Fertilizer math is fairly straightforward, but since most growers (and even agronomists) only calculate it once or twice a year (fall and spring), it is easy to forget the routine.
That math can get more confusing this season as we put together the puzzle of buying the most cost-effective fertilizer, hitting yield goals while watching expenses, and taking into account the needs of modern hybrids and varieties.
I was reminded of how confusing that math can be in an email from a reader who stated that he reads many articles about nitrogen use on corn and finds the messaging quite confusing: Farmers (and agronomists) use different terms in different regions, and compute corn's nitrogen needs quite differently depending on state recommendations and other factors. Even the way we track how many units of nitrogen are actually in a unit of fertilizer can be complicated, depending on whether the product used is a solid, liquid solution, or gas.
He wrote: "I typically put on 35 to 40 gallons of 32% split applied. One -- is this enough for 180 bu corn? Also is this 350/400 pounds of N as many articles (seem to) refer to or what is (the actual) pounds at this rate per acre. Or do I go back to units of N (per bushel). In which case at 180 bu, I would need 60 gallons of 32% for corn on corn?"
How much nitrogen to apply on corn? That is a complex question and there are many approaches. Gone is the day when we could use the old Iowa State approach of 1.2 pounds per bushel expected yield and take a credit for soybeans (1 pound per bushel soybean yield ) if in a corn/soybean rotation. That seemed to work when N was inexpensive, hybrids weren't efficient, most N was put on preplant (fall) and we didn't worry about losing nitrates into our watersheds and the Gulf of Mexico.
Today N is expensive, corn price is low, and we can't let N escape into the environment. Today's corn hybrids also are much more nitrogen efficient, and continuous corn is common. We see rates of 0.9 to 1.0 pound per bushel and some farmers are even pushing it to 0.7 to 0.8 pound per bushel yield goal. In general I believe the 1 pound per bushel expected yield works quite well, but we need to take the right credits and apply that N smartly to protect it from losses. As yields go up from 200 to 250 and now even 300 bushels per acre, it takes more N per acre and we have to be smarter on how we manage that investment.
So, how much N to apply on 180-bushel corn? I would say that 140 pounds of N to produce 180 bushels of corn (ratio of 0.78) could be enough if he applies a third to half in the spring at planting and the remaining at side-dressing. This rate would work for corn planted back after soybeans (note no continuous corn penalty) and if his organic matter is at least 2.5% or higher and he is mineralizing some additional N. He is at the low end of the recommended N rate and may want to protect what's applied with a stabilizer to make sure it stays around for the crop.
UNDERSTANDING NITROGEN MATH
Being able to calculate how much fertilizer to put on to apply the right amount of nitrogen is important.
So let's start with an easy basic -- dry urea fertilizer. Urea has an analysis of 46-0-0, which means it contains 46% N. If you want 140 pounds actual N, divide 140 pounds by 0.46 and you need to apply 304 pounds of urea per acre (140 divided by 0.46 = 304 pounds product). Really pretty simple.
The calculation gets a bit more complex if you are combining various fertilizer products. If you apply urea and MAP (monoammonium phosphate, which has an analysis of 11-52-0) together, then you need to take credit for the N in the MAP and adjust the amount of urea applied. If your yield goal calls for 30 pounds of phosphate, you would apply 58 pounds MAP (30 pounds divided by 0.52 = 57.6 rounded to 58) to meet that.
This 58-pound application also gives your crop 6 pounds of N. You can reduce the amount of urea being applied by 13 pounds (6 divided by 0.46). Your final fertilizer blend would be 291 pounds urea plus 58 pounds of MAP, for 349 pounds total product per acre.
Anhydrous ammonia is a gas but a liquid under pressure. The liquid converts back to a gas once it leaves the manifold. Ammonia is quite easy to calculate because you are dealing with weight not liquid so the calculation works the same as for urea. Ammonia has an analysis of 82-0-0. So if you want to apply 140 pounds N per acre, you apply 171 pounds of ammonia (140 pounds divided by 0.82 = 171 pounds).
Liquids get a bit more complicated because you are dealing with volume and weight. The calculation can be relatively easy if you remember that 28% UAN has 2.9 pounds N per gallon and 32% has 3.5 pounds N per gallon. How did we come up with these amounts? UAN28 weighs 10.6 pounds per gallon, so 10.6 pounds X 0.28 = 2.9 pounds. UAN32 weighs 11.06 pounds per gallon, for a calculation of 11.06 pounds x 0.32 = 3.5 pounds. So if you want to apply 140 pounds N, and use a split application, you would apply 20 gallons UAN32 in the spring (70 pounds divided by 3.5 pounds per gallon) and 24 gallons UAN28 sidedress (70 pounds divided by 2.9 pounds per gallon).
Regarding the reader's question on how much is in 350 to 400 pounds of product? The math is simple since you know the rate. Just multiple through by the analysis (%N-%P-%K) to find out how many pounds of each nutrient the product contains.
Dan Davidson can be reached at AskDrDan@dtn.com
Copyright 2016 DTN/The Progressive Farmer. All rights reserved.