Sample Manure Calculations
These are just examples of the calculations that are used to determine a balanced manure rate and supplemental nutrient needs for you crop. Because of the natural variations in soils and manures, and because of the assumptions required in making the following calculations, the rates you calculate, as in the examples on the previous page, should be considered guidelines for determining the actual rates. Actual rates should be close to or less than the maximum rates you calculate for your situation.
If a rate less than the maximum is applied, it will need to be supplemented with one or more fertilizer nutrients. The supplemental fertilization rate is determined by subtracting the nutrients applied in the manure from the crop requirement. In the following examples, applying the maximum rate would mean applying more phosphorus and potash than is recommended on the soil test. When applying an excess, check whether it will be used by other crops in the rotation. If not, a lower rate is more appropriate. If you select a lower rate, supplement it with nitrogen fertilizer and possibly phosphate and/or potash fertilizer, depending on the rate used.
Sample Calculations: Calculating Maximum Rate of Manure Required to Meet Nitrogen Needs of a Crop
Typical Manure Only
This example illustrates calculating the maximum rate of manure required to meet the nitrogen needs of a crop and the net recommendation for the field based on total analysis for N, P_{2}O_{5}, and K_{2}O and can only be used for typical, nontreated dairy, swine, other livestock, and poultry manures.
Situation
 Recommendations for 150 bu/A corn crop from the soil test report were 15050100 pounds/A of N, P_{2}O_{5} and K_{2}O, respectively.
 The liquid dairy manure analysis is 20 pounds total N, 11 pounds P_{2}O_{5}, and 21 pounds K_{2}O per 1,000 gallons.
 Manure is to be applied in the spring for this corn crop and incorporated the day after application.
 Similar manure has been applied to this field frequently in the past (two or three out of the last five years).
 The starter fertilizer program is 150 pounds per acre of 102010.
 Last year, the crop in this field was soybeans, which yielded 50 bu/A.
Net Crop Nutrient Requirement
First, all sources of nutrients must be accounted for before a manure rate is calculated by starting with the soil test recommendations and then deducting any other sources of nutrients applied or already available in the field.
Net nutrients required = soil test recommendation – residual N from past manure – residual N from a previous legume crop in the rotation – starter nutrients
Residual N from Past Applications (Table: "Manure nitrogen availability factors for use in determining manure application rates based on planning conditions", Part B)
Residual N = 20 lb N/A for a history of frequent manure applications
Residual N from a Previous Legume Crop in the Rotation (Table: "Typical crop nutrition removal for phosphorus and potassium")
Soybeans @ 50 bu/A x 1 lb residual N/bu = 50 lb N/A from the soybeans the previous year
Starter Fertilizer Nutrients That Will Be Applied Regardless of the Planned Manure Application
150 lb/A of a 102010 starter fertilizer will be used on this field. The N, P_{2}O_{5}, and K_{2}O applied in this fertilizer is calculated as follows:
150 lbs fertilizer/A x 10% N = 15 lb N/A
150 lbs fertilizer/A x 20% P_{2}O_{5} = 30 lb P_{2}O_{5}/A
150 lbs fertilizer/A x 10% K_{2}O = 15 lb K_{2}O/A
Net nutrients required 
N (lb/A) 
P_{2}O_{5} (lb/A) 
K_{2}O (lb/A) 

Soil test recommendation 
150  50  100 
 Residual manure N 
20     
 Residual legume N 
50     
 Starter nutrients 
15  30  15 
Net requirement (lbs/A) 
65  20  85 
N Balance Manure Application Rate
The N balanced manure application rate is determined by dividing the available N in the manure into the net crop N requirement calculated above.
Available N in the manure = total manure N x N availability factor (Table: "Manure nitrogen availability factors for use in determining manure application rates based on planning conditions", Part A)
N availability factor = 0.4 from Table: "Manure nitrogen availability factors for use in determining manure application rates based on planning conditions" (based on the time of application [spring], crop [corn], time until incorporation [1 day], and manure type [liquid dairy]):
Available N = 20 lb N/1,000 gal x 0.4 = 8 lb N/1,000 gal
Maximum rate per acre to meet N need = net N requirement ÷ available manure N
Rate per acre to meet N need = 65 lbs N/A ÷ 8 lbs N/1,000 gal = 8,125 gal/A
(Note: Any rate less than or equal to this rate is acceptable.)
Nutrients Applied at the Planned Rate
Based on this calculation, the farmer decided to apply 8,000 gal/A. At this application rate the following amounts of nutrients would be applied:
Nitrogen: 8,000 gal/A x 8 lb available N/1,000 gal = 64 lb available N/A
Phosphorus: 8,000 gal/A x 11 lb P_{2}O_{5}/1,000 gal = 88 lb P_{2}O_{5}/A
Potassium: 8,000 gal/A x 21 lb K_{2}O/1,000 gal = 168 lb K_{2}O/A
Final Nutrient Balance on the Field after Manure Application
The nutrient balance is simply the difference between the net crop nutrient requirement and the nutrients applied at the planned rate.
N (lb/A) 
P_{2}O_{5} (lb/A) 
K_{2}O (lb/A) 


Net crop nutrient requirement  65  20  85 
Nutrients applied at planned rate (8,000 gal/acre)  64  88  168 
Balance after manure application  1 short 
68 excess 
83 excess 
As discussed earlier, this excess P should be evaluated with the Phosphorus Index to see if it represents a potential risk to the environment. If the P Index indicates that the risk is high, manure management may have to be changed to address this concern. The excess K could result in high K levels in crops, which can, under certain circumstances, cause problems in animal feeding programs and animal health.
All Manure Including Atypical or Treated Manure
This example illustrates calculating the maximum rate of manure required to meet the nitrogen needs of a crop and the net recommendation for the field based on analysis for dry matter, total N, NH_{4}N, total P_{2}O_{5}, and K_{2}O. This calculation can be used for all manure, but it is required for atypical and/or treated manures.
Situation
 Recommendations from the soil test report for 150 bu/A corn crop were 1605040 lbs of N, P_{2}O_{5} and K_{2}O, respectively.
 Poultry manure was composted and has an analysis of 50 lbs total N; 10 lbs NH_{4}N, 40 lbs P_{2}O_{5}, and 30 lbs K_{2}O/ton; and 40 percent moisture.
 Manure is to be incorporated three days after application.
 Manure with the same analysis has been applied to this field at the rate of 10 ton/A each of the last three years.
 The starter fertilizer program is 5 gal/A of 10340 (Note: 10340 weighs 11.68 lb/gal).
Net Crop Nutrient Requirement
First all sources of nutrients must be accounted for before a manure rate is calculated by starting with the soil test recommendations and then deducting any other sources of nutrients applied or already available in the field.
Net nutrients required = soil test recommendation – residual N from past manure – residual N from a previous legume crop in the rotation – starter nutrients
Residual N from Past Applications (Table: "Factors for calculating manure nitrogen availability based on time of application, incorporation, field history, and manure analysis with ammonium and organic N fractions")
Residual N from past manure applications is estimated as follows: For each year in the last five when manure was applied, multiply the rate applied x organic N analysis x the appropriate factor from the section of Table: "Factors for calculating manure nitrogen availability based on time of application, incorporation, field history, and manure analysis with ammonium and organic N fractions" labeled “Organic N decomposed from past applications.” Note that if the organic N is not given on the manure analysis report it can be calculated by subtracting the ammonium N from the total N on the report.
In this example the manure is composted and has 50 lbs total N/ ton and 10 lbs of NH_{4+}N /ton
Organic N = total N – ammonium N = 50 lbs/ton – 10 lbs/ton = 40 lbs/ton organic N
Residual N from last year: 10 tons/A x 40 lbs organic N/ton x 0.05 = 20 lbs N/A
Residual N from 2 years ago: 10 tons/A x 40 lbs organic N/ton x 0.02 = 8 lbs N/A
Residual N from 2 years ago: 10 tons/A x 40 lbs organic N/ton x 0.01 = 4 lbs N/A
Total residual N from past applications: 32 lbs N/A
Residual N from a Previous Legume Crop in the Rotation (Table: "Typical crop nutrition removal for phosphorus and potassium")
None in this example.
Starter Fertilizer Nutrients That Will Be Applied Regardless of the Planned Manure Application
5 gal/A of a 10340 starter fertilizer will be used on this field.
First, since this is a liquid fertilizer the gal/A must be converted to lb/A. To do this multiply the number of gal/A x the weight/gal.
5 gal/A x 11.86 lbs/gal = 59.3 lbs/A
Then the N, P_{2}O_{5}, and K_{2}O applied in this fertilizer are calculated
as follows:
59.3 lbs fertilizer/A x 10% N = 6 lbs N/A
59.3 lbs fertilizer/A x 34% P_{2}O_{5} = 20 lbs P_{2}O_{5}/A
There is no K_{2}O in this fertilizer.
Net nutrients required 
N (lb/A) 
P_{2}O_{5} (lb/A) 
K_{2}O (lb/A) 

Soil test recommendation 
160  50  40 
 Residual manure N 
32     
 Residual legume N 
0     
 Starter nutrients 
6  20  0 
Net requirement (lbs/A) 
122  30  40 
N Balance Manure Application Rate
The N balanced manure application rate is determined by dividing the available N in the manure into the net crop N requirement calculated above. For treated manures the available N is the sum of the N available from both the ammonium and organic N.
Available NH4N in the manure = NH_{4}N x NH_{4}N availability factor (Table: "Factors for calculating manure nitrogen availability based on time of application, incorporation, field history, and manure analysis with ammonium and organic N fractions")
NH_{4}N availability factor = 0.4 from Table: "Factors for calculating manure nitrogen availability based on time of application, incorporation, field history, and manure anaylsis with ammonium and organic N fractions" (based on the time of application [spring], crop [corn], time until incorporation [3 days], and manure type [composted poultry])):
NH_{4}N x NH_{4}N availability factor = 10 x 0.40 = 4 lbs N/ton
Available organic in the manure = organic N x organic N availability factor (Table 1.215)
In this example, the manure is composted and has 50 lbs total N/ton and 10 lbs NH_{4}N /ton
Organic N = total N – ammonium N = 50 lbs/ton – 10 lb/ton = 40 lbs/ton organic N
Organic N availability factor = 0.1 from Table: "Factors for calculating manure nitrogen availability based on time of application, incorporation, field history, and manure anaylsis with ammonium and organic N fractions" (composted manure in this example): Organic N x Organic N availability factor = 40 x 0.10 = 4 lbs N/ton
Total available N in the manure = available NH_{4}N + available organic N
Total Available N = 4 lbs NH_{4}N/ton + 4 lbs organic N/ton = 8 lbs available N/ton
Maximum rate per acre to meet N need = net N requirement ÷ available manure N
Rate per acre to meet N need = 122 lbs N/A ÷ 8 lb N/ton = 15.25 ton/A
(Note: Any rate less than or equal to this rate is acceptable.)
(Note: This may seem to be a very high rate for poultry manure, but because it was composted, the N availability is very low compared to raw poultry manure; thus, a larger amount is required to supply adequate available N for this crop.)
Nutrients Applied at the Planned Rate
Based on this calculation, which determined that up to 15.25 tons/A could be applied, the farmer decided to apply 10 ton/A.
At this application rate the following amounts of nutrients would be applied:
Nitrogen: 10 tons/A x 8 lbs available N/ton = 80 lbs available N/A
Phosphorus: 10 tons/A x 40 lbs P_{2}O_{5}/ton = 400 lbs P_{2}O_{5}/A
Potassium: 10 tons/A x 30 lbs K_{2}O/ton = 300 lbs K_{2}O/A
Final Nutrient Balance on the Field after Manure Application
The nutrient balance is simply the difference between the net crop nutrient requirement and the nutrients applied at the planned rate.
N (lb/A) 
P_{2}O_{5} (lb/A) 
K_{2}O (lb/A) 


Net crop nutrient requirement  65  20  85 
Nutrients applied at planned rate (8,000 gal/acre)  64  88  168 
Balance after manure application  1 short 
68 excess 
83 excess 
Note that one of the consequences of applying a low N availability material like compost at a rate that comes close to matching the N needs of the crop will usually apply a large excess of P and K. This is very clear in this example even though the rate the farm chose was much lower than the rate required to supply the available N needs.
Supplemental Fertilizer Needs
In this example the N applied at the farmers planned rate will not be adequate to meet the needs of the crop. Therefore, a supplemental N fertilizer application will be needed to supply the 42 lbs N/A that he is short.