INTERNAL DISTURBING FORCES IN THE LOCOMOTIVE.
QUESTION 314. _What are the internal disturbing forces in a locomotive?_
_Answer._ They are: 1, the momentum of the parts which have a
reciprocating motion; 2, those due to the varying pressure of the
steam in the cylinder-heads; 3, those caused by the thrust of the
connecting-rods against the guide-bars; and 4, those produced by
unbalanced revolving parts.
QUESTION 315. _How can the effects of these disturbing forces be
neutralized?_
_Answer._ By putting counterweights, _A_, _A_, fig. 161, in the
driving-wheels opposite the crank-pins. The motion of these will then
be in the reverse direction to that of the weight of the parts attached
to the crank-pins, and the motion of the one will thus, to some extent,
at least, neutralize the disturbing influence of the other.
QUESTION 316. _Can the weight of these counterweights be calculated for
any locomotive?_
_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?_
_Answer._ FIND, AS BEFORE, THE CENTRES _C_, _C′_, _C″_, _C‴_, FIG.
197, OF THE SEGMENTS; DRAW _C″ C′_ AND _C‴ C_, CUTTING THE LINE _A B_;
BISECT THE INTERVAL SO INCLOSED AT _E_ FOR THE COMMON CENTRE OF GRAVITY.
[Illustration:
_Fig. 197._]
QUESTION 321. _How is the counterweight for outside-cylinder engines
with a single pair of driving-wheels calculated?_
_Answer._ FIND THE TOTAL WEIGHT, IN POUNDS, OF THE REVOLVING AND
RECIPROCATING MASSES FOR ONE SIDE, NAMELY, THE PISTON AND APPENDAGES,
CONNECTING-ROD, CRANK-PIN AND CRANK-PIN BOSS; MULTIPLY BY THE LENGTH OF
CRANK IN INCHES, AND DIVIDE BY THE DISTANCE IN INCHES OF THE CENTRE OF
GRAVITY OF THE SPACE TO BE OCCUPIED BY THE COUNTERWEIGHT. THE RESULT
IS THE COUNTERWEIGHT IN POUNDS, TO BE PLACED EXACTLY OPPOSITE TO THE
CRANK.
Question 322. _How is the counterweight for outside-cylinder engines
with coupled driving-wheels calculated?_
_Answer_. FIND THE SEPARATE REVOLVING WEIGHTS, IN POUNDS, OF CRANK-PIN,
CRANK-PIN BOSS, COUPLING-RODS AND CONNECTING-ROD, FOR EACH WHEEL, ALSO
THE RECIPROCATING WEIGHT OF THE PISTON AND APPENDAGES, AND HALF THE
CONNECTING-ROD; DIVIDE THE RECIPROCATING WEIGHT EQUALLY BETWEEN THE
COUPLED WHEELS, AND ADD THE PART, SO ALLOTTED, TO THE REVOLVING WEIGHT
ON EACH WHEEL; THE SUMS SO OBTAINED ARE THE WEIGHTS TO BE BALANCED AT
THE SEVERAL WHEELS, FOR WHICH THE NECESSARY COUNTERWEIGHT MAY BE FOUND
BY THE PRECEDING RULE.
Question 323. _How are the counterweights constructed?_
_Answer_. They are usually made of cast iron, so as to fill or partly
fill the spaces between the spokes of the wheel, as shown at _A A_ in
fig. 161. Each of the segments consists of two pieces, one of which is
put in from the outside of the wheel and the other from the inside.
The two are then bolted together with bolts, which are shown in the
above figure. In some cases the counterweights are cast solid with the
wheels, and in others, cavities are cast in the wheels which are filled
with lead, which is poured in when melted. Some locomotive builders
make the rims and spokes of the wheels hollow and fill them with lead
opposite to the crank-pins. It is difficult, however, to get the
required weight in such wheels unless they are of large diameter.
QUESTION 324. _What effect has the strain on the draw-bar, when the
engine is pulling a load, upon the distribution of the weight on the
wheels?_
_Answer_. It has the effect of pulling the back end of the engine
down, and, as it is balanced on the fulcrums of the equalizing levers,
it at the same time lifts up the front end. In this way the harder a
locomotive is pulling the greater will be the weight which is thrown
on the driving-wheels, and that on the truck will be correspondingly
diminished. The higher the draw-bar is above the rails the greater will
be the tendency to pull the engine down behind and up in front.