# A simple math question(s)

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

1. ### somegrassdudeLawnSite MemberPosts: 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

2. ### WenzelOSLLCLawnSite Senior Memberfrom Maplewood, MNPosts: 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. ### RigglePLCLawnSite Fanaticfrom Grand Rapids MIPosts: 12,097

AND...
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. ### RicLawnSite Fanaticfrom S W FloridaPosts: 11,956

.

Dude

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

.

5. ### somegrassdudeLawnSite MemberPosts: 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. ### WenzelOSLLCLawnSite Senior Memberfrom Maplewood, MNPosts: 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. ### KirilLawnSite Fanaticfrom District 9 CAPosts: 18,308

Now figure it out on an elemental basis.

8. ### KirilLawnSite Fanaticfrom District 9 CAPosts: 18,308

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?