Tips From The Pros

TURFGRASS AND SPORTS FIELD MATHEMATICS - A USEFUL TOOL TO MAKE IT EASIER

There’s an old saying that goes something like “Give a man a fish, feed him for a day. Teach a man to fish, feed him for a lifetime”. To me, this applies to turfgrass and sports field mathematics. Too often we approach this oft-despised but completely necessary subject by trying to memorize equations and conversions rather than gain a basic understanding and learn a few techniques that we can apply to so many of the daily mathematics we use in our jobs.

Turfgrass mathematics is not a stimulating subject for most folks, so pat yourself on the back for the fact that you are still reading. I see too many sports field managers struggle with the math part of the job and just leave it as kind of a blank spot in their quiver of skills. Right from the start of my career I had gained a valuable tool that really helped me throughout my career. I may be wrong, but I don’t think it is taught in many turfgrass management programs, or at least not hammered-in enough throughout the courses.

I was always good in chemistry classes in high school and university. I didn’t particularly like it, but it seems like one of those subjects that it’s hard to get a “C” in. You kind of either shine or fail. This is where I learned about “The Factor-Label Method” of converting units of measure, also sometimes called “Dimensional Analysis”. It is a tool used to convert between units and rates, which is the vast majority of turfgrass and sports field mathematics in my experiences.

Factor-Label is basic algebra, we don’t need any higher math skills than that to get through most situations in the field. You simply build an equation of fractions, with the units canceling out along the way until you arrive at the unit you desire.

Here’s the first thing I always stumbled on, flipping conversion fractions. In arithmetic, we know that does not equal . But when we add units to the fraction, it becomes a conversion, a relationship. Take the price of gasoline for example. Let’s say its $3.00 per gallon. We could express that as . But we could also then say that 1 gallon costs $3, or . When we add the units (dollars and gallons) to the fraction, it becomes a relationship and so can be expressed either way as needed to get the units to cancel in our factor-label equations.

Now let’s say your car is on empty as you pull into the gas station. You know you have a 20 gallon tank. How much will it cost to fill up the tank? Sure, you can probably figure this out in 2 seconds in your head. But let’s look at it through a factor label exercise.

Every factor-label equation uses this basic format:

What you are looking for = what you know x conversion fractions.

The first step is to identify “What you are looking for” and “What you know” to start the equation. In my example, “What you are looking for” is how many dollars it will cost to fill up, and x will be that amount. So we are looking for “x dollars”. “What you know” in my example is that we have a 20 gallon tank to fill. So the simple factor-label equation would be:

x dollars = 20 gal x conversion fractions. The idea is keep multiplying your known by conversion fractions, flipped as you need to in order to get the units on the bottom to be the same as units on top of the previous fraction. This way, the units cancel each other out. When you have the units you are looking for on the top of the conversion fraction, you are done building your factor-label equation and you simply solve for the numerical values.

You may already know the conversions, you might use any number of online conversion web calculators, or you may have to run a separate factor-label equation to get your needed conversion before plugging it into your factor-label equation.

In my example, we already know a needed conversion, gasoline is priced at . I plug it into my factor-label equation in such a way that the units cancel out.

x dollars = 20 gallons x 3 dollars/1 gallon. Notice the unit I am looking for is the unit on top of the last (and only) conversion fraction. All the units on the right side of the equation cancel out, except for the one I am looking for. So I am done building the equation and now just calculate the answer.

x dollars = 20/1 x 3/1 = (20 x 3)/(1 x 1) = 60/1 = 60 dollars to fill my tank.

Now, let’s say you want to take a drive of 600 miles. You know your car gets 25 miles per gallon, or 25 miles/1 gallon. How much will the trip cost you in gasoline?

x dollars = 600 miles x1 gallon/25 miles x 3 dollars/1 gallon = (600 x 1 x 3)/(25 x 1) = 1800/25 = 72 dollars in gas for our 600 mile trip.

Notice how I flipped the miles per gallon, and dollars per gallon conversion fractions as needed to get the units to cancel at each added conversion fraction?

When my conversion fraction has the units I am looking for on top (dollars in this example), I’m done building the equation and just do the math.

Important: Remember what your chemistry teacher always said “Always label your units in your equations!”

Turf Tips 101: Turfgrass and Sports Field Examples

Example 1:

You plan to topdress your 3 sand-based soccer fields with an appropriate sand at a recommended rate of 5 tons per acre. How many tons of sand will you need?

What you are looking for: x tons sand

What you know: 3 soccer fields

Conversions: Each field is 65,340 sq. ft. We know (or look up) that 1 acre = 43,560 sq. ft.

So:

x tons sand = 3 fields x 65,340 sq.ft/1 field x 1 acre/43,560 sq.ft. x 5 tons sand/1 acre

The units all cancel until we are left with “tons of sand” on top.

x tons sand = 3 fields x 65,340 sq.ft./1 field x 1 acre/43,560 sq.ft. x 5 tons sand/1 acre

Answer: (3 x 65,340 x 5)/43,560 = 22.5 tons of sand needed to topdress the 3 soccer fields at the recommended rate.

Example 2:

You need to fertilize a softball outfield grass area of .85 acres at a recommended rate of .75 lbs of nitrogen (N) per 1,000 sq. ft. The fertilizer analysis is 18- 26-10. How many pounds of fertilizer will you need?

We know that an 18-26-10 fertilizer is 18% nitrogen. Percent means “per 100”.

What you are looking for: x lbs fertilizer

What you know: .85 acres (1 outfield)

So:

X lbs. fertilizer = .85 acres x 43,560 sq.ft./1 acre x .75 lbs N/1,000 sq.ft. x 100 lbs fertilizer/18 lbs N = (.85 x 43,560 x .75 x 100)/(1,000 x 18) = 154.27 lbs. fertilizer needed.

Example 3:

After the fertilizer application, we realize that we overshot the target amount and used 210 lbs. of fertilizer instead of our target of 154 lbs. How many lbs. of nitrogen did we apply per 1,000 sq. ft.?

x lbs. N = 1,000 sq.ft. x 1 acre/43,560 sq.ft. x 210 lbs. fertilizer/.85 acre x 18 lbs. N/100 lbs. fertilizer = 1.02 lbs. N per 1,000 sq. ft. actually applied.

Example 4:

You are building a new natural grass football field and the plan is to seed the grass. The field designer has specified a single variety of Kentucky bluegrass to be seeded on the bare soil at 65.5 lbs. Pure Live Seed (PLS) per acre. Looking at the seed tags, you see that the purity is listed at 95% and the germination rate is 98%. You know the field and surrounds to be seeded totals 73,550 sq. ft. How much seed must you plant?

First, we have to determine the percent (PLS) in the seed, then plug that conversion into our factor-label equation. PLS% = %germination x %purity. So .98 x .95 = .93 or 93% PLS.

x lbs. seed = 73,550 sq.ft. x 1 acre/43,560 sq.ft. x 65.5 lbs.PLS/1 acre x 100 lbs. seed/93 lbs.PLS = 118.9 pounds of seed must be planted.

Resources of the Month

A good YouTube video about Dimensional Analysis/Factor Label.

There are many free online conversion web pages and apps. I like this conversion page, and also this one for Metric/English conversions.

A popular iPhone unit converter app here.

A popular android unit converter app here.

End Quote:

“Do not worry about your difficulties in Mathematics, I can assure you mine are still greater” – Albert Einstein