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Let’s proceed to solve several problems involving the r.o.f. formula.

Suppose that Gpm = 100 and that t = 2 (Lpm = 378,5 and t = 2). Substituting into the r.o.f. formula gives

Gpm x t = Gal Lpm x t = L
100 x 2 = Gal 378,5 x 2 = L

Logically if water is flowing for one minute, then flowing for two minutes gives you twice as much. Thus multiplying the two numbers makes sense. This is how the equation is solved to get the answer.

200 = Gal 757 = L

Let’s take another example with t = 30 seconds. Since r.o.f. time is “per minute” the r.o.f. time must be expressed as a fraction of a minute. Instead of 30 seconds, therefore, 0.5 of a minute must be used.

Gpm x t= Gal Lpm x t = L
100 x 0.5 = Gal 378,5 x 0.5 = L
50 = Gal 189,25 = L

The only time for which the r.o.f equals the total amount of water is when time equals one minute.

100 x 1 = 100 378,5 x 1 = 378,5

The number “1” is called the identity for multiplication since the product of any number times one equals “identically” the same number.

The remaining two problems involve division instead of multiplication. First suppose you are given a r.o.f. of Gpm = 50 and a total amount of water of Gal = 200. You are asked to find out how many minutes it would take to flow this amount of water at that rate.

Gpm x t = Gal Lpm x t = L
50 x t = 200 189,25 x t = 757

Solving this problem requires you to use the fundamental rule for simplifying equations, namely whatever you do to one side of an equation you must do to the other side. Thus we must divide both sides of the equation by 50.

Arithmetic is used to simplify the equation. This gives

(50/50) x t = 200/50 (189,25/189,25) x t = 757/189,25

Since one is the identity for multiplication, this gives us the answer of t = 4 minutes.

The second problem also requires division. You are given a value for t and for Gal (L). You must calculate the r.o.f. The logic is the same as for the preceding problem. Suppose that you are given t = 5 and Gal = 500 (L = 1.982,5)

1 x t = 4 1 x t = 4
Gpm x t = Gal Lpm x t = L
Gpm x 5 = 500 Lpm x 5 = 1.982,5

In one minute you would flow 1/5 of the total amount of water. In 5 minutes you would flow 5 times as much, or 500 gallons (1.982,5 L). So it makes sense to divide the number of gallons (L) by 5. Doing this produces the answer

Gpm = 100 Lpm = 378,5

Thus far, our r.o.f. formula has only limited use. In order to make the r.o.f. formula really useful, the “Gal” (L) must be replaced by a numerical expression that relates the amount of water used to the size of a fire in a structure. What we are after is an expression that relates the volume of a fire to the volume of the part of the structure that is burning.