Motors Part 12
Its outer end rests in a bearing plate B, of insulating material, which plate serves as the disk to hold the contact plates, 1, 2, 3, 4, to correspond with the four cylinders to which the current is to be distributed.
Wires 5, 6, 7, and 8, run to the respective spark plugs C from these contact plates. The projecting end of the shaft A carries thereon a contact finger D, which is designed to contact with the respective plates, and an insulating ring E is interposed between the shaft and finger so as to prevent short circuiting of the high tension current.
On the side of the finger is a hub F, integral therewith, and a wiper attached to a post bears against the hub so as to form continuous contact. A wire leads from the post to one terminal of the secondary coil.
[Ill.u.s.tration: _Fig. 58. Circuiting with Distributer._]
Circuiting With Distributer.--The diagram Fig. 58 shows the complete connections of a system which comprises a magneto, induction coil, condenser, and a distributer. The magneto A has on its armature shaft B two revolving disks C, D, one of which must be insulated from the shaft, and one end of the coil E of the armature is connected with one of these disks, and the other end of the coil is attached to the other disk.
Alongside of these disks is another disk F which has projecting points G to engage with and make temporary contact with a spring finger which actuates the interrupter I, this being a contact breaker which breaks the primary current at the time a spark is required.
One terminal of this interrupter is connected by a wire J with one end of the primary winding K, of the induction coil, and the other end of the primary has a wire L which runs to the disk C.
The other terminal of the interrupter has a wire M leading to a condenser N, and from the other side of the condenser is a wire O leading to the wire J before described. The wiper of the other disk D has a wire connection with the wire M.
The distributer shaft P is so mounted that it may receive its motion from the shaft of the magneto, and for this purpose the latter shaft has a gear Q one half the diameter of the gear R on the distributer shaft.
The distributer S has been described with sufficient clearness in a preceding diagram, to show how the wires T lead therefrom and connect up with the spark plugs U. One terminal of the secondary coil V is connected by a wire W with the wiper X which contacts with the hub of the distributer finger X', and the other terminal of the primary is grounded at Y, which represents the metal of the engine.
CHAPTER IX
MECHANICAL DEVICES UTILIZED IN POWER
One of the most important things in enginery is the capacity to determine the power developed. Although the method of ascertaining this appears to be somewhat complicated, it is really simple, and will be comprehended the more readily if it is constantly borne in mind that a certain weight must be lifted a definite distance within a particular time.
The Unit of Time.--The unit of time is either the second, or the minute, usually the latter, because it would be exceedingly difficult to make the calculations, or rather to note the periods as short as a second, and a very simple piece of mechanism to ascertain this, is to mount a horizontal shaft A, Fig. 59, in bearings B, B, and affix a crank C at one end.
It will be a.s.sumed that the shaft is in anti-friction bearings so that for the present we shall not take into account any loss by way of friction.
A cord, with one end attached to the shaft and the other fixed to a weight D, the latter weighing, say 550 pounds, is adapted to be wound on the shaft as it is turned by the crank.
Knowing the length of the cord and the time required to wind it up, it will be an easy matter to figure out the power exerted to lift the weight, which means, the power developed in doing it.
[Ill.u.s.tration: _Fig. 59. Ill.u.s.trating the Unit of Time._]
Suppose the cord is 100 feet long, and it requires one and a half minutes to raise the weight the full limit of the cord. It is thus raising 550 pounds 100 feet in 45 seconds.
One horse power means that we must raise 550 pounds one foot in one second of time, hence we have developed only 1/45th of one horse power.
Instead of using the crank, this shaft may be attached to the engine shaft so it will turn slowly. Then add sufficient weight so that the engine will just lift it, and wind the cord on the shaft.
You can then note the time, for, say, one minute, and when the weight is lifted, make the following calculation: Weight lifted one hundred feet in one minute of time was 825 pounds. Multiply 100 by 825, which equals 82,500. This represents _foot pounds_.
[Ill.u.s.tration: Fig. 60. The p.r.o.ney Brake.]
As there are 33,000 foot pounds in a horse power, 82,500 divided by this figure will show that 2-1/2 horse power were developed.
The p.r.o.ney Brake.--Such a device is difficult to handle, but it is ill.u.s.trated merely to show the simplicity of the calculation. As a subst.i.tute for this mechanism, a device, called the _p.r.o.ney brake_ has been devised, which can be used without rewinding of a cord. This is accomplished by frictional means to indicate the power, and by the use of weights to determine the lift.
The following is a brief description of its construction: The engine shaft A, Fig. 60, which is giving out its power, and which we want to test, has thereon a pulley B, which turns in the direction of the arrow. Resting on the upper side of the pulley is a block C, which is attached to a horizontal lever D by means of bolts E, these bolts pa.s.sing through the block C and lever D, and having their lower ends attached to the terminals of a short sprocket chain F.
Block segments G are placed between the chain and pulley B, and when the bolts E are tightened the pulley is held by frictional contact between the block C and the segments G.
The free end of the lever has a limited vertical movement between the stops H, and a swinging receptacle I, on this end of the lever, is designed to receive weights J.
The first thing to do is to get the dimensions of the pulley, its speed, and length of the lever. By measurement, the diameter of the pulley is six inches. To get the circ.u.mference multiply this by 3.1416. The distance around, therefore, is a little over 18.84 inches. The speed of the pulley being 225 times per minute, this figure, multiplied by 18.84, gives the perimeter of the pulley 4239 inches.
As we must have the figures in feet, dividing 4239 by 12, we have 353.25 feet.
The length of the lever from the center of the pulley to the suspension point of the receptacle, is 4 feet, and this divided by the radius of the pulley (which is 6 inches), gives the leverage. One half of six inches, is three inches, or 1/4 of one foot, and 4 divided by this number, is 1' 4", or 1-1/3 feet, which is the _leverage_.
Now, let us suppose the weight J is 1200 pounds. This must be multiplied by the leverage, 1-1/3 feet, which equals 1800, and this must be multiplied by the feet of travel in the pulley, namely, 353.25, which is equal to 635,850. This represents _foot pounds_.
Now, following out the rule, as there are 33,000 foot pounds in a horse power, the foregoing figure, 635,850, divided by 33,000, equals 19 horse power within a fraction.
Reversing Mechanism.--A thorough knowledge of the principles underlying the various mechanical devices, and their construction, is a part of the education belonging to motors. One of the important structures, although it is very simple, when understood, requires some study to fully master.
This has reference to reversing mechanism, which is, in substance a controllable valve motion, whereby the direction of the valve is regulated at will.
All motions of this character throw the valve to a neutral point which is intermediate the two extremes, and the approach to the neutral means a gradual decrease in the travel of the valve until the reciprocating motion ceases entirely at the neutral position.
[Ill.u.s.tration: _Fig. 61. Double Eccentric Reversing Gear._]
[Ill.u.s.tration: _Fig. 62. Reversing Gear, Neutral._]
Double Eccentric Reversing Gear.--A well known form of gear is shown in Fig. 61, in which the engine shaft A has two eccentrics B, C, the upper eccentric B being connected with the upper end of a slotted segment D by means of a stem E, and the other eccentric C is connected with the lower end of the segment by the stem F. The eccentrics B, C, are mounted on the shaft so they project in opposite directions.
The slotted segment carries therewith the pin G of a valve rod H, and the upper end of the segment has an eye I, to which eye is a rod J operated by a lever.
[Ill.u.s.tration: _Fig. 63. Reversing Gear, Reversed._]
[Ill.u.s.tration: _Fig. 64. Single Eccentric Reversing Gear._]
By this arrangement the link may be raised or lowered, and as the valve rod pin has no vertical movement, either the connecting link E or F may be brought into direct line with the valve rod H.
Fig. 61 shows the first position, in which the valve rod H is in direct line with the upper connecting rod E, actuated by the cam B.
Fig. 62 shows the neutral position. Here the pin G serves as a fulcrum for the rocking movement of the segment; whereas in Fig. 63 the valve rod H is in line with the lower connecting rod F, so that the valve is pushed to and fro by the eccentric C.
[Ill.u.s.tration: _Fig. 65. Balanced Slide Valve._]
It is more desirable, in many cases, to use a single eccentric on the engine shaft, which can be done by pivoting the segment L, Fig. 64, to a stationary support M, and connecting one end of the segment by a link N with the single eccentric O.
In this construction the valve rod P is shifted vertically by a rod Q, operated from the reversing lever, thus providing a changeable motion through one eccentric.