Now very clearly, the ratio between the number of castings, "19:13," is the same as the ratio of the weights, or 19 is to 13 as the weight for 19 castings is to the weight for 13 castings. Substituting the word "answer" in place of "weight for 13 castings," we get

Since the product of the means equals the product of the extremes, we know that

If 19 Answer =

or 1781 = 19 Answer

1781, then the answer must equal 119 of 1781.

In using proportion, the following should be kept in mind: 1. Be sure that each ratio is made up of two like terms. Thus in the problem just solved, the first ratio is made up of 19 and 13, which are both numbers of castings, and the second ratio consists of two weights.

2. Be sure that when the first term is larger than the second, the third term is also larger than the fourth, or if the first term is smaller than the second, the third is also smaller than the fourth.

3. Solve by the rule that "the product of the means equals the product of the extremes."

Examples:

1. A certain firm bought 13 used motors for a lump sum of $520. A short time later it purchased 9 more at the same rate that was paid for the first lot. What was the cost of the second lot of 9 motors?

Since they were paid for at the same rate, the costs of the two lots will be proportional to the number of motors in each lot.

By our first rule, then, one ratio will be numbers of motors and the other ratio will be costs.

Ratio of the numbers of motors is 13:9.

Ratio of the costs is 520: cost of 9 motors.

The proportion will be,

By rule 2, the fourth term must be smaller than the third term, since the second term, 9, is smaller than the first term, 13. The cost of 9 motors will

evidently be less than the cost of 13, so we know our proportion is correct. Substituting the word "answer" for "cost of 9 motors" we have

hr. and receives $3. If the other man works 11 hr., how much does he receive?

The preceding examples have all been what is termed "direct proportion," since the weights increase directly as the number of castings increase; the costs increase directly as the number of motors; and the pay increases directly as the number of hours.

61. Inverse Proportion.-There are certain cases of proportion where for a given increase of one factor, there is a corresponding decrease of another factor. Two pulleys belted together are an example of this. Our own observation tells us that the larger of the two pulleys makes the least number of r.p.m., and the more we increase the diameter of either pulley, the more will its speed in r.p.m. decrease provided the belt speed is not changed.

This is called an "inverse proportion," and the number of revolutions are said to vary inversely as the diameter; that is, in the reverse order. The same rules are observed for inverse proportions as for direct proportions.

In direct proportions, the first and third terms correspond, and the second and fourth terms correspond. Thus in the last example, 9 hr. corresponds to a pay of $3 and 11 hr. to a pay of $3.67. In inverse proportions, however, the first and fourth terms correspond and the second and third correspond.

Example:

Two pulleys of diameters 9 in. and 12 in. are belted together. The 9-in. pulley makes 270 r.p.m. How many r.p.m. does the 12-in. pulley

make?

The first two terms of the proportion will be the ratio of the diameters, 9:12.

The ratio of the r.p.m. will then be as r.p.m. of 12-in. pulley: 270, since, according to rule 2, the third term must be smaller than the fourth if the first is smaller than the second. Setting up the proportion,

It will be noted that 270 r.p.m. corresponds to the 9-in. pulley, and that these two terms are extremes.

62. Speeds and Diameters of Pulleys.-Problems involving speeds and diameters of pulleys can be worked by proportion, as just explained, or they can be worked by a set rule, based on proportion, without going through all the process of setting up the proportion.

In the previous example it was pointed out that the diameter and speed of one pulley form the extremes of the proportion, and the diameter and speed of the other form the means. Since the product of the means equals the product of the extremes, we obtain the following simple relation for pulleys belted together: The product of the diameter and revolutions of one pulley equals the product of the diameter and revolutions of the other. This gives us the following simple rule for working pulley problems.

Rule for Finding the Speeds or Diameters of Pulleys.-Take the pulley of which we know both the diameter and the r.p.m., and multiply these two numbers together. Then divide this product by the known number of the other pulley. The result is the desired number.

Examples:

1. A 36-in. pulley running 240 r.p.m. is belted to a 15-in. pulley. Find the r.p.m. of the 15-in. pulley.

36 X 240
8640 ÷ 15

=

=

8640, the product of the known diameter and revolutions. 576, the r.p.m. of the 15-in. pulley, Answer.

2. A 36-in. grindstone is to be driven at a speed of 800 f.p.m.

from a 6-in. pulley on the line shaft which is running 225 r.p.m. pulley must be put on the grindstone arbor?

What size

Explanation: First we must find the r.p.m. for the grindstone, as explained in Chap. VII. To get the required surface speed we find 85 r.p.m. necessary.

Now we have the r.p.m. and the size of the line shaft pulley. The product of these two numbers is 1350. Dividing this by the

r.p.m. of the grindstone arbor gives 16 in. as the nearest even size of pulley; so we will use that size.

PROBLEMS

116. Two pulleys are belt connected. The smaller one runs at a speed of 750 r.p.m. and the larger at 200 r.p.m. What is the ratio of their speeds?

117. The diameter of the small pulley in Problem 116 is 4 in. What is the diameter of the larger pulley?

118. Two car repairers are paid on a piece work basis. One man completes an average of 21 pieces of work per day and the other completes 10 pieces per day. What is the ratio of the first man's pay to that of the second man's?

119. Three men, each operating a drill press, are able to turn out 156 pieces a day. The foreman receives a rush order for this particular piece and puts two more drill presses at this job. How many pieces can he then obtain per day?

120. A boiler uses an average of 2 tons of coal every 51⁄2 hr. How many tons will it use in a day of 24 hr.?

121. A high-speed engine running at 300 r.p.m. drives a generator. The pulley on the engine shaft is 4 ft. in diameter and the generator pulley is 16 in. in diameter. At what speed will the generator run?

122. In the manufacture of a certain type of boiler the manufacturer uses an approximate ratio of heating surface to grate surface of 50:1. About what heating surface would a 100-hp. boiler of this type contain if the grate area was 35 sq. ft.

123. The line-shaft in Fig. 16 runs 250 r.p.m. Determine the size of line-shaft pulley to run the grinder at 1600 r.p.m., using the countershaft as shown in the figure.

124. The grinding wheel of Fig. 16 is replaced by a 6-in. emery wheel, which must be run at a surface speed of 500 ft. per minute. The large pulley on the countershaft is 15 in. in diameter, the line-shaft pulley is 3 ft. in diameter, and the speed of the line-shaft is 244 r.p.m. Calculate the required diameter of the small pulley on the countershaft.