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10. Relations between the English and metric units. The following table, which is inserted for reference, gives the relation between the most common English and metric units.

= 2.54 cm.

1 inch (in.)
1 foot (ft.) = 30.48 cm.

1 mile (mi.) = 1.609 km.

1 grain

1 oz. av.

1 lb. av.




= 64.8 mg.

= 15.44 grains

= 28.35 g.

= .0353 oz.

= .4536 kg,
42.5 lbs.

= = 2.204 lb.

The relations 1 in. 2.54 cm., 1 m. 39.37 in., 1 kilo (kg.) = 2.2 lb., 1 km. = .62 mi. should be memorized. Portions of a centimeter and of an inch scale are shown together in Fig. 2.

11. The standard unit of time. The second is taken among all civilized nations as the standard unit of time. It is 86400 part of the time from noon to noon.


12. The three fundamental units. It is evident that measurements of both area and volume may be reduced simply

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1 cm. 1 m.

1 km.


1 g.

1 g.

1 kg.




.3937 in.

1.094 yd. = 39.37 in.

= .6214 mi.



FIG. 2. Centimeter and inch scales

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to measurements of length; for an area is expressed as the product of two lengths, and a volume as the product of three lengths. For these reasons the units of area and volume are looked upon as derived units, depending on one fundamental unit, the unit of length.

Now it is found that just as measurements of area and of volume can be reduced to measurements of length, so the determination of any measurable quantities, such as the pressure in a steam boiler, the velocity of a moving train,

the amount of electricity, consumed by an electric lamp, the amount of magnetism in a magnet, etc., can be reduced simply to measurements of length, mass, and time. Hence the centimeter, the gram, and the second are considered the three fundamental units. Whenever any measurement has been reduced to its equivalent in terms of centimeters, grams, and seconds, it is said, for short, to be expressed in C.G.S. (Centimeter-Gram-Second) units.

13. Measurement of length. Measuring the length of a body consists simply in comparing its length with that of the standard meter bar kept at the International Bureau. In order that this may be done conveniently, great numbers of rods of the same length as this standard meter bar have been made and scattered all over the world. They are our common meter sticks. They are divided into 10, 100, or 1000 equal parts, great care being taken to have all the parts of exactly the same length. The method of making a measurement with such a bar is more or less familiar to everyone.

14. Measurement of mass. Similarly, measuring the mass of a body consists in comparing its mass with that of the standard kilogram. In order that this might be done conveniently, it was first necessary to construct bodies of the same mass as this kilogram, and then to make a whole series of bodies whose masses were 1 1 1 etc. of the mass of this kilogram; in other words, to construct a set of standard masses commonly called a set of weights.


2' 10' 100 1000'

With the aid of such a set of standard masses the determination of the mass of any unknown body is made by first placing the body upon the pan A (Fig. 3) and counterpoising with shot, paper, etc., then replacing the unknown body by as many of the standard masses as are required to bring the pointer back to O again. The mass of the body is equal to the sum of these standard masses. This rigorously correct method of weighing is called the method of substitution.

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If a balance is well constructed, however, a weighing may usually be made with sufficient accuracy by simply placing the unknown body upon one pan and finding the sum of the standard masses which must then be placed upon the other pan to bring the pointer again to 0. This is the usual method of weighing. It gives correct results, however, only when the knife-edge C is exactly midway between the points of support m and n of the two pans. The method of substitution, on the other hand, is independent of the position of the knife-edge. It is customary to consider that the mass of a body determined as here indicated is a measure of the quantity of matter which it contains.

FIG. 3. The simple balance


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1. The 200-meter run at the Olympic games corresponds to the 220yard run in our local games. Which is the longer and how much?

2. The French 75-mm. guns have what diameter in inches?

3. The Twentieth Century Limited runs from New York to Chicago (967 mi.) in 20 hr. Find its average speed in miles per hour.

4. Name as many advantages as you can which the metric system has over the English system. Can you think of any disadvantages?

5. What must you do to find the capacity in liters of a box when its length, breadth, and depth are given in meters? to find the capacity in quarts when its dimensions are given in feet?

6. Find the number of millimeters in 6 km. Find the number of inches in 4 mi. Which is the easier?

7. With a Vickers-Vimy biplane Captain Alcock and Lieutenant Brown completed, on June 15, 1919, the first nonstop transatlantic flight of 1890 miles from Newfoundland to Ireland in 15 hr. 57 min. How many miles per hour? How many kilometers per hour?

8. Find the capacity in liters of a box .5 m. long, 20 cm. wide, and 100 mm. deep.


15. Definition of density. When equal volumes of different substances, such as lead, wood, iron, etc., are weighed in the manner described above, they are found to have widely different masses. The term "density" is used to denote the mass, or quantity of matter, per unit volume.

Thus, for example, in the English system the cubic foot is the unit of volume and the pound the unit of mass. Since 1 cubic foot of water is found to weigh 62.4 pounds, we say that in the English system the density of water is 62.4 pounds per cubic foot.

In the C.G.S. system the cubic centimeter is taken as the unit of volume and the gram as the unit of mass. Hence we say that in this system the density of water is 1 gram per cubic centimeter, for it will be remembered that the gram was taken as the mass of 1 cubic centimeter of water. Unless otherwise expressly stated, density is now universally understood to mean density in C.G.S. units; that is, the density of a substance is the mass in grams of 1 cubic centimeter of that substance. For example, if a block of cast iron 3 cm. wide, 8 cm. long, and 1 cm. thick weighs 177.6 g., then, since there are 24 cc. in the block, the mass of 1 cc., that is, the density, is equal to 176, or 7.4 g. per cubic centimeter.


The density of some of the most common substances is given in the following table:

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Carbon bisulphide



(In grams per cubic centimeter)




Hydrochloric acid



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16. Relation between mass, volume, and density. Since the mass of a body is equal to the total number of grams which it contains, and since its volume is the number of cubic centimeters which it occupies, the mass of 1 cubic centimeter is evidently equal to the total mass divided by the volume. Thus, if the mass of 100 cubic centimeters of iron is 740 grams, the density of iron must equal 740 ÷ 100 = 7.4 grams to the cubic centimeter. To express this relation in the form of an equation, let M represent the mass of a body, that is, its total number of grams; V its volume, that is, its total number of cubic centimeters; and D its density, that is, the number of grams in 1 cubic centimeter; then



This equation merely states the definition of density in algebraic form.

17. Distinction between density and specific gravity. The term "specific gravity" is used to denote the ratio between the weight of a body and the weight of an equal volume of water.*

Thus, if a certain piece of iron weighs 7.4 times as much as an equal volume of water, its specific gravity is 7.4. But since the density of water in C.G.S. units is 1 gram per cubic centimeter, the density of iron in that system is 7.4 grams per cubic centimeter. It is clear, then, that density in C.G.S. units is numerically the same as specific gravity.

*For the present purpose the terms "weight" and "mass" may be used interchangeably. They are in general numerically equal, although an important distinction between them will be developed in § 73. Weight is in reality a force rather than a quantity of matter.

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