FACTS AND GENERAL INFORMATION RELATING TO PNEUMATIC TUBES.
We will now discuss, in an elementary way, the theory of pneumatic
tubes, in order to understand more clearly their _modus operandi_ and
the principles upon which they should be constructed. Let us begin with
the definition of a pneumatic tube.
=Definitions.=—A pneumatic tube is a tube containing air. This is
perhaps the broadest and most comprehensive definition that can be
given, but we usually associate with the idea of a pneumatic tube
the use to which it is put. If we were to embody this idea in our
definition we might define a pneumatic tube as a tube through which
material is sent by means of a current of air. This is still a very
broad definition, including all kinds of material for transportation,
for every conceivable purpose. It places no limit upon the dimensions
of the tube nor the manner of its operation. This definition would
include the toy commonly known as a putty-blower, and the pneumatic gun.
These instruments are not usually pictured in our minds when we hear
or see the term pneumatic tube used. Instead of these, we think of
the brass tubes that we have seen in the large retail stores in some
of our cities for conveying cash from the various counters to the
centrally located cashier’s desk. Again narrowing our definition to
conform more nearly with the mental picture presented, we will define a
pneumatic tube as a long tube for the purpose of transporting material
in carriers by means of a current of air in the tube. This, like all
definitions, is not entirely satisfactory, if we examine it critically,
but it will answer our present purpose.
=Intermittent and Constant Air-Current.=—Having thus defined a
pneumatic tube, there are two ways in which we may operate it to
transport our carriers containing mail, packages, or other matter. The
first method consists in storing our compressed air in a suitable tank,
or by exhausting the air from the tank; then, when we wish to despatch
a carrier we place it in the tube and connect the tube with the tank by
opening a valve. As soon as the carrier arrives at the distant end of
the tube the valve is closed and the air soon ceases to flow. When a
long interval of time elapses between the despatching of carriers, this
is the most economical method of operation, but usually carriers have
to be despatched so frequently that a great deal of time would be lost
if the air-current had to be started and stopped for each carrier.
The second and more usual method of operation consists in maintaining a
constant current of air in the tube and in having the carriers inserted
and ejected at the ends of the tube without stopping the current of air
for any appreciable length of time. It is analogous to launching boats
in a rapidly flowing stream, allowing them to float down stream and
then withdrawing them. When the boats are in the stream they present
little obstruction to the flow of water and check its speed but very
little. In order to compute the speed with which the boat will pass
from one point to another, we only have to know the speed of the stream
between those points when no boat is in it. The presence of the boat
does not change the speed appreciably. So it is with carriers in a
pneumatic tube: they are carried along with the current of air. The air
flows nearly as rapidly when a carrier is in the tube as when there
is none. The friction of the carrier against the inner surface of the
tube creates a slight drag, but it checks the speed of the air only a
little. Therefore, in order to know the speed with which a carrier will
be transported from one station to another through a pneumatic tube,
we need only to know the velocity with which the air flows through the
tube when no carrier is present. Of course there are special cases of
heavy carriers, or carriers having a large amount of friction from
their packing, or of tubes not laid horizontally, where the resistance
of the carrier must be taken into consideration, but for our present
purpose we will neglect all of these conditions.
=Laws Governing the Flow of Air in Long Tubes.=—This leads us to study
the laws governing the flow of air in long tubes, omitting for the
present the presence of a carrier. Since tubes operated intermittently
have become obsolete, we will only consider the case of a constant
current of air, this being what we have to deal with in practice.
[Illustration: FIG. 48.
PRESSURE AND VELOCITY CURVES.]
In order to make our ideas and thoughts as clear as possible let us
represent them by a diagram, Fig. 48. We will suppose that a tank, A,
is kept constantly filled with compressed air at a pressure of ten
pounds per square inch, from some source of supply. We will suppose
that the pressure of the air in this tank never changes, air being
supplied as fast as it flows away. Next, let us assume that a tube
eight inches in diameter inside and one mile long (five thousand two
hundred and eighty feet) is connected to the tank at one end and left
open to the atmosphere at the other. The air will flow in a constant
stream from the tank into the atmosphere, for the reason that air is
being supplied to the tank as fast as it flows away.
=Law of Pressure.=—First, let us consider the pressure of the air at
various points in the tube. We will, for convenience, represent the
pressure in the tank by a vertical line, D E, ten units in length,
since the pressure is ten pounds per square inch. Now let us go to a
point on the tube one quarter of a mile (one thousand three hundred
and twenty feet) from the tank, drill a hole in the tube, attach a
pressure-gauge and measure the pressure of the air at this point.
We shall find it to be about 7.91 pounds per square inch; or, 2.09
pounds below the pressure in the tank. We will represent this on our
diagram by another vertical line, F G, having a length of 7.91 units.
Again let us measure the pressure in the tube at a point one-half a
mile (two thousand six hundred and forty feet) from the tank. Here we
find it to be about 5.61 pounds per square inch, and we represent it
by the vertical line, H I, having 5.61 units of length. We note that
the pressure is 4.39 pounds below the pressure in the tank. We are at
the middle point of the tube and the pressure has fallen to nearly,
but not quite, one-half the pressure in the tank. We will now go to
a point three-quarters of a mile (three thousand nine hundred and
sixty feet) from the tank, and here the pressure is about 3.01 pounds
per square inch. We represent it by the vertical line, J K. Lastly,
we measure the pressure very near the end of the tube, one mile from
the tank, and find it to be about zero, or the same as the pressure of
the atmosphere. All of our measurements have been in pounds above the
atmospheric pressure; to express them in absolute pressure, we should
add to each the pressure of the atmosphere, which is 14.69 pounds,
nearly.
Now we will draw a smooth curve through the tops of all our vertical
lines, and we have a curve, E, G, I, K, L, representing the pressure in
the tube at every point. It falls gradually from ten pounds to zero,
but it does not fall in exact proportion to the distance from the tank.
Such a fall of pressure would be represented by the straight dash-line,
E, L. The reason why the true pressure-curve is not a straight line,
and lies above a straight line, is because air is an elastic fluid and
expands, becoming larger in volume as the pressure diminishes. The
straight dash-line represents the fall of pressure of an inelastic
fluid, like water, when flowing in the tube.
The fall of pressure along the tube is analogous to the fall of level
along a flowing stream. In fact, we frequently speak of the descent of
a stream as the “head of water” when it is used for power purposes,
and we mean by this the pressure the water would exert if it were
confined in a pipe. The descent, or change of level, in the bed of a
stream is necessary to keep the water flowing against the friction of
the banks. The descent of the water imparts energy to overcome the
friction. In a similar manner, we must have a fall of pressure along
the pneumatic tube to overcome the friction of the air against the
interior surface of the tube. We find another analogue in the flow of
the electric current along a wire; here there is a fall of potential
necessary to overcome the resistance of the wire. Since power has to be
expended to compress the air and impart to it its pressure, when this
pressure disappears we know that the air must be losing its energy or
doing work, and we look to see what becomes of it. In the present case,
we find that most of this work is expended in overcoming the friction
between the air and the surface of the tube.
=Uses of Pressure Curves.=—The pressure curve teaches us many
things. Suppose we were to establish stations on this tube at the
quarter, half, three-quarter, and mile points; we see at once that
intermediate-station or closed receivers, described in the last
chapter, must be used at all of the stations except the mile point at
the end of the tube, because the pressure in the tube is so high above
the pressure of the atmosphere that we could not open the tube to let
the carriers come out, but at the end of the tube we could use the open
receiver. In designing our sending and receiving apparatus for each
station, we look to this pressure curve to tell us the pressure which
we shall have on the pistons in our cylinders, and are thereby enabled
to make them with proper proportions for the work that they have to do.
=Law of Velocity.=—Next let us see what the velocity of the air is
in the tube. Suppose that we have some convenient means of measuring
the velocity of the air at any point, in feet per second or miles
per hour, with some form of anemometer. We will have our measurements
taken at the five points where we measured the pressure,—viz., at the
tank, one-quarter, one-half, three-quarters and one mile from the
tank. We will represent the velocities by a diagram similar to the
one used for pressures. At the tank we find the air entering the tube
with a velocity of 59.5 feet per second (40.6 miles per hour). We draw
the vertical line M N, to represent this. At the quarter mile point
the velocity is sixty-five feet per second (44.4 miles per hour) an
increase in the first quarter of a mile of 5.5 feet per second. We
construct the vertical line O P. At the half-mile point the velocity is
72.4 feet per second (49.4 miles per hour); at the three-quarter mile
point it is eighty-three feet per second (56.8 miles per hour); and at
the end of the tube, one mile from the tank, the air comes out of the
tube with a velocity of 100.4 feet per second (68.5 miles per hour),
about 1.7 times faster than it entered the tube at the tank. Drawing
all the vertical lines to represent these velocities, and drawing a
smooth curve line through the tops of our vertical lines, we have the
curve of velocities, N, P, R, T, V, for all points along the tube. It
is an increasing velocity and increases more rapidly as we approach the
end of the tube. This is shown more clearly by drawing the straight
dashed line N V.
If the fluid flowing in the tube were inelastic, like water, then the
curve of velocities would be a straight horizontal line, for the water
would not come out of the tube any faster than it went in. But we are
dealing with air, which is an elastic fluid, and, as we stated before,
it expands as the pressure is reduced and becomes larger in volume.
It is this expansion that increases its velocity as it flows along the
tube. It must go faster and faster to make room to expand. Since the
same actual quantity of air in pounds must come out of the tube each
minute as enters the tube at the other end in the same time, to prevent
an accumulation of air in the tube, and since it increases in volume as
it flows through the tube, it follows that its velocity must increase.
=Characteristics of the Velocity Curve.=—This velocity curve is both
interesting and surprising, if we have not given the subject any
previous thought. It might occur to us that the air expands in volume
in the tube, and we might reason from this fact that the velocity of
the air would increase as it flowed through the tube, but very few of
us would be able to see that the rate of increase of velocity also
increases. That is to say, it gains in velocity more rapidly as it
approaches the open end of the tube. If the velocity were represented
on the diagram by a straight horizontal line, we should know that it
was constant in all parts of the tube, which would be the case if water
were flowing instead of air. If it were represented by a straight
inclined line, like the dashed line N V, then we should know that
the velocity increased as the air flowed along the tube, but that it
increased at a uniform rate. The slope of the line would indicate the
rate of increase. Neither of these suppositions represent correctly
the velocity of the air at all points in the tube; this can only be
done by a curved line such as we have shown. The slope of the curve at
any point represents the rate of increase of velocity of the air at
that point. If the curve is nearly horizontal, then we know that the
velocity does not increase much; but if the curve is steep, then we
know that it is increasing rapidly, the actual velocity being indicated
by the vertical height of the curve above the horizontal line M U.
=Use of Velocity Curves.=—Besides being interesting, a knowledge of the
velocity of the air at all points in a tube is of much practical value.
It gives us the time a carrier will take in going from one station to
another. Usually the first questions asked, when it is proposed to lay
a pneumatic tube from station A to station B, are, How quickly can
you send a carrier between these points? How much time can be saved?
These questions are answered by constructing a velocity curve. Since
the velocity changes at every point along a tube, to get the time of
transit between two points we must know the average velocity of the
air between those points. We can find this approximately from our
curve by measuring the height of the curve above the horizontal line
M U at a large number of points, and then taking the average of all
these heights; but there is a more exact and easier method by means of
a mathematical formula. As such formula would be out of place here,
we will not give it; suffice it to say, that the average velocity of
the air between the tank and the end of the tube, in the case we have
assumed, is about seventy-three feet per second (49.7 miles per hour),
a little less than one-half the sum of the velocities at the two ends,
and a little more than the velocity at the half-mile point. Knowing
the average velocity, we can tell how long it takes for a particle
of air, and it will be nearly the same for a carrier, to travel from
the tank to the end of the tube, by dividing the distance in feet by
the average velocity in feet per second. This we find to be one minute
12.3 seconds. Since the air moves more rapidly as it approaches the
open end of the tube, a carrier will consume a greater period of time
in going from the tank to the quarter mile point than in going from
the three-quarter mile point to the open end. The last quarter of a
mile will be covered in a little more than fourteen seconds, while
the first quarter will require a little more than twenty-one seconds.
This difference is surprising, and it becomes even more marked in very
long tubes with high initial pressures. This explains why the service
between stations located near the end of the tube is more rapid than
between stations on other parts of the line.
This velocity curve shows us the velocity of the carriers at each
station along the line and enables us to regulate our time-locks
and to locate the man-holes and circuit-closers connected with each
intermediate station. It gives us the length of the “blocks” in our
“block system.” When we know the velocity and weight of our carriers,
we can compute the energy stored up in them, and from this the length
we need to make our air-cushions so as not to have the air too highly
compressed. It would be impossible to design our apparatus properly if
we did not know the laws that govern the flow of air in the tubes.
=Quantity of Air Used.=—The next important fact that we learn from
the velocity curve is the quantity of air that flows through the tube
each second or minute. If we multiply the velocity with which the
air escapes from the open end of the tube by the area of the end of
the tube in square feet, we have the number of cubic feet of air at
atmospheric pressure discharged from the tube per unit of time. The
same quantity of air must be supplied to the tank in order to maintain
a constant flow in the tube. In the present case that we have assumed,
the tube is eight inches in diameter; therefore the cross-sectional
area is 0.349 square foot. The velocity of the air as it comes out
of the end of the tube is 100.4 feet per second; therefore about
thirty-five cubic feet of air are discharged from the tube each second,
or two thousand one hundred cubic feet per minute. This same amount
must be supplied to the tank A in order to maintain the pressure
constant, but when it is compressed so that it exerts a pressure of
ten pounds per square inch, the two thousand one hundred cubic feet
will only occupy a space of one thousand two hundred and fifty cubic
feet, if its temperature does not change. This leads us to consider the
effect of temperature changes.
=Temperature of the Air.=—If the air is allowed to become heated
by compression, as is the case in practice, we have a new set of
conditions. If the air in the tank A is hot,—that is, warmer than the
surrounding atmosphere,—it will by radiation cool somewhat before it
enters the tube, and it will be still further cooled when it expands
in the tube. Again, if its temperature falls below the temperature of
the ground in which the tube is laid, it will absorb heat from the
ground, and this will tend to keep up its temperature; so in practice
we have very complicated relations between the temperature, pressure,
and volume of the air. These relations cannot be exactly expressed by
mathematical formulæ, and we will make no attempt so to express them,
but will be content with saying that in practice we find that the
temperature of the air in the tubes is nearly constant after the first
few hundred feet, so that we can without appreciable error compute the
pressures and velocities as if it were constant. Now, if the air in the
tank A is hot, we must raise the pressure a little above ten pounds
per square inch to obtain the velocities given on our diagram. When
the air cools it contracts in volume, or, if the volume cannot change,
being fixed by the limits of the containing vessel, then the pressure
is reduced, so by raising the pressure in the tank A a little above ten
pounds, we compensate the loss of pressure.
=Horse-Power.=—Having shown how the quantity of air can be computed we
are now in a position to estimate the horse-power of the air-compressor
necessary to operate the tube. I say estimate, because we have to take
into consideration the efficiency of the air-compressor, and that is
not an absolutely fixed quantity: it varies with different types of
machines and with their construction. When working with pressures of
less than ten pounds, the friction of the machine is an important
factor. The area and construction of the valves in the air-cylinders
is another very important factor. If the valves are not large enough
or do not open promptly, our cylinders will not be filled with air
at each stroke; this will reduce the efficiency of the machine. In
practice we go to a manufacturer of air-compressors and tell him how
much air must be compressed per minute, and the pressure to which it
must be compressed, with other conditions, and then he tells us what
size of machine we shall require and the horse-power of the machine
approximately. He is supposed to know the efficiency of his own
machines. We may endeavor to prove his estimate by computations of our
own. To give some idea of the horse-power required to supply the air
needed to operate our eight-inch tube one mile long, I will say that
the steam-engines of the air-compressor will have to develop in the
vicinity of one hundred and twenty-five actual horse-power. From the
horse-power of the steam-engine we can easily compute the coal that
will be consumed under boilers of the usual type.
=Efficiency.=—It will be noted that most of the power is not used
primarily in moving the carriers, but to move the air through the tube.
Very nearly as much power is used to keep the air flowing in the tube
when no carriers are in it as when carriers are being despatched. If we
should define the efficiency of a pneumatic tube as the ratio of the
power consumed in moving the carriers to the power consumed in moving
the carriers and the air, we should find this so-called efficiency to
be very low. It is analogous to pulling the carrier with a long rope
and dragging the rope on the ground. Much more power would be consumed
by the rope than by the carrier. But a business man would define the
efficiency of a pneumatic tube as the ratio of the cost of transporting
his letters, parcels, etc., to the cost of transporting them with equal
speed in any other way. Defined in this practical manner the efficiency
of a pneumatic tube is high. We do not care what becomes of the power
so long as it accomplishes our purpose.
=Pressure and Exhaust Systems.=—We have noticed that pneumatic tubes
have not always been operated by compression of the air, but that some
of the small tubes used in the telegraph service of European cities
have been operated by exhausting the air. The two systems are sometimes
distinguished by calling one a pressure system and the other an exhaust
system. These terms are very misleading, for an exhaust system is
a pressure system. The current of air is kept flowing in a tube by
maintaining a difference of pressure at the two ends, and the result is
the same whether we raise the pressure at one end above the atmospheric
or lower it at the other below the atmospheric. In either case it is
pressure that causes the air to flow. It happens that we are living in
an atmosphere of about fifteen pounds pressure per square inch, and it
is very convenient oftentimes in our computations to take the pressure
of the atmosphere as our zero, and reckon all other pressures above and
below this. If all our pressure scales read from absolute zero, the
pressure of a perfect vacuum, then all this confusion would be avoided.
We have not used the absolute zero in our diagram, because all our
gauges are graduated with their zero at atmospheric pressure, and it is
customary to speak of pressures above and below the atmospheric.
It is very natural to ask the question, why are tubes sometimes
operated by compressing the air and at other times by exhausting it?
We answer by saying that it is usually a question of simplicity and
convenience that determines which system shall be used. Some of the
cash systems in the stores use compressed air in the out-going tubes
from the cashier’s desk and exhaust the air from the return tubes. Both
ends of the tubes are then left open at the counters, no sending or
receiving apparatus being required there. The carriers are so light,
their velocity so low, and the air-pressure varies so little from the
pressure of the atmosphere that the carriers can be allowed to drop
out of the tubes on to the counters, and they can be despatched by
simply placing them at the open end of the tube into which the air is
flowing. The currents of air entering and leaving the tubes are not
so strong as to cause any special annoyance. At the cashier’s desk
some simple receiving and sending apparatus has to be used, but it is
better to concentrate all of the apparatus at one point rather than
have it distributed about the store, as would be the case if the double
system were not used. In the London pneumatic telegraph both methods
of operation have been in use. Double lines were laid, and the engines
exhausted the air from one tube and forced it into the other. The
exhausted tube was therefore used to despatch in one direction, while
the other tube, operated by the compressed air, served for despatching
in the opposite direction. So far as I know, both work equally well.
In operating large tubes, that is to say, tubes six or eight inches
in diameter, there is an advantage in using compressed, rather than
exhausted air, in the construction of the sending and receiving
apparatus, especially when the tubes are very long. With an ample
supply of compressed air always at hand, the air-cushions can be made
shorter and more effective in bringing the carriers quickly to rest.
With exhausted air the cushions are ineffective, and consequently
must be made very long in order to stop the carrier before it strikes
the closed end of the tube. This does not apply to small tubes where
the carriers are so light that they can be stopped without injury by
allowing them to strike solid buffers. Again, when compressed air is
used, we have a larger difference in pressure between the pressure in
the tube and the atmosphere to operate our mechanism by cylinders and
pistons. With an exhaust system carriers are not so easily ejected
from the tubes of the receiving apparatus; we could not use the simple
form of open receiver. Again, if the tubes are laid in wet ground, and
a leak occurs in any of the joints, water will be drawn in if air is
being exhausted from the tube, while it will be kept out if compressed
air is used.
In regard to the question of relative economy of the two systems, we
will say that when long tubes are used, requiring high pressures, or,
more strictly speaking, a large difference of pressure, to maintain the
desired velocity of air-current, there seems to be some advantage in
using an exhaust system. The reason is this: the friction of the air in
the tube, which absorbs most of the power, increases as the air becomes
heavier and more dense. When the air is exhausted from the tube, we
are using a current of rarefied air, and this moves through the tube
with less friction and, consequently, a higher velocity, for the same
difference of pressure, than the more dense compressed air. But for
short tubes that require only a small difference of pressure, this
advantage becomes very small, and is overbalanced by other advantages
of a compressed air system. So, taking everything into consideration,
there is not so much to be said in favor of an exhaust system.
=Laws Expressed in Mathematical Formulæ.=—While we have heretofore
purposely avoided all complicated mathematical formulæ, it may not
be out of place here to give a few of the more simple relations that
exist between the pressure, velocity, length and diameter of the tubes,
etc. In two tubes having the same diameter, with the same pressures
maintained at each end, but of different lengths, the mean velocities
of the air in the tubes will bear the inverse ratio to the square
roots of the lengths of the tubes. This is expressed by the following
proportion:
_u_ : U :: √L : √_l_
_u_ and U represent the mean velocities of the air in the two tubes and
_l_ and L the respective lengths of tubes.
A similar but direct ratio exists between the mean velocities and the
diameters of the tubes, thus:
_u_ : U :: √_d_ : √D
This relation, however, is only approximately true for tubes differing
greatly in diameter.
The relation of the pressure to other factors is not so simply
expressed. For example, in two tubes of the same length and diameter,
the relation between the pressures at the ends and the mean velocity of
the air may be expressed as follows:
(_pₒ_² - _p₁_²)^³/² (Pₒ² - P₁²)^³/²
_u_ : U :: ————————————— : —————————
(_pₒ_³ - _p₁_³) (Pₒ³ - P₁³)
where _u_ and U are the respective mean velocities, _p__{0} and P_{0}
the respective pressures at the initial ends of the tubes, and _p__{1}
and P_{1} the respective pressures at the final ends of the tubes,—the
pressures being measured above absolute zero.
There are other relations that can be similarly expressed, but for them
we must refer the reader to a mathematical treatise on the subject.
=Moisture in the Tubes.=—When pressures of more than five pounds per
square inch are used, it is not unusual to find some moisture on the
interior of the tube and upon the outside of the carriers when they
come out of the tube. It is seldom more than a slight dampness, or at
most a degree of wetness equal to that seen on the outside of a pitcher
of ice-water on a warm day. A slight amount of moisture in the tube is
not objectionable, for it serves as a lubricant to the carriers; but
when it is present in considerable quantity it becomes objectionable
and even annoying. This moisture is brought into the tube with the air,
and is deposited upon the walls of the tube when their temperature
is sufficiently below that of the atmosphere. The atmosphere always
contains more or less moisture in the state of vapor. The capacity of
air for water-vapor depends upon its temperature, being greater the
higher the temperature, but it is a fixed and definite quantity at any
given temperature. When the air contains all the water-vapor it can
hold at a certain temperature, it is said to be saturated. If it is
not saturated, we express the amount that it contains in per cent. of
the amount it would contain if it were saturated, and this is termed
the “relative humidity.” For example, if the air is three-fourths
saturated, we say the “relative humidity” is seventy-five; but if the
temperature changes, the “relative humidity” changes also. Suppose
the temperature to-day is seventy-five degrees Fahrenheit, and that
the “relative humidity” is eighty, a cubic foot of air then contains
0.00107 pound of water-vapor. Now suppose this air enters a pneumatic
tube and is cooled by expansion and contact with the colder walls of
the tube to sixty degrees. At this temperature a cubic foot of air can
contain only 0.00082 pound of water-vapor when it is saturated. Now,
each cubic foot of air brought into the tube brings with it 0.00107
pound of vapor, and after it is cooled down to sixty degrees it cannot
hold it all, consequently the difference, or 0.00025 pound, must be
deposited in the tube. Under these conditions one hundred thousand
cubic feet of air will deposit twenty-five pounds of water in the tube.
In the system of pneumatic tubes built and operated by the Batcheller
Pneumatic Tube Company, the presence of a large quantity of moisture
in the tube is prevented by using the same air over and over again. A
little moisture may be deposited when the tube starts into operation,
but the amount does not increase appreciably, as very little fresh air
is admitted after starting.
=Location of Obstructions in Tubes.=—In regard to the removal of
obstructions in the tubes, I have had little or no experience;
therefore under this heading I am satisfied to quote from the “Minutes
of the Proceedings of the Institution of Civil Engineers,” London,
Volume XLIII.
“Intimately connected with the working of the tubes is the removal of
obstructions which occur from time to time, causing not unfrequently
serious inconvenience and delay. The most general cause of obstruction
is a stoppage of the train arising from accident to the tube, to the
carriers or piston, or to the transmitting apparatus. In such cases the
delay is generally very brief, it being for the most part sufficient to
reverse the pressure on the train from the next station, and to drive
it back to the point it started from. If one or more of the carriers
break in the tube, reverse pressure is also generally sufficient to
remove the obstacle; but where this fails, the point of obstruction
must be ascertained. This is done by carefully observing the variations
of air pressure in the reservoir when placed in connection, first
with a line of known length, and then with the obstructed tube. By
this means the position of the obstruction can be ascertained within
one hundred feet. Or the tube may be probed with a long rod up to a
length of two hundred feet. A very ingenious apparatus, by M. Ch.
Bontemps, is shown in Figs. 49 and 50, and is employed to ascertain
the exact position of the obstruction. It acts by the reflection of
sound-waves on a rubber diaphragm. A small metal disk is cemented to
the rubber, and above this is a pointed screw, D. An electric circuit
is closed where the points C and D are brought in contact. To locate an
obstruction a pistol is fired into the tube as shown, and the resulting
wave, traversing the tube at the rate of three hundred and thirty
metres a second, strikes the obstruction and is then reflected against
the diaphragm, which in its turn reflects it to the obstacle, whence it
returns to the diaphragm. By this means indications are marked on the
recording cylinder, and if the interval of time between the first and
second indications be recorded, the distance of the obstacle from the
membrane is easily ascertained. The chronograph employed is provided
with three points; the first of these is placed in a circuit, which
is closed by the successive vibrations of the diaphragm; the second
corresponds to an electric regulator, marking seconds on the cylinder;
and the third subdivides the seconds there marked. Fig. 50 indicates a
record thus made. In this case the obstacle is situated at a distance
of sixty-two metres, and the vibration marks thirty-three oscillations
per second. The interval occupied by two successive marks from the
diaphragm on the paper corresponds to twelve oscillations, and the
distance of the obstruction is then calculated by the following formula:
D = 0.5 × 330 × 12/33 = 60 metres;
so that the distance of the obstacle is recorded within two metres.
[Illustration: FIG. 49.
OBSTRUCTION-RECORDING APPARATUS.]
[Illustration: FIG. 50.
OBSTRUCTION-RECORDING APPARATUS.]
“Amongst the special causes of accident may be mentioned the accidental
absence of a piston to the train, breaking of the piston, the freezing
up of a piston in the tube, and even forgetting the presence of a
train, which has caused the entire service to be one train late
throughout the day. Finally, the tubes themselves are sometimes broken
or disturbed during street repairs, resulting of course in a complete
cessation of traffic in the system till the damage is made good.”