ADHESION AND TRACTION.
QUESTION 307. _What is meant by the “adhesion” of a locomotive?_
_Answer._ It is the resistance which prevents or opposes the slipping
of the driving-wheels on the rails, and is due to the friction of the
former on the latter.
QUESTION 308. _On what does the amount of this friction depend?_
_Answer._ Like all friction it depends upon the weight or pressure of
the surfaces in contact, and consequently upon the load which rests
on each wheel. It also depends upon the condition of the rails, and
probably to some extent upon the material of which they and the tires
on the wheels are made.
[Illustration: _Fig. 192._
Scale ¹⁄₄ in. = 1 foot.]
QUESTION 309. _How much force is required to make the driving-wheels of
a locomotive slip on an ordinary railroad track?_
_Answer._ The force required to make them slip will, as already stated,
vary very much with the condition of the rails. If they are quite dry
and clean it will require a force equal to about one-fourth the weight
on the wheels. That is, supposing we have a wheel, _A B_, fig. 192,
attached to a frame which is fastened so that it cannot move, and
that the wheel rests on a rail and is loaded with say 10,000 pounds,
if now a rope or chain could be attached at a point, _B_, exactly at
the tread of the wheel, and carried over a pulley, _C_, then it would
require a weight, _D_, of about 2,500 pounds attached to the end of the
rope to make the wheel slip. If the rails were sanded, the adhesion
would be somewhat greater, and if they were wet or muddy or greasy,
considerably less. For ordinary circumstances the adhesion may safely
be assumed to be _one-fifth of the weight of the driving-wheels_. Of
course the total weight on all the driving-wheels must be taken in
calculating the adhesion. Thus, if a locomotive has four driving-wheels
and each one of them bears a load of 10,000 pounds, then the total
weight on the driving-wheels, or _adhesive weight_, as it is called,
will be 10,000 × 4 = 40,000 pounds, and the adhesion will be
40,000
------ = 8,000 pounds.
5
QUESTION 310. _What is meant by the tractive power of a locomotive?_
_Answer._ It is the force with which the locomotive is urged in a
horizontal direction by the pressure of the steam in the cylinders, and
which therefore tends to move the locomotive and draw the load attached
to it.
The tractive power is of course due to the pressure of steam on the
piston, and therefore its amount is dependent upon the average steam
pressure in the cylinder, on the area of the piston, and also on the
distance through which the pressure is exerted, or, in other words,
on the stroke of the piston. Thus if we have a cylinder 16 inches in
diameter and two feet stroke and an average steam pressure of 50 pounds
per square inch, then as the area of such a piston would be 201 square
inches, the average pressure on it would be 201 × 50 = 10,050 pounds,
and as each piston moves through four feet during one revolution of
the wheels, the number of foot-pounds of energy exerted by it would be
10,050 × 4 = 40,200, and for the two cylinders of a locomotive double
that amount, or 80,400 foot-pounds. Now if the driving-wheels are five
feet in diameter, their circumference will be 15.7 feet, and therefore
the locomotive will move that distance on the rails during one
revolution, if the wheels do not slip. The 80,400 foot-pounds of energy
is therefore exerted through a distance of 15.7 feet, and therefore
80,400
------ = 5,121 pounds,
15.7
which is the force exerted through each foot that the circumference of
the wheel revolves and the locomotive moves. If the wheels were only
half the diameter, or 2¹⁄₂ feet, then their circumference would be 7.85
feet and the tractive power would be
80,400
------ = 10,242 pounds,
7.85
or double what it was before. It will be seen, then, that the tractive
force of a locomotive is dependent upon (1) the average steam pressure
in the cylinders, (2) the area, (3) the stroke of the pistons, and (4)
the diameter of the driving-wheels.
QUESTION 311. _How is the tractive power of a locomotive calculated?_
_Answer._ BY MULTIPLYING TOGETHER THE AREA OF THE PISTON IN SQUARE
INCHES, THE AVERAGE STEAM PRESSURE IN POUNDS PER SQUARE INCH ON THE
PISTON DURING THE WHOLE STROKE, AND FOUR TIMES THE LENGTH OF THE STROKE
OF THE PISTON,[78] AND DIVIDING THE PRODUCT BY THE CIRCUMFERENCE OF THE
WHEEL. The result will be the tractive power exerted in pounds. The
adhesion must of course always exceed the tractive force, otherwise the
wheels will slip.
[78] This length may be taken in feet, inches or any other measure,
but in making the calculation the circumference of the wheel must be
taken in the _same_ measure as the stroke of the piston.
QUESTION 312. _How is the locomotive made to advance by causing the
wheels to revolve?_
_Answer._ The pressure of steam in the cylinders is exerted in one
direction against the piston, and in the opposite direction against the
cylinder head, as shown in fig. 192, in which the steam is represented
by the dotted shading in the back end of the cylinder, and the
direction of the pressure by the darts _s_, _s_. The pressure against
the piston is communicated by the connecting-rod to the crank-pin _E_,
and that on the cylinder-head is carried to the axle by the frame
_F F′_, and the direction of the two forces is indicated by the two
darts, _a_ and _b_. We may now regard the spokes of the wheels as
acting as levers, and assume that the fulcrum is either at the centre
_G_ of the axle, or at _B_, the point of contact of the wheel with the
rail.[79] We will assume that it is at the centre _G_ of the axle and
for the sake of even figures that the wheel is six feet in diameter and
cylinders have two feet stroke. We will also suppose that the engine is
supported so that the wheels do not touch the rails, and that a chain
or rope passing over a pulley _C_ is attached to the wheels at _B_
and with a weight at _D_. We now have a force, _a_, of 10,000 pounds
exerted on the crank-pin, or at the end of the short arm _E G_ of the
lever _E G B_. As _E G_ is one foot and _G B_ three feet long, 10,000
would be balanced by
10,000 × 1
---------- = 3,333 pounds,
3
at _B_. In other words, it would require 3,333 pounds suspended from
the chain at _D_ to resist the strain at _E_. But when this is the
case, the pressure of the axle at the fulcrum, in the direction of the
dart _c_, is equal to the pressure against the crank-pin _E_ added
to that exerted by the weight _D_ at _B_, or 10,000 + 3,333 = 13,333
pounds.
[79] The question whether the centre of the axle or the point of
contact with the rail is the fulcrum of the lever in this case has
been the subject of much animated discussion and contention. As
the word _fulcrum_ means “a point about which a lever moves,” it
is believed that the dispute is due simply to a difference in the
meaning assigned to the word fulcrum. If we regard the fulcrum as the
point which is fixed in relation to the locomotive, then it is at the
centre of the axle, but if we refer it to the surface of the earth,
then it is at the top of the rail.
As the pressure against the axle in the opposite direction, _b_, is
only 10,000 pounds, there will be an unbalanced force of 3,333 pounds
acting in the direction of the dart _c_, and tending to move it that
way. As the axle is attached to the locomotive frame, this force will
of course have a tendency to move the whole machine, and is really the
tractive force of the engine.
If, on the other hand, we regard the point of contact _B_ of the wheel
with the rail as a fulcrum, we have a force of 10,000 pounds acting
against a lever, _E G B_, four feet long. There would therefore be a
force, _c_, of
10,000 × 4
---------- = 13,333 pounds
3
exerted at _G_, and as the pressure in the direction of the dart _b_ is
only 10,000, there would be an unbalanced strain of 13,333 - 10,000 =
3,333 pounds acting against the axle in the direction of the dart _c_,
or, in other words, there is 3,333 pounds more of force pulling the
axle forward than there is pushing it backward.
When the crank-pin is below the axle, in the position shown in fig.
193, then if the centre of the axle is regarded as the fulcrum we have
a pressure of 10,000 pounds pushing against the front cylinder-head,
which is transferred to the axle by the frames, and acts in the
direction of the dart _c_, and we also have a pressure against the
crank-pin _E_ in the direction of the dart _a_. Now 10,000 pounds at
_E_ would exert a force of
10,000 × 1
---------- = 3,333 pounds
3
at _B_ and
10,000 × 2
---------- = 6,666 pounds
3
at _G_ in the direction of _b_. But the pressure on the cylinder-head
pulls against the axle _G_ in the direction c with a force of 10,000,
so that the excess of strain in the direction _b_ will be equal to
10,000 - 6,666 = 3,333 pounds. If we regard _B_ as the fulcrum, the
calculation is exactly the same, as 10,000 pounds at _E_ exerts a force
at _G_ equal to
10,000 × 2
---------- = 6,666 pounds,
3
and the difference between it and the pressure exerted by the strain
against the front cylinder-head is the force which urges the axle and
with it the locomotive forward.
It will be seen, then, that it is immaterial which point is regarded as
the fulcrum, as the result of the calculations is exactly the same.
[Illustration: _Fig. 193._
Scale ¹⁄₄ in. = 1 foot.]
It must not, however, be hastily supposed from what has been said that
the total pressure against the axle can be greater than its resistance
to the pressure. As soon as the one exceeds the other it will move.
But supposing that it requires a force equal to 3,333 pounds to draw
a train coupled to the engine, as soon as the difference between the
force exerted against the axle by the piston to move it forward and
that which presses it back exceeds 3,333 pounds, the locomotive will
move the train. If it continues to exceed it the speed of the train
will be accelerated and thus the resistance which holds the engine back
and that which pushes it forward will always be equal.
QUESTION 313. _Does the fact that the piston is working from the end of
a longer lever E G B, fig. 192, when the crank-pin is above the axle,
enable the locomotive to start a heavier train than when the crank-pin
is below the axle and the piston is working against a shorter lever, G
E B, fig. 193?_
_Answer._ No; because, as has already been shown, the pressure against
the axle is the same in both cases. It is in fact only during the
forward stroke that the pressure on the crank-pin moves the engine
forward. The forward pressure which is exerted by the crank-pin at
the axle is then greater than that exerted against the latter in the
opposite direction by the cylinder-head and frames. It is this excess
of crank pressure which moves the engine and which is the tractive
force during the forward stroke. During the backward stroke the piston
is pushing the axle backward, and the pressure against the front
cylinder-head is pulling it forward. The latter then exceeds the
former, and the difference between the two is the force which moves the
engine forward. As has been shown, this difference is the same in both
positions of the crank, and therefore the locomotive can not from this
cause pull more when the crank is above the axle than when it is below.