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Old 03-22-2012, 07:19 PM
somegrassdude somegrassdude is offline
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A simple math question(s)

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
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Old 03-22-2012, 08:26 PM
WenzelOSLLC WenzelOSLLC is online now
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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?
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Old 03-22-2012, 08:59 PM
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RigglePLC RigglePLC is offline
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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 by RigglePLC; 03-22-2012 at 09:02 PM. Reason: fix
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Old 03-22-2012, 09:28 PM
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Ric Ric is offline
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.

Dude

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



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Old 03-25-2012, 09:04 PM
somegrassdude somegrassdude is offline
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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
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Old 03-25-2012, 10:02 PM
WenzelOSLLC WenzelOSLLC is online now
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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?
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Old 03-25-2012, 10:59 PM
Kiril Kiril is offline
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Now figure it out on an elemental basis.
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Old 03-26-2012, 10:51 AM
Kiril Kiril is offline
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Quote:
Originally Posted by Kiril View Post
Now figure it out on an elemental basis.
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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