_Answer._ It can probably be calculated, but it is an exceedingly complicated problem, and one about which there is much difference of opinion. The difficulty is also increased by the fact that while counterweights may be heavy enough for one speed, they may be too heavy or too light for a slower or faster speed, and quite disproportioned when the engine is not working steam. The following rules are given in “Clark’s Railway Machinery,” and are perhaps sufficiently close to find a first approximation to the requisite position and weight of the counterweights; but the final adjustment should be made by trial. This can be done by suspending the locomotive by chains attached to the four corners of its frame, and setting the machinery in motion at the speed it is intended to run. By attaching a pencil to one or to each of the four corners of the frame, and arranging it so that it will mark on a horizontal fixed card, a diagram will be drawn, being usually an oval, which will show the amount and form of the oscillations. The counterweights can then be adjusted so that the diagram drawn by the pencil is reduced to the least possible size. When the adjustment is successful, the diameter of the diagram is reduced to about ¹⁄₁₆ of an inch.[80] Another and simpler, but less accurate, way is to place a pail or other vessel filled with water on the front of the engine and run the locomotive on a smooth track at a high speed, and adjust the counterweights so that the least amount of water will be spilled. [80] Rankine’s Treatise on the Steam Engine. QUESTION 317. _How can the centre of gravity of a counterweight in one segment be found?_ _Answer._ BY CUTTING A WOODEN TEMPLET OF UNIFORM THICKNESS TO THE FORM OF THE SURFACE, AND FREELY SUSPENDING IT BY ONE OF THE CORNERS, _a_, AS IN FIG. 194; A PLUMMET-LINE, _P a_, DROPPED FROM THE SAME POINT OF SUSPENSION IN FRONT OF THE TEMPLET WILL INTERSECT THE CENTRE LINE _b c_ AT THE CENTRE OF GRAVITY _C_. [Illustration: _Fig. 194._] QUESTION 318. _How can the centre of gravity of a counterweight in three segments be found?_ _Answer._ FIND THE CENTRE OF GRAVITY _C_, FIG. 195, OF ONE OF THE COUNTERWEIGHTS, AS ABOVE; THROUGH _C_ STRIKE AN ARC FROM THE CENTRE, _a_, OF THE WHEEL, CROSSING THE CENTRE LINES OF THE OTHER SEGMENTS AT THEIR CENTRES, _C′ C″_; DRAW _C′ C″_ MEETING _A B_ AT _D_, AND SET OFF _D E_, ONE-THIRD OF THE INTERVAL _D C_. THEN _E_ IS THE COMMON CENTRE OF GRAVITY OF THE THREE SEGMENTS. [Illustration: _Fig. 195._] QUESTION 319. _How can the centre of gravity of a counterweight in two segments be found?_ _Answer._ This is required when the crank is opposite to a spoke, as in fig. 196. FIND THE CENTRE OF GRAVITY, _C_, OF ONE SEGMENT AS BEFORE, AND BY AN ARC FIND THE OTHER CENTRE _C′_; DRAW _C C′_, CUTTING _A B_ AT _D_, WHICH IS THE COMMON CENTRE OF GRAVITY. [Illustration: _Fig. 196._] QUESTION 320. _How can the centre of gravity of a counterweight in four segments be found?_