : GC :: CG : FH = ${\tfrac {GC^{2}}{TF}}={\tfrac {GC^{2}}{T'G}}$

but, $T'G=CT'-CG$, and by similar triangles

DP : QD :: DC : CT' = ${\tfrac {DC\cdot QD}{DP}}={\tfrac {b(s-c)}{s}}$

hence, $T'G={\tfrac {b(s-c)}{s}}-CG={\tfrac {b(s-c)-s\cdot CG}{s}}$

and, $FH={\tfrac {GC^{2}\cdot s}{b(s-c)-s\cdot CG}}$ .... (D)

Again, if the length G C is not given, but the positions of the centre H, and the pin Q are fixed, let it be required to find the lengths of H F and G C. We shall have the same as above

$FH={\tfrac {GC^{2}}{T'G}}={\tfrac {(a-r)^{2}}{T'C-a+r}}$

or $r={\tfrac {(a-r)^{2}}{T'C-a+r}}$

$\therefore r(T'C-a+r)=a^{2}+r^{2}-2a^{2}$

$\therefore r(T'C+a)=a^{2}$

$\therefore r={\tfrac {a^{2}}{T'C+a}}=$

But it was shown above that $T'C={\tfrac {b(s-c)}{s}}$ hence

$r={\tfrac {a^{2}s}{as+b(s-c)}}$ .... (E)

This equation will give the length of H F, and to get G C we have

$GC=a-r$ ... (E’)

Again, suppose that the lengths of the levers H F and C G, and consequently the horizontal distance of the centre H from the main centre are given. Let it be required to find the height of the centre H and point Q

From equation E we get

$asr+b(s-c)r=a^{2}s$
$\therefore asr+brs-brc=a^{2}s$
$\therefore brc=asr+brs-a^{2}s$

sines only when ${\tfrac {1}{2}}\mathrm {vers.} \ \phi$ is indefinitely small in comparison with r; and that, by any increase of the versed sine (${\tfrac {1}{4}}s^{2}$ being always constant), the radius diminishes in a less ratio than the versed sine increases, and vice versa. Of course the converse of this is true also; that, if the radius be diminished, the versed sine increases in a greater ratio, and vice versa. When the beam is just leaving the horizontal position, the versed sine is indefinitely small in comparison of the radius; and therefore, at this point, the versed sine of the beam and radius bar will be inversely as the radii; and H F, as determined by equation (A), will be correct for this part of the stroke. At either end of the stroke, however, it ought to be equal to ${\tfrac {\mathrm {versed\ sine} \ DKGC}{\mathrm {versed\ sine} \ FL}}$. Now, as it cannot have both these values, the nearest approach to accuracy we can make is, to make its length intermediate between these two. The corrections in the Table are computed on this principle.

1. If a perpendicular H M be let fall from H on DC, MC will be the horizontal distance of the centre H, from the main centre; and since FG is parallel to PD, it is evident that FH+GC=a=MC+DN.
2. ½ versine of arc $D'DD''\times {\tfrac {s-c}{s}}$is the excess of the distance of the point D, from the line R S, over the distance of the point Q from the same line. Comparing this with the last note, the correctness of the value given above to a will be evident.