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A simple math question(s)

Discussion in 'Fertilizer Application' started by somegrassdude, Mar 22, 2012.

  1. somegrassdude

    somegrassdude LawnSite Member
    Messages: 2

    So...I have a supervisor that has me fill out chemical app sheets. He has what I consider some funky math to get AI and nutrient levels applied. And I'll add he's not very good with questions concerning his methods.

    So...lets say we apply a 14-0-14 product with .125 AI per-emergent. Lets say we apply 2200 lb on 8.8 acres. What would be the amounts of nutrients and AI applied, and could you show your math please. Thanks Thumbs Up
  2. WenzelOSLLC

    WenzelOSLLC LawnSite Senior Member
    Messages: 709

    14-0-14 is 14% or .14 so .14 X 2200= 308 lbs of each nutrient per 8.8 acres. 308/8.8= 35lbs per acre. 35/44k sqft in an acre = .79 lbs per 1000.

    Did I get that right?
  3. RigglePLC

    RigglePLC LawnSite Fanatic
    Messages: 13,727

    2200 Lbs/8.8 acres equals 250lbs per acre.
    250 lbs times .125/100% equals .3125 lbs Active ingredient per acre dithiopyr
    Last edited: Mar 22, 2012
  4. Ric

    Ric LawnSite Fanatic
    Messages: 11,969



    I taught Materials Calculation at the local College and I will occurr with the math given as being correct.

  5. somegrassdude

    somegrassdude LawnSite Member
    Messages: 2

    Wanted to say thanks for the replies. I thought my math was correct.

    Problem is telling the boss would lead to a job search. LOL
  6. WenzelOSLLC

    WenzelOSLLC LawnSite Senior Member
    Messages: 709

    So does your boss just not understand how the math works or do they TRY to fudge the numbers so they can get more money?
  7. Kiril

    Kiril LawnSite Fanatic
    Messages: 18,334

    Now figure it out on an elemental basis.
  8. Kiril

    Kiril LawnSite Fanatic
    Messages: 18,334

    Since we have no takers.

    14-0-14 = 14% N, 14% K2O

    Standard fertilizer analysis is reported as the oxides of P (P2O5) and K (K2O) unless otherwise noted on the label. N is reported as 1:1, so no special considerations are required.

    The actual amount of K being applied in the OP's case is (14% * 0.83) = 11.62% K (elemental)

    Total K (elemental) applied for the OP's case is = (0.1162 * 2200) = 255.64 lbs

    If we had 14% P2O5 then we would have (14% * 0.44) = 6.12% P (elemental)

    When determining fertilizer requirements, especially when using a soil test report to determine how much to apply, you need to know how to break a particular fertilizer down to an elemental basis if needed. For example, if a soil report calls for X pounds of K (K2O), and the fertilizer you are using is reporting K as KCl, do you know how much to apply?

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