Plumber's Licensing Exam: Math For Plumbers

This guide introduces the basic mathematical concepts and formulas that plumbers need to know to ensure that jobs go smoothly.

Objectives
- Strengthen basic math skills
- Gain confidence in math skills related to plumbing
- Learn which calculations are appropriate for certain situations
- Memorize key constants

Key Terms:
constant
offset

Basic Facts
Pipes are solid objects that need to fit in certain places in distinct ways. A few measurements and simple calculations can make any job easier.

Tape Measure Basics:
When measuring, we here in the United States use the U.S. Customary System, also known as the foot-pound second system. We use inches and feet for our basic measurements of length in the plumbing trade. We all know that there are 12 inches in a foot. The harder part is dealing with fractions of inches, which you must do on a daily basis when you are installing piping.

Example: Say that you want to find out if you can keep a drainage pipe up between the joists when you have a 32-foot run. You will want to maintain the normal 1/4-inch per foot pitch. The pipe is 2-inch PVC, with an outside diameter of approximately 21/2 inches. A 2 × 12 inch joist is in reality about 11/2 × 111/2 inches. Now, how much will the pipe drop in 32 feet to maintain the quarter-inches in an inch? Each foot in length of our 32-foot run will use up one of those quarters. So, if we divide 32 by 4, we will get how many whole inches we will need. Go ahead, do the calculation in your head or on paper. You should get 8. That means the pipe must drop 8 inches from the beginning to the end of our 32-foot run in order to maintain a normal pitch.

Tips
All of the problems in this guide are presented using a simple, step-by-step approach. Read slowly if math is difficult for you. Try the problems yourself, one step at a time, and avoid skimming the pages.

Real-Life Situation: Julio, Evan, and Pat have their work cut out for them. They wonder if Carlos, the boss, knows what he has gotten them into. Running four parallel pipes, which are three different sizes, down through the ceiling of the hallway in this octagon-shaped hospital won’t be easy. There are already two electrical conduits, a vacuum pipe, an oxygen pipe, two 2-inch diameter hot water heating pipes, and a couple of pipes about which they are clueless. There’s not much room and it all has to look perfect. Julio and Evan think they are going to make Pat, the new whiz kid, do all of the math.

So far we have only used up 8 inches of the 111/2 inches of joist depth we have to work with. But now we must consider the thickness of the pipe. If we start on the high end with the pipe tight to the underside of the floor above, the bottom of our 2-inch PVC pipe will be about 21/2 inches below the top of the joist, and the bottom of the pipe is the part we are trying to keep up within the joist space. So, we must add that 21/2 inches to the 8 inches in drop that will occur over our 32-foot distance. OK, so let’s do the calculation: 21/2 plus 8 equals 101/2 inches. Does it fit? Will it work? Yes, but be careful to keep the starting end up tight to the floor and follow the1/4-inch per foot pitch, or you will end with the bottom of the pipe below the bottom of the joist. You only have how much extra to play with? That’s right, only 1 inch.

Example:  You are installing a 3/4-inch copper pipe. The end of the pipe is 321/2 inches away from a wall. You must end your pipe 61/2 inches away from the wall. Go ahead and clean the end of the pipe and the coupling while you are thinking about this. Swab the flux on your pipe and inside the coupling and slide it on the pipe. OK, that’s enough thinking time. This is pretty easy subtraction: 1/2 minus 1/2 is zero. Now, we know our answer will be a whole number. That doesn’t happen too often, does it? We are left with 32 minus 6. You can probably do this calculation in your head, but you could also use your tape measure as a number line. Start at 32 inches and count down 6 inches: 1 inch (31), 2 inches (30), 3 inches (29), 4 inches (28), 5 inches (27), 6 inches (26). You must cut the pipe 26 inches long.

Let’s try a calculation that’s just a little harder. The end of the pipe still needs to end 61/2 inches from the wall, but this time the pipe you must put the coupling on is 211/2 inches from the wall. Last time we were able to get rid of the fraction at the start. This time, we will have a fraction left over, so let’s deal with that first. We have to subtract 1/2 from 1/8. That means we have to make the 1/8 “bigger” so we can take the 1/2 away from it. Borrow a whole 1 from the 21, and add it to the 1/8. We do this by making both denominators (that’s the bottom number) the same.

How many eighths are in one whole? That’s right, 8: 8/8 + 1/8 = 9/8

That’s big enough to subtract 1/2 from, but we still need the denominators to be the same so we can finish the problem. We have to go to the larger denominator, which is 8. We already have the 9/8 the way we want it; now, we must convert the 1/2 into eighths. We know there are eight eighths in a whole, right? So how many eighths in a half? Half of a whole would be 4/8.

Now we can subtract, since our denominators are the same and we are subtracting a smaller number from a bigger number:
9/8 – 4/8 = 5/8

We now have the fraction out of the way, so we are only left with 20 minus 6. (Remember, we borrowed 1 from 21 to make our fraction bigger.) You can do the calculation in your head or use a tape measure: 1 inch (19), 2 inches (18), 3 inches (17), 4 inches (16), 5 inches (15), 6 inches (14). The answer is 14, and when we add the fraction, we know the next piece of pipe must be 145/8 inches long.

Fitting Allowance Basics
Since you have been working in the plumbing trade for a while now, you realize that when you assemble pipes with fittings, a certain amount of length is lost because the pipe and fitting must overlap to form a watertight joint. In the field, most of us hold a fitting in the air in the place we guess it will end up, and then measure the length of pipe needed. This works most of the time, but when you have restricted space and/or angles to traverse, guessing becomes more difficult. When you must install multiple pipes in a parallel configuration, doing the math becomes essential.

Caution: Materials and labor are both very expensive. In order to win a bid, jobs are quoted with fairly thin profit margins, especially the big projects. Incorrectly installed piping and equipment can destroy any profit the plumbing company might have had. The bid is quoted intending the work to be done correctly on the first try. A few extra minutes spent with a measuring tape and a calculator can transform a marginal gain into more profit than you expected.
Take the example of the Job Connection box at the beginning of this guide The architect has drawn the pipes to be installed in the hallway ceiling as lines. It’s easy to put those pipes in with a pencil, but it’s the plumber’s responsibility to turn those lines into pipes that function the way they were designed to function.
We will begin with the first 1-inch copper water supply pipe. On the straight run, the Ace Plumbing team must install a 1 × 1 × 1 inch copper sweat tee for a supply to another area of the hospital. The architect wants the pipe centered over a doorway to allow room for some ductwork. If Julio measures to the center of the doorway and cuts the pipe at that measurement, it will be too long and the tee branch will be off center. Pat tells Julio that he must allow for the fitting allowance. Pat has brought a pipe-fitting handbook along since Pat didn’t expect Julio or Evan to be prepared properly.
The chart states that the fitting allowance for a 1-inch copper sweat tee is 3/4 inch. That means the pipe must be cut 3/4 of an inch shorter in order for the branch of the 1-inch tee to be centered over the door (Figure below).

The distance from the center of the branch (which is where the architect drew it) to the bottom of the fitting socket where the pipe will end is the fitting allowance. All other fitting patterns also have a fitting allowance. It is found in a similar way.

You should have a plumbing math book or a pipe fitter’s handbook so you have these dimensions handy.

Figure: Fitting Allowance for a Copper Tee

Sometimes a plumber must create an offset when there is an obstruction, such as a pillar that must be piped around (Figure below). If the offset can be done with 90-degree elbows, it’s a relatively simple task to figure the length of the offsetting pipe. You measure from the centerline of the existing pipe to the position where the centerline of the new pipe must be, deduct the length of the fitting allowance twice (once for each end of the pipe), and you have the length of the pipe needed.

Figure: An Offset in Piping
The next most common fitting angle used for offsets is 45 degrees. A 45-degree angled fitting has much less resistance to flow than a 90-degree angled fitting. That explains its common usage as an offset. Figuring the length of the offsetting, or middle, pipe involves only one simple additional step. When using the 45-degree angled elbow, you will multiply the distance from the centerline of the original pipe to the centerline of the new pipe by 1.414. It will always be 1.414 when you are using 45-degree elbows. This number is called a constant.

You then deduct the fitting allowance for a 45-degree elbow twice, once for each end again, and cut your pipe.

Figure: The Important Information Needed to Build an Offset

Occasionally, you know the length of the diagonal section of the piping. When you are using 45-degree elbows, the offset distance—that is, the measurement from the original pipe to the centerline of the continuation of the pipe—will be the length of the diagonal times 0.707 plus the fitting allowance. We will only touch on 45-degree offsets since they are the most common. It is unlikely that the licensing test would present you with any other angle for an offset.

Basic Calculations
After linear calculations, determining the volume of cylinders is the next most essential type of problem that must be solved by plumbers. In addition to the tank volume problem mentioned at the beginning of the guide, two typical occasions that would demand volume calculations are figuring out the water content of a hydronic heating system to determine the amount of antifreeze that needs to be added for freeze protection and to determine the size of an expansion tank for the heating system.

Simple Geometry
Let’s look at some basic shapes and their parts.
Figure below illustrates a circle’s radius, which is the distance from the center of a circle to the circle line. A diameter is the distance all the way across a circle.

Radius and diameter are measurements we often use in plumbing since we deal with round pipes, round tanks, and round fittings.

Circumference is the distance around a circle.

Figure: Circle Radius

s

Figure: Circle Diameter

Figure: Circle Circumference—the Distance around a Circle

A cylinder is a circle with depth. We are confronted with cylinders often in the form of tanks.

Figure: Cylinder

We need to know the math involved for calculating the lengths, perimeters, areas, and volumes of these shapes. I know this is starting to sound harder, but it’s not.
OK, we need to know that radius so we can figure out how much real estate the circle occupies. In other words, the area.

The formula for the area of a circle is: A = πr2.

Read it out loud: Area equals pi (3.14) multiplied by the radius squared: A = πr2.

Squared means the number is multiplied by itself. Pi is a Greek letter that stands for 3.14, which is a constant that is used for finding quantities in circles. The number really goes on indefinitely in decimals. (If you want to see how long it is, plug the formula 355 ÷ 113 into a Cray supercomputer. That will keep it busy for a while.)

Rounding to two decimal places for the final answer is close enough for our purposes. So, what is the area of a circle that has a radius of 1 1/4 inches? I’m using a fraction because that’s how we plumbers measure things. Before we can do any calculation, we have to convert the 1/4 to a decimal.

All you have to do is divide the top number by the bottom number:
1 ÷ 4 = 0.25
Let’s plug these numbers into our area formula now:

A = 3.14 × (1.25 × 1.25)2. We always do the calculation in the parentheses first, so:
1.25 × 1.25 = 1.5625

Now, we have A = 3.14 × 1.56252. Of course, it’s OK to use a calculator to do the squaring. You should now get something like A = 3.14 × 2.4414. Round it off to A = 3.14 × 2.44. I get A = 7.6616, which you can round off to A = 7.66. We always express area in square dimensions, like square feet, or square yards, or square miles. The original dimensions given to us for this problem were in inches, so we express the area of our circle with the 11/4-inch radius as 7.66 square inches.

Tips
Here are some of the constants commonly used in plumbing:
      1.414 times the distance between the pipes (the offset) in a 45-degree offset = the length of the diagonal pipe
      0.707 times the length of a diagonal pipe = the distance between the offset pipes
      231 = cubic inches in a gallon
      1 cubic foot = 1,728 cubic inches
      1 cubic foot = 7.481 gallons
      1 cubic foot of water weighs 62.426 pounds
      1 gallon = 0.1339 cubic foot
      1 gallon of water weighs 8.345 pounds

It’s nice that we have an answer, but what in the world is 0.66 of an inch? We plumbers want our dimensions in feet, inches, and parts of inches. We need to convert 0.66 to inches, which is surprisingly easy. First, decide how accurate you need to be. If 1/8 inch is accurate enough, you simply multiply the decimal number you have, in this case, it’s 0.66, by the denominator, which is 8. That will tell you how many eighths the answer is.

Let’s do the calculation:
0.66 × 8 = 5.28

That means we have 5.28 eighths. Since the 1/8 -inch tolerance is good enough in this case, we will round off 5.28 to 5. Our answer is 5/8 inch. So, if you need your measurement to the nearest 1/8 inch, 0.66 equals 5/8 inch.

Let’s say you need a higher degree of accuracy. We’ll work it to the nearest 1/16 inch. All you have to do is multiply the decimal number, 0.66 in this case, by 16, which is the denominator of the level of accuracy you need for this measurement.

Try it:
0.66 × 16 = 10.56

That means we have 10.56 sixteenths. Again, round it off to the nearest whole number, which is 11. Since we need the accuracy to the nearest sixteenth, we will now express our answer as 11/16 inch.
This simple formula works no matter what degree of accuracy you need. It will work for quarter inches as well as sixty-fourths.

Take another look at the Figure above. It shows the diameter of a circle. We use the diameter to find the circumference of, or the distance around, a circle. The formula is C = πd. This formula means that the circumference of a circle is equal to pi (π) multiplied by the diameter of the circle.

With a diameter of 8 inches, what is the circumference? Plug in the numbers and work it out:
C = 3.14 × 8
C = 25.12 inches

Here’s another decimal we must convert to a fraction of an inch so we can measure it. This time we need to calculate to the nearest 1/8 inch, so we multiply 0.12 by 8:
0.12 × 8 = 0.96
That’s how many eighths we have. Rounding 0.96 to the nearest whole number, we get 1. That gives us a circumference of 251/8 inches.

Knowing the volume of a cylinder is important in many plumbing situations. Once we find the area of one end of the cylinder by using the area formula, A = πr2, all we have to do is multiply that area by the height of the cylinder. Take a look at the formula for the volume of a cylinder: V = πr2h. Volume equals pi (3.14) multiplied by the radius squared, and then multiplied by the height of the cylinder. Run through it once using 12 inches for the radius and 48 inches for the height:
V = 3.14 × 122 × 48
V = 3.14 × 144 × 48
V = 21,703.68 cubic inches
That’s great, but you need to know how many gallons a tank holds so you can replace it with the same size tank. The supply house only stocks tanks measured in gallons, not cubic inches.

Here’s another number to memorize: 1 gallon of water has 231 cubic inches in it.

All you have to do is divide the cubic inches by 231 to get gallons. Using the cylinder example above, you get:
21,703.68 ÷ 231 = 93.96 gallons
All you have to do then is round 93.96 to 94 gallons.