This post originally was created for and also published in SportsField monitoring magazine, the official publication the the SportsTurf managers Association (STMA) in January 2021, pp. 24-25, 28.
You are watching: How much area does a 50 lb bag of fertilizer cover
What is the best method to determine just how much fertilizer is necessary to it is provided a details amount of nitrogen (or any kind of other nutrient) per 1,000 square feet?
As a component of appropriately managing and executing a sound nutrient management arrangement is making sure that friend know just how much you require to use for any type of of the “Big 3” nutrients, Nitrogen (N), Phosphorus (P) and also Potassium (K).
There are actually a number of different ways to effectively calculate these, yet one the easiest methods is together follows:
What is the best technique to convert lbs. Per 1,000 sq. Ft. Come lbs. Every acre?
This is a pretty straightforward answer. The most vital thing to recognize for this calculation is how numerous square feet space in one acre…which is 43,560. As soon as you have actually this information, all you have to do is main point the currently known lbs. Per 1,000 sq. Ft. Number by 43.56. Let’s usage our previous instance to demonstrate how the done.We currently know that we require to apply 4.5 lbs. That fertilizer per 1,000 sq. Ft., for this reason we simply multiple 4.5 X 43.56 and also come up v 196.02, i m sorry is the full pounds of fertilizer needed to sheathe one acre.
What is the best method to determine the area a bag the fertilizer deserve to cover and how many bags are needed to cover huge areas?
There are two of certain things the you require to know to determine exactly how much area one bag that fertilizer will cover. The very 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 load by the application rate and also then multiply the prize by 1,000 to get the coverage area in total sq. Ft. Again, let’s use our previous instance to illustrate how to execute this.
Once you have this information, it’s basic to determine the total variety of bags vital to cover bigger areas. Every you have to do is take it the total sq. Clip you desire to apply the fertilizer to and divide it by the coverage that one bag to figure out how countless bags you would need to have.
What other determinants should get in determining suitable application rates?
Proper application rates of any nutrients, macro or micro, will vary considerably and it’s constantly recommended to carry out soil tests at constant intervals to identify what the present levels space of every one to help guide the decision on those needed and what is not. This reports will plainly tell friend what girlfriend have an abundance of and what is missing. Eventually all fertilizers feed the soil, i beg your pardon in turn, feeding the turf. Knowing what’s happening below the surface ar is always the best way to determine what kind of fertilizer is needed and what evaluation makes the most sense for your specific site.
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Is there anything I could not have actually touched ~ above here, or anything else you would like our reader to know about fertilizer selection and application?
It’s vital to know that for ns (phosphorus) and also K (potassium), the worths expressed top top the bag are Phosphate (P2O5) and Potash (K2O) and not the element forms. For many professionals, correctly regulating P and K level in the soil calls for doing a little chemistry and math to transform the numbers to the actual quantities of every element. Here’s a brief failure on just how to make these conversions.
Here’s the basic answer that understanding exactly how much elemental phosphorus is contained in phosphate (P2O5)…43.7%. This is calculated through some very straightforward chemistry and also looking at the atomic weight of each element of the phosphate compound.Atomic weights: ns is 31 and O is 16. You deserve to see this on any type of periodic table.There are two systems of ns in phosphate, for this reason 31 X 2 = 61 gramsThere are five units that O in phosphate, therefore 16 X 5 = 80 grams61 grams + 80 grams = 141 complete grams of phosphateThe lot of p is calculation as: 61 / 141 = .437 or 43.7%.437 X 5 = 2.185 lbs of p (initial answer)185 X .5 = 1.0925 lbs the PIt is vital to main point the initial prize by .5 due to the fact that this is a 50 lb. Bag because the phosphate link is expressed together a “percentage”…meaning every 100.If this to be a 40 lb. Bag, you would certainly multiply the initial price by .4
The really same logic is offered when calculating elemental potassium. Here’s the simple answer the understanding exactly how much elemental potassium is included in potash (K2O)…83.0%.Atomic weights: K is 39 and O is 16. You deserve to see this on any type of periodic table.There are two units of K in potash, so 39 X 2 = 78 gramsThere space one units of O in potash, therefore 16 X 1 = 16 grams
78 grams + 16 grams = 94 total grams of phosphateThe lot of K is calculate as: 78 / 94 = .830 or 83.0%.830 X 10 = 8.30 lbs that K (initial answer)30 X .5 = 4.15 lbs of PIt is important to main point the initial price by .5 since this is a 50 lb. Bag since the phosphate link is expressed as a “percentage”…meaning per 100.If this to be a 40 lb. Bag, you would multiply the initial prize by .4