required, refer to the table of circular mils, and select the wire having such a number of circular mi's.

The formula is as follows:

Feet of wire x 10.7 x amperes circular mils.

Volts lost

=

By simply transposing the above terms we obtain another formula, which can be used to determine the volts lost in a given length of wire of a certain size, carrying a certain number of amperes.

The formula is as follows:

Feet of wire x 10.7 x amperes

Circular mils

=

Volts lost.

And again, by another change in the terms we obtain a formula which shows the number of amperes that a wire of given size and length will carry at a given number of volts lost:

Circular mils x volts iost

Feet of wire × 10.7

=

Amperes.

In computing the necessary size of a service or main wire, to supply current for either lamps or motors, it is necessary to know the exact number of feet from the source of supply to the center of distribution. When the distance of center of distribution is given it is well to ascertain whether it is the true center or not. It may be only the distance from a cut-out box that has been given, when it should have been the distance from the point at which the service enters the building or, perhaps, from the point at which the service is connected to the street mains. For when the size is determined it is for a certain loss which is distributed over the entire length of the wire to be installed. The transmission of additional current on the mains in the building increases the drop in volts in the main, and likewise

in the service. Most buildings are wired for a certain per cent loss in voltage, estimated from the point where the service enters the building. All additions should be estimated from that point.

In using the formula for finding the proper size wire to carry current, the first thing to be determined is the length of the wire; remember that the two wires are in parallel, and therefore the total length of the wire is twice the total distance from the commencement to the end of the circuit. If the proposed load on this circuit is given in lamps, you may reduce it to amperes, and if the proposed load is given in horsepower, you may reduce it to amperes. The voltage on the circuit is known in either case. You take the loss of the voltage and divide the product of amperes, multiplied by the length, as found, and 10.7 by it; this answer will be the size in circular mils of wire necessary to carry the amperes.

Example: What is the size of wire required for a 50 volt system, having 100 lamps at a distance of 100 ft., with a 4 per cent loss?

Answer: The load of 100 lamps on a 50 volt system is 100 amperes, and a 4 per cent loss of 50 volts is 2 volts. Multiply the total length of the wire, which is twice the distance, or 200 ft., by the 100 amperes of current; this gives us 20,000. Then muitiply this by the constant, which is 10.7; this gives us 214,000. Divide this by 2, which is the loss in volts, and you have 107,000 circular mils diameter of wire required

When determining the size of wire to be used it is always necessary to consult the table of carrying capacities, and this will very often indicate a wire much larger than that determined by the wiring formula, especially if a somewhat high loss is figured on.

When estimating the distance it is not always correct to take the total distance.

To illustrate: Suppose one lamp is roo ft. from the point at which the distance is determined, and the farthest lamp is 400 ft., the remaining lamps being distributed evenly between these two points; we would average the distances between the first and last lamp, which would be 200 ft. It is necessary to use judgment in estimating the mean or average distance, as the lamps or motors are bunched differently in each

case

In a series system the loss in voltage makes considerable difference to the power, but does not affect the quality of the light as much as in a multiple arc or parallel system. In a parallel system the lamps require a uniform pressure, and this can only be had by keeping the loss low. In a series system the lamps depend upon the constant current and the voltage varies with the resistance, in order to keep the current constant. This is accomplished by a regulator on the dynamo, which is designed to compensate for the changes of resistance in the circuit and to increase or decrease the pressure as required.

In estimating the size of wire for a series system you consider the total length of the loop. There is no average distance as the total current travels over the entire circuit. We will assume that you have an arc light circuit of a No. 6 Brown & Sharp gauge wire and want to find what loss there is in this circuit. You have the area of a No. 6 wire, which is 26,250 circular mils, and the length of the circuit, and from this we will figure the loss in this manner: Assuming the circuit to be 10,000 ft. long, and the current 10 amperes, we will multiply 10,000 ft. by 10 amperes,

and this by 1.07, which gives us 1,070,000, and divide this by 26,250. The answer is 40 volts, lost in the circuit.

Such a circuit would operate at perhaps 2,000 of 3,000 volts, and a loss of 40 volts would not be exces

sive. It would be wasting a little less energy than is required to burn one large arc lamp.

The multiple series system is a number of small wires connected in multiple, and is the same as the multiple arc or parallel system. The wire is figured in the same way as for the multiple arc system.

The series multiple system is a number of small paralel systems, and these are connected in series by the main wire. The wire is figured the same as for the series system.

The Edison three-wire system is a double multiple, and the two outside wires are the ones considered when

carrying capacity is figured. When this system is under full load or balanced, the neutral wire does not carry any current, but the blowing of a fuse in one of the outside wires may force the neutral wire to carry as much current as the outside wire and it should, therefore, be of the same size. The amount of copper needed with this system is only three-eighths of that` required for a two-wire system.

Wiring Tables. On the following pages are presented wiring tables for 110,220 and 500 volt work. These tables are used in the following manner: Suppose we wish to transmit 60 amperes a distance of 1,800 ft. at 110 volts and at a loss of 5 per cent. We take the column headed by 60 in the top row and follow it downward until we come to 1,800, or the number nearest to it. From this number we now follow horizontally to the left, and under the column headed by 5 we find the proper size of wire, which is 500,000 c. m. The same current, at a loss of 10, would require only a 0000 wire, as indicated under the column at the left, headed. by 10.

Before making selection of wire, always consult the