The parallel motion commonly employed in marine engines, working with side beams, is exactly similar in principle to that employed in land engines; but owing to the parts being inverted, it appears different. It is represented in the annexed cut. The radius shaft is usually worked by a single link, or side rod, which is attached to that radius bar and side beam which is next the vessel's side, in order to leave more space between the two engines for working the hand gear.

In constructing marine engines, it is occasionally required, that the centre of the radius rod shall be placed in some particular point; and it then becomes necessary to determine the length of the radius bar and of the parallel bar. This is usually ascertained by repeated trials with rods of different lengths, for want of a correct rule for readily determining it: we therefore give the following, which, we believe, has not before been published.

Let a b represent the beam at its horizontal position, c the centre of the cross head of the piston-rod, and c b the side rod by which it is connected to the end b of the beam a b; and let it be required that the centre of the radius bar shall be in the point d; then as the parallel bar is always parallel to the beam, and as at the horizontal position of the beam the parallel bar and radius bar will coincide or lie in the same horizontal plane, draw the line d e parallel to a b, and it will represent the position of the parallel bar and radius bar at this position of the beam; and e will be the point at which the parallel bar is attached to the side rod; and e d will be the difference between the length of the parallel bar, and that of the radius bar. Then through the point c draw the vertical line cf, to represent the path of the piston-rod, and draw a g in the highest position of the beam, making a g equal to a b; then from g with the length b c intersect the line cf in f, and draw the line f g, upon which lay off the distance b e from g to ft, then draw h i of an indefinite length and parallel to a g, and on it lay off h k equal to d e, the difference between the length of the radius bar and parallel bar; then, as the remaining portion of the parallel bar will be equal to the radius bar, and the two bars are connected at their extremity; if we draw the line k d, they will form with it an isosceles triangle, the apex of which will be at the point of junction of the two bars; therefore, to find the apex, upon k and d as centres, with any distance in the compasses describe arcs cutting in o o, through which points draw op, cutting h i inp; then draw dp and it will be the length of the radius bar; and h p will be the length of the parallel bar.

Then upon g a set off g r=h p and draw p r, and it will represent the back link, and r the point at which it is attached to the beam. 