**Look for the Merit in Metric**

Comparison of the metric system of weights and measures with those of the United States customary system is a poor way to explain the merits of the metric system. If the comparison is not done sparingly and carefully, confusion can be the result.

Initially, conversion to the metric system may seem to present problems for the fire service, but the metric system can be practical and efficient in the solution of fire science problems by greatly simplifying the mindbending mathematics firemen are accustomed to using.

It is well known by fire service hydraulics students that 1 cubic foot of liquid contains 7.481 gallons and 1 gallon of liquid contains 231 cubic inches. In the metric system, these measures would be cubic meters, liters and cubic centimeters. The terminology might be initially confusing at first, but the mathematics is simple.

A meter is roughly equivalent to a yard and a centimeter is about 2 1/2 Times smaller than an inch. Liquid measurement is by liter, roughly equivalent to a quart.

**Significant relationships**

In the metric system, the relationship between units has more significance than we are accustomed to. For example, 100 centimeters equal 1 meter and 1000 meters equal 1 kilometer. As for volume measurement, there are 1000 cubic centimeters in a liter and 1000 liters in a cubic meter. It should be evident that the metric system is easy to work with.

Equivalents can be confusing and therefore detrimental to an explanation of the metric system. The following examples are given to show that the metric system is in our ballpark.

When a fire hydraulics student has a problem involving cubic feet, he invariably multiplies by 7.481 to obtain gallons or, if he is dealing in cubic inches, divides by 231. The chances are that long division and multiplication are a necessity.

**Tank volume problem**

For example, a tank 15 x 9 x 9 feet has a volume of 1215 cubic feet. To determine the gallonage, cubic feet are multiplied by 7.481, the number of gallons in 1 cubic foot.

Even by using approximate figures, the work requires pencil and paper.

However, the metric system simplifies the operation. A 5 x 3 x 3-meter tank is 45 cubic meters in volume. Each cubic meter contains 1000 liters. The 45 cubic meters is multiplied by 1000 liters to obtain total volume in liters. Mentally, 1000 x 45 = 45,000 liters. The prefix kilo can be used to indicate thousands of, hence 45 kiloliters. Mathematically:

5 m x 3 m x 3 m = 45 m^{3}

45 m^{3} x 1000 1 = 45,000 1 = 45 kl

Obviously, answers in large denominations are not going to inhibit calculations.

**Customary method**

Presently, when we deal with the volume of a cylinder, our familiarity is with 3.1416r^{2}h or 7854D^{2}h. If we compute the capacity in gallons of 2 1/2-inch hose, 50 feet long, the figures look like this:

V = 7854D^{2}h

V = .7854 x 2.5 in x 2.5 in x 50 ft

There are some problems. We will have to decide whether to transpose the 2.5 inches to feet or 50 feet to inches. It appears as if 50 feet is most easily changed to 600 inches.

V = .7854D^{2} x 6.25 x 600

After some heavy multiplication:

V = 2945.25 cubic inches

But we’re not finished yet. To find the volume in gallons, the cubic inches must be divided by 231, the number of cubic inches in a gallon.

Is the metric system easier? We can examine it.

It has been suggested by several engineers that hose designated by diameter is not as practical as hose designated by orifice area. For example, the fact that fire service hose is 2.5 inches in diameter has very little bearing on any of our calculations. We must always compute the area to satisfy the problem.

**Proposed hose size change**

If we are going to change a system by professional management or systems analyst standards, the components of the system should be tailored to fit. The orifice area of 2 1/2-inch hose is 31.6692915 square centimeters. When we develop new measurements for fire service hose, it would be practical to make our hose orifice 32 square centimeters (32 cm^{2}) or 33 cm^{2} or 34 cm^{2}, depending on a comprehensive study by engineers.

It also has been suggested that the length of our hose, now between 15 and 16 meters, be standardized at 20 meters. The reasons are practical—a savings on couplings, ease of design and a general ease in computation. With the new lighter materials, the hose weight would not increase substantially.

What have we accomplished by hypothetically standardizing hose to an orifice area of 32 cm^{2} and a length to 20 meters? We have changed volume = .7854 D^{2}h to volume = orifice area X length.

**Metric computation**

If we compute the volume of 32-cm^{2} hose, 20 meters long, we write:

V = 32 cm^{2} x 20 m

We again face the problem of unlike terms. Shall we change centimeters to meters or meters to centimeters? Remember, in the metric system 100 centimeters equal 1 meter. Thus, 32 cm = .32 m. However, we must not forget that 32 centimeters are square centimeters for the orifice area. It is much easier to transpose 20 meters to centimeters simply by multiplying by 100: 20 m = 2000 cm. Continuing with the problem:

V = 32 cm^{2} x 2000 cm = 64,000 cm^{3}

With the knowledge that there are 1000 cubic centimeters in 1 liter, we divide 64,000 by 1000 to obtain the volume in liters. In a short time, a student is able to compute thusly:

Volume = orifice area X length

V= 32 cm2 x 2 x 103

V = 64 x 103 cm3

V = 64 x 10^{3} cm^{3} / 1 x 10^{3} cm^{3} = 64 liters

The amazing point is that all the computation is exactly as written. No other arithmetic is necessary.

This is not to say that the conversion will be without problems. There are trouble areas, especially those dealing with pressure calculation.

There is a tendency to use the kilogram (kg) incorrectly. A kilogram has weight due only to the force of gravity. All right, so does all matter. But pressure is not the result of gravity alone. The kilogram is equivalent to what the English engineering system recognizes as the slug. This is a measurement of mass. A student who has completed courses in technical math and technical physics has a working knowledge of these units. This fact points up the need for advanced education on the part of firemen.

Acceleration due to gravity and force is measured in newtons. The proper definition of a newton is: that force which gives a mass of 1 kilogram an acceleration of 1 meter per second each second (1 m/sec^{2}). Thus, a 2-kilogram object thrown to accelerate 2 meters per second each second has a force of 4 newtons behind it.

Pressure is defined as force per unit area. In the metric system, the proper terminology for pressure is newtons per square meter (N/m^{2}). This can be transposed to newtons per square centimeter (N/cm^{2}).

**Equivalent for psi**

Because we are accustomed to pounds per square inch, perhaps newtons per square centimeter, our nearest equivalent, would be more to our liking. Also, in dealing with N/m^{2} we find that the newtons are of such a high denomination in practical application that the prefix kilo must be used to denote thousands of.

For example, 700,000 N/m^{2} is approximately 100 psi. This may be written 700 kN/m^{2}. However these figures are still in areas we are least accustomed to. With the knowledge that there are 10,000 cm^{2} in a m^{2}, we divide 700,000 by 10,000 distributing the force equally among the centimeters and arrive at the newly written value 70 N/cm^{2} having the same value as 700 kN/m^{2}, or 100 psi.

In considering the metric system, we should regard it as a new frontier of study in which we will give men entering the fire service a more practical and efficient method of handling measurement problems.