Fertilizer Math

For expert insight into fertilizer application and rates, SportsField Management interviewed Christopher S. Gray, Sr., golf channel manager, professional fertilizers at LebanonTurf, and Steve Harris, manager, golf national accounts and sports turf at SiteOne. Their responses are as follows:

SportsField Management (SFM): What is the best way to determine how much fertilizer is needed to supply a specific amount of a particular nutrient per 1,000 square feet?

Gray: Part of properly managing and executing a sound nutrient management plan is making sure that you know how much you need to apply for any of the “Big 3” nutrients: nitrogen (N), phosphorus (P) and potassium (K).

There are actually a number of different ways to properly calculate these, but one the easiest ways is as follows:

• Determine what amount of N, P or K you want to apply. For this explanation, let’s say you want to apply .9# of nitrogen per 1,000 sq. ft.

• Look at the NPK analysis on the bag of fertilizer, and convert them to a more math-friendly decimal form. Let’s say you have a bag with the analysis of 20-5-10, which means there is 20% nitrogen, 5% phosphorus and 10% potassium. You’d simply convert each one to .20 nitrogen, .05 phosphorus and .10 potassium.

• Now take your desired application rate (.9# N) and divide it by the decimal form of the nitrogen (.20). It will look like this: .9 / .20 = 4.5 lbs. of fertilizer

• The calculation answer of 4.5 is the number of pounds of fertilizer that would need to be applied per 1,000 sq. ft. to achieve the .9 pounds application rate.

This same method works for any of the three nutrients.

Harris: Soil testing should always be done to determine proper nutrient levels and tailor fertilizer plans to address deficiencies and maintain desired levels. When calculating nitrogen needs, make sure to reference typical yearly plant needs and adjust based on your site-specific requirements.

SFM: What is the best method to covert lbs. per 1,000 sq. ft. to lbs. per acre?

Gray: The most important thing to know for this calculation is how many square feet are in an acre – which is 43,560. Once you have this information, all you need to do is multiply the already known lbs. per 1,000 sq. ft. number by 43.56. Let’s use our previous example to demonstrate how it’s done.

• We already know that we need to apply 4.5 lbs. of fertilizer per 1,000 sq. ft., so we simply multiply 4.5 X 43.56 and come up with 196.02, which is the total pounds of fertilizer needed to cover one acre.

Harris: There are 43,560 sq. ft., in 1 acre, therefore you would multiply the lbs. per 1,000 times 43.56 to determine rate per acre

SFM: What is the best way to determine the area a bag of fertilizer can cover and how many bags are needed to cover large areas?

Gray: There are two of specific things that you need to know to determine how much area one bag of fertilizer will cover. The first item is the weight of the bag itself and the other is the application rate of lbs. per 1,000 sq. ft.

You divide the bag weight by the application rate and then multiply the answer by 1,000 to get the coverage area in total sq. ft. Again, let’s use our previous example to illustrate how to do this:

• Most professional fertilizers come in either 40-lb. or 50-lb. bags. So, let’s assume that our 20-5-10 fertilizer comes in a 50-lb. bag for this exercise.

• We’ve already determined the application rate as 4.5 lbs. per 1,000 sq. ft.

• 50 / 4.5 = 11.111

• 11.111 X 1,000 = 11,111 total sq. ft. coverage – one bag of 20-5-10

Once you have this information, it’s easy to determine the total number of bags necessary to cover larger areas. All you need to do is take the total square footage you want to apply the fertilizer to and divide it by the coverage of one bag to figure out how many bags you would need to have.

• Let’s say you want to treat 84,000 total sq. ft. with our 20-5-10 that we just calculated will treat 11,111 sq. ft. per 50 lb. bag.

• 84,000 / 11,111 = 7.56 bags needed

Harris: Determine how much actual product is needed to apply the rate of fertilizer desired, then divide that amount into the volume of the bag (typically 50 lbs.). For example, if your desired rate of actual product is 5# per 1,000 sq. ft., a 50-lb. bag of fertilizer will cover 10,000 sq. ft.

To determine rate, you must decide how much of the nutrient you are looking to put down per 1,000 sq. ft., (most often nitrogen). To do that you must understand the analysis numbers on a bag of fertilizer. For example a bag of 25-5-15 fertilizer means that there is 25 lbs. of nitrogen, 5 lbs. of phosphorus, and 15 lbs. of potassium per 100 lbs. of material. Based on this example, a 50-lb. bag of fertilizer would contain 12.5 lbs. N, 2.5 lbs. P, and 7.5 lbs. K.

• If the desired application rate of nitrogen is 1 lb. per 1,000 sq. ft., 1 lb. (desired rate) divided by 25% (analysis of fertilizer) times 100 (100 lbs. referenced above) = 4 lbs. of material per 1,000 sq. ft. to apply 1 lb. of nitrogen

• Rates for actual phosphorus (44% available) and potassium(83% available) need to use a multiplier since not all nutrient is plant available

• Based on the example above 2.5 lbs. of P X .44 = 1.1 lbs. of actual P in the bag

• Based on the example above 7.5 lbs. of K X .83 = 6.6 lbs. of actual K in the bag

SFM: What other factors should go into determining proper application rates?

Gray: Proper application rates of any nutrients, macro or micro, will vary considerably, and it’s always recommended to perform soil tests at regular intervals to determine what the current levels are of each one to help guide the decisions on what’s needed and what is not. These reports will clearly tell you what you have an abundance of and what is missing. Ultimately, all fertilizers feed the soil, which, in turn, feed the turf. Knowing what’s happening below the surface is always the best way to determine what type of fertilizer is needed and what analysis makes the most sense for your specific site.

Harris:

• Soil type (sand, clay, loam) should be kept in mind. A sandy soil will not hold nutrients as long as a clay-based soil.

• Drainage (whether the area is typically wet, dry or in between) should be considered. An overly wet area is subject to leaching and run off resulting in contamination in low areas or creeks, ponds and lakes. Conversely a dry soil will be more difficult to get the nutrients to the plant roots.

• Type of fertilizer, whether the nutrients are all slow-release or of they are a controlled-release technology. Typically all quick-release fertilizers are applied at a lower rate per 1,000 sq. ft., to reduce the risk of burn to the plant. Controlled-release fertilizers can be applied at a higher rate since the release of the nutrient is metered.

• Plant types and time of year. Cool-season vs. warm-season fertilizer application rates and timing are different.

SFM: Is there anything else sports field managers should know regarding fertilizer selection and application?

Gray: It’s important to understand that for P (phosphorus) and K (potassium), the values expressed on the bag are Phosphate (P2O5) and Potash (K2O) and not the elemental forms. For many professionals, correctly managing P and K levels in the soil requires doing a little chemistry and math to convert the numbers to the actual amounts of each element. Here’s a brief breakdown on how to make these conversions:

Elemental phosphorus

Here’s the easy answer of understanding how much elemental phosphorus is contained in phosphate (P2O5)…43.7%. This is calculated with some very basic chemistry and looking at the atomic weight of each element of the phosphate compound.

• Atomic weights: P is 31 and O is 16. You can see these on any periodic table.

• There are two units of P in phosphate, so 31 X 2 = 61 grams

• There are five units of O in phosphate, so 16 X 5 = 80 grams

• 61 grams + 80 grams = 141 total grams of phosphate

• The amount of P is calculated as: 61 / 141 = .437 or 43.7%

With this information, you can now calculate the amount of elemental P in our 50 lb. bag example of 20-5-10 as follows:

• .437 X 5 = 2.185 lbs. of P (initial answer)

• 2.185 X .5 = 1.0925 lbs. of P

• It is necessary to multiply the initial answer by .5 because this is a 50-lb. bag because the phosphate compound is expressed as a “percentage” – meaning per 100.

• If this were a 40-lb. bag, you would multiply the initial answer by .4

Elemental potassium

The very same logic is used when calculating elemental potassium. Here’s the easy answer of understanding how much elemental potassium is contained in potash (K2O)…83.0%.

• Atomic weights: K is 39 and O is 16. You can see these on any periodic table.

• There are two units of K in potash, so 39 X 2 = 78 grams

• There are one units of O in potash, so 16 X 1 = 16 grams

78 grams + 16 grams = 94 total grams of phosphate

• The amount of K is calculated as: 78 / 94 = .830 or 83.0%

With this information, you can now calculate the amount of elemental K in our 50 lb. bag example of 20-5-10 as follows:

• .830 X 10 = 8.30 lbs. of K (initial answer)

• 8.30 X .5 = 4.15 lbs. of P

• It is necessary to multiply the initial answer by .5 because this is a 50 lb. bag because the phosphate compound is expressed as a “percentage”…meaning per 100.

• If this were a 40 lb. bag, you would multiply the initial answer by .4