Four Steps in Fertilizer Calculations

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Four Steps in Fertilizer Calculations

• Determine the area
• Determine the rate per area
• Determine the amount of nutrient (area X rate)
• Convert amount of nutrient to amount of
  fertilizer

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Two Basic Types of Calculations(nutrient vs.
fertilizer)

• You have a given amount of fertilizer and you
  need to calculate how much nutrient is in it
• You have a given amount of a nutrient to apply
  and you need to calculate how much fertilizer to
  use

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Memorize and Understand This!

• Weight of fertilizer X of Nutrient Weight of
  nutrient
• of urea X 46 N of N
• Weight of fertilizer (Weight of nutrient)/( of
  nutrient)
• of urea ( of N)/(46 N)

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Converting From Fertilizer to Nutrient

• How many pounds of N are in a 50 bag of ammonium
  nitrate (35-0-0)?
• 50 fert X 0.35 17.5 N
• 50 fert X (0.35N/fert) 17.5 N

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Converting From Nutrient to Fertilizer

• How many pounds of ammonium nitrate (35-0-0) do
  you need to give you 5 of N?
• x fert X 0.35 5 N
• x fert X (0.35N/fert) 5 N
• x fert (5 N)/(0.35N/fert) 14.3
  fertilizer

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Example 1

• The area of a home lawn is 15,000 ft2
• You want to apply 1N per 1000 ft2 using ammonium
  sulfate (21-0-0)
• Total amount of N (1N/K) X (15K)
  15N
• Total amount of fertilizer 15N/0.21 71
  ammonium sulfate

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Example 1, more difficult

• You want to apply 1N per 1000 ft2 to the same
  lawn, using 75 IBDU and 25 urea
• 15N X 75 11.25N from IBDU (31N)
• x IBDU (11.25N)/(0.31N/IBDU) 36.3 IBDU
• 15N X 25 3.75N from urea (46N)
• x urea (3.75N)/(0.46N/urea) 8.2 urea

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Example 2, Acres

• 1 acre approx. 43,000 ft2 43K
• 1N/K X (43K/acre) 43N/acre
• How much ammonium nitrate (33-0-0) would you buy
  for 75 acres if you want to apply a total of
  4N/K during the next year, in 4 separate
  applications? It comes in 50 bags. How many
  bags would you order?

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Continued

• 1N/K 43N/acre
• 4N/K ?N/acre 172N/acre
• For 75 acres, need 172N/acre X 75 acres
  12900N
• x AN (12900N)/(0.33N/fert) 39091 AN
• 39091 AN/(50 AN/bag) 782 bags

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Example 3

• You are growing 25 acres of tall fescue and plan
  on applying a total of 4 N on the following
  schedule 2 N as MU (39-0-0) on May 1 1 N
  as 26-4-8 on Mar. 1 1 N as 26-4-8 on Oct. 1
• Question 1 How much of each fertilizer do you
  need to order?

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Question 1 Calculations

• Will apply 2/K of both fertilizers
• (2N/K)(43K/acre)(25 acres) 2150N
• (2150 N)/(0.39 N/ MU) 5513 MU
• (2150 N)/(0.26 N/ 26-4-8) 8269 fert

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Example 3 continued

• You also want to apply 1.5 actual P/1000 and 3
  actual K/1000 ft2.
• Question 2 How much actual P and K are applied
  in the 26-4-8 applications?

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Question 2 calculations

• There are a couple ways to approach this.
• Figure out how much fertilizer is applied per
  1000 ft2
• (8269 fert)/(25 acres) 331 fert/acre
  (331 fert/acre)/(43K ft2/acre) 7.7 fert/K
• Now calculate how much actual P and K are applied
  in 7.7 fert/1000 ft2.

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Continued...

• 7.7 fert/1000 ft2 (26-4-8)
• For P (7.7 fert)(.04 P2O5/fert) 0.3
  P2O5/1000 ft2
• (0.3 P2O5)(0.44 P/P2O5) 0.14 P/1000
• To apply a total of 1.5 actual P/1000, need 1.36
  P/1000 (1.5-0.14) from some other source. Lets
  use super phosphate (0-20-0).

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Continued...

• Need 1.36 P/1000 from 0-20-0 for 25 acres
• How much total P? (1.36P/K)(43K/acre)(25
  acres) 1462 P
• How much 0-20-0 is needed? First calculate the
  amount of P2O5 equal to 1462 P (1462P)/(0.44P2
  O5/P) 3323 P2O5
• How much 0-20-0 is needed? (3323 P2O5)/(0.20
  P2O5 / fert) 16614 0-20-0

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Continued...

• Now figure out how much K must be applied
• Remember we applied 7.7 26-4-8/1000
• How much actual K is applied in 7.7 fert/1000
  ft2? (7.7fert/1000)(0.08K2O/fert) 0.62 K2O
• (0.62 K2O)(0.83P/ K2O) 0.51 K/1000

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Continued...

• To apply a total of 3 actual K/1000, need 2.5
  K/1000 (3-0.5) from some other source. Lets use
  KCl (0-0-60)
• How much extra K is needed? (2.5K/1000)(43K/acre)
  (25 acres) 2688 K
• Convert from actual K to K2O (2688K)/(0.83K2O/
  K) 3238 K2O

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Continued...

• How much 0-0-60 is needed? (3238K2O)/(0.60K2O/
  fert) 5397 of KCl