Exercise 3.  Interpreting Soil
and Manure Test Results

Exercise 3a.  Interpreting Soil Tests

Soil test results for lower slope position - nutrient values are in ppm

Sample
Depth
Nitrate
N
Phosphate Potassium Sulphates Iron Copper Zinc Boron Manganese pH
(acidity)
EC
(salinity)
0-6" 8 53 267 >20 156 1.35 4.4 2.55 12.8 6.7 (normal) 0.59 (OK)
6-12" 6

>20




7.0 0.63
Total (lbs/acre)

27 105 533 >80
Estimated Available (lbs/acre)

38 105 533 113
Micronutrient Levels:
Iron - adequate
Copper - adequate
Zinc - adequate
Boron - adequate
Manganese - adequate
Recommendations:
Barley - Malt
Lbs. Nitrogen Applied
Growing Condition Yield (Bu/Ac) N P2O5 K2O S 40 50 60 70
Excellent 80 60 21 0 0 76 79 80 81
Average 62 52 14 0 0 60 62 63 64
Comments:
This recommendation is made for Black soil
Previous crop: canola
On high K soils, crops may respond to 0-0-60 due to chloride or to decreased root disease
Recommended application rates are based on seed placed or banded fertillizer efficiencies unless otherwise indicated.
The method of application, however, is left to your discretion.  The total amount can not necessarily be placed with the seed.

 

Soil test results for upper slope position - nutrient values are in ppm

Sample
Depth
Nitrate
N
Phosphate Potassium Sulphates Iron Copper Zinc Boron Manganese pH
(acidity)
EC
(salinity)
0-6" 11 19 173 8 165 1.09 4.1 2.10 15.3 6.3 (normal) 0.33 (OK)
6-12" 11

9




6.7 0.63
Total (lbs/acre)

44 38 346 33
Estimated Available (lbs/acre)

62 39 346 47
Micronutrient Levels:
Iron - adequate
Copper - adequate
Zinc - adequate
Boron - adequate
Manganese - adequate
Recommendations:
Barley - Malt
  Lbs. Applied Nitrogen
Growing
Conditions
Yield
(Bu/Ac)
N P2O5 K2O S 40 50 60 70
Excellent 80 52 33 25 0 78 80 81 81
Average 62 44 25 17 0 62 63 64 64
Comments:
This recommendation is made for Black soil
Previous crop: canola
On high K soils, crops may respond to 0-0-60 due to chloride or to decreased root disease
Recommended application rates are based on seed placed or banded fertillizer efficiencies unless otherwise indicated.
The method of application, however, is left to your discretion.  The total amount can not necessarily be placed with the seed.

The following questions pertain to the two preceeding soil tests representing an upper and lower slope position. Fifty percent of the field is characteristic of lower slopes while the other 50% represents upper slopes. Combine the soil test results to answer the following questions.

  1. Convert ppm to lbs/acre for the 0-6 inch sample depth, then determine the amount present in the 0-2 inch depth. Provide the 0-2" lbs/acre equivalent for both nitrogen and phosphorus.

     

  2. If there is 8 ppm of nitrate-N in the 0-6" sample depth and 6 ppm of nitrate-N in the 6-12" sample depth, how many ppm of nitrate-N are there in the 12-24" depth to approximate 38 lbs/acre of available nitrate-N?

     

  3. For an 80 bushel malt barley crop, how many lbs/acre of nitrogen would you need? How many lbs/acre of P2O5 is required?

     

  4. What is the conversion factor for P to P2O5?

     

  5. What is the conversion factor for K to K2O?

     

  6. What is the pH of the combined soil sample and is it a problem? What is a normal acceptable range of surface soil pH?

     

  7. What is the E.C. of the soil? Are the E.C. levels of these soil samples acceptable? What related problem can occur with frequent manure application?

     

  8. What other analyses would you recommend for monitoring manure application onto soil? Why? (Try to provide at least 3 possible analyses)

 

 

Exercise 3b. Interpreting Manure Test Results

Manure Test results (on a dry matter basis):

Test Result Test Result
Moisture @ 60 oC 71.70% Kjeldahl Nitrogen (Total) 2.83%
Total P2O5 4.95% Total K2O 2.8%
Available Nitrogen 1.92% Total Carbon 52.9%
E.C. ms/cm 51.80 C:N Ratio 26.99
SAR 6.9 Sodium 0.31%
Available Carbon 38.16% Ash 14.72%
Calcium 1.00% AvailableOrganic Matter 76.33%
Available C:N Ratio 28.7 Organic Matter 85.3%
pH 7.3 Magnesium 0.33%

 

Manure nutrient conversions from dry basis to as-is:

Dry basis to as-is
Amount of nutrient in manure x (%DM/100)

As-is to dry basis:

  • Amount of nutrient in manure x (100/%DM)

Dry Matter (DM) = 100 - moisture content of manure

Example 1:
1.33% available N on a dry basis. The moisture content of the manure is 71.7%. Calculate the amount of available N on an as-is basis.

  • = 1.33 x {(100-71.7) / (100)}
  • = 0.376 % available N on an as-is basis
Example 2:
0.45% available N on an as-is basis. Dry matter content is 40%. Calculate the amount of available N on a dry matter basis.
  • = 0.45 x {(100) / (40)}
  • = 1.125% available N on a dry matter basis
Manure nutrient conversion from % to lbs/ton.

Example: 0.5% available N on an as-is basis.

0.5
100
=   x  
2000
2000 lbs in 1 ton
  • = 0.5 x 2000 / 100 = 10 lbs N / ton of manure
  • 0.5% is a unitless ratio, and since we know there is 2000 lbs in a ton, cross-multiplying gives us the lbs/ton equivalent.

 

Using the manure test results and the above information answer the following three questions:

  1. Determine the amount (lbs/ton) of total nitrogen as-is in the manure.

     

     

  2. Determine the amount (lbs/ton) of P2O5 as-is in the manure.

     

     

  3. Determine the amount (lbs/ton) of K20 as-is in the manure.