by Donald Proctor

To understand a concept clearly, most people can organize their thoughts in words, rather than in codes or symbols. While using symbols is usually faster, correct answers are a lot more important than quick answers. For the picnic table problem in the last installment of Logical Math (see "The Importance of Math in our Vocational World"), you logically decided that you needed to multiply length times width to determine the surface area of the table.

Your mind probably did not just logically decide that A= l × w, even though it is easier to write. In the case of the picnic table, both words and symbols are pretty clear, but even when you write A = l × w, your mind should be saying that area equals length times width.

I urge you to write down the relationship in words first and then use symbols as a next step. At the least, this extra step makes understanding the solution to your problem that much easier for someone else to follow.

Vocational Vocabulary

This may strike you as a silly directive, but if we do not know exactly what the words we use mean, we cannot communicate very well. We’ll examine a lot of words and their meanings (see table below), but the list will be far from complete.

As you encounter new terms in your job, be sure that you understand precisely what they mean. If there is any doubt about what a word means, don’t hesitate to ask for a clear definition. In some cases, two different terms may have almost identical meanings. Also, always bear in mind that you will encounter words or terms used incorrectly and you may or may not be able to figure out the intended meaning. Unfair? Yes! But that’s life.

Let’s look at an example problem to see how words or terms can be used to set up a problem solution.

You are told to purchase paint to cover the north wall of the Blower Building, which is 18 ft high and 42 ft long. The paint specifications suggest that the coverage factor for this paint is 210 ft2/gal. The paint costs $7.60/gal or $31.00 per 5-gal pail. How much paint should you buy?

One of several possible approaches to a solution might involve the following reasoning:

   Paint volume needed = (volume of paint in a container ÷ area that volume will cover) ×
              (height of area to be covered × length of area to be covered),
                                                   or
            Paint volume needed = (1.0 gal ÷ 210 ft2) × (18 ft × 42 ft) = 3.6 gal.

Next you must decide whether it is best to buy four 1-gal pails or one 5-gal pail:

     Four 1-gal pails × $7.60/pail = $30.40, or one 5-gal pail × $31.00/pail = $31.00.

Now you must decide if it’s wise to save $0.60 and take a chance that you can get by with only 0.4 gal of “wiggle room” in your estimate or spend the extra money and have 1.4 gal extra. (I probably would buy the 5-gal pail — and I wonder why I always have so many old, dried-up cans of paint in my garage.)

Notice that the words used to set up the solution involve an inverted use of the coverage factor: 1.0 gal per 210 ft2 rather than 210 ft2/gal. Seeing the reasoning expressed in words adds a lot of clarity and understanding to someone else reviewing your solution. As other problem-solving techniques are presented, we usually will try to set the solution up in word form as a first step.

Solution to Lickety Split Nitrogen Loading Problem Let’s recall the problem from the end of the last installment of Logical Math (see article "The Importance of Math in our Vocational World") and apply some of the good habit ideas we have discussed. The question asked you to determine the maximum solids loading rate for liquid digested sludge onto a bamboo cane field. A maximum of 50 lb (23 kg) of nitrogen per acre per year could be applied, spread over at least three applications. Typical sludge is 5.0% solids with 47% volatile solids. There are 3.2 lb (1.5 kg) of total nitrogen per 100 lb (45 kg) of volatile solids. The city’s sludge truck has a 2800-gal (11,000-L) tank, and the city produces an average of 4700 gal/d (18,000 L/d) digested sludge. (Metric conversions will be left out for the rest of the solution to avoid cluttering the explanation.)   As a first step, let’s determine which factors will affect the magnitude of the answer — that is, affect the application depth limit. If the limit would allow more nitrogen, we logically could put on deeper layers of liquid sludge, so the 50 lb/ac•yr sludge limit must logically be a multiplying factor. Requiring the annual application to be divided into three applications means that three applications per year must be a dividing factor, or we can multiply by the inverse, one year divided by three applications. The three items of information about the sludge composition (percent solids, percent volatile solids, and amount of nitrogen per unit of volatiles) each indicate that as any one of those factors increase, so does the amount of nitrogen per unit volume of liquid sludge. To hold the nitrogen application rate down to the approved limit, each of these factors must either be used as dividing factors or inverted for multiplication. The limitation of the nitrogen application would not change whether you hauled the sludge in a bigger truck or in a little-bitty wheelbarrow, so that tank volume should not enter into our solution. Finally, the problem does not even say that all of the sludge will be applied to the cane pole rearing site. The daily (or annual) volume of sludge might well affect the acreage needed but does not impact the application depth, so we don’t need that piece of data. Let’s see what our solution looks like so far: Amount of solids per application = (50 lb N/ac •yr) × (1 yr ÷ 3 applications) × (100 lb sludge ÷ 5 lb total solids) × (100 lb total solids ÷ 47 lb volatile solids) × (100 lb volatile solids ÷ 3.2 lb N) = 22,163 lb sludge/ac. Now we must apply conversion factors to get the desired units for the answer: Depth of application = 22,163 lb sludge × (1 gal ÷ 8.34 lb) × (1 ft3 ÷ 7.48 gal) × (1 ac ÷ 43,560 ft2) × (12 in. ÷ 1 ft) = 0.097 in. For all practical purposes, the limit is 0.1 in. We could have used 1.0 ft3 ÷ 62.4 lbs as a single conversion factor in place of both 1 gal ÷ 8.34 lb and 1 ft3 ÷ 7.48 gal. Also, as the problem was worded, we could have left the answer in feet of depth rather than in inches, but it’s a bit hard to visualize a liquid depth of only 0.008 ft. Take time to understand and appreciate the way logical thought was applied to the given factors to determine how to set up the solution — that is, which are logically multipliers and which are dividing factors or have an inverse impact on the answer. Take special note that you did not need to know how to arrange all factors at once to solve for an answer — you only needed to determine the significance, or lack thereof, of one factor at a time. You also didn’t need to memorize or look up a formula for the solution. Plain logical thought (“horse sense”) is quite portable and gains in strength the more you exercise it. Notice also that the explicit units of measure carried through with the numerical values help to assure that no necessary conversion factors were left out. Was the answer reasonable? An application of only 0.1 in. of liquid sludge may not seem like much at first glance, but 0.1 in. of syrup on your pancakes is a lot. How’s that for a memorable but disgusting comparison?