Whenever we write the denominator of a decimal instead of using the decimal point, we have a common fraction.

We can reduce the fraction to one having a smaller denominator whenever the numerator and denominator can both be divided by the same number. Thus the numerator and denominator of we can both be divided by 5. Dividing both terms of the fraction by 5 we have 15400 = 26.

In the last example above 125 and 1000 can both be divided by 25. Dividing we get

125/4000

=

Both terms of can be divided by 5.

5/40-1/8.

=

Therefore, 0.125 = 1254000 = %.

To reduce a common fraction to a decimal we divide the numerator by the denominator.

Example:

Reduce to a decimal.

% means 3 divided by 5. We divide in this way.

5) 3.0
.6

The dividend is 3 units or 30 tenths, and 30 tenths divided by 5 equals 6 tenths.

To reduce a common fraction to a decimal, annex zeros to the numerator, placing the decimal point before the zeros. Then divide by the denominator, as in division of decimals.

It may happen that even after annexing zeros the numerator can not be exactly divided by the denominator. In this case we carry the division as far as we need to and omit the rest of the decimal.

Example:

3 of an inch equals how many hundredths of an inch?

3) 1.00
.33+

There is a remainder as far as we carry the division, but as we are measuring only to hundredths of an inch we carry the division only to hundredths. The sign + is sometimes placed after such a decimal to show that the division might have been carried farther.

13/20.

Examples:

1.

Reduce 0.7 to a common fraction.-Answer, 0. 2. Reduce 0.65 to a common fraction.--Answer,

3. Reduce % to a decimal.-Answer, .4.

4. Five-eighths of a dollar equals how many cents. -Answer, $0.625, or 621⁄2 cents.

5.

Reduce 15% to a decimal.-Answer, 15.375. 6. The diameter of No. 0000 wire B. & S. gauge is 0.46 inch. What is the diameter in common fractions? Answer, 400, or 20 inch.

7. The diameter of a pipe is found to be 37% inches. What is the diameter in decimals?-Answer, 3.875 inches.

Percentage.

Percentage is a particular case of decimal fractions. In percentage the denominator is always one hundred.. When we find a certain per cent of a number, we find that many hundredths of the number. Thus six per cent means six hundredths. To find six per cent of a number we multiply the number by 0.06.

Examples:

1. Twenty per cent of a number is what fraction of the number?-Answer, 2100, or %.

2. One-eighth of a number is what per cent of the number? Reduce % to hundredths.-Answer, 12% per cent.

3. Thirty-three and one-third per cent of a number is what fraction of the number?-Answer 1.

4. Find 6 per cent of $1,000.-Answer, $60.00.

5. If an engine wastes in friction 25 per cent of the power it receives, and it receives 80 horse-power, how many horse-power does it waste?-Answer, 20 horsepower.

6. If an electric motor gives out 65 per cent of the electrical power it receives, and receives 12 horsepower, how many horse-power does it give out?Answer, 7.8 horse-power.

7. If a dynamo gives out in electrical horse-power 90 per cent of the power it receives, and it receives. 5,000 horse-power, how many electrical horse-power does it give out?-Answer, 4,500 electrical horsepower.

8. The efficiency of a motor is 80 per cent, that is, it gives out in useful power 80 per cent of the power it receives. It receives 101⁄2 horse-power. What power does it give out?-Answer, 8.4 horse-power.

9. If the power factor of an alternating current is 86.6, that is, the true power is 86.6 per cent of the apparent power, what is the true power if the apparent power is 4,000 kilowatts? Multiply 4,000 by 0.866.Answer, 3464 kilowatts.

10. What is the true power if the apparent power is 10,500 kilowatts and the power factor 93.97?— Answer, 9866.85 kilowatts. In the last example we multiply 10,500 by 0.9397.

Finding the Horse-power of an Engine.

In finding the horse-power of an engine, the area of the piston must first be found. The end of the piston

is a circle and you have learned how to find the area of a circle. The area must be found in square inches. Measure the diameter of the piston in inches and find the area in square inches.

Next the pressure per square inch must be found. It will not do to take the pressure as shown by the pressure gauge, for this is the pressure of the steam in the boiler. We want the pressure of the steam on the piston, and this is shown by the indicator card. The indicator card shows that the pressure changes, being greatest near the beginning of the stroke and dropping after the cut-off. Now, since there are so many different pressures during a stroke, which pressure shall we take? The answer is we must take the average pres

sure.

To find the average pressure, we divide the indicator diagram into a number of parts (say ten), as shown in Fig. 5, and draw vertical lines to the curve. These vertical lines are the pressure lines for the different parts of the stroke. They should be drawn in the middle of the division. They should begin on the line which represents the back pressure, that is, the lower line of the indicator diagram.

Two vertical lines should be drawn just touching the extreme ends of the diagram. The space between these two lines should be divided into ten or twenty equal parts and from the middle point of each of these parts the pressure lines should be drawn as described above. In Fig. 5, A C and B D are the vertical lines just touching the ends of the diagram. The space A B is divided into ten equal parts. From the middle of each part, and perpendicular to the line A B a pressure line is drawn. In the figure the pressure lines are dotted. The part of the pressure line within the curve represents the difference between the forward pressure and the back pressure. This is the pressure that counts in measuring power. We must measure the length of

each pressure line in inches and multiply its length by the scale of the spring.

Fig. 5 is an indicator-card from the crank end of the engine. The average pressure for this card should be found, also the average pressure for a card taken from the head end. The average of these two pressures is the average effective pressure for one revolution.

For example, if we are using a 40-pound spring, each inch represents 40 pounds of pressure and we multiply by 40. Thus, if the length of a pressure line at a certain point is 1.1 inches, the pressure at that point is 1.1 times 40, which equals 44 pounds. That is to say, 44 pounds is the pressure per square inch when the piston is at that particular point in the stroke. If we find in this way the pressure shown by each pressure line, add all these pressures together and divide by the number of pressure lines, we have the average pressure per square inch for the entire stroke. The greater the number of pressure lines used, the more accurate the result will be, but it is seldom necessary to use more than twenty lines.